Buffer - Introductory Microcomputer Interfacing Laboratory - Solved Exam, Exams of Microcomputers

Main points of this past exam are: Buffer, Tri-State, Fourier Convolution Theorem, Transition Voltages, Converter, Infinite Impulse, Response Digital Filter

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2012/2013

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145M Final Exam Solutions page 1 May 13, 2006 S. Derenzo
UNIVERSITY OF CALIFORNIA
College of Engineering
Electrical Engineering and Computer Sciences Department
145M Microcomputer Interfacing Lab
Final Exam Solutions May 13, 2006
1.1 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.
1.2 Fourier Convolution Theorem: The Fourier transform of the convolution of two
functions is the simple product of the Fourier transforms of the two functions
[2 points off for stating the Fourier Frequency Convolution Theorem]
1.3 Transition Voltages (A/D Converter): Specific input voltages at which the output
switches between one output number and the next.
1.4 Infinite Impulse Response Digital Filter: A digital filter whose output depends on
previous outputs
Note: The above definitions were taken directly from the textbook.
2.1
r=a/br/a=1 / br/b=a/b2
σ
r
2=
σ
a
2(f/a)2+
σ
b
2(f/b)2=
σ
a
2(1/ b2)+
σ
b
2(a2/b4)
σ
r
2=104a2(1 / b2)+104b2(a2/b4)=2×104(a2/b2)
σ
r=0.01414( a/b)=0.01414 r
3.1 A periodic waveform is the convolution of a time-limited waveform and a comb of delta
functions. The Fourier transform of a periodic waveform is the simple product of the
Fourier transform of the time-limited waveform and the Fourier transform of a comb of delta
functions. The latter is a comb of delta functions in the frequency domain so the simple
product is nonzero only at discrete frequencies. See Figure 5.26 of the textbook.
[7 points off if the periodic function is not described as the convolution of a time-limited
function and a train of delta functions]
3.2 Sampling in the time domain is equivalent to a simple product of the waveform and a comb
of delta functions in the time domain. The Fourier transform of the sampled data is the
convolution of the Fourier transform of the wavefunction and the Fourier transform of the
comb of delta functions. Since the latter is a comb of delta functions in the frequency
domain, overlap will occur unless the original waveform is not frequency limited. This
overlap causes aliasing and can be eliminated by increasing the sampling frequency or by
bandwidth limiting the waveform so that the highest frequencies persent are sampled at least
twice per cycle. See Figure 5.29 of the textbook.
[10 points off for only stating the Nyquist Theorem]
[5 points off for stating that aliasing is caused by frequency overlap but not stating that
sampling is equivalent to multiplying by a train of delta function in the time domain]
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UNIVERSITY OF CALIFORNIA

College of Engineering Electrical Engineering and Computer Sciences Department 145M Microcomputer Interfacing Lab Final Exam Solutions May 13, 2006 1.1 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. 1.2 Fourier Convolution Theorem: The Fourier transform of the convolution of two functions is the simple product of the Fourier transforms of the two functions [2 points off for stating the Fourier Frequency Convolution Theorem] 1.3 Transition Voltages (A/D Converter): Specific input voltages at which the output switches between one output number and the next. 1.4 Infinite Impulse Response Digital Filter: A digital filter whose output depends on previous outputs Note: The above definitions were taken directly from the textbook. 2.1 r = a / b ∂r / ∂a = 1 / b ∂r / ∂b = −a / b^2 σr^2 = σ (^) a^2 (∂f / ∂a)^2 + σb^2 (∂f / ∂b)^2 = σ (^) a^2 ( 1 / b^2 ) + σb^2 (a^2 / b^4 ) σr^2 = 10 −^4 a^2 ( 1 / b^2 ) + 10 −^4 b^2 (a^2 / b^4 ) = 2 × 10 −^4 (a^2 / b^2 ) σr = 0.01414(a / b) = 0.01414 r 3.1 A periodic waveform is the convolution of a time-limited waveform and a comb of delta functions. The Fourier transform of a periodic waveform is the simple product of the Fourier transform of the time-limited waveform and the Fourier transform of a comb of delta functions. The latter is a comb of delta functions in the frequency domain so the simple product is nonzero only at discrete frequencies. See Figure 5.26 of the textbook. [7 points off if the periodic function is not described as the convolution of a time-limited function and a train of delta functions] 3.2 Sampling in the time domain is equivalent to a simple product of the waveform and a comb of delta functions in the time domain. The Fourier transform of the sampled data is the convolution of the Fourier transform of the wavefunction and the Fourier transform of the comb of delta functions. Since the latter is a comb of delta functions in the frequency domain, overlap will occur unless the original waveform is not frequency limited. This overlap causes aliasing and can be eliminated by increasing the sampling frequency or by bandwidth limiting the waveform so that the highest frequencies persent are sampled at least twice per cycle. See Figure 5.29 of the textbook. [10 points off for only stating the Nyquist Theorem] [5 points off for stating that aliasing is caused by frequency overlap but not stating that sampling is equivalent to multiplying by a train of delta function in the time domain]

3.3 Multiplying by a function that tapers to zero with zero slope at the edges of the sampling window reduces the discontinuities that produce unwanted frequency components. This is also equivalent to convolving the frequency spectrum of the untruncated waveform with the Fourier transform of the windowing function, which is designed to have only low frequency components. 3.4 Not using a windowing function is equivalent to multiplying the waveform with a rectangular time function which is equivalent to convolving the frequency spectrum of the untruncated waveform with the Fourier transform of the square wave, which has long range frequency components. 4.1 Required elements: 256 sensor circuits with 16-bit digital outputs 256 digital deglitching circuits, each consisting of 16 delays, 16 exclusive OR circuits, one 16- input OR circuit, and one16-bit transparent latch. (Similar to Figure 3.37 of the textbook) 256 tri-state buffers, each with 16 inputs and 16 outputs one address decoder with input connected to 8 bits of output port and output used to select among the 256 tri-state buffers 32-bit I/O port, with 16 bits set to input and 8 bits set to output computer 4.2 When one of the sensor circuit output changes the changed bits will generate different inputs to the corresponding exclusive OR circuits and these output will be combined in the OR circuit to put the transparent latch into hole mode during the glitch. [10 points off if glitch rejection not described] 4.3 (1) Set 16 bits to input and 8 bits to output, set m = 0 (2) Wait until the seconds counter changes (3) set n = 0 (4) select tri-state n to switch from high impedance to transparent mode (5) read 16-bit input bus (6) select the milliseconds timer and read the 16 bits (7) select the minutes timer and read the 16 bits (8) store the data and time into an array[n, m] (9) set n to n + 1 (10) loop back to (4) until n = 256 (11) set m to m + 1 and loop back to (2) 5.1 f 1 = 20 kHz G 1 = 0. choose n = 8 f 1 /fc = 0.784 and fc = 25.5 kHz G 2 = 0.001 f 2 /fc = 2.371 f 2 = 60.5 kHz fs > f 1 + f 2 = 80.5 kHz [5 points off for an incorrect or missing anti-aliasing filter, or using a sampling frequency of 40 kHz]

145M Final Exam Grades: Problem 1 2 3 4 5 Total Average 38.0 16.8 31.8 39.2 32.9 158. rms 3.0 4.2 10.4 7.3 7.2 19. Maximum 40 20 45 50 45 200 145M Numerical Grades: Short labs Long labs Lab Partic. Midterm #1 Midterm #2 Final Total Average 77.5 394.4 100 93.8 85.1 158.7 912. rms 1.7 7.9 0 3.9 14.0 19.3 30. Maximum 105 400 100 100 100 200 1005 Note: The average of labs 1, 8, 10, 22, and 24a was 1 point per lab higher than the average of labs 2, 9, 21, 23, and 24b (omitting the lowest long lab grade). This was due to the nature of the labs and small differences in grading standards. One bonus point was added to the long lab total for each of labs 2, 9, 21, 23, and 24b. 145M Letter Grade Distribution Letter Grade Course Totals (1000 max) A+ 961 A 958 A– 921, 922 B+ 902, 908, 912 B 891, 894 B- C+ 857 C C– D+ D