Fourier Series â€¢ Fourier Transform â€¢ Laplace Transform â€¢ Applications of Laplace...

Basics in Systems and Circuits Theory

Michael E.Auer 01.11.2011 BSC04

• Fourier Series • Fourier Transform • Laplace Transform • Applications of Laplace Transform • Z-Transform

BSC Modul 4: Advanced Circuit Analysis



• The Fourier series of a periodic function f(t) is a representation that resolves f(t) into a dc component and an ac component comprising an infinite series of harmonic sinusoids.

• Given a periodic function f(t) = f(t+nT) where n is an integer and T is the period of the function.

where ω0=2π/T is called the fundamental frequency in radians per second.

ac

nn

dc

tnbtnaatf ∑∞

=

++=1

0000 )sincos()( ωω

Trigonometric Fourier Series (1)



• in alternative form of f(t)

where

• and an and bn are as follow

∫=

T

on dttntfT

a0

)cos()(2 ω

)(tan , 1n

22

n

nnnn a

bbac −−=+= φ

∫=T

on dttntfT

b0

)sin()(2 ω

ac

nnn

dc

tncatf ∑∞

=

++=1

00 )cos(()( φω

Trigonometric Fourier Series (2)

(Inverse tangent or arctangent)



)2()( and 21 ,010 ,1

)( +=

<<<<

= tftftt

tf

==

==

==

∫

∫

evenn ,0oddn,/2

)sin()(2

and 0)cos()(2

0 0

0 0

πω

ω

ndttntf

Tb

dttntfT

a

T

n

T

n

∑∞

=

−=+=1

12 ),sin(1221)(

kkntn

ntf π

π

==°−

=

==

=

evenn ,0 oddn,90

evenn ,0 oddn,/2

n

n

nA

φ

π

Truncating the series at N=11

a) Amplitude and b) Phase spectrum

Fourier Series Example

Determine the Fourier series of the waveform shown right. Obtain the amplitude and phase spectra.



Three types of symmetry

1.Even Symmetry : a function f(t) if its plot is symmetrical about the vertical axis.

In this case,

)()( tftf −=

0

)cos()(4

)(2

2/

0 0

2/

00

=

=

=

∫

∫

n

T

n

T

b

dttntfT

a

dttfT

a

ω

Typical examples of even periodic function

Symmetry Considerations (1)



2.Odd Symmetry : a function f(t) if its plot is anti-symmetrical about the vertical axis.

In this case,

)()( tftf −=−

∫=

=2/

0 0

0

)sin()(40

T

n dttntfT

b

a

ω

Typical examples of odd periodic function




3.Half-wave Symmetry : a function f(t) if

)()

2( tfTtf −=−

=

=

=

∫

∫

evenan for , 0

oddn for , )sin()(4

evenan for , 0

oddn for , )cos()(40

2/

0 0

2/

0 0

0

T

n

T

n

dttntfTb

dttntfTa

a

ω

ω

Typical examples of half-wave odd periodic functions




Example 1

Find the Fourier series expansion of f(t) given below.

∑∞

=

−=

1 2sin

2cos112)(

ntnn

ntf ππ

πAns:




Example 2

Determine the Fourier series for the half-wave cosine function as shown below.

∑∞

=

−=−=1

22 12 ,cos1421)(

kknnt

ntf

πAns:




Steps for Applying Fourier Series

1.Express the excitation as a Fourier series.

2.Transform the circuit from the time domain to the frequency domain.

3.Find the response of the dc and ac components in the Fourier series.

4.Add the individual dc and ac responses using the superposition principle.

Circuit Applications (1)



Example

Find the response v0(t) of the circuit below when the voltage source vs(t) is given by

( ) 12 ,sin1221)(

1−=+= ∑

∞

=

kntnn

tvn

s πωπ




Amplitude spectrum of the output voltage

Solution

Phasor of the circuit

For dc component, (ωn=0 or n=0), Vs = ½ => Vo = 0

For nth harmonic,

In time domain,

s0 V25

2Vπ

πnj

nj+

=

)5

2tan(c4254)(

1

1

220 ∑∞

=

−−+

=k

ntnosn

tv πππ

s22

1

0 V425

5/2tan4V ,902Vππ

π nn

nS+

−∠=°−∠=

−




Given:

The average power is

The rms value is

∑∑∞

=

∞

=

−+=−+=1

Im0mdc1

0ndc )cos(II)( and )cos(VV)(mn

Vn tmtitntv φωφω

)(1

2220 ∑

∞

=

++=n

nnrms baaF

∑∞

=

−+=1

nndcdc )cos(IV21IVP

nnn φθ

Average Power and RMS Values (1)



Example

Determine the average power supplied to the circuit shown below if i(t)=2+10cos(t+10°)+6cos(3t+35°) A

Answer: 41.5W

Average Power and RMS Values (2)



• The exponential Fourier series of a periodic function f(t) describes the spectrum of f(t) in terms of the amplitude and phase angle of ac components at positive and negative harmonic.

• The plots of magnitude and phase of cn versus nω0 are called the complex amplitude spectrum and complex phase spectrum of f(t) respectively.

∫ == −T tjnn Tdtetf

Tc

0 0 /2 where,)(10 πωω∑

∞

−∞=

=n

tjnn

oectf ω)(

Exponential Fourier Series (1)



• The complex frequency spectrum of the function

f(t)=et, 0<t<2π with f(t+2π)=f(t)

(a) Amplitude spectrum; (b) phase spectrum

Exponential Fourier Series (2)



•Filter are an important component of electronics and communications system.

•This filtering process cannot be accomplished without the Fourier series expansion of the input signal.

•For example,

(a) Input and output spectra of a lowpass filter, (b) the lowpass filter passes only the dc component when ωc << ω0

Application – Filter (1)



(a) Input and output spectra of a bandpass filter, (b) the bandpass filter passes only the dc component when Β << ω0

Application – Filter (2)



• It is an integral transformation of f(t) from the time domain to the frequency domain F(ω)

• F(ω) is a complex function; its magnitude is called the amplitude spectrum, while its phase is called the phase spectrum.

Given a function f(t), its Fourier transform denoted by F(ω), is defined by

∫∞

∞−

−= )()( dtetfF tjωω

Definition of Fourier Transform (1)



Example 1:

Determine the Fourier transform of a single rectangular pulse of wide τ and height A, as shown below.


2sin

22

2/2/

)(

2/2/

2/

2/

ωττ

ω

ττ

ω

ω

ωτωτ

ω

τ

τ

ω

cA

jeeA

ejA

dtAeF

jj

tj

tj

=

−=

−−=

=

−

−

−∫Solution:

Amplitude spectrum of the rectangular pulse



Example 2:

Obtain the Fourier transform of the “switched-on” exponential function as shown.


Solution:

ω

ω

ω

ωω

ja

dte

dteedtetfF

etuetf

tja

tjjattj

atat

+=

=

==

<>

==

∫∫∫

∞

∞−

+−

∞

∞−

−−∞

∞−

−

−−

1

)()(

Hence,0 t,00 t,

)()(

)(



[ ] )()()()( 22112211 ωω FaFatfatfaF +=+

Linearity:

If F1(ω) and F2(ω) are, respectively, the Fourier Transforms of f1(t) and f2(t)

Example:

[ ] ( ) ( )[ ] [ ])()(21)sin( 000

00 ωωδωωδπω ωω −−+=−= − jeFeFj

tF tjtj

Properties of Fourier Transform (1)



[ ] constant a is ,)(1)( aa

Fa

atfF ω=

Time Scaling:

If F (ω) is the Fourier Transforms of f (t), then

If |a|>1, frequency compression, or time expansion

If |a|<1, frequency expansion, or time compression




[ ] )()( 00 ωω FettfF tj−=−

Time Shifting:


Example:

[ ]ω

ω

jetueF

jt

+=−

−−−

1)2(

2)2(




[ ] )()( 00 ωωω −= FetfF tj

Frequency Shifting (Amplitude Modulation):


Example:

[ ] )(21)(

21)cos()( 000 ωωωωω ++−= FFttfF




)()( sFjtudtdfF ω=

Time Differentiation:

If F (ω) is the Fourier Transforms of f (t), then the Fourier Transform of its derivative is

Example:

( )ωja

tuedtdF at

+=

− 1)(




)()0()()( ωδπωω F

jFdttfF

t=

∫ ∞−

Time Integration:

If F (ω) is the Fourier Transforms of f (t), then the Fourier Transform of its integral is

Example:

[ ] )(1)( ωπδω+=

jtuF




• Fourier transforms can be applied to circuits with non-sinusoidal excitation in exactly the same way as phasor techniques being applied to circuits with sinusoidal excitations.

• By transforming the functions for the circuit elements into the frequency domain and take the Fourier transforms of the excitations, conventional circuit analysis techniques could be applied to determine unknown response in frequency domain.

• Finally, apply the inverse Fourier transform to obtain the response in the time domain.

Y(ω) = H(ω)X(ω)

Circuit Application (1)



Example:

Find v0(t) in the circuit shown below for

vi(t)=2e-3tu(t)

Circuit Application (2)

Solution:

)()(4.0)( gives ansformFourier tr inverse theTaking

)5.0)(3(1)(V

Hence,21

1)(V)(V)( iscircuit theoffunction transfer The

3

2)(V is signalinput theof ansformFourier tr The

35.00

0

i

0

i

tueetv

jj

jH

j

tt −− −=

++=

+==

+=

ωωω

ωωωω

ωω



• It is an integral transformation of f(t) from the time domain to the complex frequency domain F(s)

• Given a function f(t), its Laplace transform denoted by F(s), is defined by

• Where the parameter s is a complex number

[ ] ∫∞ −⋅==

0 )()()( dtetftfLsF st

Definition of Laplace Transform

ωσ js += σ, ω – real numbers



• When one says "the Laplace transform" without qualification, the unilateral or one-sided transform is normally intended. The Laplace transform can be alternatively defined as the bilateral Laplace transform or two-sided Laplace transform by extending the limits of integration to be the entire real axis. If that is done the common unilateral transform simply becomes a special case of the bilateral transform.

• The bilateral Laplace transform is defined as follows:

Bilateral Laplace Transform

[ ] ∫∞

∞−

−⋅== )()()( dtetftfLsF st

http://en.wikipedia.org/wiki/Two-sided_Laplace_transform



Determine the Laplace transform of each of the following functions shown:

Examples of Laplace Transforms (1)

a) The Laplace Transform of unit step, u(t) is given by

[ ] ∫∞ − ===

0

11)()(s

dtesFtuL st



b) The Laplace Transform of exponential function, e-at u(t), a>0 is given by

[ ] ∫∞ −

+===

0

1)()(α

α

sdteesFtuL stt


c) The Laplace Transform of impulse function, δ(t) is given by

[ ] ∫∞ − ===

01)()()( dtetsFtuL stδ




ssF 1)( =

α+=

ssF 1)( 1)( =sF



Table of Selected Laplace Transforms (1)



[ ] )()()()( 22112211 sFasFatfatfaL +=+

Linearity:

If F1(s) and F2(s) are, respectively, the Laplace Transforms of f1(t) and f2(t)

Example:

[ ] ( ) 22)(21)()cos(

ωω ωω

+=

+= −

sstueeLtutL tjtj

Properties of Laplace Transform (1)



[ ] )(1)(asF

aatfL =

Scaling:

If F (s) is the Laplace Transforms of f (t), then

Example:

[ ] 22 42)()2sin(

ωωω+

=s

tutL




[ ] )()()( sFeatuatfL as−=−−

Time Shift:


Example:

[ ] 22)())(cos(ω

ω+

=−− −

sseatuatL as




[ ] )()()( asFtutfeL at +=−

Frequency Shift:


Example: [ ] 22)()()cos(

ωω

+++

=−

asastuteL at




)0()()( −−=

fssFtu

dtdfL

Time Differentiation:

If F (s) is the Laplace Transforms of f (t), then the Laplace Transform of its derivative is


Time Integration:

If F (s) is the Laplace Transforms of f (t), then the Laplace Transform of its integral is

)(1)(0

sFs

dttfLt

=

∫



)(lim)0( ssFfs ∞→

=

Initial and Final Values:

The initial-value and final-value properties allow us to find f(0) and f(∞) of f(t) directly from its Laplace transform F(s).

Initial-value theorem

)(lim)(0

ssFfs→

=∞ Final-value theorem




In principle we could recover f(t) from F(s) via

But, this formula isn’t really useful.

sesFj

tfj

j

std)(21)( ∫

∞+

∞−

⋅=σ

σπ

The Inverse Laplace Transform (1)

∫= xF 1



Suppose F(s) has the general form of

The finding the inverse Laplace transform of F(s) involves two steps:

1.Decompose F(s) into simple terms using partial fraction expansion.

2.Find the inverse of each term by matching entries in Laplace Transform Table.

)()()(

sDsNsF =


numerator polynomial

denominator polynomial



Example

Find the inverse Laplace transform of

Solution: 4

61

53)( 2 ++

+−=

ssssF

0 t),()2sin(353(4

61

53)( 2111

≥+−=

++

+−

=

−

−−−

tutes

Ls

Ls

Ltf

t




• The Laplace transform is useful in solving linear integro-differential equations.

• Each term in the integro-differential equation is transformed into s-domain.

• Initial conditions are automatically taken into account.

• The resulting algebraic equation in the s-domain can then be solved easily.

• The solution is then converted back to time domain.

Application to Integro-differential Equations (1)



Example:

Use the Laplace transform to solve the differential equation

Given: v(0) = 1; v’(0) = -2

)(2)(8)(6)(2

2

tutvdt

tdvdt

tvd=++




Solution:

Taking the Laplace transform of each term in the given differential equation and obtain

[ ] [ ]

)()21(41)(

Transform, Laplace inverse By the42

)(2424)()86(

have we,2)0(';1)0( ngSubstituti

2)(8)0()(6)0(')0()(

42

41

21

412

2

2

tueetv

ssssV

sss

sssVss

vvs

sVvssVvsvsVs

tt −− ++=

++

++=⇒

++=++=++

−==

=+−+−−




Steps in Applying the Laplace Transform:

1.Transform the circuit from the time domain to the s-domain

2.Solve the circuit using nodal analysis, mesh analysis, source transformation, superposition, or any circuit analysis technique with which we are familiar

3.Take the inverse transform of the solution and thus obtain the solution in the time domain.

Circuit Element Models (1)



Assume zero initial condition for

the inductor and capacitor,

Resistor : V(s)=RI(s)

Inductor: V(s)=sLI(s)

Capacitor: V(s) = I(s)/sC

The impedance in the s-domain is defined as Z(s) = V(s)/I(s)

The admittance in the s-domain is defined as Y(s) = I(s)/V(s)

Time-domain and s-domain representations of passive elements under zero initial conditions.





Non-zero initial condition for the inductor and capacitor,

Resistor : V(s)=RI(s)

Inductor: V(s)=sLI(s) + LI(0)

Capacitor: V(s) = I(s)/sC + v(0)/s



Introductory Example

Charging of a capacitor v

V(0) = 0



Example 1:

Find v0(t) in the circuit shown below, assuming zero initial conditions.

Circuit Element Models Examples (1)



Solution:

Transform the circuit from the time domain to the s-domain:

s

L

tu

3sC1 F

31

ss H 1s1 )(

=⇒

=⇒

⇒

Apply mesh analysis, on solving for V0(s):

220 )2()4(2

23)(V

++=

ss 0 V, )2sin(

23)( 4

0 ≥= − ttetv t

Inverse transform




Example 2:

Determine v0(t) in the circuit shown below, assuming zero initial conditions.

V )()21(8 :Ans 22 tutee tt −− −−




Example 3:

Find v0(t) in the circuit shown below. Assume v0(0)=5V .

V )()1510()( v:Ans 20 tueet tt −− +=




Example 4:

The switch shown below has been in position b for a long time. It is moved to position a at t=0. Determine v(t) for t > 0.

where0, t ,I)IV()( v:Ans 0/

00 RCReRt t =>+−= − ττ




• Circuit analysis is relatively easy to do in the s-domain.

•By transforming a complicated set of mathematical relationships in the time domain into the s-domain where we convert operators (derivatives and integrals) into simple multipliers of s and 1/s.

•This allow us to use algebra to set up and solve the circuit equations.

•In this case, all the circuit theorems and relationships developed for dc circuits are perfectly valid in the s-domain.

Circuit Analysis



Example:

Consider the circuit below. Find the value of the voltage across the capacitor assuming that the value of vs(t)=10u(t) V and assume that at t=0, -1A flows through the inductor and +5V is across the capacitor.

Circuit Analysis Example (1)



Solution:

Transform the circuit from time-domain (a) into s-domain (b) using Laplace Transform. On rearranging the terms, we have

By taking the inverse transform, we get

230

135V1 +

−+

=ss

V )()3035()(v 21 tueet tt −− −=




Example: The initial energy in the circuit below is zero at t=0. Assume that vs=5u(t) V. (a) Find V0(s) using the Thevenin theorem. (b) Apply the initial- and final-value theorem to find v0(0) an v0(∞). (c) Obtain v0(t).

Ans: (a) V0(s) = 4(s+0.25)/(s(s+0.3)) (b) 4,3.333V, (c) (3.333+0.6667e-0.3t)u(t) V.




• The transfer function H(s) is the ratio of the output response Y(s) to the input response X(s), assuming all the initial conditions are zero.

h(t) is the impulse response function.

• Four types of gain:

1. H(s) = voltage gain = V0(s)/Vi(s)

2. H(s) = Current gain = I0(s)/Ii(s)

3. H(s) = Impedance = V(s)/I(s)

4. H(s) = Admittance = I(s)/V(s)

)()()(

sXsYsH =

Transfer Functions



Example: The output of a linear system is y(t)=10e-tcos4t when the input is x(t)=e-tu(t). Find the transfer function of the system and its impulse response.

Solution: Transform y(t) and x(t) into s-domain and apply H(s)=Y(s)/X(s), we get

Apply inverse transform for H(s), we get

16)1(44010

16)1()1(10

)()()( 22

2

++−=

+++

==ss

ssXsYsH

)()4sin(40)(10)( tutetth t−−= δ

Transfer Functions Example (1)



Example:

The transfer function of a linear system is

Find the output y(t) due to the input e-3t·u(t) and its impulse response.

62)(+

=s

ssH

)(12e-(t)2 0; t,42 :Ans -6t63 tuee tt δ≥+− −−

Transfer Functions Example (2)



In continuous systems Laplace transforms play a unique role. They allow system and circuit designers to analyze systems and predict performance, and to think in different terms - like frequency responses - to help understand linear continuous systems. Z-transforms play the role in sampled systems that Laplace transforms play in continuous systems. In continuous systems, inputs and outputs are related by differential equations and Laplace transform techniques are used to solve those differential equations. In sampled systems, inputs and outputs are related by difference equations and Z-transform techniques are used to solve those differential equations.

Introduction



Fourier, Laplace and Z-Transforms

For right-sided signals (zero-valued for negative time index) the Laplace transform is a generalization of the Fourier transform of a continuous-time signal, and the z-transform is a generalization of the Fourier transform of a discrete-time signal.

Fourier Transform

Laplace Transform Z-Transform

generalization

continuous-time signals discrete-time signals



The Z-transform converts a discrete time-domain signal, which is a sequence of real or complex numbers, into a complex frequency-domain representation. It can be considered as a discrete-time equivalent of the Laplace transform. There are numerous sampled systems that look like the one shown below.

discrete-time signals

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

http://en.wikipedia.org/wiki/Time-domain



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

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

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

http://en.wikipedia.org/wiki/Frequency-domain



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

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



Let us assume that we have a sequence, yk. The subscript "k" indicates a sampled time interval and that yk is the value of y(t) at the kth sample instant.

yk could be generated from a sample of a time function. For example: yk = y(kT), where y(t) is a continuous time function, and T is the sampling interval. We will focus on the index variable k, rather than the exact time kT, in all that we do in the following.

[ ] k

kkk zyyZ −

∞

=∑=

0

Definition of the Z-Transform



kk ayy ⋅= 0

Z-Transform Example

Given the following sampled signal:

[ ]az

z

zaz

azaaZk

kk

kkk

−=

−=

==⋅ ∑∑

∞

=

∞

=

−

1

1100

We get the Z-Transform for y0 = 1



Z-Transform of Unit Impulse and Unit Step

Given the following sampled signal Dk:

Dk is zero for k>0, so all those terms are zero. Dk is one for k = 0, so that

[ ] 1=kDZ

[ ]1

...1 321

−=+++= −−−

zzzzzuZ k

Given the following sampled signal uk:

uk is one for all k.



More Complex Example of Z-Transform

Given the following sampled signal fk:

( ) )sin(bkTekTff akTk

−==

[ ] k

k

akT

k

kkk zbkTezffZ −

∞

=

−∞

=

− ∑∑ == )sin(00

Finally:

[ ]

−+

−= *2

1cz

zcz

zj

fZ k jbTaTec +−=where



S- and Z-Plane Presentation



Inverse Z-Transform

The inverse z-transform can be obtained using one of two methods:

a) the inspection method, b) the partial fraction method.

In the inspection method each simple term of a polynomial in z, H(z), is substituted by its time-domain equivalent. For the more complicated functions of z, the partial fraction method is used to describe the polynomial in terms of simpler terms, and then each simple term is substituted by its time-domain equivalent term.

Fourier Series â€¢ Fourier Transform â€¢ Laplace Transform â€¢ Applications of Laplace...

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