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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.

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Answer #1

Butterworth Notch filter It 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. The given bandwidth is 400 Hz Thus, The given center frequency of the filter is 5 kHz That 1S, fLfH-5000 Consider the expression Now on substituting the values Un+左)2-4002 +4(25×106) In + fr = 10008 Solve equation (1), and (2) fe = 5204 Hz, and f, = 4804 Hz 2Consider the transfer function of 2nd order Butterworth low pass filter with a gain of 1 and oc equal to1 H,(s Scale the filter to q-2t( 4804)-96087 rads. s +1.414s +1 2 5+1.414+1 960玩) +1.414|-. S 9.11x108 12 96087 96087 HL (s) s +42680.8s +9.11x10Consider the transfer function of 2nd order Butterworth high pass filter with a gain of 1 and oc equal to 1. H()-+1414s+ ScalTherefore, the transfer function of the Butterworth Notch filter is H(s) H(s)+H2(s) H(s) 9.11x10 s2 42680.8s 9.11x108 46234.5Consider the transfer function of the RLC low pass filter. LC H1 (s) L LC Compare equation (3), and (5) 42680.8 9.11x 108 LC Choose Put L 23.4 mH 1k - 9.11x 108 - 42680.8 (23.4m)C 1k 42680.8 L=23.4 mH | 260(91110)23.4m) C 46.8 nFNow consider the transfer function of 2nd order Butterworth High pass filter, that is, Recall (4) H2(s) Consider the transfer function of the RLC high pass filter. s2 46234.5s +10.69 x108 R 1 L LC Compare equation (4), and (6) 10.69 × 108 LC - 46234.5 Choose Put L 21.6 mH IK 46234.5 (21.6 m)C 1k mic 10.698 108 1k 46234.5 10.698108)(21.6 m) L=21.6 rnH | C = 43.2 nFDraw the filter circuit 1kΩ 23.4 mH 46.8 nF 43.2 nF 1 kS2 Summing Amplifier 21.6 mH IkΩ 23.4 mil Vi 46.8 nF 43.2 nF 1 k2 vo VTo sketch the Bode magnitude plot. Recall (3), and (4) H, (s) 9,11 x108 s2 +42680.8s 9.11 x108 4 s2 + 46234.5s +10.69 x108 Consider the MATLAB code. syms s w 20E3:0.5:40E3; H2-s.2/ (s. 2+46234.5. *s+10.69E8); H-H1+H2; Hmag-abs (H1+H2) Hm-20. *loq10 (Hmag); plot (w, Hm) xlabel w yl ae HmangThe following is the bode magnitude plot 5 -10 -15 20 -25 -30 4.4 10 4.5 10 4.6 10

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