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Analog Network Signal Processing:

Please answer problem 6.18 below for both part A & part B . Please show all work. Please show graph and MATLAB code for part B.

***MATLAB MUST BE USED FOR PART B THIS PROBLEM***

Part A:

6.18 Determine the transfer function of a second-order (2 poles) Notch Butterworth filter with a center frequency of 5 kHz, passband gain-1, and a bandwidth of 400 Hz. Design it using the Multiple Feedback filter configuration, find all the values of the circuit components, and draw the circuit.

Part B: Also plot bodemag using Matlab.

engineering   Electrical-Engineering   
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Answer #1

Butterworth Notch filter is designed by passing the same input to a low pass and a high pass filter and the output of the two filters is combined using a summing amplifier Band width is 400 Hz JH_fL = 400 NR5000 f 25x106 fH + 10008 Solve the two equations to get the two frequencies H5204 Hz f 4804 Hz Consider the transfer function of 2md order Butterworth low pass filter with a gain of 1 and cac equal to 1 H(s) 2 +1.414s +1 Scale the filter to a,-2π(4804) 96087 rad/s 1.414+1 +1.414 9608T H, (s) 9.11x10 s2 +42680.8s +9,1lx10 Consider the transfer function of 2nd order Butterworth high pass filter with a gain of 1 and ac equal to 1 H, (s)5 Scale the filter to ah-2n(5204)-104087 rad/s +1.414s +1 104087 H, (s) 1.414+1 S 1 104087 +1.4 1 4110407 H,(s) 2 +46234.5s +10.69 x10 Thus, the transfer function of the Butterworth Notch filter is, H(s)-H(s)+H,(s) 9.11x10 s2 +42680.8s +9.11x108 s2 46234.5s +10.69x108

9.11x108 2 +42680.8s +9.11x10 Consider Hi(s)- Consider the transfer function of the RLC low pass filter. H(s)-C R 1 L LC -9.11x10 LC 42680.8 Choose 42680.8 L-23.4 mH 9.11x108L C =46.8 nF Consider H, (s) Consider the transfer function of the RLC high pass filter. H.(s) s2 +46234.5s +10.69x10 L LC -10.69 x10 LC 46234.5 Choose 46234.5 L=21,6 mH 10.698 101 C 43.2 nF

Draw the filter circuit. 1 k 23.4 mH 46.8 nF l k243.2 nF Amplifier 21.6 mH 46,8 nF 43.2 nF 1 k2 Li 21.6 mH Summing Amplifier Write the MATLAB code to plot the Bode magnitude plot syms S W-20E3:0.5:40E3 H1=(9. 11E8) . / (s. ^2+42680.8. *S+9. 11E8) ; H2=s,2/ (s.2+46234.5. * s+10. 6988) ; Hmag-abs (Hl+H2); Hm-20.log10 (Hmag); plot (w, Hm) xlabel w ylabel Hmag 10 15 20 25 30 10 10 10

MATLAB code:

syms s

w=20E3:0.5:40E3;

s=i.*w;

H1=(9.11E8)./(s.^2+42680.8.*s+9.11E8);

H2=s.^2/(s.^2+46234.5.*s+10.69E8);

H=H1+H2;

Hmag=abs(H1+H2);

Hm=20.*log10(Hmag);

plot(w,Hm)

xlabel w

ylabel Hmag

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