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Main points of this past exam are: Hold Amplifier, Data-Acquisition, Analog Waveform, Transfer, Filtering, Oscillating, Frequency Indices
Typology: Exams
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College of Engineering Electrical Engineering and Computer Sciences Department
present must < one-half the sampling frequency.
components of a periodically sampled waveform.
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]
frequency and thereby prevent aliasing.
actuator such as a speaker or heater. [2 points off if high power or current output not mentioned]
156V (p-p) 60 Hz 15R
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]
(^6) samples (S = 50 sec, N = 5 x 10^5 samples also acceptable)
6
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]
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]
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
where n = 100f and N = 10^6.
(^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]
y 0 = 0 y 1 = 1 y 2 = – y 3 = 0 y 4 = 0
y 0 = 0 y 1 = 1 y 2 = – y 3 = 1 y 4 = 0
y 0 = 0 y 1 = 0.100 y 2 = 0. y 3 = 0.081 y 4 = 0.
First derivative — a Low pass filter — c
a — FIR b — FIR c — IIR
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)