The Laplace transform of the causal periodic square wave of period T shown in the figure below is It <img src="/images/question-image/electronics-and-communications-engineering/signal-processing/1681538348-the-laplace-transform-of-the-causal-periodic.jpg" title="Signal Processing mcq question image" alt="Signal Processing mcq question image">

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

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

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

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

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

Laplace transform of the function f(t) is given by $${\text{F}}\left( {\text{s}} \right) = {\text{L}}\left\{ {{\text{f}}\left( {\text{t}} \right)} \right\} = \int_0^\infty {{\text{f}}\left( {\text{t}} \right){{\text{e}}^{ - {\text{st}}}}{\text{dt}}{\text{.}}} $$ Laplace transform of the function shown below is given by

$$\frac{{1 - {{\text{e}}^{ - 2{\text{s}}}}}}{{\text{s}}}$$

$$\frac{{1 - {{\text{e}}^{ - {\text{s}}}}}}{{2{\text{s}}}}$$

$$\frac{{2 - 2{{\text{e}}^{ - {\text{s}}}}}}{{\text{s}}}$$

$$\frac{{1 - 2{{\text{e}}^{ - {\text{s}}}}}}{{\text{s}}}$$

One period (0, T) each of two periodic waveforms W₁ and W₂ are shown in the figure. The magnitudes of the nth Fourier series coefficients of W₁ and W₂, for n ≥ 1, n is odd, are respectively proportional to

|n|-3 and |n|-2

|n|-1 and |n|-3

|n|-1 and |n|-2

|n|-4 and |n|-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

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

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

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

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 magnitude and phase of the complex Fourier series coefficient a_k of a periodic signal x(t) are shown in the figure. Choose the correct statement from the four choices given. Notation: C is the set of complex number, R is the set of purely real numbers, and P is the set of purely imaginary numbers.

$$x\left( t \right) \in R$$

$$x\left( t \right) \in P$$

$$x\left( t \right) \in \left( {C - R} \right)$$

The information given is not sufficient to draw any conclusion about x(t)

The Laplace transform of a function f(t) u(t), where f(t) is periodic with period T, is A(s) times the Laplace transform of its first period. Then

A(s) = s

$$A\left( s \right) = {1 \over {\left( {1 - \exp \left( { - {T_s}} \right)} \right)}}$$

$$A\left( s \right) = {1 \over {\left( {1 + \exp \left( { - {T_s}} \right)} \right)}}$$

A(s) = exp(Ts)

The pictorial view of the frustum of a square pyramid is shown in figure 'X'. Its top view, when viewed in the direction of the arrow, will look like which of the given alternatives 1, 2, 3 and 4?

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

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

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

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