1
GATE ECE 1999
MCQ (Single Correct Answer)
+1
-0.3
The input to a channel is a band pass signal. It is obtained by linearly modulating a sinusoidal carrier with a signal- tone signal. The output of the channel due to this input is given by y(t) = (1/100) cos$$(100t - {10^{ - 6}})\,$$ cos$$({10^6}t - 1.56)$$. The group delay $$({t_g})$$ and the phase delay $$({t_p})$$, in seconds, of the channel are
A
$${t_g} = {10^{ - 6}},\,{t_p} = 1.56$$
B
$${t_g} = 1.56,\,\,{t_p} = {10^{ - 6}}$$
C
$${t_g} = \,\,{10^{ - 8}},\,\,{t_p} = 1.56 \times {10^{ - 6}}$$
D
$${t_g} = {10^{ - 8}},\,{t_p} = 1.56$$
2
GATE ECE 1999
MCQ (Single Correct Answer)
+2
-0.6
The input to a matched filter is given by $$s(t) = \left\{ {\matrix{ {10\sin (2\pi \times {{10}^6}t),} & {0 < \left| t \right| < {{10}^{ - 4}}\sec } \cr 0 & {Otherwise} \cr } } \right.$$

The peak amplitude of the filter output is

A
10 Volts
B
5 Volts
C
10 millivolts
D
5 millivolts
3
GATE ECE 1999
MCQ (Single Correct Answer)
+2
-0.6
The Nyquist sampling frequency (in Hz) of a signal given by $$16 \times {10^{4\,}}\,\sin {c^2}(400t)*{10^6}\,\sin {c^3}(100t)$$ is
A
200
B
300
C
500
D
1000
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