1
GATE ECE 2008
MCQ (Single Correct Answer)
+2
-0.6
In the following network (Fig .1), the switch is closed at t = 0- and the sampling starts from t = 0. The sampling frequency is 10 Hz. GATE ECE 2008 Signals and Systems - Discrete Time Signal Z Transform Question 11 English
The expression and the region of convergence of the z-transform of the sampled signal are
A
$${{5z} \over {z - {e^{^{ - 5}}}}},\left| z \right| < {e^{ - 5}}$$
B
$${{5z} \over {z - {e^{^{ - 0.05}}}}},\left| z \right| < {e^{ - 0.05}}$$
C
$${{5z} \over {z - {e^{^{ - 0.05}}}}},\left| z \right| > {e^{ - 0.05}}$$
D
$${{5z} \over {z - {e^{^{ - 5}}}}},\left| z \right| < {e^{ - 5}}$$
2
GATE ECE 2008
MCQ (Single Correct Answer)
+2
-0.6
In the following network (Fig.1), the switch is closed at t = 0 and the sampling starts from t=0. The sampling frequency is 10 Hz. GATE ECE 2008 Signals and Systems - Discrete Time Signal Z Transform Question 12 English

The samples x (n) (n=0, 1, 2,...........) are given by

A
5(1-$${e^{ - 0.05n}}$$)
B
$$5{e^{ - 0.05n}}$$
C
$$5(1 - {e^{ - 5n}})$$
D
$$5{e^{ - 5n}}$$
3
GATE ECE 2008
MCQ (Single Correct Answer)
+1
-0.3
The impulse response h(t) of a linear time-invariant continuous time system is described by $$h\left( t \right) = \,\,\exp \left( {\alpha t} \right)u\left( t \right)\,\,\, + \,\,\exp \left( {\beta t} \right)u\left( { - t} \right),$$ where u(t) denotes the unit step function, and $$\alpha $$ and $$\beta $$ are real constants. This system is stable if
A
$$\alpha $$ is positive and $$\beta $$ is positive
B
$$\alpha $$ is negative and $$\beta $$ is negative
C
$$\alpha $$ is positive and $$\beta $$ is negative
D
$$\alpha $$ is negative and $$\beta $$ is positive
4
GATE ECE 2008
MCQ (Single Correct Answer)
+1
-0.3
The input and output of a continuous system are respectively denoted by x(t) and y(t). Which of the following descriptions corresponds to a causal system?
A
$$y\left( t \right) = x\left( {t - 2} \right) + x\left( {t + 4} \right)$$
B
$$y\left( t \right) = \left( {t - 4} \right)x\left( {t + 1} \right)$$
C
$$y\left( t \right) = \left( {t + 4} \right)x\left( {t - 1} \right)$$
D
$$y\left( t \right) = \left( {t + 5} \right)x\left( {t + 5} \right)$$
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