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a) Assume that you are only given a sampling of x(t), specifically a sampling obtained by taking 6000 samples per second (samples equally spaced in time). Can one process this sampling in such a way that a band-limited interpolation of the processed (output) DT signal would be the same as y(t)? Answer yes/no. If you answered yes, explain how. If you answered no, explain why not. | a) Assume that you are only given a sampling of x(t), specifically a sampling obtained by taking 6000 samples per second (samples equally spaced in time). Can one process this sampling in such a way that a band-limited interpolation of the processed (output) DT signal would be the same as y(t)? Answer yes/no. If you answered yes, explain how. If you answered no, explain why not. | ||
| − | :Answer: Yes. Since the sampling frequency is 6000 Hz, there will be no aliasing when we sample | + | :Answer: Yes. Since the sampling frequency is 6000 Hz, there will be no aliasing when we sample a signal which has a maximum frequency of 1400 Hz (fs > 2*fm). The sampled signal Xd(w) has a cutoff frequency <math>7\pi/15 (= 1400*2\pi/6000)</math>. If we have a discrete-time LPF with a cutoff frequency o<math>4\pi/15 (= 800*2\pi/6000)</math>, we can have a signal reconstructed same as y(t). |
b) Now assume that the sampling from Part a) is downsampled by a factor 2. Can one process this downsampled signal in such a way a band-limited interpolation of the processed (output) DT signal would be the same as y(t)? Answer yes/no. If you answered yes, explain how. If you answered no, explain why not. | b) Now assume that the sampling from Part a) is downsampled by a factor 2. Can one process this downsampled signal in such a way a band-limited interpolation of the processed (output) DT signal would be the same as y(t)? Answer yes/no. If you answered yes, explain how. If you answered no, explain why not. | ||
| − | :Answer: Yes. Since the sampling frequency is 3000 Hz, there will be no aliasing when we sample | + | :Answer: Yes. Since the sampling frequency is 3000 Hz, there will be no aliasing when we sample a signal which has a maximum frequency of 1400 Hz (fs > 2*fm). The sampled signal Xd(w) has a cutoff frequency <math>14\pi/15 (= 1400*2\pi/3000)</math>. If we have a discrete-time LPF with a cutoff frequency <math>8\pi/15 (= 800*2\pi/3000)</math>, we can have a signal reconstructed same as y(t). |
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Revision as of 21:01, 19 October 2015
Homework 4 Solution, ECE438, Fall 2015, Prof. Boutin
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Question
A continuous-time signal x(t) is such that its CTFT X(f) is zero when when |f|>1,400 Hz. You would like to low-pass-filter the signal x(t) with a cut off frequency of 800Hz and a gain of 7. Let's call this desired filtered signal y(t).
a) Assume that you are only given a sampling of x(t), specifically a sampling obtained by taking 6000 samples per second (samples equally spaced in time). Can one process this sampling in such a way that a band-limited interpolation of the processed (output) DT signal would be the same as y(t)? Answer yes/no. If you answered yes, explain how. If you answered no, explain why not.
- Answer: Yes. Since the sampling frequency is 6000 Hz, there will be no aliasing when we sample a signal which has a maximum frequency of 1400 Hz (fs > 2*fm). The sampled signal Xd(w) has a cutoff frequency $ 7\pi/15 (= 1400*2\pi/6000) $. If we have a discrete-time LPF with a cutoff frequency o$ 4\pi/15 (= 800*2\pi/6000) $, we can have a signal reconstructed same as y(t).
b) Now assume that the sampling from Part a) is downsampled by a factor 2. Can one process this downsampled signal in such a way a band-limited interpolation of the processed (output) DT signal would be the same as y(t)? Answer yes/no. If you answered yes, explain how. If you answered no, explain why not.
- Answer: Yes. Since the sampling frequency is 3000 Hz, there will be no aliasing when we sample a signal which has a maximum frequency of 1400 Hz (fs > 2*fm). The sampled signal Xd(w) has a cutoff frequency $ 14\pi/15 (= 1400*2\pi/3000) $. If we have a discrete-time LPF with a cutoff frequency $ 8\pi/15 (= 800*2\pi/3000) $, we can have a signal reconstructed same as y(t).
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