The impulse response and the excitation function of a linear time invariant causal system are shown in figure (a) and (b) respectively. The output of the system at t = 2 sec is equal to <img src="/images/question-image/electronics-and-communications-engineering/signal-processing/1681542224-the-impulse-response-and-the-excitation.jpg" title="Signal Processing mcq question image" alt="Signal Processing mcq question image">

$$\frac{1}{2}$$

$$\frac{3}{2}$$

Consider a system whose input r and output y are related by the equation $$y\left( t \right) = \int\limits_{ - \infty }^\infty {x\left( {t - \tau } \right)} h\left( {2\tau } \right)d\tau $$

Where h(t) is shown in the graph
Which of the following four properties are possessed by the system? BIBO: Bounded input gives a bounded output
Causal: The system is causal.
LP : The system is low pass.
LTI: The system is linear and time-invariant.

Causal, LP

BIBO, LTI

BIBO, Causal, LTI

LP, LTI

Letx(t) be the input and y(t) be the output of a continuous time system. Match the system properties P₁, P₂ and P₃ with system relations R₁, R₂, P₃, P₄.
Properties
P₁ : Linear but NOT time-invariant
P₂ : Time-invariant but NOT linear
P₃ : Linear and time-invariant
Relations
R₁ : y(t) = t²x(t)
R₂ : y(t) = t |x(t)|
R₃ : y(t) = |x(t)|
R₄ : y(t) = x(t - 5)

(P1, R1), (P2, R3), (P3, R4)

(P1, R2), (P2, P3), (P3, R4)

(P1, R3), (P2, R1), (P3, R2)

(P1, R1), (P2, R2), (P3, R3)

The Laplace transform of the causal periodic square wave of period T shown in the figure below is It

$$F\left( s \right) = \frac{1}{{1 + {e^{ - \frac{{sT}}{2}}}}}$$

$$F\left( s \right) = \frac{1}{{s\left( {1 + {e^{ - \frac{{sT}}{2}}}} \right)}}$$

$$F\left( s \right) = \frac{1}{{s\left( {1 - {e^{ - \frac{{sT}}{2}}}} \right)}}$$

$$F\left( s \right) = \frac{1}{{1 - {e^{ - sT}}}}$$

The function x(t) is shown in the figure. Even and odd parts of a unit-step function u(t) are respectively,

$$\frac{1}{2},\frac{1}{2}x\left( t \right)$$

$$ - \frac{1}{2},\frac{1}{2}x\left( t \right)$$

$$\frac{1}{2}, - \frac{1}{2}x\left( t \right)$$

$$ - \frac{1}{2}, - \frac{1}{2}x\left( t \right)$$

For the discrete-time system shown in the figure, the poles of the system transfer function are located at

2, 3

$$\frac{1}{2},3$$

$$\frac{1}{2},\frac{1}{3}$$

$$2,\frac{1}{3}$$

Consider the system shown in the figure below. The transfer function $$\frac{{Y\left( z \right)}}{{X\left( z \right)}}$$ of the system is

$$\frac{{1 + a{z^{ - 1}}}}{{1 + b{z^{ - 1}}}}$$

$$\frac{{1 + b{z^{ - 1}}}}{{1 + a{z^{ - 1}}}}$$

$$\frac{{1 + a{z^{ - 1}}}}{{1 - b{z^{ - 1}}}}$$

$$\frac{{1 - b{z^{ - 1}}}}{{1 - a{z^{ - 1}}}}$$

The unit impulse response of a linear time invariant system is the unit step function u(t). For t > 0, the response of the system to an excitation e^-at u(t), a > 0 will be

ae-at

$$\left( {{1 \over a}} \right)\left( {1 - {e^{ - at}}} \right)$$

a(1 - e-at)

1 - e-at

Consider the signal x(t) shown in the figure. Let h(t) denote the impulse response of the filter matched to x(t), with h(t) being non-zero only in the interval 0 to 4 sec. The slope of h(t) in the interval 3

$$\frac{1}{2}{\sec ^{ - 1}}$$

-1 sec-1

$$ - \frac{1}{2}{\sec ^{ - 1}}$$

1 sec-1

Referring this circuit, determine the maximum output voltage when a single pulse is applied as shown. The total resistance is 60 Ω.

2.73 V

27.33 V

30 V

2.67 V

The impulse response functions of four linear systems S₁, S₂, S₃, S₄ are given respectively by
$${h_1}\left( t \right) = 1,$$ $${h_2}\left( t \right) = u\left( t \right),$$ $${h_3}\left( t \right) = \frac{{u\left( t \right)}}{{t + 1}},$$ $${h_4}\left( t \right) = {e^{ - 3t}}u\left( t \right)$$
Where u(t) is the unit step function. Which of these systems is time invariant, causal, and stable?

S1

S2

S3

S4