1. 1 EENGR 3810 Chapter 3 Analysis and Transmission of Signals

2. 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

3. 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

4. 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

5. Active Filter Problem Make a Bode Plot of the Active Filter shown below using MATLAB. 5

6. Discrete Fourier Transform (DFT) Problem 1 Estimate the continuous Fourier transform of the signal: g(t) = 4e -6t. 6

7. Discrete Fourier Transform (DFT) Problem 2 Estimate the continuous Fourier transform of the signal: g(t) = 6te -3t 7

8. Inverse FFT Problem Estimate the inverse Fourier transform of G(  ) = 6/(3+J  ) 8

9. 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. 10 Example 1 of Direct Fourier Transform Find the Fourier transform of the signal g(t) shown below:

11. 11 Example 2 of Direct Fourier Transform Find the Fourier transform of the signal g(t) shown below:

12. 12 Example 1 of Inverse Fourier Transform Find the inverse Fourier transform of the signal G(  ) shown below:

13. 13 Example 2 of Inverse Fourier Transform Find the inverse Fourier transform of the signal G(  ) shown below:

14. 14 Unit Gate Function rect (x)

15. 15 Unit Gate Function rect (x/  )

16. 16 Unit Triangle Function

17. 17 Interpolating Function sinc (x)

18. 18 Gate Pulse and Its Fourier Spectrum

19. 19 What is the Fourier transform of a (T-6) Gate pulse

20. 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. 21 Time Shifting This shows that delaying a signal by t 0 seconds does not change its magnitude. The phase spectrum is changed by -  t 0.

22. 22 Time Shifting (From Page 92) From Table 3.1 = G(  )

23. 23 Frequency Shifting The signal’s spectrum is shifted by  =  0

24. 24 Frequency Shifting G(  ) = 2 g(t) cos  0 t = g(t) /2 = G(  ) /2

25. 25

26. Exponential Form of Fourier Series 26

27. 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

28. Calculating the Fourier Transform F(  ) If f(t) = 0, t  0 and f(t) = te -at, t  0, a  0 Find: F(  ) 28

29. 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

30. 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

31. 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

32. FFT Example 1 To 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

33. Example 1 FFT Plot 33

34. 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

35. Example 2 FFT Plot 35

36. Inverse FFT Example 1 Estimate 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') diary 36

37. Inverse FFT Example 1 37

38. 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

39. Frequency Bands. 39

40. 40 R-L Low-Pass Filter (Cut-off Frequency)  C = 2  f C = R/L f C = Cut-off Frequency f C = R / (2  L) X L = j  C

41. 41 R-L Low-Pass Filter A plot of the magnitude of V o versus the frequency results in the above curve f C = R / (2  L)

42. 42 R-L Low-Pass Filter (Transfer Function - H(s)) Vo(s) = (R / R +sL)Vi H(s) = Vo / Vi = R / (R + sL) H(s) = (R/L) / (s + RL)  C = R/L H(s) =  C / (s +  C ) s

43. 43 R-C Low-Pass Filter (Cut-off Frequency) X C = 1/j  C  C = 2  f C = 1/RC

44. 44 R-C Low-Pass Filter A plot of the magnitude of V o versus the frequency results in the above curve

45. 45 R-C Low-Pass Filter (Transfer Function – H(s)) V 0 = (1/sC) / (R + 1/sC)V i H(s) = V 0 / V i = (1/sC) / (R + 1/sC) H(s) = 1 / (sRC + 1) H(s) = (1 / RC) / [s + (1/RC)]  C = 1/RC H(s) =  C / (s +  C ) 1/s

46. 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. 47 R-L High-Pass Filter (Cut-off Frequency)  C = 2  f C = R/L f C = Cut-off Frequency f C = R / (2  L)

48. 48 R-L High-Pass Filter A plot of the magnitude of V o versus the frequency results in the above curve. f C = R / (2  L)

49. 49 R-L High-Pass Filter (Transfer Function – H(s)) sL V 0 = (sL) / (R + sL)V i 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. 50 RC High-Pass Filter (Cut-off Frequency)  C = 2  f C = 1/RC

51. 51 R-C High-Pass Filter A plot of the magnitude of V o versus the frequency results in the above curve.

52. 52 R-C High-Pass Filter (Transfer Function – H(s)) V 0 = R / (R + 1/sC)V i H(s) = Vo / Vi = R / (R + 1/sC) H(s) = RsC / RsC + 1 H(s) = s / [s + (1/RC)] H(s) = s / (s +  C )  C = 1/RC

53. 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. 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. 55 Pass-Band Filters Pass-band filters can be constructed using a low-pass and a high-pass filter in series.

56. 56 Pass-Band Filters

57. 57 RC Pass-Band Filters

58. 58

59. 59 Series Resonant Pass-band Filter

60. 60 Series Resonant Pass-band Filter 00

61. 61 Series Resonant Pass-band Filter

62. 62 Series Band-pass Filter H(s) =  s / (s 2 +  s +  0 2 ) +V0_+V0_ H(s) = (R/Ls) / [s 2 + (R/Ls) + (l/LC)]  = R/L  0 2 = 1/LC

63. 63 Parallel Resonant Pass-band Filter

64. 64 Parallel Pass-band Filter +V0_+V0_ 1/ss H(s) = (s/RC) / [s 2 + (s/RC) + (1/LC)] H(s) =  s / s 2 +  s +  0 2  = 1/RC  0 2 = 1/LC

65. 65 Series or Parallel Pass-band Filter H(s) =  s / s 2 +  s +  0 2

66. 66 Band-reject Filter

67. 67 Band-reject Filter

68. 68 Band-reject Filter ( using a series resonant circuit) Band-reject Filter

69. 69 Series Band-reject Filter +V0_+V0_ s 1/s H(s) = [s 2 + (1/LC)] / [s 2 + (R/Ls) + (l/LC)] H(s) = (s 2 +  0 2 ) / (s 2 +  s +  0 2 )  = R/L  0 2 = 1/LC

70. 70 Band-reject Filter (Using a Parallel Resonant Network) Band-reject Filter

71. 71 Parallel Band-reject Filter H(s) = [s 2 + (1/LC)] / [s 2 + (R/Ls) + (l/LC)] H(s) = (s 2 +  0 2 ) / (s 2 +  s +  0 2 )  = R/L  0 2 = 1/LC

72. 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 P 2 and P 1 in decibels:

73. 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 P 2 and P 1 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. 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. 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. 76 Bode Plots Bode plots are straight-line segments because the dB change per decade is constant. At ƒ = ƒ c, the actual response curve is 3dB down from Bode plots are straight-line segments because the dB change per decade is constant. At ƒ = ƒ c, the actual response curve is 3dB down from the idealized Bode plot, whereas at ƒ = 2ƒ c and ƒ c / 2, the actual response curve is 1 dB down from the asymptotic response. The idealized Bode plot, whereas at ƒ = 2ƒ c and ƒ c / 2, the actual response curve is 1 dB down from the asymptotic response.

77. 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));grid title(‘Bode Plot’) xlabel (‘w (Rad/sec)’) ylabel(‘Decibels (dB)’) H(s) = 2 / (s + 2)

78. 78 Bode Plot

79. 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. 80 Semi-log Paper

81. 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. 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));grid title('Bode Plot') xlabel(‘w (Rad/sec)') ylabel('Decibels (dB)')

83. 83 Bode Plot

84. 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. 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));grid title('Bode Plot') xlabel(‘w (Rad/sec)') ylabel('Decibels (dB)')

86. 86 Bode Plot

87. Active Filters Circuits 87

88. 88 General Operational Amplifier Circuit ZfZf ZiZi H(s) = -Z f / Z i = -K  C / (s +  C )  C = 1 / R 2 C

89. 89 First-order low-pass Filter H(s) = -Z f / Z i = -K  C / (s +  C ) K = R 2 / R 1 (Gain)  C = 1 / R 2 C

90. 90 First-order low-pass Filter H(s) = -K  C / (s +  C ) K = 1/1 = 1  C = 1 / R 2 C = 1 H(s) = -1 / (s+1)

91. 91 First-order low-pass Filter H(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. 92 First-order low-pass Filter H(s) = -1 /(s+1)

93. 93 First-order High-pass Filter H(s) = -Z f / Z i = -K s / (s +  C ) K = R 2 / R 1 (Gain)  C = 1 / R 1 C

94. 94 First-order High-pass Filter H(s) = -K s / (s +  C ) K = 200k / 20k = 10  C = 1 / R 1 C = 1 / (20K)(0.1uF) = 500 H(s) = -10s / (s+500)

95. 95 First-order High-pass Filter H(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. 96 First-order High-pass Filter H(s) = -10s /(s+50)

97. 97 Scaling Factors (k f and k m ) Scaling in Magnitude: R‘ = k m R L‘ = k m L C‘ = C / k m Scaling in Frequency: R‘ = R L‘ = L / k f C‘ = C / k f Scaling in both Magnitude and Frequency: Magnitude scaling factor k m Frequency scaling factor k f

98. 98 Pass-Band Filters Low-pass FilterHigh-pass FilterInverting Amp vivi v0v0 H(s) = v 0 / v i = - K  C2 s / [(s +  C1 )(s +  C2 )] H(s) = -  C2 / (s+  C2 ) H(S) = -s / (s +  C1 ) H(s) = -R f / R i = K H(s) = - K  C2 s / [(s 2 + (  C1 +  C2 )s +  C!  C2 ]  C2 = 1 / R L C L  C1 = 1 / R H C H

99. 99 Pass-Band Filters H(s) = - K  C2 s / [(s 2 + (  C1 +  C2 )s +  C!  C2 ]  C2 = 1 / R L C L = 62500  C1 = 1 / R H C H = 628.3 K = 2k/1k = 2 H(s) = - (2 )(62500) s / [(s 2 + (628.3 + 62500)s + (628.3)(62500)] H(s) = -125000s / s 2 + 63128.3s + 39268750

100. 100 Pass-Band Filters n=[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. 101 Pass-Band Filters  = f C2 - f C1 = 10,000 – 100 = 9,900 Hz f C1 = 100 Hz f C2 = 10,000 Hz fsfs H(s) = -125000s / s 2 + 63128.3s + 39268750

102. 102 Band-reject Filter  C1 = 1 / R L C L  C2 = 1 / R H C H H(s) = K[(s 2 + (  C1  C2 )s +  C!  C2 ] / [(s +  C1 )(s +  C2 )] Low-pass Filter High-pass Filter Inverting Amp vivi v0v0 H(s) = K[(s 2 +  C1 s +  C!  C2 ] / [s 2 + (  C1 +  C2 )s +  C1  C2 ]

103. 103 Band-reject Filter H(s) = -K[(s 2 +  C1 s +  C!  C2 ] / [s 2 + (  C1 +  C2 )s +  C1  C2 ]  C1 = 1 / R L C L = 1 / (20k)(0.5uF) = 100 K = 3k / 1k = 3  C2 = 1 / R H C H = 1 / (1k)(0.5uF) = 2000 H(s) = 3[s 2 + 100s + 200000] / [s 2 + 2100s + 200000] H(s) = [3s 2 + 300s + 600000] / [s 2 + 2100s + 200000]

104. 104 Band-reject Filter H(s) = [3s 2 + 300s + 600000] / [s 2 + 2100s + 200000]  =  C2 -  C1 = 2000 – 100 = 1900 Rads/s

105. 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 by multiplying individual transfer functions: H(s) = (-1) n / (s + 1) n A low-pass frequency scale factor (k f ) is used to place the cutoff frequency at any value of  c desired for a n th-order unity-gain low-pass filter : k f =  c /  cn  cn = [(2) 1/n – 1] 1/2

106. 106 4 th-order Unity-gain Low-pass Filter (500 Hz Cutoff Frequency and Gain of 10)  c4 = [(2) 1/4 – 1] 1/2 = 0.435 rads/s K f = 500/0.435 = 7222.39 H(s) = -10 [(7222.39) 4 / (s + 7222.39) 4 ]

107. 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) = -10a 4 / (s 4 + 4as 3 + 6a 2 s 2 + 4a 3 s + a 4 ) H(s) = -27.2 x 10 15 / (s 4 + 2.89 x 10 4 s 3 + 3.13 x 10 8 s 2 + 1.51 x 10 12 s+ 2.72 x 10 15 )

108. 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. 109 Bode Plot

110. 110 Second-order Butterworth Filters H(s) = 1/ s + b 1 s + 1 b 1 = 2 / C 1 1 = 1 / C 1 C 2 Order of a Butterworth Filter: n = -0.05A s / log 10 (  s /  p ) Gain (K) = 20 log 10 [1 / (1 + (  /  C ) 2n ]

111. 111 Forth-order Butterworth Filters From Table 15.1 H(s) = 1/ (s 2 + 0.765s + 1)(s 2 + 1.848s + 1) Normalizes values for  C = 1 rad/s: C 1 = 2 / b 1 = 2 / 0.765 = 2.61 F C 2 = 1 / C 1 = 1 / 2.61 = 0.38 F C 3 = 2 / b 2 = 2 / 1.848 = 1.08 F C 4 = 1 / C 3 = 1 / 1.08 = 0.924 F

112. 112 Fourth-order Butterworth Filters Scaling factors for f C = 500 Hz: k f = 2  f C =  C =(6.2832)(500) = 3141.6 K m = 1000 if resistors = 1k  K f k m = 3.1416 x 10 6 C = C n / k f k m C 1 = 2.61 / 3.1416 x 10 6 = 831 nF C 2 = 0.38 / 3.1416 x 10 6 = 121 nF C 3 = 1.08 / 3.1416 x 10 6 = 3.44 nF C 4 = 0.924 / 3.1416 x 10 6 =294 nF

113. 113 Fourth-order Butterworth Filters Scalled up H(s): H(s) = -10 / [(s\k f ) 2 + 0.765(s/k f ) + 1)(s/k f ) 2 + 1.848(s/k f ) + 1)] = -10 / [ (s 2 /9.87 x10 6 ) + = -10 / [(1 x10 -7 s 2 + 2.44 x 10 -4 s +1)(1 x 10 -7 s 2 + 5.88 x 10 -4 + 1)] = -10 / (1 x 10 -14 s 2 + 8.32 x 10 -11 s 3 + 3.33 x 10 -7 s 2 + 8.32 x 10 -4 s + 1

114. 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. 115 Bode Plot