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Main points of this past exam are: Tristate Buffer, Transition Voltages, Converter, Digital Filter, Anti-Aliasing Filter, Fourier Frequency Convolution Theorem, Nyquist Sampling Theorem
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College of Engineering Electrical Engineering and Computer Sciences Department 145M Microcomputer Interfacing Lab Final Exam Solutions May 17, 2002
1a Tri-State buffer: Digital circuit with input and output lines, and a select line. When the select line is in one logic state, output = input. Otherwise the output is in a high-impedance state and neither drives nor loads any circuit connected to it. 1 b Transition voltages (of an A/D converter): Specific input voltages at which the output switches between one output number and the next. 1 c Digital filter: Filter whose digital output is a linear combination of previous digital input and output values. The finite impulse response (FIR) filter does not depend on previous output values. The infinite impulse response (IIR) depends on previous output values. 1 d Anti-Aliasing Filter: Analog filter designed to eliminate high frequencies above the Nyquist sampling limit (one-half the sampling frequency) that would otherwise appear as erroneous lower frequencies in the sampled data. 1 e Fourier frequency convolution theorem: The Fourier transform of the simple product of two functions is the convolution of the Fourier transforms of the two functions. [2 points off for giving Fourier convolution theorem] [OK for giving both] 1 f Nyquist sampling theorem: To recover a waveform from its sampled values, the highest frequency present must be less than one-half the sampling frequency.
2a f = a + b σ (^) f f σ (^) a σ (^) b σ (^) a σ b a
f b
2
2 2
2 = ∂^2 2 ∂
σ (^) f = σ (^) a^2 +σ b^2
2 b f N
ai i
=
1
σ (^) f σ σ σ i i
N a a i
N a
f a N
2
1
2 2 2 1
= =
σ (^) f = σ a / N
3a TR, FL, 1B 3 b SA, DS, HF 3 c FL 3 d SA, HF, 1B [DS would require 20 kHz x 2^12 = 1.3 GHz clock speed] 3 e DS, 1B 3 f SA, HF [DS takes a time average and does not need a steady input] [1 point off for each wrong answer] [1 point off for each missing correct answer]
Problem 4 Since 5 square waves were sampled, we expect non-zero magnitudes at frequency indexes at odd multiples of 5 and 128 minus odd multiples of 5. These are n = 5 and 123 (first harmonic at 15 Hz), n = 15 and 113 and (third harmonic at 45 Hz), n =25 and 103 (fifth harmonic at 75 Hz), etc. The additional large magnitudes at n = 20 and 108 corresponds to a pure 60 Hz harmonic, which must be due to power line interference in the lab. If magnitude at n = 20 were due to distortion generating even harmonics, additional magnitudes at n = 10, 30, 40, 50, etc. would also be expected. [3 points off if odd harmonics not identified] [3 points off if magnitudes above index 64 not explained] [4 points off if 60 Hz electromagnetic interference not identified]
5a Multiplying the data with a square window convolves the true frequency spectrum with the Fourier transform of the square wave, which produces long-range spectral leakage that falls off as 1/freqency. [5 points off for not explaining what is happening in the frequency domain] 5 b To recover the original square-wave-windowed time samples, just use the inverse transform. Then multiply the time samples with the raised cosine window and take the fft. [10 points off for convolving the original fft with the fft of the raised cosine and not removing effects of the square window] 5 c Multiplying the data with the raised cosine window convolves the true frequency spectrum with the Fourier transform of the raised cosine window, which produces short-range spectral leakage (nearest-neighbor frequency amplitudes) but very small long-range spectral leakage. [5 points off for not explaining what is happening in the frequency domain]
6a Essential features for the circuit design:
Digital I/O requirements
[5 points off for not stopping timers before reading final values- this can cause erroneous data when read occurs when bits are changing] [5 points off if touch plate switches are not connected to input port and there is no workable plan to detect when the swimmers finish- this design causes many redundant numbers to be written to the screen and to disk] [no deduction for having the computer start the counters rather than the pistol switch, but using the computer for this purpose is a poor design due to unpredictable system interrupt delays]
145M Final Exam Grades:
Problem 1 2 3 4 5 6 Total Average 28.8 24.2 24.2 11.9 29.4 56.6 175. rms 1.1 2.0 3.7 2.2 7.9 4.4 11. Maximum 30 25 30 15 40 60 200
145M Numerical Grades:
Lab total Lab Partic. Midterm #1 Midterm #2 Final Total Average 446.0 89.9 82.5 86.8 175.0 8 8 0. 2 rms 35.0 5.9 12.7 10.2 11.4 4 5. 4 Maximum 500 100 100 100 200 1 0 0 0
145M Letter Grade Distribution
Letter Grade Course Totals (1000 max) A+ 932 A 914, 930 A– 891, 898, 906. B+ 867, 872, 872.5, 875 B B- 839. C+ C 765 C– D+ D