1 EENGR 3810 Chapter 3 Analysis and Transmission of Signals.
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Transcript of 1 EENGR 3810 Chapter 3 Analysis and Transmission of Signals.
1
EENGR 3810 Chapter 3
Analysis and Transmission of Signals
2
Chapter 3 Homework
3.1-4b, 3.1-5b, 3.1-6b, 3.1-7b, 3.2-2,
3.3-4b, 3.3-7b
Chapter 3 Homework Continued
RL Low-Pass Filter Problem:
a. Determine the transfer function of the filter.
b. Determine the Matrix for Bode Plot.
c. Build the Bode Plot Program (MATLAB Software).
d. Make the Bode Plot.
3
L = 10mH
R = 1k
Chapter 3 Homework Continued
RC High-Pass Filter Problem:
a. Determine the transfer function of the filter.
b. Build the Bode Plot Program (MATLAB Software).
c. Make the Bode Plot.
4
Active Filter Problem
Make a Bode Plot of the Active Filter shown below using MATLAB.
5
Discrete Fourier Transform (DFT) Problem 1
Estimate the continuous Fourier transform of the signal: g(t) = 4e-6t.
6
Discrete Fourier Transform (DFT) Problem 2
Estimate the continuous Fourier transform of the signal: g(t) = 6te-3t
7
Inverse FFT Problem
Estimate the inverse Fourier transform of G() = 6/(3+J)
8
9
APeriodic Signals Representation by Fourier Integral
• Let G() = the direct Fourier transform of g(t) and g(t) be the inverse Fourier transform, we have the following:
10
Example 1 of Direct Fourier Transform Find the Fourier transform of the signal g(t) shown below:
11
Example 2 of Direct Fourier Transform• Find the Fourier transform of the signal g(t) shown below:
12
Example 1 of Inverse Fourier Transform
• Find the inverse Fourier transform of the signal G() shown below:
13
Example 2 of Inverse Fourier TransformFind the inverse Fourier transform of the signal G() shown below:
14
Unit Gate Functionrect (x)
15
Unit Gate Functionrect (x/)
16
Unit Triangle Function
17
Interpolating Function sinc (x)
18
Gate Pulse and Its Fourier Spectrum
19
What is the Fourier transform of a (T-6) Gate pulse
20
Time Scaling
The scaling property states that time compression of a signal results in its spectral expansion, and time expansion of the signal results in a spectral compression.
21
Time Shifting
This shows that delaying a signal by t0 seconds does not change its magnitude. The phase spectrum is
changed by -t0.
22
Time Shifting
(From Page 92)
From Table 3.1
= G()
23
Frequency Shifting
The signal’s spectrum is shifted by = 0
24
Frequency Shifting
G() = 2 g(t) cos 0t
= g(t) /2
= G() /2
25
Exponential Form of Fourier Series
26
Where
f(t)e dt = (an – jbn) / 2 = An/2-n-jn0t
jn0t
Cn = 1/T
e
e
-jn0te
= cos 0t + jsin t
= cos 0t - jsin t
sin 0t =
cos 0t =jn0te -jn0t+ e
2
jn0te -jn0t- e
2j
f(t) = Cnejn0t
Fourier Symbols
• G() Is used in this book and F() is used in most Books. Thus, F() = G() for these problems.
• g(t) Is used in this book and f(t) is used in most Books. Thus, f(t) = g(t) for these problems.
27
Calculating the Fourier Transform F()
If f(t) = 0, t 0 and f(t) = te-at , t 0, a 0
Find: F()
28
Calculating the Fourier Transform F()
If f (t) = -A, -/2 t 0 and f (t) = A, 0 t /2
Calculate the Fourier Transform F().
29
Calculating the Inverse Fourier Transform
If F() = 0, - -3; F() = 4, -3 -2; F() = 1, , -2 2;
F() = 4, 2 3; and F() = 0, 3 0 Find the Inverse Fourier Transform f(t).
30
Numerical Computation of Fourier Transform:Discrete Fourier Transform (DFT)
• We have to use the samples of g(t) to compute G(), the Fourier transform of g(t).– G() must be at some finite number of
frequencies.– Thus, we can only compute samples of G().
• DFT can be computed by the FFT Algorithm – Developed by Tukey and Cooley in 1965.– Known as the Fast Fourier Transform (FFT)
31
FFT Example 1To Illustrate the FFT, consider the problem of estimating the continuous Fourier transform of the signal: g(t) = 2e-3t.
Analytically, the Fourier Transform is: G() = 2/(3+J)
% This program computes the Fourier Transform Approximation of g(t) = 2e-3t
diary EENG3810.dat
N= 128; % choose a power of 2 for speed
t = linspace(0,3,N); % time points for function evaluation
Fa = 2*exp(-3*t); % evaluate function, minimize aliasing: f(3)- 0
Ts = t(2)- t(1); % the sample period
Ws = 2*pi/Ts ; % the sampling frequency in rad/sec
F = fft(Fa) ; % compute the fft
Fc = fftshift(F)*Ts ; % shift the scale
W = Ws*(-N/2 : (N/2) -1)/N; % frequency axis
fa = 2./(3 + j*W) ; % analytical Fourier Transform
plot(W,abs(Fa),W,abs(Fc),'o') % generate plot, 'o' marks fft
xlabel('Frequency,Rads/s')
ylabel('F(\omega)')
title('Fourier Transform Approximation’)
diary
32
Example 1 FFT Plot
33
FFT Example 2
Estimate the continuous Fourier transform of the signal: g(t) = 4te-3t.
% This program computes the Fourier Transform Approximation
diary EENG3810b.dat
N= 128; % choose a power of 2 for speed
t = linspace(0,3,N); % time points for function evaluation
Fa = 4*t.*exp(-3*t); % evaluate function, minimize aliasing: f(3)- 0
Ws = 2*pi; % the sampling frequency in rad/sec
F = fft(Fa) ; % compute the fft
Fc = fftshift(F)*Ts ; % shift the scale
W = Ws*(-N/2 : (N/2) -1)/N; % frequency axis
plot(W,abs(Fa),W,abs(Fc),'o') % generate plot, 'o' marks fft
xlabel('Frequency,Rads/s')
ylabel('F(\omega)')
title('Fourier Transform Approximation')
diary
34
Example 2 FFT Plot
35
Inverse FFT Example 1Estimate the inverse Fourier transform of G() = 2/(3+J)
% This program computes the Inverse Fourier Transform Approximation
diary EENG3810.dat
N= 128; % choose a power of 2 for speed
t = linspace(0,3,N); % time points for function evaluation
Fa = 2./(3 + j*W) ; % evaluate function, minimize aliasing: f(3)- 0
Ts = t(2)- t(1); % the sample period
Ws = 2*pi/Ts ; % the sampling frequency in rad/sec
F = ifft(Fa) ; % compute the fft
W = Ws*(-N/2 : (N/2) -1)/N; % frequency axis
plot(W,abs(Fa)) % gnerate plot, 'o' marks fft
xlabel('Frequency,Rads/s')
ylabel('F(t)')
title('Inverse Fourier Transform Approximation')
diary36
Inverse FFT Example 1
37
38
Passive Filters• Any combination of passive (R, L, and C) and or
active (transistor or amplifier) elements designed to select or reject a band of frequencies is called a filter
• There are two classifications of filters– Passive – composed of series or parallel
combinations of R, L, and C elements– Active – employ active devices such as
transistors and operational amplifiers in combination with R, L, and C elements
• Four broad categories of filters are: low-pass, high-pass, pass-band, and band-reject
Frequency Bands.
39
40
R-L Low-Pass Filter (Cut-off Frequency)
C = 2fC = R/L
fC = Cut-off Frequency
fC = R / (2L)
XL = jC
41
R-L Low-Pass Filter
A plot of the magnitude of Vo versus the frequency results in the above curve
fC = R / (2L)
42
R-L Low-Pass Filter(Transfer Function - H(s))
Vo(s) = (R / R +sL)ViH(s) = Vo / Vi = R / (R + sL)H(s) = (R/L) / (s + RL)
C = R/LH(s) = C / (s + C)
s
43
R-C Low-Pass Filter (Cut-off Frequency)
XC = 1/jC
C = 2fC = 1/RC
44
R-C Low-Pass Filter
A plot of the magnitude of Vo versus the frequency results in the above curve
45
R-C Low-Pass Filter(Transfer Function – H(s))
V0 = (1/sC) / (R + 1/sC)Vi
H(s) = V0 / Vi = (1/sC) / (R + 1/sC)H(s) = 1 / (sRC + 1)H(s) = (1 / RC) / [s + (1/RC)]C = 1/RCH(s) = C / (s + C)
1/s
46
Any Low-Pass Filter (Transfer Function – H(s))
H(s) = C / (s + C)
For R-C Low-Pass Filter C = 1/RC
For R-L Low-Pass Filter C = R/L
47
R-L High-Pass Filter (Cut-off Frequency)
C = 2fC = R/L
fC = Cut-off Frequency
fC = R / (2L)
48
R-L High-Pass Filter
A plot of the magnitude of Vo versus the frequency results in the above curve.
fC = R / (2L)
49
R-L High-Pass Filter(Transfer Function – H(s))
sL
V0 = (sL) / (R + sL)Vi
H(s) = Vo / Vi = (sL) / (R + sL)H(s) = s / (R/L +s)H(s) = s / s + R/L)H(s) = s / (s + C)
C = R/L
50
RC High-Pass Filter (Cut-off Frequency)
C = 2fC = 1/RC
51
R-C High-Pass Filter
A plot of the magnitude of Vo versus the frequency results in the above curve.
52
R-C High-Pass Filter(Transfer Function – H(s))
V0 = R / (R + 1/sC)Vi
H(s) = Vo / Vi = R / (R + 1/sC)H(s) = RsC / RsC + 1H(s) = s / [s + (1/RC)]H(s) = s / (s + C)
C = 1/RC
53
Any High-Pass Filter (Transfer Function – H(s))
For R-C High-Pass Filter C = 1/RC
For R-L High-Pass Filter C = R/L
H(s) = s / (s + C)
54
Band-pass and Band-reject Filters
Band-pass and band-reject filters have two cut off frequencies (C1 and C2), a center frequency 0, a bandwidth ,and a quality factor Q.
These quantities are defined as:
0 = [(C1)(C2)]1/2
= C2 - C1
Q = 0 /
Also:
0 = (1/LC)1/2
C1 = - (R/2L) + [(r/2L)2 + (1/LC)]1/2
C2 = (R/2L) + [(R/2L)2 + (1/LC)]1/2
55
Pass-Band Filters
Pass-band filters can be constructed using a low-pass and a high-pass filter in series.
56
Pass-Band Filters
57
RC Pass-Band Filters
58
59
Series Resonant Pass-band Filter
60
Series Resonant Pass-band Filter
0
61
Series Resonant Pass-band Filter
62
Series Band-pass Filter
V(s)
sL
L1
1/sC
C1
R1
H(s) = s / (s2 + s + 02)
+
V0
_
H(s) = (R/Ls) / [s2 + (R/Ls) + (l/LC)]
= R/L 02 = 1/LC
63
Parallel Resonant Pass-band Filter
64
Parallel Pass-band Filter
0Vi
R
C L +
V0
_
1/s s
H(s) = (s/RC) / [s2 + (s/RC) + (1/LC)]
H(s) = s / s2 + s + 02
= 1/RC 02 = 1/LC
65
Series or Parallel Pass-band Filter
H(s) = s / s2 + s + 02
66
Band-reject Filter
67
Band-reject Filter
68
Band-reject Filter(using a series resonant circuit)
Band-reject Filter
69
Series Band-reject Filter
Vi
R
C
L
+
V0
_
s
1/s
H(s) = [s2 + (1/LC)] / [s2 + (R/Ls) + (l/LC)]
H(s) = (s2 + 02 ) / (s2 + s + 0
2)
= R/L 02 = 1/LC
70
Band-reject Filter (Using a Parallel Resonant Network)
Band-reject Filter
71
Parallel Band-reject Filter
Vi R
C
L
H(s) = [s2 + (1/LC)] / [s2 + (R/Ls) + (l/LC)]
H(s) = (s2 + 02 ) / (s2 + s + 0
2)
= R/L 02 = 1/LC
72
Decibels• The bel, defined as:
• To provide a unit of measure of less magnitude, a decibel is defined:
• The result is the following important equation, which compares power levels P2 and P1 in decibels:
73
Decibels• Voltage Gain
– Decibels are also used to provide a comparison between voltage levels. By substituting the basic power equation into the equation which compares levels of P2 and P1 in decibels
• Modern VOMs and DMMS have a dB scale designed to provide and indication of power ratios referenced to a standard level of 1mW at 600.
74
Bode Plots
• The curves obtained from the magnitude and/or phase angle versus frequency are called Bode plots– Through the use of straight-line segments
called idealized Bode plots, the frequency response of a system can be found efficiently and accurately
75
Bode Plots
• High-Pass R-C Filter– Two frequencies separated by a 2:1 ratio are said to be an octave
apart– For Bode plots, a change in frequency by one octave will result in
a 6-dB change in gain– Two frequencies separated by a 10:1 ratio are said to be a decade
apart– For bode plots, a change in frequency by one decade will result in
a 20-dB change in gain
76
Bode Plots
Bode plots are straight-line segments because the dB change per decade is constant.At ƒ = ƒc , the actual response curve is 3dB down from Bode plots are straight-line segments because the dB change per decade is constant.At ƒ = ƒc , the actual response curve is 3dB down from the idealized Bode plot, whereas at ƒ = 2ƒc and ƒc/ 2, the actual response curve is 1 dB down from the asymptotic response.The idealized Bode plot, whereas at ƒ = 2ƒc and ƒc/ 2, the actual response curve is 1 dB down from the asymptotic response.
77
MATLAB Software – Bode Plot Template
n=[0 2];d=[1 2];bode (n,d);w=logspace (-1, 2, 200);[mag,pha]=bode (n,d,w);semilogx (w,20*log10 (mag));gridtitle(‘Bode Plot’)xlabel (‘w (Rad/sec)’)ylabel(‘Decibels (dB)’)
H(s) = 2 / (s + 2)
78
Bode Plot
10-1
100
101
102
-20
-15
-10
-5
0
5
10Bode Plot
w(Rad/sec)
Dec
ibel
s (d
B)
X: 2.026Y: -3.066
79
MATLAB Software For Semi-log Paper
n=[1];
d=[1];
bode (n,d);
w=logspace (0, 4, 200);
[mag,pha]=bode (n,d,w);
semilogx (w,20*log10 (mag));grid
title(‘Bode Plot’)
xlabel (‘w (Rad/sec)’)
xlabel(‘Decibels (dB)’)
80
Semi-log Paper
10-1
100
101
102
103
104
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1Bode Plot
w(Rad/sec)
Dec
ibel
s (d
B)
81
RC Low-Pass Filter
F(s) = (1 / RC) / [s + (1/RC)]
F(s) = (83333) / [s + (83333)]
Matrix for Bode Plot:
n=[0 83333];
d=[1 83333];
82
Bode Plot Program (MATLAB Software)
%%%n=[0 83333];d=[1 83333];bode(n,d);w=logspace(0,6,200);[mag,pha]=bode(n,d,w);semilogx(w,20*log10(mag));gridtitle('Bode Plot')xlabel(‘w (Rad/sec)')ylabel('Decibels (dB)')
83
Bode Plot
100
101
102
103
104
105
106
-20
-15
-10
-5
0
5
10Bode Plot
w(Rad/sec)
Dec
ibel
s (d
B)
X: 8.805e+004Y: -3.256
84
RL High-Pass Filter
R = 5K and L = 3.5 mH
F(s) = s / s + R/L
F(s) = s / s + 1428571
n = [1 0];
d = [ 1 1428571];
85
Bode Plot Program(MATLAB Software)
%%%n=[1 0];d=[1 1428571];bode(n,d);w=logspace(5,10,200);[mag,pha]=bode(n,d,w);semilogx(w,20*log10(mag));gridtitle('Bode Plot')xlabel(‘w (Rad/sec)')ylabel('Decibels (dB)')
86
Bode Plot
105
106
107
108
109
1010
-25
-20
-15
-10
-5
0
5
10Bode Plot
w(Rad/sec)
Dec
ibel
s (d
B)
X: 1.431e+006Y: -3.002
Active Filters Circuits
87
88
General Operational Amplifier Circuit
+
Vo
_
X1
Vi
Zf
Zi
H(s) = -Zf / Zi = -K C / (s + C)
C = 1 / R2C
89
First-order low-pass Filter
+
Vo
_
X1
Vi
R1
R2
C
H(s) = -Zf / Zi = -K C / (s + C)
K = R2 / R1 (Gain)
C = 1 / R2C
90
First-order low-pass Filter
+
Vo
_
X1
Vi
1
R11
R2
1F
C
H(s) = -K C / (s + C)
K = 1/1 = 1
C = 1 / R2C = 1
H(s) = -1 / (s+1)
91
First-order low-pass FilterH(s) = -1 /(s+1)
MATLAB Bode Plot Program
n=[0 -1];
d=[1 1];
bode (n,d);
w=logspace (-1, 1, 200);
[mag,pha]=bode (n,d,w);
semilogx (w,20*log10 (mag));grid
title('Bode Plot')
xlabel ('w(Rad/sec)')
ylabel('Decibels (dB)')
92
First-order low-pass FilterH(s) = -1 /(s+1)
10-1
100
101
-25
-20
-15
-10
-5
0
5
10Bode Plot
w(Rad/sec)
Dec
ibel
s (d
B)
X: 1.012Y: -3.061
93
First-order High-pass Filter
+
Vo
_
X1
Vi
1
R11
R2
1F
C
H(s) = -Zf / Zi = -K s / (s + C)
K = R2 / R1 (Gain)
C = 1 / R1C
94
First-order High-pass Filter
+
Vo
_
X1
Vi
20k
R1200k
R2
0.1uF
C
H(s) = -K s / (s + C)
K = 200k / 20k = 10
C = 1 / R1C = 1 / (20K)(0.1uF) = 500
H(s) = -10s / (s+500)
95
First-order High-pass FilterH(s) = -10s / (s+50)
n=[-10 0];
d=[1 50];
bode (n,d);
w=logspace (0, 4, 200);
[mag,pha]=bode (n,d,w);
semilogx (w,20*log10 (mag));grid
title('Bode Plot')
xlabel ('w(Rad/sec)')
ylabel('Decibels (dB)')
96
First-order High-pass FilterH(s) = -10s /(s+50)
100
101
102
103
104
-15
-10
-5
0
5
10
15
20
25
30Bode Plot
w(Rad/sec)
Dec
ibel
s (d
B)
X: 51.11Y: 17.08
97
Scaling Factors (kf and km)
Scaling in Magnitude:
R‘ = kmR L‘ = kmL C‘ = C / km
Scaling in Frequency: R‘ = R L‘ = L / kf C‘ = C / kf
Scaling in both Magnitude and Frequency:
Magnitude scaling factor km
Frequency scaling factor kf
98
Pass-Band Filters
Low-pass Filter High-pass Filter Inverting Ampvi v0
H(s) = v0 / vi = - KC2s / [(s + C1)(s + C2)]
H(s) = -C2 / (s+C2) H(S) = -s / (s + C1) H(s) = -Rf / Ri = K
H(s) = - KC2s / [(s2 + (C1 + C2)s + C!C2]
C2 = 1 / RLCL C1 = 1 / RHCH
+
Vo
_
XH
R2
RH1
RH2
CHXi
XL
Vi
RL1 x
RL2
C1
R1
99
Pass-Band Filters
+
Vo
_
XH 2k
R2
7958
RH17958
RH2
0.2uF
CHXi
XL
Vi
80
RL180
RL2
0.2uFCL
1k
Ri
H(s) = - KC2s / [(s2 + (C1 + C2)s + C!C2]
C2 = 1 / RLCL= 62500 C1 = 1 / RHCH = 628.3 K = 2k/1k = 2
H(s) = - (2 )(62500) s / [(s2 + (628.3 + 62500)s + (628.3)(62500)]
H(s) = -125000s / s2 + 63128.3s + 39268750
100
Pass-Band Filtersn=[0 -125000 0];
d=[1 63128.3 39268750];
bode (n,d);
w=logspace (0, 7, 200);
[mag,pha]=bode (n,d,w);
f=w/6.283
semilogx (f,20*log10 (mag));grid
title('Bode Plot')
xlabel ('Frequency (Hz)')
ylabel('Decibels (dB)')
101
Pass-Band Filters
= fC2 - fC1 = 10,000 – 100 = 9,900 Hz
fC1 = 100 Hz fC2 = 10,000 Hz
101
102
103
104
105
-3
-2
-1
0
1
2
3
4
5
6
7Bode Plot
Frequency (Hz)
Dec
ibel
s (d
B)
X: 103.7Y: 3.166
fs
H(s) = -125000s / s2 + 63128.3s + 39268750
102
Band-reject Filter
+
Vo
_
XH
R2
RH1
RH2
CH
Xi
XL
Vi
RL1
RL2
CL
Ri1
Ri2
C1 = 1 / RLCL C2 = 1 / RHCH
H(s) = K[(s2 + (C1C2)s + C!C2] / [(s + C1)(s + C2)]
Low-pass Filter
High-pass Filter
Inverting Ampvi
v0
H(s) = K[(s2 + C1s + C!C2] / [s2 + (C1 + C2)s +C1C2 ]
103
Band-reject Filter
+
Vo
_
XH
3k
Rf
1k
RH11k
RH2
0.5uF
CH
Xi
XL
Vi
20k
RL120k
RL20.5uF
CL
1k
Ri1
1k
Ri2
H(s) = -K[(s2 + C1s + C!C2] / [s2 + (C1 + C2)s +C1C2 ]
C1 = 1 / RLCL = 1 / (20k)(0.5uF) = 100 K = 3k / 1k = 3 C2 = 1 / RHCH = 1 / (1k)(0.5uF) = 2000
H(s) = 3[s2 + 100s + 200000] / [s2 + 2100s + 200000]
H(s) = [3s2 + 300s + 600000] / [s2 + 2100s + 200000]
104
Band-reject Filter
H(s) = [3s2 + 300s + 600000] / [s2 + 2100s + 200000]
100
101
102
103
104
105
-20
-15
-10
-5
0
5
10
15Bode Plot
w(Rad/sec)
Dec
ibel
s (d
B)
= C2 - C1 = 2000 – 100 = 1900 Rads/s
105
Higher Order Op Amp Filters
n -order operational amplifier filters will provide: a. A sharper transition from pass-band to stop-band. b. A slope of 20n dB/decade.
H(s) for cascaded low-pass filters can be calculated bymultiplying individual transfer functions:
H(s) = (-1)n / (s + 1)n
A low-pass frequency scale factor (kf) is used to place the cutoff
frequency at any value of c desired for a n th-order unity-gain low-pass filter :
kf = c / cn
cn = [(2)1/n – 1]1/2
106
4 th-order Unity-gain Low-pass Filter(500 Hz Cutoff Frequency and Gain of 10)
+
Vo
_
1384.6
R1
X1
138.46
R2138.46
R3
1uFC1
X2
138.46
R4
138.46
R4
138.46
R3
V1
X2
138.46
R2138.46
R1
1uFC1
X1
1uFC2
X3
138.46
R5138.46
R6
1uFC3
c4 = [(2)1/4 – 1]1/2 = 0.435 rads/s
Kf = 500/0.435 = 7222.39
H(s) = -10 [(7222.39)4 / (s + 7222.39)4]
107
4 th-order Unity-gain Low-pass Filter(500 Hz Cutoff Frequency and Gain of 10)
H(s) = -10 [(7222.39)4 / (s + 7222.39)4]
Let a = 7222.39 Thus, H(s) = -10 [(a)4 / (s + a)4]
H(s) = -10 [(a)4 / (s + a)4]
H(s) = -10a4 / (s4 + 4as3 + 6a2s2 + 4a3s + a4)
H(s) = -27.2 x 1015 / (s4 + 2.89 x 104s3 + 3.13 x 108s2+ 1.51 x 1012s+ 2.72 x 1015)
108
Bode Plot
n=[0 0 0 -27.2*10^15];
d=[1 2.89*10^4 3.13*10^8 1.51*10^12 2.72*10^15];
bode (n,d);
w=logspace (2, 4, 200);
[mag,pha]=bode (n,d,w);
f=w/6.283
semilogx (f,20*log10 (mag));grid
title('Bode Plot')
xlabel ('Frequency (Hz)')
ylabel('Decibels (dB)')
109
Bode Plot
101
102
103
104
10
12
14
16
18
20
22Bode Plot
Frequency (Hz)
Dec
ibel
s (d
B)
X: 500.4Y: 16.96
110
Second-order Butterworth Filters
+
Vo
_
X1
V1
R1 R2
C1
1u
C2
H(s) = 1/ s + b1s + 1
b1 = 2 / C1 1 = 1 / C1C2
Order of a Butterworth Filter: n = -0.05As / log10 (s / p)
Gain (K) = 20 log10 [1 / (1 + ( / C)2n]
111
Forth-order Butterworth Filters
From Table 15.1H(s) = 1/ (s2 + 0.765s + 1)(s2 + 1.848s + 1)
Normalizes values for C = 1 rad/s:C1 = 2 / b1 = 2 / 0.765 = 2.61 FC2 = 1 / C1 = 1 / 2.61 = 0.38 FC3 = 2 / b2 = 2 / 1.848 = 1.08 FC4 = 1 / C3 = 1 / 1.08 = 0.924 F
+
Vo
_
X1
Vi
R1 R2
C1
1u
C2
X2X3
R3 R4
C3 2k
R5
1u
C4
1k
R6
112
Fourth-order Butterworth Filters
Scaling factors for fC = 500 Hz:
kf = 2fC = C =(6.2832)(500) = 3141.6
Km = 1000 if resistors = 1k
Kfkm = 3.1416 x 106
C = Cn / kfkm
C1 = 2.61 / 3.1416 x 106 = 831 nFC2 = 0.38 / 3.1416 x 106 = 121 nFC3 = 1.08 / 3.1416 x 106 = 3.44 nFC4 = 0.924 / 3.1416 x 106 =294 nF
113
Fourth-order Butterworth Filters
+
Vo
_
X1
Vi
1k
R1
1k
R2
831nF C1
121nF
C2
X2X3
1k
R3
1k
R4
344nF C3 10k
R5
294nF
C4
1k
R6
Scalled up H(s):H(s) = -10 / [(s\kf)2 + 0.765(s/kf) + 1)(s/kf)2 + 1.848(s/kf) + 1)] = -10 / [ (s2/9.87 x106) + = -10 / [(1 x10-7s2 + 2.44 x 10-4s +1)(1 x 10-7s2 + 5.88 x 10-4 + 1)] = -10 / (1 x 10-14s2 + 8.32 x 10-11s3 + 3.33 x 10-7s2 + 8.32 x 10-4s + 1
114
Bode Plot
n=[0 0 0 0 -10];
d=[1*10^-14 8.32*10^-11 3.43*10^-7 8.32*10^-4 1];
bode (n,d);
w=logspace (2, 4, 200);
[mag,pha]=bode (n,d,w);
f=w/6.283
semilogx (f,20*log10 (mag));grid
title('Bode Plot')
xlabel ('Frequency (Hz)')
ylabel('Decibels (dB)')
115
Bode Plot
101
102
103
104
-10
-5
0
5
10
15
20
Bode Plot
Frequency (Hz)
Dec
ibel
s (d
B)
X: 500.4Y: 16.99