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Computer System - Introductory Microcomputer Interfacing Laboratory - Solved Exam, Exams of Microcomputers

AMET UniversityMicrocomputers

Main points of this past exam are: Computer System, Sampling Frequency, Number of Samples, Butterworth, Anti-Aliasing Filter, Hanning, Frequency Range

Typology: Exams

2012/2013

Uploaded on 03/22/2013

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145M April 12, 2006 page 1 S. Derenzo
Solutions for Midterm #2 - EECS 145M Spring 2006
1.1 Filter gain >0.99 for frequencies <78,400 Hz
1.2 Filter gain <0.01 for frequencies >177,800 Hz
1.3 S = M โˆ†t = M/fs = 216/218 Hz = 0.25 s
1.4 H0 corresponds to 0 Hz (d.c.); H1 corresponds to 1/S = 4 Hz
1.5 The FFT produces coefficients Hn, where n = 0 to Mโ€“1. Therefore, the coefficient with the
highest index is HM-1 or H65,535, which corresponds to 4 Hz.
[2 points off for HM and 0 Hz] [3 points off for HM and 218 Hz]
1.6 The FFT coefficient that corresponds to the highest frequency is HM/2 or H32,768. The
corresponding frequency is (M/2)/S = 131,072 Hz
1.7 For a 4,000 Hz sinewave, the primary FFT coefficients are H1000 and HM-1000. Additional
neighboring coefficients H999, H1001, HM-999, and HM-1001 are non-zero (actually half the value
of the primary coefficients) due to the side lobes produced by the Hann window.
[2 points off for omitting side lobes] [2 points off for omitting HM-999, HM-1000, and HM-1001]
1.8 For a 4,000 Hz symmetric square wave, a sequence of harmonics will appear at odd
multiples of the 4,000 Hz fundamental. So Hk1000 and HM-k1000 would be non-zero, and the
Hann side lobes would be at Hk1000-1, Hk1000+1, HM-k1000-1, and HM-k1000+1.
[1 point off for omitting side lobes] [4 points off for omitting harmonics]
1.9 For a 4,002 Hz sinewave, H1000, H1001, HM-1000, and HM-1001 would be non-zero and of equal
magnitude, and the Hann side lobes would appear at H999, H1002, HM-999 and HM-1001.
[2 points off for omitting side lobes] [2 points off for omitting HM-1000 and HM-1001]
[4 points off for stating that all coefficients are non-zero]
1.10 The primary 4,000 Hz sinewave would produce non-zero values at H999, H1000, and H1001.
A second smaller sinewave of slightly higher frequency 4,000 + 4m Hz would produce non-
zero values at H1000+m-1, H1000+m, and H1000+m+1 (there are also complex conjugate coefficients
at HM-1000, etc.). For the smaller sinewave to appear as a separate peak, there must be a
valley between the coefficient H1001 and the coefficient at H1000+m, which requires 1000 + m
> 1002, or m >2. The smallest value of m we can have is 3, which corresponds to a
frequency 12 Hz above 4,000 Hz.
[4 points off for 4 Hz] [3 points off for 8 Hz] [both 12 Hz and 16 Hz were accepted]
1000 1002 1004
1.11 A sinewave of frequency 4M โ€“ 84,000 Hz = 178,144 Hz will produce non-zero
coefficients at H20999, H21000, H21001, HM-20999, HM-21000, and HM-21001.
pf3

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Solutions for Midterm #2 - EECS 145M Spring 2006

1.1 Filter gain >0.99 for frequencies <78,400 Hz 1.2 Filter gain <0.01 for frequencies >177,800 Hz 1.3 S = M โˆ†t = M/fs = 216 /2^18 Hz = 0.25 s 1.4 H 0 corresponds to 0 Hz (d.c.); H 1 corresponds to 1/S = 4 Hz 1.5 The FFT produces coefficients Hn, where n = 0 to Mโ€“1. Therefore, the coefficient with the highest index is HM-1 or H65,535, which corresponds to 4 Hz. [2 points off for HM and 0 Hz] [3 points off for HM and 2^18 Hz] 1.6 The FFT coefficient that corresponds to the highest frequency is HM/2 or H32,768. The corresponding frequency is (M/2)/S = 131,072 Hz 1.7 For a 4,000 Hz sinewave, the primary FFT coefficients are H 1000 and HM-1000. Additional neighboring coefficients H 999 , H 1001 , HM-999, and HM-1001 are non-zero (actually half the value of the primary coefficients) due to the side lobes produced by the Hann window. [2 points off for omitting side lobes] [2 points off for omitting HM-999, HM-1000, and HM-1001] 1.8 For a 4,000 Hz symmetric square wave, a sequence of harmonics will appear at odd multiples of the 4,000 Hz fundamental. So Hk1000 and HM-k1000 would be non-zero, and the Hann side lobes would be at Hk1000-1, Hk1000+1, HM-k1000-1, and HM-k1000+1. [1 point off for omitting side lobes] [4 points off for omitting harmonics] 1.9 For a 4,002 Hz sinewave, H 1000 , H 1001 , HM-1000, and HM-1001 would be non-zero and of equal magnitude, and the Hann side lobes would appear at H 999 , H 1002 , HM-999 and HM-1001. [2 points off for omitting side lobes] [2 points off for omitting HM-1000 and HM-1001] [4 points off for stating that all coefficients are non-zero] 1.10 The primary 4,000 Hz sinewave would produce non-zero values at H 999 , H 1000 , and H 1001. A second smaller sinewave of slightly higher frequency 4,000 + 4m Hz would produce non- zero values at H1000+m-1, H1000+m, and H1000+m+1 (there are also complex conjugate coefficients at HM-1000, etc.). For the smaller sinewave to appear as a separate peak, there must be a valley between the coefficient H 1001 and the coefficient at H1000+m, which requires 1000 + m

1002, or m >2. The smallest value of m we can have is 3, which corresponds to a frequency 12 Hz above 4,000 Hz. [4 points off for 4 Hz] [3 points off for 8 Hz] [both 12 Hz and 16 Hz were accepted] 1000 1002 1004 1.11 A sinewave of frequency 4M โ€“ 84,000 Hz = 178,144 Hz will produce non-zero coefficients at H 20999 , H 21000 , H 21001 , HM-20999, HM-21000, and HM-21001.

M = 216 = 65,536. M โ€“ 21,000 = 44,536.

A sinewave of frequency 84,000 Hz will produce non-zero coefficients at exactly the same frequency indexes. This is an example of how a higher frequency can alias to a lower frequency. However, the 84,000 Hz sinewave will be only slightly reduced by the anti- aliasing filter (gain >0.90, while the 178,144 Hz sinewave will be greatly reduced (gain โ‰ˆ0.01). So the coefficients will be about 100 times smaller for the 178,144 Hz sinewave. [3 points off for not stating the non-zero coefficients] [1 point off for omitting H 20999 , H 21001 , HM-20999, and HM-21001] [3 points off for stating that the magnitudes are the same for sampling 178,144 and 84, Hz] 1.12 To reduce the answer to 1.10 by a factor of two (i.e. to 6 Hz), sample for twice as long. [2 points off for doubling the sampling frequency, which increases the number of Fourier coefficients but not the frequency spacing โˆ†f = 1/S] [2 points off for reducing the sampling frequency] 2.1 v = 3 m/s f = 100 kHz (1 + 3/300) = 101,000 Hz v = 30 m/s f = 100 kHz (1 + 30/300) = 110,000 Hz v = 60 m/s f = 100 kHz (1 + 60/300) = 120,000 Hz The exact calculation gives 101,010; 111,111; 111,235; and 125,000 Hz; both were OK. 2.2 โˆ†f = 100 Hz, so the minimum length of the sampling window must be S = 1/โˆ†f = 0.01 s 2.3 Sample for a whole number of 100 kHz cycles. Since a windowing function was not used, this completely eliminates spectral leakage for the 100 kHz primary signal. The much smaller echo signal would still have spectral leakage, but the peak would not be obscured. [2 points off for increasing the sampling window S, this helps somewhat, but for a non- integer number of 100 kHz cycles, the spectral leakage will be severe because the primary has 100 times the amplitude of the echo signal] [4 points off for using an anti-aliasing filter since aliasing is only a problem for the white noise] [4 points off for reducing the sampling frequency, which does not reduce spectral leakage] [4 points off for using the Hann window, which was specifically excluded in the problem statement] 2.4 For n = 8 and G 1 = 0.9, f 1 /fc = 0.913. At 60 m/s f 1 = 120 kHz, and fc = 135.4 kHz 2.5 For n = 8 and G 2 = 0.01, f 2 /fc = 1.778, and f 2 = 233.7 kHz The minimum sampling frequency fs = f 1 + f 2 = 353.7 kHz 2.6 The minimum number of samples is the product of the minimum sampling window (S = 0.01 s from part 2.2) and the minimum sampling frequency (fs = 353.7 kHz from part 2.5) = 3537 samples. The next power of two is 4096 samples. 2.7 The echo is 100 times smaller than the 100 kHz tone and the white noise is 10 times smaller than the peak echo. The low pass filter falls off as 1/f^8. [2 points off if no FFT frequency index] [2 points off if no frequency scale in Hz] [2 points off if white noise not shown] [4 points off if primary tone or echo not shown] [3 points off for blank vertical scale] [2 points if vertical scale is not numbered] [2 points off if effect of the low-pass filter not shown]