Matlab Manual

EC 45

IV Semester

Rajesh N

Dept of Electronics and Communication Engg Nitte Meenakshi Institute of Technology

Yelahanka, Bangalore – 560 064

MAT LAB MANUAL DEPT OF ELECTRONICS AND COMMUNICATION ENGG

Syllabus

Part - A ( 8 Lab Sessions of 3Hrs Each)

(Perform the experiments using MatLab)

1. Vector and matrix operations, solving non linear equations.

2. a) Representation of basic signals, Basic operations on continuous/discrete time

signals: Add/sub of two signals, scaling, shifting, time reversal , signal energy /

power calculation

b) Verification of Shannon sampling theorem.

3. Convolution and Correlation of continuous/ discrete time signals.

4. Fourier analysis of signals: Computing CTFS/CTFT and DTFS/DTFT, properties of

DTFT

5. The z and Inverse z-Transform: Properties of z-Ttransform. Determination of

system function and the unit sample response of a linear time invariant systems

described by difference equation.

6. Laplace Transform analysis of linear systems.

7. PID design.

8. Frequency response analysis: Bode Plots , Nyquist Plot. Root Locus

9. Simulink model of first /second order systems.

NITTE M EENAKSHI INSTITUTE OF TECHNOLOGY, BANGALORE. 2


EXPERIMENT No.1

Vector and matrix operations, solving non linear equations.

Aim: Study Vector and matrix operations and solving non linear equations.

Refer Appendix for vector and matrix operations.

Example:1. Solve the given quadratic equation

x2 + 5x +6 = 0

roots = -3 and -2

program: b=[1 5 6];

roots(b)

Result:

ans =

-3.0000

-2.0000

Example:2. Solving given difference equation

Theory:

• A difference equation with constant coefficients describes a LTI system. For example

the difference equation y[n] + 0.8y[n-2] + 0.6y[n-3] = x[n] + 0.7x[n-1] + 0.5x[n-2]

describes a LTI system of order 3. The coefficients 0.8, 0.7, etc are all constant i.e.,

they are not functions of time (n). The difference equation y[n]+0.3ny[n-1]=x[n]

describes a time varying system as the coefficient 0.3n is not constant.

• The difference equation can be solved to obtain y[n], the output for a given input x[n]

by rearranging as y[n] = x[n] + 0.7x[n-1]+0.5x[n-2]- 0.8y[n-2]- 0.6y[n-3] and solving.

• The output depends on the input x[n]

o With x[n]= δ[n], an impulse, the computed output y[n] is the impulse

response.

o If x[n]=u[n], a step response is obtained.

o If x[n] = cos(wn) is a sinusoidal sequence, a steady state response is obtained

(wherein y[n] is of the same frequency as x[n], with only an amplitude gain

and phase shift-refer Fig.7.3).

o Similarly for any arbitrary sequence of x[n], the corresponding output

response y[n] is computed.



• The difference equation containing past samples of output, i.e., y[n-1], y[n-2], etc

leads to a recursive system, whose impulse response is of infinite duration (IIR). For

such systems the impulse response is computed for a large value of n, say n=100 (to

approximate n=∞). The MATLAB function filter is used to compute the impulse

response/ step response/ response to any given x[n]. Note: The filter function

evaluates the convolution of an infinite sequence (IIR) and x[n], which is not possible

with conv function (remember conv requires both the sequences to be finite).

• The difference equation having only y[n] and present and past samples of input (x[n],

x[n-k]), represents a system whose impulse response is of finite duration (FIR). The

response of FIR systems can be obtained by both the ‘conv’ and ‘filter’ functions. The

filter function results in a response whose length is equal to that of the input x[n],

whereas the output sequence from conv function is of a longer length (xlength +

hlength-1).

Algorithm: 1. Input the two sequences as a and b representing the coefficients of y and x.

2. If IIR response, then input the length of the response required (say 100, which can be

made constant).

3. Compute the output response using the ‘filter’ command.

4. Plot the input sequence & impulse response (and also step response, etc if required).

MATLAB Implementation: MATLAB has an inbuilt function ‘filter’ to solve difference equations numerically, given

the input and difference equation coefficients (b,a).

y=filter(b,a,x)

where x is the input sequence, y is the output sequence which is of same length as x.

Given a difference equation a0y[n]+a1y[n-1]+a y[n-2]=b x[n]+b2 0 2x[n-2], the coefficients are

written in a vector as b=[b 0 b ] and a=[a a a0 2 0 1 2]. Note the zero in b (x[n-1] term is missing).

Also remember a , the coefficient of y[n] should always be 1. 0

For impulse response the number of zeros = the length of the IIR response

required (say 100 implemented by function zeros(1,99)).

,....0,0,0,1][↑

=nx

For step response the number of ones = the length of the IIR response

required-1 (say 100 implemented by function ones(1,100).

,....1,1,1,1,1][↑

=nx

Similarly for any given x[n] sequence, the response y[n] can be calculated.



MATLAB Program: clear all, close all ; %y(n)-0.1y(n-1)=x(n) %x(n)=u(n) %initial condition y(-1)=2 b=input('Enter the numerator co-effiecients') a=input('Enter the denominator co-effiecients') N=input('Enter the length of the input sequence') n=0:1:N-1; inp=1.^n; X=2; XIC=filtic(b,a,X); RES1=filter(b,a,inp,XIC); figure(1); stem(n,RES1); grid on; xlabel('n'); ylabel('amplitude'); title('step response output of difference equation'); %x(n) is an impulse function IMP=[1,zeros(1,50)]; RES2=filter(b,a,IMP,XIC); n=0:1:length(RES2)-1; figure(2); stem(n,RES2); grid on; xlabel('n'); ylabel('amplitude'); title('impulse response output of difference equation');

RESULTS:

Enter the numerator co-effiecients[1]

b = 1

Enter the denominator co-effiecients[1 -0.1]

a = 1.0000 -0.1000

Enter the length of the input sequence50

N = 50

RES1=

1.2000, 1.1200, 1.1120, 1.1112, 1.1111, 1.1111, 1.1111, 1.1111, 1.1111,

1.1111, 1.1111, 1.1111, 1.1111, 1.1111, 1.1111, 1.1111, 1.1111, 1.1111,

1.1111, 1.1111 , 1.1111, 1.1111, 1.1111, 1.1111, 1.1111, 1.1111, 1.1111,

1.1111, 1.1111, 1.1111, 1.1111, 1.1111, 1.1111, 1.1111, 1.1111, 1.1111,

1.1111, 1.1111, 1.1111, 1.1111, 1.1111, 1.1111, 1.1111, 1.1111, 1.1111,

1.1111, 1.1111, 1.1111, 1.1111, 1.1111



RES2=

1.2000, 0.1200, 0.0120, 0.0012, 0.0001, 0.0000, 0.0000, 0.0000, 0.0000 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000

Questions: 1)Difference equation y(n) - y(n-1) + 0.9y(n-2) = x(n);

Calculate impulse response h(n) and also step response at n=0,…..,100

2)Plot the steady state response y[n] to x[n]=cos(0.05πn)u(n), given y[n]-0.8y[n-1]=x[n]



*note- Use [h,t]=impz(b,a,N) to find the impulse response and find the difference between y~=h’ (=0). Using impz, we need not specify the impulse input x[n], just the coefficients and length N that we desire (refer experiment 2). Example: Find the first 8 samples of the complete response y[n] for the difference equation

[ ] ];2[9025.0]1[95.0]2[]1[][31][ −−−+−+−+= nynynxnxnxny n≥0; where

][3

cos][ nunnx ⎟⎠⎞

⎜⎝⎛=π and the initial conditions are y[-1]=-2; y[-2]=-3; x[-1]=1; x[-2]=1.

Solution: Use MATLAB function filter(b,a,x,xic), where xic are the initial conditions computed as xic=filtic(b,a,Y,X), where Y=[y(-1) y(-2)…y(N-1)] & X=[x(-1) x(-2)…x(-M)] are the initial condition arrays. If x[n]=0 for n<0, then X need not be specified. The sample program for the above problem is: b=[1/3 1/3 1/3]; a=[1 -0.95 0.9025]; %difference equation coefficient arrays Y=[-2 -3]; X=[1 1]; %initial condition arrays xic=filtic(b,a,Y,X); n=0:7; %For 8 output samples using filter function, the input should also be of size 8 x=cos(pi*n/3); %input sequence y=filter(b,a,x,xic); %output response Answer: y = 1.8075 4.3555 2.8398 -1.5664 -4.7176 -3.4014 1.3596 5.0281

Exercises:

1. Implement all the exercises given in chapter-2: “S&S”, M.J.Roberts.



EXPERIMENT No.2

Representation of basic signals Program to represent some basic signals: %Unit step function defined as 0 for input argument values %Less than zero, ½ for input argument values equal to zero, %And one for input argument values greater than zero. function y=u(t) zero=(t==0) ;pos =(t>0) ;y=zero/2 + pos ; % function to compute the triangle function.uses the definition % of the triangle functionin terms of the ramp function.works % for vectors and scalars equally well. function y=tri(t) y=ramp(t+1) -2*ramp(t) + ramp(t-1) ; % function to compute sinc(t) defined as sin(pi*t)/(pi*t) . % works for vectors or scalars equally well.this function % may be intrinsic in some versions of MATLAB. function y=sinc(t) zero=(t==0) ; % indicate the locations of zeros in t. num=(~zero).*sin(pi*t) + zero ; den = (~zero).*(pi*t) +zero ; y=num./den ; % rectangle function . uses the definition of the rectangle % function in terms of the unit step function.works on % vectors or scalars equally well. function y=rect(t) y=u(t+0.5) - u(t-0.5) ; % Function to compute the ramp function defined as zero for % Values of the argument less than zero and the value of % The argument for arguments greater than or equal to zero. % Works for vectors and scalars equally well. function y=ramp(t) y=t.*(t >=0) ; % Function to generate the discrete-time impulse function defined as one % for input integer arguments equal to zero and zero otherwise . % Returns- NaN for noninteger arguments .Works for vectors and % Scalars equally well .



function y=impDT(n) y=(n==0) ; % impulse is one where argument % is zero and zero otherwise. ss=find(round(n)~=n) % Find noninteger values of n.. y(ss)==NaN ; % Set corresponding outputs to NaN. % function to compute values of the dirichlet function. % Works for vectors or scalars equally well. % x= sin(N*pi*t)/(N*sin(pi*t)) function x =drcl(t,N) x=diric(2*pi*t,N) ; % program to demonstrate some basic CT signals %-------------------------------------------------------------------------------- clc; disp('Select the finction') %----Next part of the program accepts choice of function -------------------- disp('1-Sine 2-Exponential 3-Ramp 4-Step 5-Rectangle 6-Sinc '); Choice = input('Enter choice of function = '); t=-10:0.1:10; switch(Choice) case1 % Sine plot(t,sin(t)); grid on case2 % Exponential plot(t,exp(t)); grid on case3 % Ramp plot(t,(t.*(t >=0))); grid on case4 % Step plot(t,(t>=0)); grid on case5 % Rectangle plot(t,(t>=-1&t<=1)); grid on case6 % Sinc plot(t,((~(t==0)).*sin(pi*t) + (t==0))./ ((~(t==0)).*(pi*t) +(t==0))); grid on otherwise subplot(321); plot(t,sin(t)); grid on; title('Sine Wave') subplot(322); plot(t,exp(t)); grid on ; title('Exponential') subplot(323); plot(t,(t.*(t >=0))); grid on ; title('Ramp') subplot(324); plot(t,(t>=0)); grid on ; title('Step') subplot(325); plot(t,(t>=-1&t<=1)); grid on; title('Rectangle') subplot(326); plot(t,((~(t==0)).*sin(pi*t) + (t==0))./ ((~(t==0)).*(pi*t) +(t==0))); grid on;title('Sinc') end %------------------------ end of program ----------------------------------------



Continuous time Signals

Discrete time Signals

--------------------------------------------------------------------------------------------------------------- % program to plot some demonstrations of CT function combination. % x1(t) = exp(-t).sin(20.pi.t)+exp(-t/2)sin(19.pi.t) % x2(t) = sinc(t)cos(20.pi.t). t=0:1/120:6; x1 =exp(-t).*sin(20*pi*t) + exp(-t/2).*sin(19*pi*t) ; subplot(2,1,1); p=plot(t,x1,'k'); set(p,'LineWidth',2) ; xlabel('\itt'); ylabel('x_1(\itt)') ; t=-4:1/60:4; x2 =sinc(t).*cos(20*pi*t); subplot(2,1,2); p=plot(t,x2,'k'); set(p,'LineWidth',2) ; xlabel('\itt'); ylabel('x_2(\itt)') ;



g.m function y=g(t) %calculate the functional variation for each range of t. y1=-4 -2*t ;y2= -4 +3*t ;y3=16 -2*t ; %splice together the different functional variations in their respective ranges of validity. y=y1.*(-2<t & t<=0) +y2.*(0<t & t<=4) +y3 .*(4<t & t<=8) ; Script file: % program to plot the function ,g(t) and then to % plot 3*g(t+1) ,g(3*t)/2 ,and -2*g(t-1)/2 . % Write a function for g(t) and store. tmin =-4 ; tmax =20; % set the time range for the plot . dt=0.1 ; % set the time between points t=tmin:dt:tmax; % set the vector of times for the plot. g0=g(t) ; % compute the original g(t). g1=3*g(t+1) ; % compute the first transformation . g2=g(3*t) /2 ; % compute the second transformation . g3=-2*g((t-1)/2) ; % compute the third transformation . % Find the maximum and minimum g values in all the transformed % Functions and use them to scale all plots the same . gmax=max([max(g0) ,max(g1) ,max(g2) ,max(g3)]) ; gmin=min([min(g0) ,min(g1) ,min(g2) ,min(g3)]) ; % plot all four functions in a 2 by 2 arrangement . % plot them all on the same scale using the axis command . % plot grid lines, using the grid command ,to aid in reading values. Subplot (2,2,1); p=plot(t,g0,'k'); set(p,'linewidth',2) ; xlabel('t');ylabel('g(t)'); title('Original Function, g(t)'); axis([tmin,tmax,gmin,gmax]);grid; subplot(2,2,2); p=plot(t,g1,'k') ; set (p,'LineWidth',2); xlabel('t'); ylabel('3g(t+1)'); title ('First Transformation'); axis([tmin,tmax,gmin,gmax] ) ;grid; subplot(2,2,3);p=plot(t,g2,'k'); set(p,'LineWidth',2); xlabel('t');ylabel('g(3t)/2');title('Second Transformation'); axis([tmin,tmax,gmin,gmax] );grid; subplot(2,2,4); p=plot(t,g3,'k'); set(p,'LineWidth',2); xlabel('t'); ylabel('-2g((t-1)/2)');title('Third Transformation'); axis([tmin,tmax,gmin,gmax] );grid ;



--------------------------------------------------------------------------------------------------------------------------- % program to illustrate some of MATLAB s built-in functions % close all ; t= -20:1/20:20 ; x1=chirp(t,1/20,20,1/3) ; subplot(4,2,1) ; p=plot(t,x1, 'k' ) ; axis([-20,20,-1.5,1.5] ) ; title( 'chirp - A "chirped" cosine') ; xlabel ( '\itt') ; ylabel( 'x_1(\itt )') ; x2=diric(t,5) ; subplot (4,2,2) ; p=plot(t,x2, 'k') ; axis([-20,20, -1.5, 1.5] ) ; title ( 'diric -The MATLAB "Dirichlet" function') ; xlabel( ' \ itt') ; ylabel( 'x_2(\itt)') ; x3=sawtooth(t) ; subplot(4,2,3) ;p=plot(t,x3, 'k') ; axis([-20,20 -1.5 ,1.5] ) ; title( 'sawtooth -A Periodic Sawtooth') ; xlabel( '\ itt') ; ylabel( 'x_3(\itt)') ; x4=square(t) ;subplot(4,2,4) ; p=plot(t,x4, 'k') ; axis([-20,20, -1.5, 1.5] ) ;title( 'square - A Square Wave') ; xlabel( '\itt') ; ylabel( 'x_4(\itt )') ; x5=rectpuls(t/10) ; subplot(4,2,5) ; p =plot(t,x5, 'k') ; axis([ -20 ,20 ,-1.5 ,1.5] ) ; title( 'rectpuls - A Rectangular Pulse Wave') ; xlabel( '\itt') ; ylabel ( 'x_5 (\itt)' ); x6=tripuls(t/10) ; subplot (4,2,6) ; p=plot(t,x6, 'k') ; axis([ -20 ,20 , -1.5 ,1.5] ) ; title( 'tripuls -A Triangular Pulse Wave') ; xlabel ( '\itt') ;ylabel( 'x_6(\itt)') ; x7=sinc(t/2) ; subplot(4,2,7) ; p=plot(t,x7, 'k') ; axis([ -20 ,20 ,-1.5 ,1.5]) ; title( 'sinc(t)') ; xlabel ( '\itt') ;ylabel( 'x_7(\itt)' ); x8=sign(t/2) ; subplot(4,2,8) ; p=plot(t,x8, 'k') ; axis([ -20 ,20 -1.5 ,1.5] ) ; title( 'sign - The Signum Function') ; xlabel( '\itt') ; ylabel( 'x_8(\itt )') ; figure; x24=x2.*x4 ; subplot(2,2,1) ; plot(t,x24, 'k') ; axis([ -20 ,20 ,-1.5 ,1.5] ) ; title( 'x_2*x_4 - Even*odd') ; xlabel( '\itt') ; ylabel( 'x_2_4( \itt)') ;



x34=x3 .*x4 ; subplot(2,2,2) ; plot(t,x34 , 'k') ; axis([ -20 ,20 ,-1.5 ,1.5] ) ; title( 'x_3*x_4 -odd*odd') ; xlabel( '\itt') ; ylabel( 'x_3_4(\itt)') ; x26=x2 .*x6 ; subplot(2,2,3) ; plot(t,x26 , 'k') ; axis([ -20 ,20 ,-1.5 ,1.5]) ; title( 'x_2*x_6 - Even*Even') ; xlabel( '\itt') ;ylabel( 'x_2_6(\itt)') ; x37=x3 .*x7 ; subplot(2,2,4) ; plot(t, x37 , 'k') ; axis([ -20 ,20 , -1.5 ,1.5] ) ; title( 'x_3*x_7 - odd*Even') ; xlabel( '\itt') ; ylabel( 'x3_7(\itt)') ; xle= (x1 + x1(end: -1 :1)) /2 ; xlo= (x1 - x1(end : -1 :1))/2 ; subplot(2,1,1) ; plot(t,xle, 'k') ; axis([ -20 ,20 , -1.5 ,1.5] ) ; title( 'Even Part of x_1') ; xlabel( '\itt') ; ylabel( 'x_1_e(\itt)') ; subplot(2,1,2) ; plot(t ,xlo , 'k') ; axis( [ -20 ,20 ,-1.5 ,1.5] ) ; title( 'odd Part of x_1') ; xlabel( '\itt') ; ylabel ( 'x_1_o(\itt)') ;

% pole zero plot from coefficients of a difference equation. % This Matlab program accepts the coefficients of the difference equation % and generates pole zero plot against the unit circle. % % The format of the difference equation is as follows..



% y(n) = -[a1*y(n-1) + a2*y(n-2) + a3*y(n-3) + ....] % +b0*x(n) + b1*x(n-1) + b2*x(n-2) + .... % the above equation can also be written as follows, % a0*y(n) + [a1*y(n-1) + a2*y(n-2) + a3*y(n-3) + .... % = b0*x(n) + b1*x(n-1) + b2*x(n-2) + .... % here a0 = 1 and the coefficients are to be entered as, % b = [b0 b1 b2 b3 . . .] % a = [a0 a1 a2 a3 . . .] %----- Next part accepts coefficients of the difference equation ---------------- clc; disp(' Pole zero plot'); disp(' '); b = input('enter [b0 b1 b2...] = '); % enter the coefficients of x(n) a = input('enter [a0 a1 a2...] = '); % enter the coefficients of y(n) %------ Next part plots the pole zero plot -------------------------------------- zplane(b,a); % plot the pole zero plot title('Pole zero plot'); %------------------------ end of program ----------------------------------------

Verification of Sampling theorem. Aim: To verify Sampling theorem for a signal of given frequency. Theory:

• Sampling is a process of converting a continuous time signal (analog signal) x(t) into

a discrete time signal x[n], which is represented as a sequence of numbers. (A/D

converter)

)(tx)• Converting back x[n] into analog (resulting in ) is the process of reconstruction.

(D/A converter)



• Techniques for reconstruction-(i) ZOH (zero order hold) interpolation results in a

staircase waveform, is implemented by MATLAB plotting function stairs(n,x), (ii)

FOH (first order hold) where the adjacent samples are joined by straight lines is

implemented by MATLAB plotting function plot(n,x),(iii) spline interpolation, etc.

)(tx)• For to be exactly the same as x(t), sampling theorem in the generation of x(n)

from x(t) is used. The sampling frequency fs determines the spacing between samples.

• Aliasing-A high frequency signal is converted to a lower frequency, results due to

under sampling. Though it is undesirable in ADCs, it finds practical applications in

stroboscope and sampling oscilloscopes.

The Sampling Theorem: If a signal xa(t) has a bandlimited Fourier transform Xa(jΩ) such that Xa(jΩ)=0 for Ω≥2πfN,

then xa(t) can be uniquely reconstructed from equally spaced samples xa(nT), -∞<n<∞, if

1/T≥2 fN (f ≥2fS N).

Time domain sequence and corresponding spectrum.



N

NNN

FTorTorT

2/12/2/2

:need aliasing avoid To

>Ω>Ω>Ω− ππ

occurs aliasing,2/1 wherecase NFT <

t

x

t

fs >= 2 * highest freq in x

fs < highest freq in x



Plots of a sampled cosine wave of 200Hz

Algorithm:

1. Input the desired frequency fd (for which sampling theorem is to be verified).

2. Generate an analog signal xt of frequency fd for comparison.

3. Generate oversampled, Nyquist & under sampled discrete time signals.

4. Plot the waveforms and hence prove sampling theorem.

MATLAB Program: close all, clear all;; f=5; fs=2*f; n4=-50:1:50; % for f >> fs m n=0.01:0.01:1; %a=1+sin(2*pi*f*n)+(1+sin(2*pi*2*f*n))/2+(1+sin(2*pi*3*f*n))/3; a=(1+cos(2*pi*f*n)); figure(1); subplot(4,1,1);plot(a);grid on; subplot(4,1,2);stem(a);grid on; b=fft(a,101); b=fftshift(b); subplot(4,1,3); stem(n4,abs(b));grid on; s=abs(ifft(b)); s=s(1:100); subplot(4,1,4); plot(s); grid on; % for f = fs m

n2=0:1/(fs/2):1; %a1=1+sin(2*pi*f*n2)+(1+sin(2*pi*2*f*n2))/2+(1+sin(2*pi*3*f*n2))/3; a1=(1+cos(2*pi*f*n2)); figure(2); subplot(4,1,1);plot(a);grid on; subplot(4,1,2);stem(a1);grid on; b=fft(a1,101);



b=fftshift(b); subplot(4,1,3); stem(n4,abs(b));grid on; g1=abs(ifft(b,101)); subplot(4,1,4); plot(g1(1:length(n2)),'r'); grid on; % for f =2* fs m

n1=0:1/(fs):1; %a2=1+sin(2*pi*f*n1)+(1+sin(2*pi*2*f*n1))/2+(1+sin(2*pi*3*f*n1))/3; a2=(1+cos(2*pi*f*n1)); figure(3); subplot(4,1,1);plot(n*100,a);grid on; subplot(4,1,2);stem([a2]);grid on; b=fft(a2,101); b=fftshift(b); subplot(4,1,3); stem(n4,abs(b));grid on; g=abs(ifft(b,101)); subplot(4,1,4); plot(g(1:length(n1)),'r'); grid on; % for f =2*(2* f ) s m

n3=0:1/(2*fs):1; %a2=1+sin(2*pi*f*n3)+(1+sin(2*pi*2*f*n3))/2+(1+sin(2*pi*3*f*n3))/3; a2=(1+cos(2*pi*f*n3)); figure(4); subplot(4,1,1);plot(n*100,a);grid on; subplot(4,1,2);stem(n3*100,a2);grid on; b=fft(a2,101); b=fftshift(b); subplot(4,1,3); stem(n4,abs(b));grid on; g=abs(ifft(b,101)); subplot(4,1,4); plot(n3*100, g(1:length(n3)),'r'); grid on; Result: fs>>fm

Fig: a) 5Hz sinusoidal time domain signal b) Sampled sequence at high sampling rate. c) Spectrum. d)

constructed signal. Re



fs=fm

Fig: a) 5Hz sinusoidal time domain signal b) Sampled sequence at half the Nyquist rate. c) Spectrum. d) Reconstructed signal. fs=2*fm

Fig: a) 5Hz sinusoidal time domain signal b) Sampled sequence at Nyquist rate. c) Spectrum. d) Reconstructed signal. fs=2*(2*fm)

Fig: a) 5Hz sinusoidal time domain signal b) Sampled sequence at double the Nyquist rate. c) Spectrum. d) Reconstructed signal.



Note: Mark the labeling for above graphs.

Inference:From the under sampling plot observe the aliasing effect. The analog signal is of

5Hz. The reconstructed (from under sampled plot) signal is of a lower frequency.

At Nyquist rate plot shows a reconstructed signal almost similar to that of the analog signal.

Using low pass filtering the wave form can be further smoothened.

The over sampled plot shows a reconstructed signal similar to that of the analog signal. Questions: 1. Generate sine/cosine waveforms of multiple frequencies, say x1=sin(2*pi*fd1*n1)+

sin(2*pi*fd2*n2); where fd1, fd2 are 2 different frequencies.

2. Sample a band limited continuous time signal band limited to fm = 3400 Hz under the

following conditions

a. Nyquist Rate b. Twice the Nyquist Rate c. Half the Nyquist rate.

Impulse response of a given system

Aim: To find the impulse response h(n) of the given LTI system whose response y(n) to an input x(n) is given. Theory:

Fig.2.1: A LTI system

• A discrete time LTI system (also called digital filters) as shown in Fig.2.1 is

represented by o A linear constant coefficient difference equation, for example,

];2[]1[][]2[]1[][ 21021 −+−+=−−−+ nxbnxbnxbnyanyany o A system function H(z) (obtained by applying Z transform to the difference

equation). 22

11

22

110

1)()()( −−

−−

++++

==zazazbzbb

zXzYzH

o A frequency response function obtained by applying DTFT on the impulse response h[n] (or by replacing z with ejΩ

in H(z)) to get

Ω−Ω−

Ω−Ω−

Ω

ΩΩ

++++

== jj

jj

j

jj

eaeaebebb

eXeYeH 2

21

2210

1)()()(

• Given the difference equation or H(z), the impulse response of the LTI system is

computed using filter or impz MATLAB functions. This is elaborated in experiment

no.7 - Solving a given difference equation. If the difference equation contains past

samples of output, i.e., y[n-1], y[n-2], etc , then its impulse response is of infinite



duration (IIR). For such systems the impulse response is computed for a large value of

n, say n=100 (to approximate n=∞).

• Given only the input sequence x[n] and the output sequence y[n], we can find the

impulse function h[n] by using the inverse operation deconv. (The conv operation

convolves 2 sequences x[n] and h[n] to obtain the output y[n]. for convolution

operation refer experiment no 3: Linear convolution of two given sequences). If both

y[n] and x[n] are finite then the impulse response is finite (FIR).

• The deconvolution operation is valid only if the LTI system is ‘invertible’.

Given Problem 1) Find the impulse response h(n) of the given LTI system whose response y(n)=[1 2 3 2 1 2 ]

to an input x(n)=[1 2 3] is given. Also verify the result using conv operation. (FIR)

2) A LTI system is described by the difference equation y(n) - y(n-1) + 0.9y(n-2) = x(n);

Find its impulse response h(n) at n=0,…..,50. (IIR)

Algorithm for Problem1:

1. Input the two sequences as x and y.

2. Use deconv to get impulse response h.

3. Plot the sequences.

4. Verify using conv(x,h) to get y back.

Algorithm for Problem2:

1. Input the two coefficient arrays as b and a.

2. Input the length of the impulse response required as N.

3. Use impz unction to get impulse response h.

4. Plot the impulse response.

MATLAB Implementation: h=deconv(y,x): ‘y’ the output sequence should be of longer length than x. DECONV -

Deconvolution and polynomial division. [Q,R] = DECONV(B,A) deconvolves vector A out

of vector B. The result is returned in vector Q and the remainder in vector R such that B =

conv(A,Q) + R. If A and B are vectors of polynomial coefficients, deconvolution is

equivalent to polynomial division. The result of dividing B by A is quotient Q and remainder

R. (Remember conv is polynomial multiplication).

Note: Valid y and x should be given, else the conv verification will result in a slightly

different y (because of the remainder generated by deconv).



h=impz(b,a,N), returns the impulse response h[n], where b and a are coefficient arrays

obtained from difference equation, N = length of the response required. (refer expt.7). For the

given difference equation b=[1]; a=[1 -1 0.9].

IMPZ Impulse response of digital filter. [H,T] = IMPZ(B,A) computes the impulse response

of the filter B/A choosing the number of samples for you, and returns the response in column

vector H and a vector of times (or sample intervals) in T (T = [0 1 2 ...]').

[H,T] = IMPZ(B,A,N) computes N samples of the impulse response.

MATLAB Programs: Program 1: y= input('The output sequence y(n) of the system='); x=input('the input sequence of the system='); h=deconv(y,x); disp('the impulse response of the system is='); disp(h); %graphical display part N=length(h); n=0:1:N-1; stem(n,h); xlabel('Time index n'); ylabel('Amplitude'); title('impulse response of a system') %Verification yv=conv(x,h); disp('the verified output sequence is'); disp(yv) Result: The output sequence y(n) of the system = [1 2 1 1 0 -1] the input sequence of the system = [1 1 1 1] the impulse response of the system is = 1 1 -1 the verified output sequence is 1 2 1 1 0 -1

Inference: The impulse response h[n] is of finite duration. The verified convolution output sequence is the same as the given y[n].



Program 2 disp('Enter the sequences follows'); disp('Numerator coefficients: b=[b0 b1 b2 ...bn]'); disp('Denominator coefficients: a=[a0 a1 a2 ...an]'); b=input('Enter the numerator coefficients'); a=input('Enter the denominator coefficients'); N=50; h = impz(b,a,N); disp('Impulse response is:'); n=0:1:N-1; disp(h); stem(n,h); xlabel('n'); ylabel('h(n)'); grid on; title('IMPULSE RESPONSE USING DIFFERENTIAL EQUATION'); Result: Enter the sequences as follows Numerator coefficients: b=[b0 b1 b2 ...bn] Denominator coefficients: a=[a0 a1 a2 ...an] Enter the numerator coefficients[1] Enter the denominator coefficients>> [1 -1 0.9] Impulse response is: 1.0000, 1.0000, 0.1000, -0.8000, -0.8900, -0.1700, 0.6310, 0.7840, 0.2161 -0.4895, -0.6840, -0.2434, 0.3722, 0.5912, 0.2563, -0.2758, -0.5065, -0.2583 0.1976, 0.4300, 0.2522, -0.1348, -0.3618, -0.2405, 0.0852, 0.3016, 0.2249 -0.0465, -0.2489, -0.2071, 0.0169, 0.2033, 0.1881, 0.0051, -0.1642, -0.1688 -0.0210, 0.1309, 0.1498, 0.0320, -0.1028, -0.1316, -0.0391, 0.0794, 0.1145 0.0431, -0.0600, -0.0988, -0.0448, 0.0441



Inference: The impulse response h[n] is of infinite duration. h[n] is stable (as h[n] is absolutely

summable , decaying exponential). ∑+∞

−∞=

∞<n

nh ][

EXPERIMENT NO 3:

Linear convolution of two given sequences.

Aim: To obtain convolution of two finite duration sequences. Theory:

• The output y[n] of a LTI (linear time invariant) system can be obtained by convolving

the input x[n] with the system’s impulse response h[n].

∑∑+∞

−∞=

+∞

−∞=

−=−=∗=kk

khknxknhkxnhnxny ][][][][][][][ • The convolution sum is

• x[n] and h[n] can be both finite or infinite duration sequences.

( ) ][9.0][ nunh n=• Even if one (or both) of the sequences is infinite (say, ), we can

analytically evaluate the convolution formula to get a functional form (closed form

solution) of y[n].

• If both the sequences are of finite duration, then we can use the MATLAB function

‘conv’ to evaluate the convolution sum to obtain the output y[n]. Convolution is

implemented as polynomial multiplication (refer MATLAB help).

• The length of y[n] = xlength + hlength -1.

• The conv function assumes that the two sequences begin at n=0 and is invoked by

y=conv(x,h).

• Even if one of the sequences begin at other values of n, say n=-3,or n=2; then we need

to provide a beginning and end point to y[n] which are ybegin=xbegin+hbegin and

yend=xend+hend respectively.

Algorithm: 1. Input the two sequences as x1, x2

2. Convolve both to get output y.


MATLAB Implementation:

MATLAB recognizes index 1 to positive maximum. Index 0 is also not recognized. The

timing information for a sequence is provided by another vector, say n=-3:5; creates a

vector with values from -3 to 5 with an increment of 1.



During plotting, the time vector size and the sequence size should be the same, i.e., the

number of elements in the sequence x1 and in its time vector n1 should be the same.

Similarly for x2 and y.

MATLAB Program: % linear convolution clear all; close all; x=input('Enter the signal x(n)=') h=input('Enter the impulse response sequence h(n)=') lx=length(x); lh=length(h); len=lx+lh-1; t=0:1:len-1; for n=1:len y(n)=0; for k=1:lx if((n-k)>=0&(n-k)<lh) y(n)=y(n)+x(k).*h(n-k+1); end end end % to dispaly the result. disp('System response Output') disp(y) % Compare the result with inbuilt function. yc=conv(x,h); disp('Convolution from inbuilt function') disp(yc); t1=0:1:lx-1; t2=0:1:lh-1; % Plot subplot(3,1,1), stem(t1,x); grid on; ylabel('----> x(n)'); xlabel('-----------> n'); title('Linear Convolution of two sequences'); subplot(3,1,2), stem(t2,h); grid on; ylabel('----> h(n)'); xlabel('-----------> n'); subplot(3,1,3), stem(t,y); grid on; ylabel('----> y(n)'); xlabel('-----------> n');



Result: Enter the signal x(n)=[1 2 3 4] x = 1 2 3 4 Enter the impulse response sequence h(n)=[1 1 1 1] h = 1 1 1 1 System response Output: 1 3 6 10 9 7 4 Convolution from inbuilt function: 1 3 6 10 9 7 4

Inferences: ………….. Questions: 1. Write a program to compute the convolution of two sided sequences. 2. Write a program to compute the convolution using FT

Autocorrelation of a given sequence and verification of its properties.

Aim: To obtain autocorrelation of the given sequence and verify its properties. Theory:

• Correlation is mathematical technique which indicates whether 2 signals are related

and in a precise quantitative way how much they are related. A measure of similarity

between a pair of energy signals x[n] and y[n] is given by the cross correlation

sequence rxy[l] defined by ∑∞

−∞=

±±=−=n

xy llnynxlr ,...2,1,0];[][][ .

• The parameter ‘l’ called ‘lag’ indicates the time shift between the pair.

∑∞

−∞=

±±=−=n

xx llnxnxlr ,...2,1,0];[][][ • Autocorrelation sequence of x[n] is given by

• Some of the properties of autocorrelation are enumerated below



o The autocorrelation sequence is an even function i.e., ][][ lrlr xxxx −=

o At zero lag, i.e., at l=0, the sample value of the autocorrelation sequence has

its maximum value (equal to the total energy of the signal εx) i.e.,

. ∑∞

−∞=

==≤n

xxxxx nxrlr ][]0[][ 2ε

• This is verified in Fig. 5.1, where the autocorrelation of the rectangular pulse (square)

has a maximum value at l=0. All other samples are of lower value. Also the maximum

value = 11 = energy of the pulse [12+12+12..].

o A time shift of a signal does not change its autocorrelation sequence. For

example, let y[n]=x[n-k]; then ryy[l] = rxx[l] i.e., the autocorrelation of x[n] and

y[n] are the same regardless of the value of the time shift k. This can be

verified with a sine and cosine sequences of same amplitude and frequency

will have identical autocorrelation functions.

o For power signals the autocorrelation sequence is given by

∑−=

∞→±±=−

+=

k

knkxx llnxnxk

lr ,...2,1,0];[][12

1lim][ and for periodic signals with

period N it is ∑−

=

±±=−=1

0,...2,1,0];[][1][

N

nxx llnxnx

Nlr and this rxx[l] is also

periodic with N. This is verified in Fig. 5.3 where we use the periodicity

property of the autocorrelation sequence to determine the period of the

periodic signal y[n] which is x[n] (=cos(0.25*pi*n)) corrupted by an additive

uniformly distributed random noise of amplitude in the range [-0.5 0.5]

Algorithm:

1. Input the sequence as x.

2. Use the ‘xcorr’ function to get auto correlated output r.


MATLAB Implementation: MATLAB has the inbuilt function XCORR(A), when A is a vector, is the auto-

correlation sequence. If A is of length M vector (M>1), then the xcorr function returns the

length 2*M-1 auto-correlation sequence. The zeroth lag of the output correlation is in the

middle of the sequence at element M.

XCORR(...,MAXLAG) computes the (auto/cross) correlation over the range of lags:

-MAXLAG to MAXLAG, i.e., 2*MAXLAG+1 lags. If missing, default is MAXLAG = M-1.

[C,LAGS] = XCORR(...) returns a vector of lag indices (LAGS).



MATLAB Programs

%r = conv(x, fliplr(x)); %Energy signals r=xcorr(x); disp('autocorrelation sequence r='); %simple sequence disp(r); x2=[3,-1,2,1]; >> xcorr(x2) %plot the sequences ans = 3 5 -3 15 -3 5 3 subplot(2,1,1) stem(n,x); %Computation of Autocorrelation of a title('square sequence'); % rectangular Sequence subplot(2,1,2) n = -5:5; k = -N:N; N=10; stem(k, r); % Generate the square sequence title('autocorrelation output'); x = ones(1,11); xlabel('Lag index'); ylabel('Amplitude'); % Compute the correlation sequence

Result: The output plot is in Fig.5.1 autocorrelation sequence r = 1.0000 2.0000 3.0000 4.0000 5.0000 6.0000 7.0000 8.0000 9.0000 10.0000 11.0000 10.0000 9.0000 8.0000 7.0000 6.0000 5.0000 4.0000 3.0000 2.0000 1.0000 Inference: Following Autocorr properties are verified 1) Max peak of 11 (=energy of the pulse) at zero lag (l=0) 2)the autocorr seq is an even function

Fig.5.2

Fig. 5.3



Inference 2: The plot of noise corrupted signal y[n] (Fig. 5.3) does not show any noticeable

periodicity, but its autocorrelation sequence has distinct peaks at lags that are multiples of

8,indicating thaty[n] is a periodic with N=8 (cos(0.25*pi*n)=cos(wn) implies w=2*pi*f; N=

1/f = 2/0.25 =8). Also notice the strong peak at zero lag.

From Fig.5.2, the autocorrelation of the noise component has a very strong peak only at zero

lag. The amplitudes are considerably smaller at other values of the lag as the sample values

(of noise generated by rand function) are uncorrelated with each other.

Cross correlation of a given sequence and verification of its properties. Aim: To obtain cross correlation of the given sequence and verify its properties. Theory:

• Cross Correlation has been introduced in the last experiment. Comparing the

equations for the linear convolution and cross correlation we find that

. i.e., convolving the

reference signal with a folded version of sequence to be shifted (y[n]) results in cross

correlation output. (Use ‘fliplr’ function for folding the sequence for correlation).

][][)]([][][][][ lylxnlynxlnynxlrnn

xy −∗=−−=−= ∑∑∞

−∞=

∞

−∞=

• The properties of cross correlation are 1) the cross correlation sequence sample values

are upper bounded by the inequality yxyyxxxx rrlr εε=≤ ]0[]0[][

2) The cross correlation of two sequences x[n] and y[n]=x[n-k] shows a peak at the

value of k. Hence cross correlation is employed to compute the exact value of the

delay k between the 2 signals. Used in radar and sonar applications, where the

received signal reflected from the target is the delayed version of the transmitted

signal (measure delay to determine the distance of the target).

3) The ordering of the subscripts xy specifies that x[n] is the reference sequence that

remains fixed in time, whereas the sequence y[n] is shifted w.r.t x[n]. If y[n] is the

reference sequence then ][][ lrlr xyyx −= . Hence r [l] is obtained by time reversing the yx

sequence rxy[l].

Program:



%Verification of the properties %Small MATLAB code xi=[1,2,3,-1,0,0]; x2=[0,0,1,2,3,-1]; %=xr[n-2],i.e., k=2 >> xcorr(xi,x2)

0,0,0,0,1,1,5,15,5,1,1 −−↑

ans =

Inference: Strong peak of 15 at lag = -2 implies the delay between xi and x2 is 2. Also peak =15=energy of xi (1+22 2+3 +(-1)2) implies that both xi and x2 are strongly correlated. %consider the below sequences xr=[1,2,3,-1]; x2=[3,-1,1,2]; xcorr(xr,x2) ans = 3,10,2,2,7,5,2 −

↑

xcorr(x2,xr) ans = 2,5,7,2,2,10,3↑

−

Inference: Strong peak of 10 at lag = 2 implies the delay between xr and x2 is 2, but since 10<15, it implies that xr and x2 are uncorrelated slightly (may be due to noise,etc). is verified. ][][ lrlr xyyx −= Algorithm:

1. Input the sequence as x and y.

2. Use the ‘xcorr’ function to get cross correlated output r.


MATLAB Implementation: MATLAB has the inbuilt function XCORR: Say C = XCORR(A,B), where A and B are

length M vectors (M>1), returns the length 2*M-1 cross-correlation sequence C. If A and B

are of different length, the shortest one is zero-padded. Using convolution to implement

correlation, the instruction is FLIPLR Flip matrix in left/right direction. FLIPLR(X)

returns X with row preserved and columns flipped in the left/right direction. X = 1 2 3

becomes 3 2 1.

MATLAB Program:

% Computation of Cross-correlation Sequence using folded sequence and convolution

x = input('Type in the reference sequence = '); x = input('Type in the reference sequence = '); y = input('Type in the second sequence = '); y = input('Type in the second sequence = '); % Compute the correlation sequence % Compute the correlation sequence n1 = length(y)-1; n1 = length(y)-1; n2 = length(x)-1; n2 = length(x)-1; r = xcorr(x,y); r = conv(x,fliplr(y)); %[r,lag]=xcorr(x,y); disp(‘Cross correlation output is =’); disp('Cross correlation output is ='); disp(r); disp(r); k = (-n1):n2; %time vector for plotting N=max(n1,n2) ; stem(k,r); k = (-N):N; xlabel('Lag index'); ylabel('Amplitude'); stem(k,r); %stem(lag,r); xlabel('Lag index'); ylabel('Amplitude'); % Computation of Cross-correlation Sequence

%using xcorr function



Result: Type in the reference sequence = [1 3 -2 1 2 -1 4 4 2] Type in the second sequence = [2 -1 4 1 -2 3] Cross correlation output is = 3 7 -11 14 13 -15 28 6 -2 21 12 12 6 4 Verification: with xcorr(x,y), the output is -0.0000 , -0.0000 , 0.0000, 3.0000, 7.0000, -11.0000, 14.0000, 13.0000, -15.0000, 28.0000, 6.0000, -2.0000, 21.0000, 12.0000, 12.0000, 6.0000 , 4.0000 Note: For sequences with different lengths, the length of the output using xcorr is 2*max(n1,n2)-1 = 2* 9-1=17 whereas using conv program, the length of output is n1+n2-1.

Fig.6.1 Cross correlation output Fig.6.2 Cross correlation output using xcorr using Convolution

FREQUENCY RESPONSE: The frequency response is a representation of the system's response to sinusoidal inputs at varying frequencies. The output of a linear system to a sinusoidal input is a sinusoid of the same frequency but with a different magnitude and phase. Any linear system can be completely described by how it changes the amplitude and phase of cosine waves passing through it. This information is called the system's frequency response.Since both the impulse response and the frequency response contain complete information about the system, there must be a one-to-one correspondence between the two. Given one, you can calculate the other.The relationship between the impulse response and the frequency response is one of the foundations of signal processing:A system's frequency response is the Fourier Transform of its impulse response Since h [ ] is the common symbol for the impulse response, H [ ] is used for the frequency response. Discrete Time Fourier Transform:


The discrete-time Fourier transform (DTFT) X(ejω) of a sequence x[n] is defined

In general X(ejω) is a complex function of the real variable ω and can be written as

where Xre(ejω) and Xim(ejω) are, respectively, the real and imaginary parts of X(ejω), and are real functions of ω. X(ejω) can alternately be expressed in the form


The quantity |X(ejω)| is called the magnitude function and the quantity θ(ω) is called the phase function In many applications, the Fourier transform is called the Fourier spectrum and, likewise, |X(ejω)| and θ(ω) are referred to as the magnitude spectrum and phase spectrum, respectively. The DTFT X(ejω) is a periodic continuous function in ω with a period 2π. The DTFT satisfies a number of useful properties that are often uitilized in a number of applications. MATLAB COMMANDS: For complex Z, the magnitude R and phase angle theta are given by:

R = abs (Z) Theta = angle (Z)

Y = fft(X) returns the discrete Fourier transform of vector X, computed with a fast Fourier transform (FFT) algorithm.

Y = fft(X) Y = fft(X,n) returns the n-point FFT.

Y = fft(X, n) 01: Compute the discrete Fourier transform of the following function analytically and Then plot the magnitude and phase:

Its DTFT is given as:

MATLAB Code: w = [0:500]*pi/500; z = exp(-j*w);

x = 3*(1-0.9*z).^(-1);

a = abs(x);

b = angle(x)*180/pi;

subplot(2,1,1);

plot(w/pi,a);

subplot(2,1,2);

plot(w/pi,b);



OUTPUT:

#2: Evaluate the DTFT of the given coefficients.

num=[2 1] den=[1 –0.6]

• Plot real and imaginary parts of Fourier spectrum. • Also plot the magnitude and phase spectrum.

MATLAB CODE: % Evaluation of the DTFT title('Imaginary part of H(e^j\omega)')

clc; xlabel('\omega /\pi');

% Compute the frequency samples of the

DTFT

ylabel('Amplitude');

subplot(2,2,3)

w = -4*pi:8*pi/511:4*pi; plot(w/pi,abs(h));

num = [2 1]; grid on;

den = [1 -0.6]; title('Magnitude Spectrum |H(e^j\omega)|')

h = freqz(num, den, w); xlabel('\omega /\pi');

% Plot the DTFT ylabel('Amplitude');

subplot(2,2,1) subplot(2,2,4)

plot(w/pi,real(h)); plot(w/pi,angle(h));

grid on; grid on;

title('Real part of H(e^j\omega)') title('Phase Spectrum arg[H(e^j\omega)]')

xlabel('\omega /\pi'); xlabel('\omega /\pi');

ylabel('Amplitude'); ylabel('Phase, radians');

subplot(2,2,2)

plot(w/pi,imag(h));

grid on;

NITTE M EENAKSHI INSTITUTE OF TECHNOLOGY, BANGALORE.

33


OUTPUT

Fast Fourier Transform: The Fast Fourier Transform (FFT) is just a computationally fast way to calculate the DFT. 03: Determine the Fourier transform of the following sequence. Use the FFT (Fast Fourier Transform) function.

x (n) = 4 6 2 1 7 4 8

MATLAB Code:

grid on n = 0:6; title('Magnitude Response'); x = [4 6 2 1 7 4 8]; subplot(2,1,2); a = fft(x); plot(pha); mag = abs(a); grid on pha = angle(a); title('phase Response'); subplot(2,1,1);

plot(mag); OUTPUT:


34


Time-shifting property of DTFT:

If ‘x[n]’ be the sequence ,and X(ejw) be its fourier transform then,

x[n] ----F----- X(ejw)

then, for the time-shifted sequences, a simple transformation of the index of summation in the discrete-time Fourier transform yields

Fx[n-nd] ---- ----- e-jwnd jwX(e )

#04: Observe the time-shifting property of DTFT.

• Plot and compare the magnitude spectrum of the original input with time delayed input. • Plot and compare the phase spectrum of the original input with delayed input. MATLAB CODE: % Time-Shifting Properties of DTFT clc; w = -pi:2*pi/255:pi; D = 10; num = [1 2 3 4 5 6 7 8 9]; h1 = freqz(num, 1, w); h2 = freqz([zeros(1,D) num], 1, w); subplot(2,2,1) plot(w/pi,abs(h1)); grid on; title('Magnitude Spectrum of Original Sequence') subplot(2,2,2) plot(w/pi,abs(h2)); grid on; title('Magnitude Spectrum of Time-Shifted Sequence') subplot(2,2,3) plot(w/pi,angle(h1)); grid on; title('Phase Spectrum of Original Sequence') subplot(2,2,4) plot(w/pi,angle(h2)); grid on; title('Phase Spectrum of Time-Shifted Sequence')


35


Frequency-shifting property of the DTFT: #05: Observe the frequency shifting property of DTFT.

• Plot and compare the magnitude spectrum of the original input with frequency shifted

input. • Plot and compare the phase spectrum of the original input with frequency shifted input..

MATLAB CODE: % Frequency-Shifting Properties of DTFT clc; w = -pi:2*pi/255:pi; wo = 0.4*pi; num1 = [1 3 5 7 9 11 13 15 17]; L = length(num1); h1 = freqz(num1, 1, w); n = 0:L-1; num2 = exp(wo*i*n).*num1; h2 = freqz(num2, 1, w); subplot(2,2,1); plot(w/pi,abs(h1)); grid on; title('Magnitude Spectrum of Original Sequence') subplot(2,2,2) plot(w/pi,abs(h2)); grid on; title('Magnitude Spectrum of Frequency-Shifted Sequence') subplot(2,2,3) plot(w/pi,angle(h1)); grid on; title('Phase Spectrum of Original Sequence') subplot(2,2,4) plot(w/pi,angle(h2)); grid on; title('Phase Spectrum of Frequency-Shifted Sequence')


36


OUTPUT

Questions: 1. Determine the Fourier transform of the following sequence. Use the FFT (Fast Fourier Transform) function,

x(n) = 4 3 2 1 1 2 3 4

2. Compute the discrete Fourier transform of the following function analytically and then Plot the magnitude and phase

x(n) = 2(0.8)n+2 u(n-2)

3. Determine H(w) and plot its magnitude and phase for the following system.

Y(n) = 2x(n) + x(n-1) – 0.25y(n-1) + 0.25y(n-2) 4. Given below is the frequency response of the system,

H(ejw -jw) = 1+e

1-e-jw -j2w +0.9e

a) Write a MATLAB code that manually implements the above system. b) Compare with the spectrum obtained from freqz command

• Plot the magnitude and frequency spectrum. • Also plot real and imaginary plots


37


Z-transform and Inverse Z-transform Analysis Objective: To study the Z-transform and Inverse Z-transform practically using MATLAB. Description: In mathematics and signal processing, the Z-transform converts a discrete time-domain signal, which is a sequence of real or complex numbers, into a complex frequency-domain representation.

The Z-transform, like many other integral transforms, can be defined as either a one-sided or two-sided transform.

Bilateral Z-transform: The bilateral or two-sided Z-transform of a discrete-time signal x[n] is the function X(z) defined as

.

Unilateral Z-transform : Alternatively, in cases where x[n] is defined only for n ≥ 0, the single-sided or unilateral Z-transform is defined as

In signal processing, this definition is used when the signal is causal.

As analog filters are designed using the Laplace transform, recursive digital filters are developed with a parallel technique called the z-transform. The overall strategy of these two transforms is the same: probe the impulse response with sinusoids and exponentials to find the system's poles and zeros. The Laplace transforms deals with differential equations, the s-domain, and the s-plane. Correspondingly, the z-transform deals with difference equations, the z-domain, and the z-plane. However, the two techniques are not a mirror image of each other; the s-plane is arranged in a rectangular coordinate system, while the z-plane uses a polar format. Recursive digital filters are often designed by starting with one of the classic analog filters, such as the Butterworth, Chebyshev, or elliptic. A series of mathematical conversions are then used to obtain the desired digital filter. The Z transform of a discrete time system X[n] is defined as Power Series. Rational Z-transform to factored Z-transform: Example:

Let the given transfer function be in the rational form, 2z4 3 2+16z +44z +56z+32

G(z)= -------------------------------- 3z4 3+3z -15z2+18z-12

It is required to convert it into factored form, so that we can find the poles and zeros mathematically by applying quadratic equation. Matlab command required for converting rational form to factored form be


38

http://en.wikipedia.org/wiki/Signal_processing

http://en.wikipedia.org/wiki/Causal_system


‘Zp2sos’ The factored form of G(z) as evaluated by ‘zp2sos’ be,

G(z)= ( 0.6667 + 0.4z-1 -2 -1 -2 + 0.5333 z ) (1.000 + 2.000 z +2.000 z ) (1.000 + 2.000z-1 -2 -1 -2-4.000z )(1.000 - 1.000 z + 1.000 z )

Factored Z-transform / zeros,poles to rational Z-transform: It is the inverse of the above case, when the transfer function is given in factored form and it is required to convert in rational form then a single ‘matlab’ command can serve the purpose. Example: Lets use the above result i-e;transfer function in factored for,

G(z)=( 0.6667 + 0.4z-1 + 0.5333 z-2) (1.000 + 2.000 z-1 +2.000 z-2) (1.000 + 2.000z-1 -4.000z-2 )(1.000 - 1.000 z-1 + 1.000 z-2)

For building up transfer function in rational form we find the poles and zeros of above system simply by using matlab ‘root’ command or by hand. Or simply we have poles and zeros of the given system we can find the transfer function in factored form. Matlab command that converts poles and zeros of the system in to transfer function is ‘zp2tf’ . Rational Z-transform to partial fraction form: This technique is usually used , while taking the inverse Z-transform and when the order ‘H(z)’ is high so that it is quite difficult to solve it mathematically. Example: Consider the transfer function in the rational form i-e; 18z3

G(z)= ------------------ 18z3+3z2-4z-1 We can evaluate the partial fraction form of the above system using matlab command. The partial fraction form be,

G(z)= 0.36__ + __0.24__ + _0.4____ 1 – 0.5z-1 1+0.33 z-1 (1+0.33 z-1)

Matlab command that converts rational z-transform in to partial fraction form is ‘residuez’. Partial fraction form to Z-transform: This technique is used when it is required to convert partial fraction expression in to rational Z-transform. Example: Take the partial fraction form of above ,

G(z)= 0.36__ + __0.24__ + _0.4____ 1 – 0.5z-1 1+0.33 z-1 (1+0.33 z-1)

Matlab command that converts partial fraction form into rational z-transform is ‘residuez’

Zplane: Zero-pole plot zplane(b,a) This function displays the poles and zeros of discrete-time systems.


39


Pole Zero Diagrams For A Function In Z Domain: Z plane command computes and display the pole-zero diagram of Z function. The Command is Zplane(b,a) To display the pole value, use root(a) To display the zero value, use root(b)

Matlab Code: b=[0 1 1 ] a= [1 -2 +3] roots(a) roots(b) zplane(b,a); ans = 1.0000 + 1.4142i 1.0000 - 1.4142i ans= -1

Frequency Response: The Freqz function computes and display the frequency response of given Z- Transform of the function freqz(b,a,Fs) b= Coeff. Of Numerator, a= Coeff. Of Denominator, Fs= Sampling Frequency Matlab Code: b=[2 5 9 5 3] a= [5 45 2 1 1] freqz(b,a);


40


Example: Plot the magnitude and phase of the frequency response of the given digital filter Using freqz function: y(n) = 0.2x(n) + 0.52y(n-1) – 0.68(y(n-2) Matlab Code: b = [0.2]; a= [1, -0.52, 0.68]; w = [0:1:500]*pi/500; H=freqz(b,a,w); magH = abs(H); phaH = angle(H)*180/pi; subplot(2,1,1); plot(w/pi,magH); title('Magnitude Response'); xlabel('frequency in pi units'); ylabel('H'); subplot(2,1,2); plot(w/pi,phaH); title('Phase Response'); xlabel('frequency in pi units'); ylabel('Degrees');

Example: Assume we have a transfer function in the z-domain given by

( )375.025.0375.025.01 221

1

−−=

−−= −−

−

zzz

zzzzX .

Factoring this, it’s partial fraction expansion can be found to be

( ) ( )( ) ( ) ( ) ( ) ( )5.08.0

75.08.0

5.075.05.075.021

+−

+−

=+

+−

=+−

=z

zz

zz

zcz

zczz

zzX .

The inverse z-transform of this is thus


41


( ) ( )( ) ( )( )[ ] ( )nunx nn 5.08.075.08.0 −−+= There are several MATLAB functions that could assist with calculating and analyzing these results. We can find the roots of the denominator polynomial using >> den = [1 -0.25 -0.375]; >> roots(den) ans = 0.7500 -0.5000 We can then plot the zeros and poles either with

(1) zeros and poles in column vectors >> z = [0] z = 0 >> p = [0.75; -0.5] p = 0.7500 -0.5000 >> zplane(z,p)

numerator and denominator coefficients in row vectors >> num = [0 1 0] num = 0 1 0 >> den = [1 -0.25 -0.375] den = 1.0000 -0.2500 -0.3750 >> zplane(num,den)

-1 -0.5 0 0.5 1

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Real Part

Imag

inar

y P

art

We can find the impulse response (or inverse z-transform) of the polynomial based on the power series expansion method using the “impz” function h = impz(num,den,N) where N is the number of terms or coefficients to compute >> h = impz(num,den,10)


42


h = 0 1.0000 0.2500 0.4375 0.2031 0.2148 0.1299 0.1130 0.0770 0.0616 MATLAB can also be used to find the partial fraction expansion Assume you have a polynomial of the form

( ) ( )( ) N

N

MM

zazazaazbzbzbb

zAzBzH −−−

−−−

++++++++

==L

L2

21

10

22

110

This can be turned into a partial fraction expansion of the form ( )( )( )( ) K

pzzR

pzzR

pzzR

pzzR

zAzB

Kzp

Rzp

Rzp

Rzp

RzAzB

n

n

n

n

+−

++−

+−

+−

=

+−

++−

+−

+−

= −−−−

L

L

3

3

2

2

1

1

113

31

2

21

1

1

1111

where we can get the residues (coefficients), poles, and direct terms. Thus for our example we have

( ) 21

1

2 375.025.01375.025.0 −−

−

−−=

−−=

zzz

zzzzX

so in MATLAB we would enter >> num = [0 1]; >> den = [1 -0.25 -0.375]; >> [R,P,K]=residuez(num,den) R = 0.8000 -0.8000 P = 0.7500 -0.5000


43


K = [] which corresponds to a partial fraction expansion of

( )

( ) ( )( ) ( )( )[ ] ( )nunxz

zz

zzX

nn 5.08.075.08.0

05.0

8.075.0

8.0

−−+=

++

−+

−=

We can also plot the frequency response of a particular polynomial with freqz

( )375.025.02 −−

=zzzzX we would enter For

>> num = [0 1 0]; >> den = [1 -0.25 -0.375]; >> freqz(num, den, 512)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-200

-150

-100

-50

0

Normalized Frequency (×π rad/sample)

Pha

se (d

egrees

)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-5

0

5

10

Normalized Frequency (×π rad/sample)

Mag

nitude

(dB)

Here we have used the form freqz(num,den,N) where ‘num’ are the numerator coefficients, ‘den’ are the denominator coefficients, and ‘N’ is the number of points to use in the plot which goes from 0 to π. An alternative form is to use [H,f] = freqz(num,den,N,Fs) plot(f, abs(H)) ‘num’ and ‘den’ are the same. ‘Fs’ is the sampling frequency. ‘N’ values between 0 and Fs are calculated. The response data versus frequency are stored in H. >> num = [0 1 0]; >> den = [1 -0.25 -0.375];


44


>> [H,f] = freqz(num,den,512,8000); >> plot(f,abs(H))

0 500 1000 1500 2000 2500 3000 3500 40000.5

1

1.5

2

2.5

3

Question: Task#1: Express the following z-transform in factored form , plot its poles and zeros,and then determine its ROCs.

2z4 3 2+16z +44z +56z+32 G(z)= --------------------------------

3z4 3+3z -15z2+18z-12 Task#2: Determine the partial fraction expansion of the z-transform G(z) given by

18z3

G(z)= ------------------ 18z3 2+3z -4z-1

LAPLACE TRANSFORMS

You can compute Laplace transform using the symbolic toolbox of MATLAB. MATLAB forms Laplace transform symbolically. Thus, you need to first define the variable t as a "symbol".

SYMS Command

Whenever you want to use symbolic mathematics, you must use the syms operator to alert MATLAB that you are using a symbolic variable, and that it does not have a specific value Example 1.

Solution: >> syms t >> f=5*exp(-2*t);


45


>> L=laplace(f) L= 5/(s+2)

>> ilaplace(1/s-2/(s+4)+1/(s+5))

Matlab result:

Ans : = 1-2*exp(-4*t)+exp(-5*t)

Examples: 1. You can compute Laplace transform using the symbolic toolbox of MATLAB. If you want to compute the Laplace ace transform of x(t) = t , you can use the following MATLAB program. >> f=t; >> syms f t >> f=t; >> laplace(f) ans =1/s^2 where f and t are the symbolic variables, f the function, t the time variable. 2. The inverse transform can also be computed using MATLAB. If you want to compute the inverse Laplace transform of

You can use the following command lines. >> syms F S >> F=24/(s*(s+8)); >> ilaplace(F) ans = 3-3*exp(-8*t) 3. We can also do inverse Laplace transform using partial fraction expansion, and MATLAB can help you with that. If you want to find the partial-fraction expansion of

the following MATLAB program gives you the coefficients in the expansion. You write the coefficients of the numerator and the denominator in separate vectors and MATLAB gives you the coefficients with the corresponding poles in the expansion. >> n=[0 0 4 4 4]; >> d=[1 3 2 0 0]; >> [r,p,k]=residue(n,d)


r = -3 4 -1 2

p = -2 -1 0 0

Therefore, the partial fraction expansion is:

Calculating the Laplace F(s) transform of a function f(t) is quite simple in Matlab. First you need to specify that the variable t and s are symbolic ones. This is done with the command >> syms t s Next you define the function f(t). The actual command to calculate the transform is >> F=laplace(f,t,s) To make the expression more readable one can use the commands, simplify and pretty. here is an example for the function f(t),

>> syms t s >> f=-1.25+3.5*t*exp(-2*t)+1.25*exp(-2*t); >> F=laplace(f,t,s) F = -5/4/s+7/2/(s+2)^2+5/4/(s+2) >> simplify(F) ans = (s-5)/s/(s+2)^2 >> pretty(ans)

s - 5 ----------

2s (s + 2) which corresponds to F(s),

Alternatively, one can write the function f(t) directly as part of the laplace command:

>>F2=laplace(-1.25+3.5*t*exp(-2*t)+1.25*exp(-2*t))

Inverse Laplace Transform The command one uses now is ilaplace. One also needs to define the symbols t and s. Lets calculate the inverse of the previous function F(s),

>> syms t s


>> F=(s-5)/(s*(s+2)^2); >> ilaplace(F) ans = -5/4+(7/2*t+5/4)*exp(-2*t) >> simplify(ans) ans = -5/4+7/2*t*exp(-2*t)+5/4*exp(-2*t) >> pretty(ans) - 5/4 + 7/2 t exp(-2 t) + 5/4 exp(-2 t)

Which corresponds to f(t)

Alternatively one can write

>> ilaplace((s-5)/(s*(s+2)^2)) Here is another example.


>> F=10*(s+2)/(s*(s^2+4*s+5)); >> ilaplace(F) ans = -4*exp(-2*t)*cos(t)+2*exp(-2*t)*sin(t)+4

Which gives f(t),

making use of the trigonometric relationship,

Matlab often gives the inverse Laplace Transform in terms of sinhx and coshx. Using the following definition one can rewrite the hyperbolic expression as a function of exponentials:

Also, you may find the “Heaviside(t) function which corresponds to the unit step function u(t): thus the function H(t) = heaviside(t) =0 for t<0 and H(t) = heaviside(t)=1 for t>0. As an example, suppose that Matlab gives you the following result for the inverse Laplace transform: 2 heaviside(t-10) exp(-5/2t+25) sinh(1/2t-5) This can be re-written, using the definition of the sinh(x) function:


This last expression is closer to what your hand calculations will give you for the inverse Laplace Transform. SIMULINK: • To begin your SIMULINK session open first MATLAB ICON by clicking mouse twice and

then type

»simulink

You will now see the Simulink block library.

• Browse through block libraries. E.g., if you click Continuous, you will see the following:



Observe the description of the integrator beside the block that will be used for integrator, 1s . This

operator notation comes from linear systems domain, where s is Laplace variable. Roughly, s

corresponds to derivative operator,


ddt , and its inverse

1s , to integration, z.

• Math Operations library is shown below

After browsing through other block libraries, we are now ready to start generating a simple

Simulink diagram.

• Choose in menu selection File, then New and Model. This file will be called untitled.


Untitled file, where the SIMULINK configuration is constructed using different blocks

in the library. With a mouse you can configure your model

into the empty space. EXAMPLE 1: Input signal is sin t . It is fed into an amplifier with gain 2. Simulate the output of the amplifier. Solution: Pick first the amplifier. This you can find under Math and it is called Gain. Move the cursor on top of Gain, keep it down and move the Gain block to untitled file and release it. If you fail, try again. The result is shown below.

The gain has only one parameter, which has value of 1. You can change it by moving the cursor close to 1 and clicking once and then operating as you would with any word processing system. Once finished, click OK. The numerical value inside the block will now change to 2. Next find the input signal block. This is under Sources and is called Sine Wave. Again move the cursor on top of it, keep pressing the mouse while you move the block to the untitled file and then release the mouse. In order to see the result, you need to install a sink from Sinks library. In the beginning, the easiest sink to use is scope. Move that block in the same way as the others to the untitled file. The result is shown below.



The only thing missing of the system is to connect the peaces together. This is done with the mouse. Take the cursor to the output of Sine Wave block. You’ll see a hairline cursor, when you are close enough. Now press the mouse down. Keep it down and move it close to the input of gain. You’ll see a line forming, while you drag your mouse. Once you reach the input another hairline cursor can be seen and you can release the mouse.

Your simulation system is now complete. Before simulation you should check that the parameter values in the Sine Wave block are correct. Open it by placing the cursor on top and click twice.



For this example Amplitude is 1 and frequency (rad/s) is also 1, so default values are OK. There is no phase shift and sampling is not issue here. You can begin simulation by choosing Start simulation from Simulation menu or by clicking the start button.

Simulation menu

Start button

EXERCISE 2: Solve the differential equation

2 1, 0

( 0 ) 0 .

d x x td tx

= − + >

=

.

Note: The input is 1 after t> 0. This can be taken as a step function from the Sources block library. Note however, that the stepping time is not t=0 but t=1. Solution: If you are inexperienced with differential equations, you can use differential operator D.

Let dDdt

= . Then the differential equation becomes

(1) 2Dx x= − +1

1or

2 ( 2)Dx x D x+ = + = .

This further leads to NITTE M EENAKSHI INSTITUTE OF TECHNOLOGY, BANGALORE. 53


12

xD

=+

(2)

You can use either (1) or (2) for SIMULINK configuration. We will use the first one:

2 1Dx x= − + . 1s

Laplace transform operator s is almost the same as D, except for the initial conditions. In SIMULINK

means integration (see the block below). Input to the integrator is dxDxdt

= and output x. Thus in

configuration you set up the right hand side and connect the everything to the input of the integrator.

The solution is found by choosing under Simulation menu Start simulation or by clicking Start button. The result is shown above. If you study the equation and the result, it is clear (why?) that we have made a mistake in configuration. The sign in summation is wrong. Correct it as shown below. Now the result corresponds to our expectations (if any).



If you want to see both the input and output at the same time, use Mux (multiplexer) block, which you can find under Signals and Systems block library. Set up the system as shown below.

Double-click

Initial condition

As seen above you can change the initial value to e.g. –3. The default time for simulation is 10. If you wish to simulate longer you have to change it. To do that open Simulation menu and choose Simulation parameters.



You can change the Stop time according to your needs. Complete the exercise. Run the simulation from SIMULINK and also from Command window.

DAMPED OSCILLATOR : Solve the damped oscillator problem

2

2 5 9 (d x dx )x u tdt dt

+ + =

(0) 2

(0) 2

dx xdtx

= = −

=

&

Assume that u(t) = 0, that is, there is no input.

PURPOSE: To illustrate how to configure a SIMULINK diagram for a higher order differential equation and how to introduce initial conditions into it. SOLUTION: Solve equation first with respect to the highest order derivative to obtain

2

2 5 9d x dx xdt dt

= − −

To set up the right-hand side two integrators are needed:

d xdt

2

2

dxdtThe input to the first integraror is the second derivative and its output is . The latter is the innput to the

second integrator producing x(t) at its output. In this way we have constructed the left-hand side of the equation.



d xdt

2

2Since the second derivative is equal to the right hand side, we collect it term by term. In order to do that we

need dxdt from the output of the first integrator, x(t) from the output of the second integrator and u(t), the step input

must be generated. Here x(t) must also be multiplied by 9, so a gain is required. All these items are to be summed up so a sum block is also needed. The final configuration is given below. The initial values are added to the integrators. The resulting configuration is given below.

Next, set up the initial conditions by clicking the integrators one at a time and making appropriate changes.

dxdtThe solution x(t) and are shown in Fig. below. The first figure is SIMULINK scope and the

second is the result from Command window simulation.



The sharpness of the lower curve around t = 0.4 s is not real, it should be smooth. First you might suspect numerical difficulties (there are none) due to too large a step size. This is not the case. It is due to display graphics, i.e., not enough points have been saved to have a smooth presentation.

dxdtThe damping factor can be changed by changing the coefficient 5 in front of . If the coefficient

is zero (no damping), the result is a sinusoidal. Increasing the damping will result in damping oscillations. Complete the study to obtain the following responses.

Let us also plot a phase plane plot (x vs dx/dt). Note that here time has been eliminated. To see the effect better, start with less damping. Change the coefficient 5 to 1.

The result is shown below.



XY Graph does not adjust the scales automatically. In order to see the whole picture, click the XY Graph open and adjust the scales. Adjusting also the Sample time results in smooth picture.

CONCLUSION: A stable system. It converges towards the origin. Physical interpretation: In origin both position and velocity are zero.



REVIEW:

1. Finding the roots of a polynomial using ‘roots’ »a=[1 10 35 50 24]; »r=roots(a) r = -4.0000 -3.0000 -2.0000 -1.0000

2. Multiplying Polynomials using ‘conv’ »a=[1 2 1];b=[1 4 3]; »c=conv(a,b) c = 1 6 12 10 3

3. Adding Polynomials. Only polynomials with the same length can be added together. m=length(x);n=length(y); if m>=n z=x+[zeros(1,m-n),y]; else z=y+[zeros(1,n-m),x]; end

4. Evaluating Polynomials using ‘polyval’ »a=[1 2 1]; »polyval(a,[1:3]) ans = 4 9 16

5. For typical systems the transfer function can be expressed as a rational function, such as

61161

)3)(2)(1(1

)()()( 23

22

++++

=+++

+==

ssss

ssss

sAsBsH

6. Partial Fraction Expansion using ‘residue’ »b=[1 0 1]; % B(s) »a=[1 6 11 6]; % A(s) »[gamma,alpha,k]=residue(b,a)

gamma =

5.0000 -5.0000 1.0000 alpha = -3.0000 -2.0000 -1.0000 k = []

7. Inverse Laplace Transform using ‘ilaplace’ (this was not introduced in EE 311)

» syms F s » F=(s^2+1)/(s^3+6*s^2+11*s+6); »ilaplace(F) ans = 5*exp(-3*t)-5*exp(-2*t)+exp(-t)

8. Plotting Complex Frequency Response of H(s) using ‘freqs’

» b=[1 0 1];



» a=[1 6 11 6]; » freqs(b,a) % The output of this is seen in Figure One below.

1 0 0 1 0 1-1 0 0

-5 0

0

5 0

1 0 0

F re q u e n c y ( ra d / s )

Pha

se (d

egre

es)

1 0 0 1 0 11 0 - 4

1 0 - 2

1 0 0

F re q u e n c y ( ra d / s )

Mag

nitu

de

Figure One Output of ‘freqs’ 9. Create Bode Diagrams using ‘bode’ (frequency response with dB magnitude plot)

» sys=tf([1 0 1],[1 6 11 6]); » bode(sys) % This is shown in Figure Two below.

B o d e D i a g r a m

F r e q u e n c y ( r a d / s e c )

Pha

se (d

eg)

Mag

nitu

de (d

B)

- 2 0 0

- 1 5 0

- 1 0 0

- 5 0

0

1 0- 1

1 00

1 01

1 02

- 9 0

- 4 5

0

4 5

9 0

Figure Two Bode plot of the same system from Figure One

SIMULINK model : Now we wish to use SIMULINK to simulate the system, e.g., see the output when the input is a step function (see the Figure Three below). SIMULINK is a powerful simulation tool provided by MATLAB. It allows for analysis/simulation of interconnections of dynamic systems (both continuous-time and discrete-time).


s +12

s +6s +11s+63 2

T ransfer Fcn

sim out

T o Workspace

Step Scope

Figure Three SIMULINK diagram for the step response of the system used in Figures One and Two


Root Locus Closed-Loop Poles

The root locus of an (open-loop) transfer function H(s) is a plot of the locations (locus) of all possible closed loop poles with proportional gain k and unity feedback:

)(1)(

)()(

sKHsKH

sRsY

+=The closed-loop transfer function is:

and thus the poles of the closed loop system are values of s such that 1 + K H(s) = 0.

If we write H(s) = b(s)/a(s), then this equation has the form:

Let n = order of a(s) and m = order of b(s) [the order of a polynomial is the highest power of s that appears in it].

We will consider all positive values of k. In the limit as k -> 0, the poles of the closed-loop system are a(s) = 0 or the poles of H(s). In the limit as k -> infinity, the poles of the closed-loop system are b(s) = 0 or the zeros of H(s).

No matter what we pick k to be, the closed-loop system must always have n poles, where n is the number of poles of H(s). The root locus must have n branches, each branch starts at a pole of H(s) and goes to a zero of H(s). If H(s) has more poles than zeros (as is often the case), m < n and we say that H(s) has zeros at infinity. In this case, the limit of H(s) as s -> infinity is zero. The number of zeros at infinity is n-m, the number of poles minus the number of zeros, and is the number of branches of the root locus that go to infinity (asymptotes).

Since the root locus is actually the locations of all possible closed loop poles, from the root locus we can select a gain such that our closed-loop system will perform the way we want. If any of the selected poles are on the right half plane, the closed-loop system will be unstable. The poles that are closest to the imaginary axis have the greatest influence on the closed-loop response, so even though the system has three or four poles, it may still act like a second or even first order system depending on the location(s) of the dominant pole(s).

Plotting the root locus of a transfer function

Consider an open loop system which has a transfer function of


62


How do we design a feed-back controller for the system by using the root locus method? Say our design criteria are 5% overshoot and 1 second rise time. Make a MATLAB file called rl.m. Enter the transfer function, and the command to plot the root locus:

num=[1 7]; den=conv(conv([1 0],[1 5]),conv([1 15],[1 20])); sys=tf(num,den); rlocus(sys) axis([-22 3 -15 15])

Choosing a value of K from the root locus

The plot above shows all possible closed-loop pole locations for a pure proportional controller. Obviously not all of those closed-loop poles will satisfy our design criteria. To determine what part of the locus is acceptable, we can use the command sgrid(Zeta,Wn) to plot lines of constant damping ratio and natural frequency. Its two arguments are the damping ratio (Zeta) and natural frequency (Wn) [these may be vectors if you want to look at a range of acceptable values]. In our problem, we need an overshoot less than 5% (which means a damping ratio Zeta of greater than 0.7) and a rise time of 1 second (which means a natural frequency Wn greater than 1.8). Enter in the MATLAB command window:

zeta=0.7; Wn=1.8; sgrid(zeta, Wn)

On the plot above, the two dotted lines at about a 45 degree angle indicate pole locations with Zeta = 0.7; in between these lines, the poles will have Zeta > 0.7 and outside of the lines Zeta < 0.7. The semicircle indicates pole locations with a natural frequency Wn = 1.8; inside the circle, Wn < 1.8 and outside the circle Wn > 1.8.


http://www.library.cmu.edu/ctms/ctms/extras/mfile.htm

http://www.library.cmu.edu/ctms/ctms/extras/plot.htm#axis


Going back to our problem, to make the overshoot less than 5%, the poles have to be in between the two white dotted lines, and to make the rise time shorter than 1 second, the poles have to be outside of the white dotted semicircle. So now we know only the part of the locus outside of the semicircle and in between the two lines are acceptable. All the poles in this location are in the left-half plane, so the closed-loop system will be stable.

From the plot above we see that there is part of the root locus inside the desired region. So in this case we need only a proportional controller to move the poles to the desired region. You can use

command in MATLAB to choose the desired poles on the locus: rlocfind

[k,poles] = rlocfind(sys) Click on the plot the point where you want the closed-loop pole to be. You may want to select the points indicated in the plot below to satisfy the design criteria.

Note that since the root locus may has more than one branch, when you select a pole, you may want to find out where the other pole (poles) are. Remember they will affect the response too. From the plot above we see that all the poles selected (all the "+" signs) are at reasonable positions. We can go ahead and use the chosen k as our proportional controller.

Closed-loop response : In order to find the step response, you need to know the closed-loop transfer function. You could compute this using the rules of block diagrams, or let MATLAB do it for you

sys_cl= feedback(k*sys,1)

The two arguments to the function feedback are the numerator and denominator of the open-loop system. You need to include the proportional gain that you have chosen. Unity feedback is assumed.

If you have a non-unity feedback situation, look at the help file for the MATLAB function , which can find the closed-loop transfer function with a gain in the feedback loop. feedback

Check out the step response of your closed-loop system: step(sys_cl)

As we expected, this response has an overshoot less than 5% and a rise time less than 1 second. NITTE M EENAKSHI INSTITUTE OF TECHNOLOGY, BANGALORE. 64

http://www.library.cmu.edu/ctms/ctms/extras/step.htm


PID Controller

Introduction This tutorial will show you the characteristics of the each of proportional (P), the integral (I), and the derivative (D) controls, and how to use them to obtain a desired response. In this tutorial, we will consider the following unity feedback system:

Plant: A system to be controlled Controller: Provides the excitation for the plant; Designed to control the overall system behavior

The three-term controller The transfer function of the PID controller looks like the following:

• Kp = Proportional gain • KI = Integral gain • Kd = Derivative gain

First, let's take a look at how the PID controller works in a closed-loop system using the schematic shown above. The variable (e) represents the tracking error, the difference between the desired input value (R) and the actual output (Y). This error signal (e) will be sent to the PID controller, and the controller computes both the derivative and the integral of this error signal. The signal (u) just past the controller is now equal to the proportional gain (Kp) times the magnitude of the error plus the integral gain (Ki) times the integral of the error plus the derivative gain (Kd) times the derivative of the error.

This signal (u) will be sent to the plant, and the new output (Y) will be obtained. This new output (Y) will be sent back to the sensor again to find the new error signal (e). The controller takes this new error signal and computes its derivative and its integral again. This process goes on and on.



The characteristics of P, I, and D controllers

A proportional controller (Kp) will have the effect of reducing the rise time and will reduce but never eliminate the steady-state error. An integral control (Ki) will have the effect of eliminating the steady-state error, but it may make the transient response worse. A derivative control (Kd) will have the effect of increasing the stability of the system, reducing the overshoot, and improving the transient response. Effects of each of controllers Kp, Kd, and Ki on a closed-loop system are summarized in the table shown below.

CL RESPONSE RISE TIME OVERSHOOT SETTLING TIME S-S ERRORKp Decrease Increase Small Change Decrease Ki Decrease Increase Increase Eliminate Kd Small Change Decrease Decrease Small Change

Note that these correlations may not be exactly accurate, because Kp, Ki, and Kd are dependent on each other. In fact, changing one of these variab can change the effect of the other two. For this reason, the table should only be used as a reference when you are determining the values for Ki, Kp and Kd.

Example Problem

Suppose we have a simple mass, spring, and damper problem.

The modeling equation of this system is

(1)

Taking the Laplace transform of the modeling equation (1), we get

The transfer function between the displacement X(s) and the input F(s) then becomes


http://www.library.cmu.edu/ctms/ctms/extras/ess/ess.htm


Let

• M = 1kg • b = 10 N.s/m • k = 20 N/m • F(s) = 1

Plug these values into the above transfer function

The goal of this problem is to show you how each of Kp, Ki and Kd contributes to obtain

• Fast rise time • Minimum overshoot • No steady-state error

Open-loop step response

Let's first view the open-loop step response. Create a new m-file and add in the following code:

num=1; den=[1 10 20]; plant=tf(num,den); step(plant)

Running this m-file in the MATLAB command window should give you the plot shown below.

The DC gain of the plant transfer function is 1/20, so 0.05 is the final value of the output to an unit step input. This corresponds to the steady-state error of 0.95, quite large indeed. Furthermore, the rise time is about one second, and the settling time is about 1.5 seconds. Let's design a controller that will reduce the rise time, reduce the settling time, and eliminates the steady-state error.

Proportional control From the table shown above, we see that the proportional controller (Kp) reduces the rise time, increases the overshoot, and reduces the steady-state error. The closed-loop transfer function of the above system with a proportional controller is:





Let the proportional gain (Kp) equal 300 and change the m-file to the following:

Kp=300; contr=Kp; sys_cl=feedback(contr*plant,1); t=0:0.01:2; step(sys_cl,t)

Running this m-file in the MATLAB command window should give you the following plot.

Note: The MATLAB function called feedback was used to obtain a closed-loop transfer function directly from the open-loop transfer function (instead of computing closed-loop transfer function by hand).

The above plot shows that the proportional controller reduced both the rise time and the steady-state error, increased the overshoot, and decreased the settling time by small amount.

Proportional-Derivative control

Now, let's take a look at a PD control. From the table shown above, we see that the derivative controller (Kd) reduces both the overshoot and the settling time. The closed-loop transfer function of the given system with a PD controller is:

Let Kp equal 300 as before and let Kd equal 10. Enter the following commands into an m-file and run it in the MATLAB command window.

Kp=300; Kd=10;

NITTE M EENAKSHI INSTITUTE OF TECHNOLOGY, BANGALORE. 68contr=tf([Kd Kp],1);




sys_cl=feedback(contr*plant,1); t=0:0.01:2; step(sys_cl,t)

This plot shows that the derivative controller reduced both the overshoot and the settling time, and had a small effect on the rise time and the steady-state error.

Proportional-Integral control

Before going into a PID control, let's take a look at a PI control. From the table, we see that an integral controller (Ki) decreases the rise time, increases both the overshoot and the settling time, and eliminates the steady-state error. For the given system, the closed-loop transfer function with a PI control is:

Let's reduce the Kp to 30, and let Ki equal 70. Create an new m-file and enter the following commands.

Kp=30; Ki=70; contr=tf([Kp Ki],[1 0]); sys_cl=feedback(contr*plant,1); t=0:0.01:2; step(sys_cl,t)

Run this m-file in the MATLAB command window, and you should get the following plot.






We have reduced the proportional gain (Kp) because the integral controller also reduces the rise time and increases the overshoot as the proportional controller does (double effect). The above response shows that the integral controller eliminated the steady-state error.

Proportional-Integral-Derivative control

Now, let's take a look at a PID controller. The closed-loop transfer function of the given system with a PID controller is:

After several trial and error runs, the gains Kp=350, Ki=300, and Kd=50 provided the desired response. To confirm, enter the following commands to an m-file and run it in the command window. You should get the following step response.

Kp=350; Ki=300; Kd=50; contr=tf([Kd Kp Ki],[1 0]); sys_cl=feedback(contr*plant,1); t=0:0.01:2; step(sys_cl,t)




Now, we have obtained a closed-loop system with no overshoot, fast rise time, and no steady-state error.

General tips for designing a PID controller When you are designing a PID controller for a given system, follow the steps shown below to obtain a desired response.

1. Obtain an open-loop response and determine what needs to be improved 2. Add a proportional control to improve the rise time 3. Add a derivative control to improve the overshoot 4. Add an integral control to eliminate the steady-state error 5. Adjust each of Kp, Ki, and Kd until you obtain a desired overall response. You can always

refer to the table shown in this "PID Tutorial" page to find out which controller controls what characteristics.

Lastly, please keep in mind that you do not need to implement all three controllers (proportional, derivative, and integral) into a single system, if not necessary. For example, if a PI controller gives a good enough response (like the above example), then you don't need to implement a derivative controller on the system. Keep the controller as simple as possible. Code: % This Matlab code simulates a PID controller for a fisrt order time delay system % G(s) = (K*e(-theta*s))/(T*s+1) clear all; clc; T=360; K=14.9; theta= 80; % System Parameters refe = 10; % Reference to follow Tf = 1000; % Simulation time yold = 0;yold1=0; yp = []; ys_prime = []; er = refe; % Error (Initial error = Reference) er1 = 0; % First derivative of error er2 = 0; % Second derivative of error eold = refe; eold2 = 0; dt = 1; for i=1:dt:Tf dtspan = [i i+dt]; eold2 = eold ; eold = er; er = refe - yold; er2 = er + eold2 - 2*eold; er1 = er - eold; init_con = [yold ; (yold-yold1)]; % Initial conditions for the diffirential equations options = []; [t,y] = ode45(@pid_ctrl,dtspan,init_con,options,er,er1,er2); yold1 = yold; ys = y(length(y),1); if i <= theta ys_prime = [0 ys_prime]; else ys_prime = [ys ys_prime]; end

NITTE M EENAKSHI INSTITUTE OF TECHNOLOGY, BANGALORE. 71 yold = ys_prime(1);


yp = [yp yold]; end plot(yp); xlabel('Time'); ylabel('Output'); title('Output of the system with PID Controller'); grid on;

pid_ctrl.m function yprime = pid_ctrl(t,y,er,er1,er2) T=360; K=14.9; % System parametrs Kp = 0.5; Ki = 0.05; Kd = 3; % PID Controller parametrs yprime = [y(2); ((1/T)*(-y(2)+ K*Kd*er2 + K*Kp*er1 + K*Ki*er))];

Bode Plots As noted above, a Bode plot is the representation of the magnitude and phase of G(j*w) (where the frequency vector w contains only positive frequencies). To see the Bode plot of a transfer function, you can use the MATLAB command. For example, bode

num = 50; den = [1 9 30 40]; sys = tf(num,den); bode(sys)

displays the Bode plots for the transfer function:

50 ----------------------- s^3 + 9 s^2 + 30 s + 40

Please note the axes of the figure. The frequency is on a logarithmic scale, the phase is given in degrees, and the magnitude is given as the gain in decibels.

Note: a decibel is defined as 20*log10 ( |G(j*w| )

Gain and Phase Margin Let's say that we have the following system:



where K is a variable (constant) gain and G(s) is the plant under consideration. The gain margin is defined as the change in open loop gain required to make the system unstable. Systems with greater gain margins can withstand greater changes in system parameters before becoming unstable in closed loop.

Keep in mind that unity gain in magnitude is equal to a gain of zero in dB.

The phase margin is defined as the change in open loop phase shift required to make a closed loop system unstable.

The phase margin also measures the system's tolerance to time delay. If there is a time delay greater than 180/Wpc in the loop (where Wpc is the frequency where the phase shift is 180 deg), the system will become unstable in closed loop. The time delay can be thought of as an extra block in the forward path of the block diagram that adds phase to the system but has no effect the gain. That is, a time delay can be represented as a block with magnitude of 1 and phase w*time_delay (in radians/second).

For now, we won't worry about where all this comes from and will concentrate on identifying the gain and phase margins on a Bode plot.

The phase margin is the difference in phase between the phase curve and -180 deg at the point corresponding to the frequency that gives us a gain of 0dB (the gain cross over frequency, Wgc). Likewise, the gain margin is the difference between the magnitude curve and 0dB at the point corresponding to the frequency that gives us a phase of -180 deg (the phase cross over frequency, Wpc).

One nice thing about the phase margin is that you don't need to replot the Bode in order to find the new phase margin when changing the gains. If you recall, adding gain only shifts the magnitude plot up. This is the equivalent of changing the y-axis on the magnitude plot. Finding the phase margin is simply the matter of finding the new cross-over frequency and reading off the phase margin. For example, suppose you entered the command bode(sys). You will get the following bode plot:



You should see that the phase margin is about 100 degrees. Now suppose you added a gain of 100, by entering the command bode(100*sys). You should get the following plot (note we changed the axis so the scale would be the same as the plot above, your bode plot may not be exactly the same shape, depending on the scale used):

As you can see the phase plot is exactly the same as before, and the magnitude plot is shifted up by 40dB (gain of 100). The phase margin is now about -60 degrees. This same result could be achieved if the y-axis of the magnitude plot was shifted down 40dB. Try this, look at the first Bode plot, find where the curve crosses the -40dB line, and read off the phase margin. It should be about -60 degrees, the same as the second Bode plot.

We can find the gain and phase margins for a system directly, by using MATLAB. Just use the margin command. This command returns the gain and phase margins, the gain and phase cross over frequencies, and a graphical representation of these on the Bode plot. Let's check it out:

margin(sys)



Bandwidth Frequency The bandwidth frequency is defined as the frequency at which the closed-loop magnitude response is equal to -3 dB. However, when we design via frequency response, we are interested in predicting the closed-loop behavior from the open-loop response. Therefore, we will use a second-order system approximation and say that the bandwidth frequency equals the frequency at which the open-loop magnitude response is between -6 and - 7.5dB, assuming the open loop phase response is between -135 deg and -225 deg. For a complete derivation of this approximation, consult your textbook.

In order to illustrate the importance of the bandwidth frequency, we will show how the output changes with different input frequencies. We will find that sinusoidal inputs with frequency less than Wbw (the bandwidth frequency) are tracked "reasonably well" by the system. Sinusoidal inputs with frequency greater than Wbw are attenuated (in magnitude) by a factor of 0.707 or greater (and are also shifted in phase).

Let's say that we have the following closed-loop transfer function representing a system:

1 --------------- s^2 + 0.5 s + 1

First of all, let's find the bandwidth frequency by looking at the Bode plot: num = 1; den = [1 0.5 1]; sys = tf(num,den); bode (sys)

Since this is the closed-loop transfer function, our bandwidth frequency will be the frequency corresponding to a gain of -3 dB. looking at the plot, we find that it is approximately 1.4 rad/s. We can also read off the plot that for an input frequency of 0.3 radians, the output sinusoid should have a magnitude about one and the phase should be shifted by perhaps a few degrees (behind the input). For an input frequency of 3 rad/sec, the output magnitude should be about -20dB (or 1/10 as large as the input) and the phase should be nearly -180 (almost exactly out-of-phase). We can use the


lsim command to simulate the response of the system to sinusoidal inputs.

First, consider a sinusoidal input with a frequency lower than Wbw. We must also keep in mind that we want to view the steady state response. Therefore, we will modify the axes in order to see the steady state response clearly (ignoring the transient response).

w = 0.3; num = 1;

http://www.library.cmu.edu/ctms/ctms/extras/lsim.htm


den = [1 0.5 1]; sys = tf(num,den); t = 0:0.1:100; u = sin(w*t); [y,t] = lsim(sys,u,t); plot(t,y,t,u) axis([50,100,-2,2])

Note that the output (blue) tracks the input (purple) fairly well; it is perhaps a few degrees behind the input as expected.

However, if we set the frequency of the input higher than the bandwidth frequency for the system, we get a very distorted response (with respect to the input):

w = 3; num = 1; den = [1 0.5 1]; sys = tf(num,den); t = 0:0.1:100; u = sin(w*t); [y,t] = lsim(sys,u,t); plot(t,y,t,u) axis([90, 100, -1, 1])

Again, note that the magnitude is about 1/10 that of the input, as predicted, and that it is almost exactly out of phase (180 degrees behind) the input. Feel free to experiment and view the response for several different frequencies w, and see if they match the Bode plot.

Closed-loop performance In order to predict closed-loop performance from open-loop frequency response, we need to have several concepts clear:

• The system must be stable in open loop if we are going to design via Bode plots. • If the gain cross over frequency


is less than the phase cross over frequency (i.e. Wgc < Wpc), then the closed-loop system will be stable.

http://www.library.cmu.edu/ctms/ctms/extras/plot.htm

http://www.library.cmu.edu/ctms/ctms/extras/plot.htm#axis

http://www.library.cmu.edu/ctms/ctms/freq/freq.htm#Wgc

http://www.library.cmu.edu/ctms/ctms/freq/freq.htm#Wgc


• For second-order systems, the closed-loop damping ratio is approximately equal to the phase margin divided by 100 if the phase margin is between 0 and 60 deg. We can use this concept with caution if the phase margin is greater than 60 deg.

• A very rough estimate that you can use is that the bandwidth is approximately equal to the natural frequency.

Let's use these concepts to design a controller for the following system:

Where Gc(s) is the controller and G(s) is:

10 ---------- 1.25s + 1

The design must meet the following specifications:

• Zero steady state error. • Maximum overshoot must be less than 40%. • Settling time must be less than 2 secs.

There are two ways of solving this problem: one is graphical and the other is numerical. Within MATLAB, the graphical approach is best, so that is the approach we will use. First, let's look at the Bode plot. Create an m-file with the following code:

num = 10; den = [1.25,1]; sys = tf(num,den); bode(sys)

There are several several characteristics of the system that can be read directly from this Bode plot. First of all, we can see that the bandwidth frequency is around 10 rad/sec. Since the bandwidth frequency is roughly the same as the natural frequency (for a first order system of this type), the rise time is 1.8/BW=1.8/10=1.8 seconds. This is a rough estimate, so we will say the rise time is about 2 seconds.

The phase margin for this system is approximately 95 degrees. The relation damping ratio = pm/100 only holds for PM < 60/ Since the system is first-order, there should be no overshoot.



The last major point of interest is steady-state error. The steady-state error can be read directly off the Bode plot as well. The constant (Kp, Kv, or Ka) is found from the intersection of the low frequency asymptote with the w=1 line. Just extend the low frequency line to the w=1 line. The magnitude at this point is the constant. Since the Bode plot of this system is a horizontal line at low frequencies (slope = 0), we know this system is of type zero. Therefore, the intersection is easy to find. The gain is 20dB (magnitude 10). What this means is that the constant for the error function it 10. The steady-state error is 1/(1+Kp)=1/(1+10)=0.091. If our system was type one instead of type zero, the constant for the steady-state error would be found in a manner similar to the following

Let's check our predictions by looking at a step response plot. This can be done by adding the following two lines of code into the MATLAB command window.

sys_cl = feedback(sys,1); step(sys_cl)

As you can see, our predictions were very good. The system has a rise time of about 2 seconds, has no overshoot, and has a steady-state error of about 9%. Now we need to choose a controller that will allow us to meet the design criteria. We choose a PI controller because it will yield zero steady state error for a step input. Also, the PI controller has a zero, which we can place. This gives us additional design flexibility to help us meet our criteria. Recall that a PI controller is given by:


http://www.library.cmu.edu/ctms/ctms/extras/ess/ess.htm



K*(s+a) Gc(s) = -------

s The first thing we need to find is the damping ratio corresponding to a percent overshoot of 40%. Plugging in this value into the equation relating overshoot and damping ratio (or consulting a plot of this relation), we find that the damping ratio corresponding to this overshoot is approximately 0.28. Therefore, our phase margin should be at least 30 degrees. We must have a bandwidth frequency greater than or equal to 12 if we want our settling time to be less than 1.75 seconds which meets the design specs.

Now that we know our desired phase margin and bandwidth frequency, we can start our design. Remember that we are looking at the open-loop Bode plots. Therefore, our bandwidth frequency will be the frequency corresponding to a gain of approximately -7 dB.

Let's see how the integrator portion of the PI or affects our response. Change your m-file to look like the following (this adds an integral term but no proportional term):

num = 10; den = [1.25 1]; plant = tf(num,den); numPI = 1; denPI = [1 0]; contr = tf(numPI,denPI); bode(contr * plant, logspace(0,2))

Our phase margin and bandwidth frequency are too small. We will add gain and phase with a zero. Let's place the zero at 1 for now and see what happens. Change your m-file to look like the following:

num = 10; den = [1.25 1]; plant = tf(num,den); numPI = [1 1]; denPI = [1 0]; contr = tf(numPI,denPI); bode(contr * plant, logspace(0,2))



It turns out that the zero at 1 with a unit gain gives us a satisfactory answer. Our phase margin is greater than 60 degrees (even less overshoot than expected) and our bandwidth frequency is approximately 11 rad/s, which will give us a satisfactory response. Although satisfactory, the response is not quite as good as we would like. Therefore, let's try to get a higher bandwidth frequency without changing the phase margin too much. Let's try to increase the gain to 5 and see what happens. This will make the gain shift and the phase will remain the same.

num = 10; den = [1.25 1]; plant = tf(num,den); numPI = 5*[1 1]; denPI = [1 0]; contr = tf(numPI,denPI); bode(contr * plant, logspace(0,2))

That looks really good. Let's look at our step response and verify our results. Add the following two lines to your m-file:

sys_cl = feedback(contr * plant,1); step(sys_cl)



Sample Bode Plots

Let's say that we have the following system:

We can view the open loop Bode plot of this system by looking at the Bode plot of Gc(s)G(s). However, we can also view the Bode plots of G(s) and of Gc(s) and add them graphically. Therefore, if we know the frequency response of simple functions, we can use them to our advantage when we are designing a controller:

bode(1, [1 0]) 1 --- s

bode(1, [1 1])

1 ------ s + 1

bode([1 0], 1)



s

bode([1 1], 1)

s+1

Note that the location of the pole or zero(a) will greatly affect the frequency response. If we have a pole close to the origin, the response at lower frequencies will undergo the most drastic changes. If we have a pole or zero farther from the origin, the response at higher frequencies will undergo the most drastic changes. Also remember that a change in gain will shift the entire gain response up or down and will not affect the phase. Let's see a couple of examples:

bode(1,[ 1, 0.1]) 1

------ s + 0.1

bode(1,[ 1, 10])

1

------ s + 10



bode([1, 0.1],1)

s + 0.1

bode([1, 10],1)

s + 10

Modifying the location of these poles or zeros, we can manipulate the frequency response of the combined system to something that will suit our purposes.

The Nyquist Diagram

The Nyquist plot allows us to predict the stability and performance of a closed-loop system by observing its open-loop behavior. The Nyquist criterion can be used for design purposes regardless of open-loop stability (remember that the Bode design methods assume that the system is stable in open loop). Therefore, we use this criterion to determine closed-loop stability when the Bode plots display confusing information. NITTE M EENAKSHI INSTITUTE OF TECHNOLOGY, BANGALORE. 83


The Nyquist diagram is basically a plot of G(j* w) where G(s) is the open-loop transfer function and w is a vector of frequencies which encloses the entire right-half plane. In drawing the Nyquist diagram, both positive and negative frequencies (from zero to infinity) are taken into account. We will represent positive frequencies in red and negative frequencies in green. The frequency vector used in plotting the Nyquist diagram usually looks like this (if you can imagine the plot stretching out to infinity):

However, if we have open-loop poles or zeros on the jw axis, G(s) will not be defined at those points, and we must loop around them when we are plotting the contour. Such a contour would look as follows:

Please note that the contour loops around the pole on the jw axis. As we mentioned before, the MATLAB nyquist command does not take poles or zeros on the jw axis into account and therefore produces an incorrect plot. To correct this, please download and use nyquist1.m. If we have a pole on the jw axis, we have to use nyquist1. If there are no poles or zeros on the jw-axis, or if we have pole-zero cancellation


, we can use either the command or nyquist1.mnyquist .

The Cauchy criterion The Cauchy criterion (from complex analysis) states that when taking a closed contour in the complex plane, and mapping it through a complex function G(s), the number of times that the plot of G(s) encircles the origin is equal to the number of zeros of G(s) enclosed by the frequency contour minus the number of poles of G(s) enclosed by the frequency contour. Encirclements of the origin are counted as positive if they are in the same direction as the original closed contour or negative if they are in the opposite direction.

When studying feedback controls, we are not as interested in G(s) as in the closed-loop transfer function:

G(s) --------- 1 + G(s)

http://www.library.cmu.edu/ctms/ctms/extras/nyquist1.htm

http://www.library.cmu.edu/ctms/ctms/extras/pz.htm

http://www.library.cmu.edu/ctms/ctms/extras/nyquist1.htm


If 1+ G(s) encircles the origin, then G(s) will enclose the point -1. Since we are interested in the closed-loop stability, we want to know if there are any closed-loop poles (zeros of 1 + G(s)) in the right-half plane. More details on how to determine this will come later.

Therefore, the behavior of the Nyquist diagram around the -1 point in the real axis is very important; however, the axis on the standard nyquist diagram might make it hard to see what's happening around this point. To correct this, you can add the lnyquist.m function to your files. The lnyquist.m command plots the Nyquist diagram using a logarithmic scale and preserves the characteristics of the -1 point.

To view a simple Nyquist plot using MATLAB, we will define the following transfer function and view the Nyquist plot:

0.5 ------- s - 0.5

sys=tf(0.5,[1 -0.5]); nyquist(sys)

Now we will look at the Nyquist diagram for the following transfer function:

s + 2 ----- s^2

Note that this function has a pole at the origin. We will see the difference between using the , and commands with this particular function. nyquist, nyquist1 lnyquist

num = [1 2]; den = [1 0 0]; sys = tf(num,den); nyquist(sys)

nyquist1(sys)

lnyquist(sys)


http://www.library.cmu.edu/ctms/ctms/extras/lnyquist.htm


Note that the plot is not the correct one, the nyquist nyquist1 plot is correct, but it's hard to see what happens close to the -1 point, and the plot is correct and has an appropriate scale. lnyquist

Closed Loop Stability

Consider the negative feedback system:

Remember from the Cauchy criterion that the number N of times that the plot of G(s)H(s) encircles -1 is equal to the number Z of zeros of 1 + G(s)H(s) enclosed by the frequency contour minus the number P of poles of 1 + G(s)H(s) enclosed by the frequency contour (N = Z - P). Keeping careful track of open- and closed-loop transfer functions, as well as numerators and denominators, you should convince yourself that:

• the zeros of 1 + G(s)H(s) are the poles of the closed-loop transfer function • the poles of 1 + G(s)H(s) are the poles of the open-loop transfer function.

The Nyquist criterion then states that:

• P = the number of open-loop (unstable) poles of G(s)H(s) • N = the number of times the Nyquist diagram encircles -1 • clockwise encirclements of -1 count as positive encirclements • counter-clockwise (or anti-clockwise) encirclements of -1 count as negative encirclements

• Z = the number of right half-plane (positive, real) poles of the closed-loop system

The important equation which relates these three quantities is: Z = P + N

Note: This is only one convention for the Nyquist criterion. Another convention states that a positive N counts the counter-clockwise or anti-clockwise encirclements of -1. The P and Z variab remain the same. In this case the equation becomes Z = P - N. Throughout these tutorials, we will use a positive sign for clockwise encirclements. It is very important (and somewhat tricky) to learn how to count the number of times that the diagram encircles -1.

Another way of looking at it is to imagine you are standing on top of the -1 point and are following the diagram from beginning to end. Now ask yourself: How many times did I turn my head a full 360 degrees? Again, if the motion was clockwise, N is positive, and if the motion is anti-clockwise, N is negative.



Knowing the number of right-half plane (unstable) poles in open loop (P), and the number of encirclements of -1 made by the Nyquist diagram (N), we can determine the closed-loop stability of the system. If Z = P + N is a positive, nonzero number, the closed-loop system is unstable.

We can also use the Nyquist diagram to find the range of gains for a closed-loop unity feedback system to be stable. The system we will test looks like this:

where G(s) is :

s^2 + 10 s + 24 --------------- s^2 - 8 s + 15

This system has a gain K which can be varied in order to modify the response of the closed-loop system. However, we will see that we can only vary this gain within certain limits, since we have to make sure that our closed-loop system will be stable. This is what we will be looking for: the range of gains that will make this system stable in the closed loop.

The first thing we need to do is find the number of positive real poles in our open-loop transfer function:

roots([1 -8 15])

ans = 5 3The poles of the open-loop transfer function are both positive. Therefore, we need two anti-clockwise (N = -2) encirclements of the Nyquist diagram in order to have a stable closed-loop system (Z = P + N). If the number of encirclements is less than two or the encirclements are not anti-clockwise, our system will be unstable.

Let's look at our Nyquist diagram for a gain of 1:

sys = tf([ 1 10 24], [ 1 -8 15]); nyquist(sys)

There are two anti-clockwise encirclements of -1. Therefore, the system is stable for a gain of 1. Now we will see how the system behaves if we increase the gain to 20:

nyquist(20*sys)



The diagram expanded. Therefore, we know that the system will be stable no matter how much we increase the gain. However, if we decrease the gain, the diagram will contract and the system might become unstable. Let's see what happens for a gain of 0.5:

nyquist(0.5*sys)

The system is now unstable. By trial and error we find that this system will become unstable for gains less than 0.80. We can verify our answers by zooming in on the Nyquist plots as well as by looking at the closed-loop steps responses for gains of 0.79, 0.80, and 0.81.

Gain Margin We already defined the gain margin as the change in open-loop gain expressed in decibels (dB), required at 180 degrees of phase shift to make the system unstable. Now we are going to find out where this comes from. First of all, let's say that we have a system that is stable if there are no Nyquist encirclements of -1, such as :

50

----------------------- s^3 + 9 s^2 + 30 s + 40

Looking at the roots, we find that we have no open loop poles in the right half plane and therefore no closed-loop poles in the right half plane if there are no Nyquist encirclements of -1. Now, how much can we vary the gain before this system becomes unstable in closed loop? Let's look at the following figure:


http://www.library.cmu.edu/ctms/ctms/freq/exa1sol.htm


The open-loop system represented by this plot will become unstable in closed loop if the gain is increased past a certain boundary. The negative real axis area between -1/a (defined as the point where the 180 degree phase shift occurs...that is, where the diagram crosses the real axis) and -1 represents the amount of increase in gain that can be tolerated before closed-loop instability.

If we think about it, we realize that if the gain is equal to a, the diagram will touch the -1 point:

Therefore, we say that the gain margin is 'a' units. However, we mentioned before that the gain margin is usually measured in decibels. Hence, the gain margin is :

GM = 20*log10(a) [dB] We will now find the gain margin of the stable, open-loop transfer function we viewed before. Recall that the function is:

50

----------------------- s^3 + 9 s^2 + 30 s + 40

and that the Nyquist diagram can be viewed by typing: sys = tf(50, [1 9 30 40 ]); nyquist(sys)

As we discussed before, all that we need to do to find the gain margin is find 'a', as defined in the preceding figure. To do this, we need to find the point where there is exactly 180 degrees of phase shift. This means that the transfer function at this point is real (has no imaginary part). The numerator is already real, so we just need to look at the denominator. When s = j*w, the only terms in the denominator that will have imaginary parts are those which are odd powers of s. Therefore, for G(j*w) to be real, we must have:

-j w^3 + 30 j w = 0 which means w=0 (this is the rightmost point in the Nyquist diagram) or w=sqrt(30). We can then find the value of G(j*w) at this point using polyval:

w=sqrt(30); polyval(50,j*w)/polyval([1 9 30 40],j*w)

You should get the following output from MATLAB ans = -0.2174

The answer is: -0.2174 + 0i. The imaginary part is zero, so we know that our answer is correct. We can also verify by looking at the Nyquist plot again. The real part also makes sense. Now we can proceed to find the gain margin.

We found that the 180 degrees phase shift occurs at -0.2174 + 0i. This point was previously defined as -1/a. Therefore, we now have 'a', which is the gain margin. However, we need to express the gain margin in decibels,

-1/a = -0.2174



=> a = 4.6 => GM = 20*log10( 4.6) = 13.26 dB

We now have our gain margin. Let's see how accurate it is by using a gain of a = 4.6 and zooming in on the Nyquist plot:

a = 4.6 nyquist(a*sys)

The plot appears to go right through the -1 point. We will now verify the accuracy of our results by viewing the zoomed Nyquist diagrams and step responses for gains of 4.5, 4.6, and 4.7.

Phase Margin We have already discussed the importance of the phase margin. Therefore, we will only talk about where this concept comes from. We have defined the phase margin as the change in open-loop phase shift required at unity gain to make a closed-loop system unstable. Let's look at the following graphical definition of this concept to get a better idea of what we are talking about.

Let's analyze the previous plot and think about what is happening. From our previous example we know that this particular system will be unstable in closed loop if the Nyquist diagram encircles the -1 point. However, we must also realize that if the diagram is shifted by theta degrees, it will then touch the -1 point at the negative real axis, making the system marginally stable in closed loop. Therefore, the angle required to make this system marginally stable in closed loop is called the phase margin (measured in degrees). In order to find the point we measure this angle from, we draw a circle with radius of 1, find the point in the Nyquist diagram with a magnitude of 1 (gain of zero dB), and measure the phase shift needed for this point to be at an angle of 180 deg.


http://www.library.cmu.edu/ctms/ctms/freq/exa2sol.htm

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