University of Illinois at Chicago Spring ECE 412 Introduction to Filter Synthesis Homework #2 Solutions. Problem 1

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1 Problem 1 (a) Magnitude (impedance) scale the circuit so that all resistors are 1kΩ. Solution: Since all of the resistors in the circuit are 1Ω, we need to magnitude scale by k m = 1000; therefore, we should multiply the resistors and inductors by k m and divide the capacitors by k m. The resulting magnitude-scaled circuit is shown below. 1kΩ 4kH v 1 (t) 1 2 mf 1 3 mf 1kΩ v 2 (t) (b) Determine the frequency response of the circuit in (a) by using a circuit simulator (for example, LTspice). If you simulate the circuit from 0.01 rad/s to 100 rad/s, the plots should look similar to Homework 1, Problem 1(c). Solution: One thing to keep in mind while setting up the simulation is that the frequency range in Hz is found by dividing the radian frequencies by 2π; therefore, the circuit simulation should run from 1.59 mhz 15.9 Hz. The LTspice schematic and the resulting frequency response are shown below. in R1 {1*km} L1 {4*km} out HW 2, 1(b) V1 AC 1 0 C1 {1/2/km} C2 {1/3/km} R2 {1*km}.ac dec 100 {0.01/(2*pi)} {100/(2*pi)}.param km= mHz, Hz, dB mHz 100mHz 1Hz 10Hz Dr. Vahe Caliskan 1 of 8 Posted: February 8, 2013

2 Hz, mHz 100mHz 1Hz 10Hz -270 (c) From the simulation results in (b), determine the magnitude (in db) and the phase (in degrees) at a frequency of 1 Hz. Solution: From the simulation results at 1 Hz, the magnitude is 45.5dBand the phase is 226. (d) From the simulation results in (b), determine the frequency (in Hz) at which the magnitude of the gain is 20 db. Solution: When the gain is 2, the frequency is 286mHz = 0.286Hz. (e) Frequency scale the circuit in (b), so that the frequency at which the gain is 20 db is shifted to 100,000 rad/s (= 100 krad/s). Solution: From part (d) we know that the gain crosses 2 at approximately 0.286Hz which equals 1.797rad/s. In order to move the 2 crossing to 100, 000rad/s, we need a frequency scaling factor of k f = 100,000/ ,649. This means that we should start with the magnitude-scaled circuit of part (a), and divide the capacitors and inductors by k f. The resulting frequency-scaled circuit is shown below. 1kΩ 72mH v 1 (t) 9nF 6nF 1kΩ v 2 (t) (f) Simulate the circuit in (e) from 1000 rad/s (= 1 krad/s) to 10,000,000 rad/s (= 10 Mrad/s) to verify that the gain at 100 krad/s is indeed 20 db. Solution: From the simulation results (shown below) we see that the the gain crosses 2 at approximately 15.9kHz which equals 100, 000rad/s. Dr. Vahe Caliskan 2 of 8 Posted: February 8, 2013

3 Dr. Vahe Caliskan 3 of 8 Posted: February 8, 2013

4 Problem 2 (a) Magnitude (impedance) scale the circuit so that all capacitors are 100µF. Solution: Since all of the capacitors are 1F, we need to magnitude scale by k m = 1F/100µF = 10,000; therefore, we should multiply the resistors and the inductor by k m and divide the capacitors by k m. The resulting magnitude-scaled circuit is shown below. 10kΩ 100µF 100µF v 1 (t) 5kH 20kΩ v 2 (t) (b) Determine the frequency response of the circuit in (a) by using a circuit simulator. If you simulate the circuit from 0.01 rad/s to 100 rad/s, the plots should look similar to Homework 1, Problem 2(c). Solution: As in Problem 1, the circuit simulation should run from 1.59 mhz 15.9 Hz. The LTspice schematic and the resulting frequency response are shown below. in R1 C1 C2 out {1*km} V1 AC 1 0 {1/km} {1/km} L1 {km/2} R2 {2*km} HW 2, 2(b).ac dec 100 {0.01/(2*pi)} {100/(2*pi)}.param km=10k mHz, mHz 100mHz 1Hz 10Hz Dr. Vahe Caliskan 4 of 8 Posted: February 8, 2013

5 mHz 100mHz 1Hz 10Hz -420 (c) From the simulation results in (b), determine the frequency (in Hz) at which the magnitude of the gain is 10 db. Solution: From the gain frequency response, we see that when the gain is 1, the frequency is 137.6mHz = Hz. (d) Frequency scale the circuit in (b), so that the frequency at which the gain is 10 db is shifted to 100,000 rad/s (= 100 krad/s). Solution: From part(c) we know that the gain crosses 1 at approximately Hz which equals rad/s. In order to move the 1 crossing to 100, 000rad/s, we need a frequency scaling factor of k f = 100,000/ ,664. This means that we should start with the magnitude-scaled circuit of part (a), and divide the capacitors and inductors by k f. The resulting frequency-scaled circuit is shown below. 10kΩ 865pF 865pF v 1 (t) 43.2mH 20kΩ v 2 (t) (e) Simulate the circuit in (e) from 1000 rad/s (= 1 krad/s) to 10,000,000 rad/s (= 10 Mrad/s) to verify that the gain at 100 krad/s is indeed 10 db. Solution: From the simulation results (shown below) we see that the the gain crosses 1 at approximately 15.9kHz which equals 100, 000rad/s. Dr. Vahe Caliskan 5 of 8 Posted: February 8, 2013

6 in R1 {1*km} C1 {1/km/kf} C2 {1/km/kf} out HW 2, 2(e) V1 AC 1 0 L1 {km/2/kf} R2 {2*km}.ac dec 100 {1k/(2*pi)} {10meg/(2*pi)}.param km=10k, kf= KHz, KHz 10KHz 100KHz 1MHz Dr. Vahe Caliskan 6 of 8 Posted: February 8, 2013

7 Problem 3 Draw the asymptotic magnitude Bode plots for the following transfer functions. Be sure to indicate important horizontal frequencies (in rad/s), vertical gain levels (in db) and slopes (in db/dec). Solution: First simplify the transfer functions to put them in a more appropriate form for drawing the asymptotic Bode plots. (a) H(s) = (s1,000)(s8,000) (s4,000)(s40,000) = 0.05(1s/1,000)(1s/8,000) (1s/4,000)(1s/40,000) Solution: The low freqency gain is dB. The vertical difference between 1000 and 4000 rad/s is computed by 2/dec log(4000/1000)dec = 14dB(this is the slope times the number of decades). Similarly the vertial difference between 8000 and rad/s is given by 20 log(40000/8000) = 14dB. Using these differences and the starting point of 26dB the horizontal levels of 14dB and can be computed as shown. 2/dec 14dB 14dB 2/dec 26dB 12dB rad/s (b) H(s) = (s40,000)(s1,000)2 (s400)(s10,000) 2 = (1s/40,000)(1s/1,000)2 (1s/400)(1s/10,000) 2 Solution: The low freqency gain is 1. The vertical difference between 1000 and 400 rad/s is computed by 2/dec log(1000/400)dec = 8dB; therefore, the gain at 1000 rad/s is 8dB lower than or just 8dB. The vertical difference between and 1000 rad/s is just 2 since the frequency difference is 1 decade. The vertical difference between and rad/s is computed by 20log(40000/10000)= 12dB; therefore, the gain at rad/s is 12dB lower than 12dB or just. 12dB 2/dec 2/dec 2/dec 2 8dB 8dB rad/s Dr. Vahe Caliskan 7 of 8 Posted: February 8, 2013

8 (c) H(s) = 10(s5,000)4 (s20,000) 4 (s500) 4 (s200,000) 4 = 10(1s/5,000)4 (1s/20,000) 4 (1s/500) 4 (1s/200,000) 4 Solution: The low freqency gain is The vertical difference between 500 and 5000 rad/s (also and rad/s) is one decade; therefore the vertical differences are 8 since the magnitude of the slopes are 8/dec /dec 8/dec rad/s (d) H(s) = 500s 2 (s100) 2 (s10,000) 2 = s 2 (1s/100) 2 (1s/10,000) 2 Solution: The low freqency behavior is proportional to s 2 which is a line with slope of 4/decwith gainof 186dBat 1rad/s. Going from1to 100rad/s(2 decades), the gainincreases by 8 due to the 4/dec slope. As a result, the slope flattens out to 106dB at 100rad/s. At 10000rad/s, the slope starts falling off at 4/dec due to the double pole. 106dB 4/dec 8 4/dec 186dB rad/s Dr. Vahe Caliskan 8 of 8 Posted: February 8, 2013