1
GATE ECE 1999
MCQ (Single Correct Answer)
+1
-0.3
$$If\,\,L\left[ {f\left( t \right)} \right]\, = \,F\left( s \right),$$ then $$L\left[ {f\left( {t - T} \right)} \right]$$ is equal to
A
$${e^{sT}}F\left( s \right)\,$$
B
$${e^{ - sT}}\,F\left( s \right)\,\,$$
C
$${{F\left( s \right)} \over {1 + {e^{sT}}}}\,$$
D
$${{F\left( s \right)} \over {1 - {e^{ - sT}}}}$$
2
GATE ECE 1999
MCQ (Single Correct Answer)
+1
-0.3
The z-transform F(z) of the function f(nT) = $${a^{nT}}$$ is
A
$${z \over {z - {a^T}}}$$
B
$${z \over {z + {a^T}}}$$
C
$${z \over {z - {a^{ - T}}}}$$
D
$${z \over {z + {a^{ - T}}}}$$
3
GATE ECE 1999
MCQ (Single Correct Answer)
+2
-0.6
The z-transform of a signal is given by c(z)=$${1 \over 4}{{{z^{ - 1}}(1 - {z^{ - 4}})} \over {{{(1 - {z^{ - 1}})}^2}}}$$. Its final value is
A
1/4
B
zero
C
1.0
D
infinity
4
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
$${t_g} = {10^{ - 8}},\,{t_p} = 1.56$$
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