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