Hold Amplifier - Introductory Microcomputer Interfacing Laboratory - Solved Exam, Exams of Microcomputers

Main points of this past exam are: Hold Amplifier, Data-Acquisition, Analog Waveform, Transfer, Filtering, Oscillating, Frequency Indices

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2012/2013

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UNIVERSITY OF CALIFORNIA
College of Engineering
Electrical Engineering and Computer Sciences Department
145M Microcomputer Interfacing Lab
Final Exam Solutions May 18, 1992
1A Nyquist Theorem: To recover a waveform from its sampled values, the highest frequency
present must < one-half the sampling frequency.
1B Discrete Fourier Transform: Transform for determining the amplitude of the frequency
components of a periodically sampled waveform.
1C Differential Linearity Error (of an A/D converter): Difference between the spacing of neigh-
boring transition voltages and their average spacing. [4 points off if step size or transition volt-
age not mentioned] [4 points off if absolute or relative accuracy was defined]
1D Anti-Aliasing Filter: Low-pass filter used to block frequencies above one-half the sampling
frequency and thereby prevent aliasing.
1E Power Amplifier: Amplifier having high power or current output, and required to drive an
actuator such as a speaker or heater. [2 points off if high power or current output not
mentioned]
1F Digital Filter: Filter whose output is a linear combination of previous input and output values.
2A 110V (rms)
156V (p-p)
60 Hz
15R
R
Low Pass Filter
fc 4 kHz
Data
acquisition
circuit
Micro-
computer
Screen
display
Disk
Keyboard
data
data
ready
10V p-p
[3 points off if 156 V p-p sent directly into acquisition circuit]
[3 points off if low-pass filter omitted]
2B f = 0.01 Hz, S = 1/f = 100 sec, N = 100 sec x 10 kHz = 106 samples
(S = 50 sec, N = 5 x 105 samples also acceptable)
2C F0 corresponds to 0 Hz or dc.
2D 60 Hz corresponds to F6000 and FN-6000 where N = 106
2E Since the distortion has a period of 60 Hz, only multiples of 60 Hz will be present.
The highest harmonic nmax that can pass the anti-aliasing filter is 5000 Hz/ 60Hz 80.
The nth harmonic will be at F6000n and F N – 6000n
[3 points off if only frequencies given]
[6 points off if only F6000 and FN– 6000 given]
[6 points off if answer says that all Fourier amplitudes are non-zero]
2F 1) Data acquisition circuit samples waveform, digitizes, and sets data ready line
2) When program detects data ready line, it reads data, stores data, and resets data ready line
3) Loop back to step 1 until 106 values taken
4) Multiply values by Hanning window
5) Compute the FFT
[2 points off for each step missing]
145M Final Exam Solutions (B) page 1 May 18, 1992 S. Derenzo/L. Bushnell
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UNIVERSITY OF CALIFORNIA

College of Engineering Electrical Engineering and Computer Sciences Department

145M Microcomputer Interfacing Lab

Final Exam Solutions May 18, 1992

1A Nyquist^ Theorem:^ To recover a waveform from its sampled values, the highest frequency

present must < one-half the sampling frequency.

1B Discrete^ Fourier^ Transform:^ Transform for determining the amplitude^ of^ the^ frequency

components of a periodically sampled waveform.

1C Differential Linearity Error^ (of an A/D converter):^ Difference between the spacing of neigh-

boring transition voltages and their average spacing. [4 points off if step size or transition volt- age not mentioned] [4 points off if absolute or relative accuracy was defined]

1D Anti-Aliasing^ Filter:^ Low-pass filter used to block frequencies above one-half the sampling

frequency and thereby prevent aliasing.

1E Power^ Amplifier:^ Amplifier having high power or current output, and required to drive an

actuator such as a speaker or heater. [2 points off if high power or current output not mentioned]

1F Digital Filter:^ Filter whose output is a linear combination of previous input and output values.

2A 110V (rms)

156V (p-p) 60 Hz 15R

R

Low Pass Filter fc ≈ 4 kHz

Data acquisition circuit

Micro- computer

Screen display

Disk

Keyboard

data

data ready

≈10V p-p

[3 points off if 156 V p-p sent directly into acquisition circuit] [3 points off if low-pass filter omitted]

2B ∆f = 0.01 Hz, S = 1/∆f = 100 sec, N = 100 sec x 10 kHz = 10

(^6) samples (S = 50 sec, N = 5 x 10^5 samples also acceptable)

2C F^0 corresponds to 0 Hz or dc.

2D 60 Hz corresponds to F^6000 and FN-6000^ where N = 10

6

2E Since the distortion has a period of 60 Hz, only multiples of 60 Hz will be present.

The highest harmonic nmax that can pass the anti-aliasing filter is ≈ 5000 Hz/ 60Hz ≈ 80. The nth harmonic will be at F6000n and FN – 6000n [3 points off if only frequencies given] [6 points off if only F 6000 and FN– 6000 given] [6 points off if answer says that all Fourier amplitudes are non-zero]

2F 1) Data acquisition circuit samples waveform, digitizes, and sets data ready line

  1. When program detects data ready line, it reads data, stores data, and resets data ready line
  2. Loop back to step 1 until 10^6 values taken
  3. Multiply values by Hanning window
  4. Compute the FFT [2 points off for each step missing]

3A The first harmonic could be anywhere in the 59.9 to 60.1 Hz range, which corresponds to 20

potentially non-zero Fourier coefficients (40 filters) from F 5990 to F 6010. The range of the 80th harmonic will involve 40 x 80 = 3200 filters. The average number of filters per harmonic = 1600. The total number of filters is then 1600 x 80 = 128,000 (much less than the total number of 500,000 real plus 500,000 imaginary components) [4 points off for an answer of 2-5 filters] [3 points off for answers of 10-40 filters or 10^6 filters] [2 points off for an answer from 3000 to 7000 filters] [2 points off if the 59.9 to 60.1 Hz range was neglected]

3B The real FIR filter is given by

Fn = fk cos(2πnk/ N) k = 0

N− 1

where n = 100f and N = 10^6. [2 points off if frequency f missing]

The real IIR filter is given by

Fn(t + ∆ t) = [F n (t) − f 0 + fN] cos(2 πn /N)

where n = 100f and N = 10^6.

3C^1 Take 10

(^6) samples (f 0 to fN–1 ) and store in memory 2 Compute all needed coefficients in a loop 3 Start conversion and read new sample when ready 4 Delete oldest value (f 0 ), store new value (fN ) after last value (fN–1 ), and shift all fk to fk– 5 Loop back to step 2 [3 points off for each missing step]

4A

y 0 = 0 y 1 = 1 y 2 = – y 3 = 0 y 4 = 0

4B

y 0 = 0 y 1 = 1 y 2 = – y 3 = 1 y 4 = 0

4C

y 0 = 0 y 1 = 0.100 y 2 = 0. y 3 = 0.081 y 4 = 0.

4D Second derivative — b

First derivative — a Low pass filter — c

a — FIR b — FIR c — IIR

5A^1 With an input impulse, sample the output waveform h(t) and perform the FFT.

2 Perform the FFT of the desired output y(t) 3 Compute the required input u(t) using

u(t) = FFT−^1

FFT(y) FFT(h)