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Chapter 8 IIR Filter Design 1. Problem P 8.1 Analog Butterworth lowpass filter design: Ω p = 30 rad/s, R p = 1 dB, Ωs = 40 rad/s, As = 30 dB. M ATLAB Script: 111 112 S OLUTIONS M ANUAL FOR DSP USING M ATLAB A PRIL 98 The system function is given by Ha (s) = 2:8199 1022 1122 s + 57:338s + 984:84 s + 50:777s + 984:84 11 2 + 41:997s + 984:842 + 31:382s + 984:84ss 11 s2 + 19:395s + 984:84s2 + 6:5606s + 984:84 1 s2 + 61:393s + 984:84 1 s + 31:382 The filter design plots are given in Figure 8.1. 2. Problem P 8.2 Analog Elliptic lowpass filter design: Ω p = 10 rad/s, R p = 1 dB, Ωs = 15 rad/s, As = 40 dB. M ATLAB Script: A PRIL 98 S OLUTIONS M ANUAL FOR DSP USING M ATLAB 113 Analog Butterworth Lowpass Filter Design Plots in P 8.1 Magnitude Response 0 1 0.89 Magnitude 30 Decibels Log−Magnitude Response 0 0 30 40 Analog frequency in rad/sec 100 100 0 30 40 Analog frequency in rad/sec 100 Phase Response 0 0.1 Impulse Response Phase in pi units 0.05 −3.47 −4.89 ha(t) 0 0 30 40 Analog frequency in rad/sec 100 −0.05 0 0.5 1 t (sec) 1.5 2 Figure 8.1: Analog Butterworth Lowpass Filter Plots in Problem P 8.1 114 S OLUTIONS M ANUAL FOR DSP USING M ATLAB A PRIL 98 The system function is given by Ha (s) = 46978s4 + 220:07s2 + 2298:5 s5 + 9:23s4 + 184:71s3 + 1129:2s2 + 7881:3s + 22985 : The filter design plots are given in Figure 8.2. Analog Elliptic Lowpass Filter Design Plots in P 8.2 Magnitude Response 1 0.89 log−Magnitude in dB Magnitude 0 Log−Magnitude Response 40 0 0 1015 Analog frequency in rad/sec 30 100 0 1015 Analog frequency in rad/sec 30 Phase Response 0 3 2 ha(t) −1.26 −1.64 1 0 −1 −2 Impulse Response Phase in pi units 0 1015 Analog frequency in rad/sec 30 0 2 4 t (sec) 6 8 10 Figure 8.2: Analog Elliptic Lowpass Filter Design Plots in P 8.2 3. Problem P 8.3 A PRIL 98 S OLUTIONS M ANUAL FOR DSP USING M ATLAB 115 The filter passband must include the 100 Hz component while the stopband must include the 130 Hz component. To obtain a minimum-order filter, the transition band must be as large as possible. This means that the passaband cutoff must be at 100 Hz while the stopband cutoff must be at 130 Hz. Hence the analog Chebyshev-I lowpass filter specifications are: Ω p = 2π (100) rad/s, R p = 2 dB, Ωs = 2π (130) rad/s, As = 50 dB. M ATLAB Script: 116 S OLUTIONS M ANUAL FOR DSP USING M ATLAB A PRIL 98 The system function is given by Ha (s) = 7:7954 1022 1 1 s2 + 142:45s + 51926 s2 + 75:794s + 301830 The magnitude response plots are given in Figure 8.3. 1 s2 + 116:12s + 168860 11 2 + 26:323s + 388620ss + 75:794 Analog Chebyshev−I Lowpass Filter Design Plots in P 8.3 Magnitude Response 1 0.79 Magnitude 0 0 100130 Analog frequency in Hz 200 Log−Magnitude Response 0 Decibels 50 100 0 100130 Analog frequency in Hz 200 Figure 8.3: Analog Chebyshev-I Lowpass Filter Plots in Problem P 8.3 4. Problem P 8.4 Analog Chebyshev-II lowpass filter design: Ω p = 2π (250) rad/s, R p = 0:5 dB, Ωs = 2π (300) rad/s, As = 45 dB. M ATLAB Script: A PRIL 98 S OLUTIONS M ANUAL FOR DSP USING M ATLAB 117 118 S OLUTIONS M ANUAL FOR DSP USING M ATLAB A PRIL 98 The filter design plots are given in Figure 8.4. Digital Butterworth Filter Design Plots in P 8.7 Log−Magnitude Response 0 Decibel 50 0 0.4 Frequency in Hz 0.6 1 Impulse Response 0.1 ha(t) 0 −0.1 0 10 20 30 40 5060 time in seconds 70 80 90 100 Figure 8.4: Analog Chebyshev-II Lowpass Filter Plots in Problem P 8.4 5. Problem P 8.5 M ATLAB function afd. m: 6. Problem P 8.6 A PRIL 98 S OLUTIONS M ANUAL FOR DSP USING M ATLAB 119 Digital Chebyshev-1 Lowpass Filter Design using Impulse Invariance. M ATLAB script: (a) Part (a): T = 1. M ATLAB script: The filter design plots are shown in Figure 8.5. (b) Part (b): T = 1=8000. M ATLAB script: 120 S OLUTIONS M ANUAL FOR DSP USING M ATLAB A PRIL 98 Filter Design Plots in P 8.6a Log−Magnitude Response 0 Decibel 40 60 0 1500 2000 Frequency in Hz 4000 Impulse Response 0.3 0.2 0.1 0 −0.1 −0.2 h(n) 0 10 20 30 40 50 n 60 70 80 90 100 Figure 8.5: Impulse Invariance Design Method with T =1 in Problem P 8.6a The filter design plots are shown in Figure 8.6. (c) Comparison: The designed system function as well as the impulse response in part 6b are similar to those in part 6a except for an overall gain due to Fs = 1=T = 8000. This problem can be avoided if in the impulse invariance design method we set h (n) = T ha (nT ) 7. Problem P 8.7 Digital Butterworth Lowpass Filter Design using Impulse Invariance. MATLAB script: