GATE ECE 2015 Set 3 | GATE ECE Year Wise Previous Years Questions

GATE ECE 2015 Set 3

MCQ (Single Correct Answer)

-0.6

A realization of a stable discrete time system is shown in the figure. If the system is excited by a unit step sequence input x[n ] , the response y[ n] is GATE ECE 2015 Set 3 Signals and Systems - Discrete Time Signal Z Transform Question 23 English

GATE ECE 2015 Set 3 Signals and Systems - Discrete Time Signal Z Transform Question 23 English

$$4{\left( { - {1 \over 3}} \right)^n}u\left[ n \right] - 5{\left( { - {2 \over 3}} \right)^n}u\left[ n \right]$$

$$5{\left( { - {2 \over 3}} \right)^n}u\left[ n \right] - 3{\left( { - {1 \over 3}} \right)^n}u\left[ n \right]$$

$$5{\left( {{1 \over 3}} \right)^n}u\left[ n \right] - 5{\left( {{2 \over 3}} \right)^n}u\left[ n \right]$$

$$5{\left( {{2 \over 3}} \right)^n}u\left[ n \right] - 5{\left( {{1 \over 3}} \right)^n}u\left[ n \right]$$

GATE ECE 2015 Set 3

MCQ (Single Correct Answer)

-0.6

Suppose x $$\left[ n \right]$$ is an absolutely summable discrete-time signal. Its z-transform is a rational function with two poles and two zeroes. The poles are at z = ± 2j. Which one of the following statements is TRUE for the signal x=$$\left[ n \right]$$ ?

It is a finite duration signal.

It is a causal signal.

It is a non-causal signal.

It is a periodic signal.

GATE ECE 2015 Set 3

MCQ (Single Correct Answer)

-0.3

Consider a four-point moving average filter defined by the equation $$y[n] = \sum\limits_{i = 0}^3 {{\alpha _i}x[n - i]} $$. The condition on the filter coefficients that results in a null at zero frequency is

$${\alpha _1} = {\alpha _2} = 0;\,{\alpha _0} = - {\alpha _3}$$

$${\alpha _1} = {\alpha _2} = 1;\,{\alpha _0} = - {\alpha _3}$$

$${\alpha _0} = {\alpha _3} = 0;\,{\alpha _1} = {\alpha _2}$$

$${\alpha _1} = {\alpha _2} = 0;\,{\alpha _0} = {\alpha _3}$$

GATE ECE 2015 Set 3

MCQ (Single Correct Answer)

-0.6

The complex envelope of the bandpass signal $$x(t)\, = \, - \sqrt 2 \left( {{{\sin (\pi t/5)} \over {\pi t/5}}} \right)\sin \left( {\pi t - {\pi \over 4}} \right),$$ centered about f = $${1 \over {2\,}}\,Hz,$$ is

$$\left( {{{\sin (\pi t/5)} \over {\pi t/5}}} \right){e^{j{\pi \over 4}}}$$

$$\left( {{{\sin (\pi t/5)} \over {\pi t/5}}} \right){e^{ - j{\pi \over 4}}}$$

$$\sqrt 2 \left( {{{\sin (\pi t/5)} \over {\pi t/5}}} \right){e^{j{\pi \over 4}}}$$

$$\sqrt 2 \left( {{{\sin (\pi t/5)} \over {\pi t/5}}} \right){e^{ - j{\pi \over 4}}}$$

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