HW4

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Texas Tech University *

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3303

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Electrical Engineering

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Dec 6, 2023

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HW 4-1 (30 pts ). Use an RC circuit to realize a low-pass filter that can attenuate a signal at 1.5kHz by 35dB. 1) Determine the cut-off frequency of the low pass filter 2) Given that R=1kΩ, find the value of C 3) Use one of the following circuit simulators to produce a bode diagram for the magnitude response of the RC circuit. (Note: please take a screenshot of Bode diagram and incorporate it into your work in a single PDF or WORD file). LTspice: AC sweep analysis https://www.youtube.com/watch?v=fziUQaVQxA4 PSpice: AC sweep analysis https://www.youtube.com/watch?v=4qe9dIOsDfg Multisim live: AC sweep analysis https://www.multisim.com/help/simulation/grapher/ac-sweep/ Solution: 1) The signal’s angular frequency ω = 2πx1500 = 9.42x10 3 rad/s. To attenuate the signal by 35 dB, using the low-pass filter’s Bode diagram, the following equation is obtained: 20 x log 10 (ω/ ω c ) = 35 dB, (10 pts) in which ω c represents the cut-off frequency of the low-pass filter. By solving the equation, ω c = 167.6 rad/s, which is 26.7 Hz. (5 pts) 2) The cut-off frequency of the RC low-pass filter is 1/(RC). Therefore, C = 1/(Rω c ) = 5.97 µF (5 pts) 3) Multisim Live schematic and AC sweep result: (10 pts)

Cursor 1 is at the cut-off frequency (26.7 Hz), and cursor 2 is at the signal’s frequency. Note the dB value: -3 dB and -35 dB, resp. HW 4-2 (30 pts ). Using the butterfly diagram (Fig 7-28), compute the 4-point DFTs of each of the following signals. Then, compare your results with those obtained by using the fft() function of Matlab (Note: please include the Matlab code and result) 1) {16 , 8, 12, 4} Solution: 1) Following the butterfly diagram for 4-point fft: ( 10 pts ).

The result is {40, 4-4j, 16, 4+4j} ( 10 pts ). Matlab result (1 0 pts ). “x = [16, 8, 12, 4]; fft(x)” ans = 40.0000 + 0.0000i 4.0000 - 4.0000i 16.0000 + 0.0000i 4.0000 + 4.0000i HW 4-3 (40 pts ). For the continuous-time signal v ( t ) in HW2-Q3, 1) Determine the Nyquist rate by considering signal’s 95% energy bandwidth. 2) Assume the above Nyquist rate is used to sample v ( t ), find the DTFT of the discrete-time signal obtained by sampling v ( t ). 3) Find the DFT for the first 10 samples. 4) Use Matlab to plot the magnitude spectra of DTFT and DFT. Further, plot and collate the magnitude spectrum of v(t) with those obtained by DTFT and DFT. (Note: please take a screenshot of the plot and incorporate it into your work in a single PDF or WORD file) 5) Explain how the discrepancy in the magnitude spectra is caused, and discuss how to mitigate such discrepancy. Solution: 1) The signal’s 95% energy bandwidth is given by: B = 1000 × tan ( 95% × π 2 ) = 1.27 × 10 4 rad/s Thus, the Nyquist rate is given by: 2 B /( 2 π )= 4.04 × 10 3 Hz. ( 5 pts ) 2) The sampling rate as f s = 4.04 × 10 3 Hz, and the sampling period T s = 2.47 × 10 − 4 s . Then, the samples of v(t) is given by:

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v [ n ] =− 2 e − 1000 nT s u [ n ] The DTFT of v(t) is given by: − 2 e − 1000 nT s e − jΩn = ¿ − 2 ∑ 0 + ∞ ( e −( 1000 T s + j Ω ) ) n v [ n ] e − jΩn = ¿ ∑ 0 + ∞ ¿ V ( Ω ) = ∑ − ∞ + ∞ ¿ Then, using the formula for the sum of geometric series, V ( Ω ) = − 2 1 − e −( 1000 T s + jΩ ) ( 5 pts ) 3) The first N=10 samples are v [ n ] =− 2 e − 1000 nT s ,n = 0,1 ,…, 9 Apply 10-point DFT to v [ n ] , yielding: − 2 e − 1000 nT s e − jk Ω 0 n = ¿ − 2 ( 1 − e − 10 ( 1000 T s + jk Ω 0 ) ) 1 − e −( 1000 T s + jk Ω 0 ) V [ k ]= ∑ n = 0 9 ¿ ( 5 pts ) 4) The magnitude of V ( Ω ) and V [ k ] are illustrated in the figure below. Further, the FT of v ( t ) is given by: V ( ω ) = − 2 1000 + jω In order to collate its magnitude spectral plot to those of DTFT and DFT, V ( ω ) is scaled by f s and ω is replaced by Ω = ωT s . The result is shown in the figure below. Discrepancy could be observed among the magnitude spectra by FT, DTFT, and DFT.

5) The discrepancy between the magnitude spectra of DTFT and FT is caused by sampling ( 5 pts ), and it can be mitigated by utilizing a larger sampling rate ( 5 pts ). The discrepancy between the magnitude spectra of DTFT and FT is caused by windowing ( 5 pts ), and it can be mitigated by adopting a larger window size ( 5 pts ). With the above insight, if the signal’s 99% energy bandwidth is considered for sampling, and the first N=100 samples are used to compute the DFT, the discrepancy between the magnitude spectra of FT, DTFT, and FT could be significantly mitigated, as illustrated in the figure below: (5 pts )

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