6) A continuous-time signal x(t) is sampled at 1000 samples/second with negligib
ID: 2268631 • Letter: 6
Question
6) A continuous-time signal x(t) is sampled at 1000 samples/second with negligible aliasing. You use its samples to compute its spectrogram using the FFT at various time shifts. You then obtain the graph on the side, where the x-axis represent time in seconds and the y-axis represent the normalized frequency of each FFT 3 output between 0 and 2T. 6 2 Among the signals below, which signal x() was most likely sampled? 0.1 0.2 0.3 0.4 0.5 cos (27-50), for 0 t 0.1s cos(2. 200t), for 0.2sts0.3s cos(27-50), for 0.4 t 0.5s cos(2.t.100t), for 0ts0.1s cos(27-300), for 0.2 0.3s | cos(2n-100t), for 0.4 0.5s n(t) = otherwise otherwise cos ( 27-50t ) , for 0 t cos(27-100t), for 0.2 cos ( 27-50), for 0.45 t 0.1s cos ( 27-100t ) , for 0 t 0.1s cos (2-.400), for 0.2 t 0.3s cos(2.T-100r), for 0.4 sts0.55 0.3s 0.5s x2 (t)= otherwise otherwiseExplanation / Answer
Along X axis time is considered, and along y axis normalised frequency is considered.
And as FFT is used, there is a shift in frequency happens (low frequency will map to high value of index, for example when an fft of signal of length 1024 is computed then -fs/2 will be at 513th sample and the D.C. com[onent will be at 1024, and fs will be at 512).
The frequency here is 0-2 pi i.e 0- 500 is mapped to 2 pi-pi (sampling frequency) and -500-0 is mapped to pi to 0.
Here all the signals conidered are cosines hence one can observe two frequency components f0, -f0 at each time symmetric around pi.
Now between 0 to 0.1 a frequency near to 6 is shown. so f =500 * (6.28-5.7)/3.14 =92 which is near to 100Hz. and it is also the frequency component between 0.4 to 0.5.
Now between 0.2-0.3 there is a frequency component near 4, so f =500*(6.28-3.9)/3.14=378 which is near to 400.
Hence the given signal is x2(t).
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