GATE EE 2022 | Linear Time Invariant Systems Question 8 | Signals and Systems

GATE EE 2022

MCQ (Single Correct Answer)

-0.67

Let a causal LTI system be governed by the following differential equation $$y(t) + {1 \over 4}{{dy} \over {dt}} = 2x(t)$$, where x(t) and y(t) are the input and output respectively. Its impulse response is

$$2{e^{ - {1 \over 4}t}}u(t)$$

$$2{e^{ - 4t}}u(t)$$

$$8{e^{ - {1 \over 4}t}}u(t)$$

$$8{e^{ - 4t}}u(t)$$

GATE EE 2022

MCQ (Single Correct Answer)

-0.67

Consider the system as shown below:

GATE EE 2022 Signals and Systems - Linear Time Invariant Systems Question 8 English

where y(t) = x(e^t). The system is

linear and causal.

linear and non-causal.

non-linear and causal.

non-linear and non-causal.

GATE EE 2015 Set 2

MCQ (Single Correct Answer)

-0.6

For linear time invariant systems, that are Bounded Input Bounded stable, which one of the following statement is TRUE?

The impulse response will be integral, but may not be absolutely integrable.

The unit impulse response will have finite support.

The unit step response will be absolutely integrable.

The unit step response will be bounded.

GATE EE 2013

MCQ (Single Correct Answer)

-0.6

The impulse response of a continuous time system is given by h(t) = $$\delta$$(t − 1) + $$\delta$$(t − 3). The value of the step response at t = 2 is

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