1
GATE ECE 2014 Set 1
+2
-0.6
Let x $$\left[ n\right]$$= $${\left( { - {1 \over 9}} \right)^n}\,u(n) - {\left( { - {1 \over 3}} \right)^n}u( - n - 1).$$ The region of Convergence (ROC) of the z-tansform of x$$\left[ n \right]$$
A
is $$\left| z \right| > {1 \over 9}$$
B
is $$\left| z \right| < {1 \over 3}$$
C
is $${1 \over 3} > \left| z \right| > {1 \over 9}$$
D
does not exist.
2
GATE ECE 2008
+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.
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
$${{5z} \over {z - {e^{^{ - 5}}}}},\left| z \right| < {e^{ - 5}}$$
3
GATE ECE 2008
+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.

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
$$5{e^{ - 5n}}$$
4
GATE ECE 2007
+2
-0.6
The z-transform X (z) f a sequence x$$\left[ n \right]$$ is given by = $${{0.5} \over {1 - 2{z^{ - 1}}}}$$ . It is given that the region of convergence of X$$\left[ z \right]$$ includes the unit circle. The value of x$$\left[ 0 \right]$$ is
A
-0.5
B
0
C
0.25
D
0.5
EXAM MAP
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