1
GATE ECE 2015 Set 3
Numerical
+2
-0
Consider a continuous-time signal defined as $$x(t) = \left( {{{\sin \,(\pi t/2)} \over {(\pi t/2)}}} \right)*\sum\limits_{n = - \infty }^\infty {\delta (t - 10n)} $$ Where ' * ' denotes the convolution operation and t is in seconds. The Nyquist sampling rate (in samples/sec) for x(t) is __________________.
Your input ____
2
GATE ECE 2010
MCQ (Single Correct Answer)
+2
-0.6
The Nyquist sampling rate for the signal $$s(t) = {{\sin \,(500\pi t)} \over {\pi \,t}} \times {{\sin \,(700\pi t)} \over {\pi \,t}}$$ is given by
3
GATE ECE 2006
MCQ (Single Correct Answer)
+2
-0.6
The minimum sampling frequency (in samples /sec) required to reconstruct the following signal from its samples without distortion $$x(t) = 5{\left( {{{\sin \,\,2\,\pi \,1000\,t)} \over {\pi \,t}}} \right)^3} + 7{\left( {{{\sin \,\,2\,\pi \,1000\,t} \over {\pi \,t}}} \right)^2}$$
would be
4
GATE ECE 2006
MCQ (Single Correct Answer)
+2
-0.6
A signal m(t) with bandwidth 500 Hz is first multiplied by a signal g(t) where $$g(t)\, = \,\,\sum\limits_{k = - \infty }^\infty {{{( - 10)}^k}\,\delta (t - 0.5x{{10}^{ - 4}}k)} $$
The resulting signal is then passed through an ideal low pass filter with bandwidth 1 kHz. The output of the low pass filter would be
The resulting signal is then passed through an ideal low pass filter with bandwidth 1 kHz. The output of the low pass filter would be
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