1
GATE EE 2007
+2
-0.6
$$X\left( z \right) = 1 - 3\,\,{z^{ - 1}},\,\,Y\left( z \right) = 1 + 2\,\,{z^{ - 2}}$$ are $$Z$$-transforms of two signals $$x\left[ n \right],\,\,y\left[ n \right]$$ respectively. A linear time invariant system has the impulse response $$h\left[ n \right]$$ defined by these two signals as $$h\left[ n \right] = x\left[ {n - 1} \right] * y\left[ n \right]$$ where $$*$$ denotes discrete time convolution. Then the output of the system for the input $$\delta \left[ {n - 1} \right]$$
A
has $$Z$$-transforms $${z^{ - 1}}X\left( z \right)Y\left( z \right)$$
B
equals
$$\delta \left[ {n - 2} \right] - 3\delta \left[ {n - 3} \right] + 2\delta \left[ {n - 4} \right] - 6\delta \left[ {n - 5} \right]$$
C
has $$Z$$-transform $$1 - 3\,{z^{ - 1}} + 2\,{z^{ - 2}} - 6\,{z^{ - 3}}$$
D
does not satisfy any of the above three.
2
GATE EE 2007
+2
-0.6
A signal is processed by a causal filter with transfer function $$G(s).$$ For a distortion free output signal waveform, $$G(s)$$ must
A
provide zero phase shift for all frequencies
B
provide constant phase shift for all frequencies
C
provide linear phase shift that is proportional to frequency
D
provide a phase shift that is inversely proportional to frequency
3
GATE EE 2007
+2
-0.6
Consider the discrete-time system shown in the figure where the impulse response of $$G\left( z \right)$$ is
$$g\left( 0 \right) = 0,\,\,g\left( 1 \right) = g\left( 2 \right) = 1,\,g\left( 3 \right) = g\left( 4 \right) = .... = 0$$

This system is stable for range of values of $$K$$

A
$$\left[ { - 1,1/2} \right]$$
B
$$\left[ { - 1,1} \right]$$
C
$$\left[ { - 1/2,1} \right]$$
D
$$\left[ { - 1/2,2} \right]$$
4
GATE EE 2007
+2
-0.6
A signal is processed by a causal filter with transfer function $$G(s).$$ For a distortion free output signal waveform, $$G(s)$$ must.

$$G\left( z \right) = a{z^{ - 1}} + \beta \,\,{z^{ - 3}}$$ is a low-pass digital filter with a phase characteristic same as that of the above question if

A
$$\alpha = \beta$$
B
$$\alpha = - \beta$$
C
$$\alpha = {\beta ^{\left( {1/3} \right)}}$$
D
$$\alpha = {\beta ^{ - \left( {1/3} \right)}}$$
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