Signals and Systems MCQ Quiz in தமிழ் - Objective Question with Answer for Signals and Systems - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

Duality property of Fourier transform:

If X(ω) is the Fourier transform of x(t),

$X\left(t\right)\leftrightarrow 2\pi x\left(-\omega \right)$

Calculation:

$e^{-a\left|t\right|}\stackrel{F.T}{\leftrightarrow }\frac{2a}{a^2+{\omega }^2}$

Step 2:

$\frac{1}{2\pi \left(t^2+\frac{1}{4}\right)}\to e^{-\frac{\left|\omega \right|}{2}}$

$\frac{2}{\pi \left(1+4t^2\right)}\leftrightarrow e^{-\frac{\left|\omega \right|}{2}}$

Concept:

The Fourier Transform of a continuous-time signal x(t) is given as:

$X\left(\omega \right)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t)e^{-j\omega t}dt$

Analysis:

Given:

x(t) = e^j10t  defined from t = -1 to 1. 

$X\left(\omega \right)=\underset{-1}{\overset{1}{\int }}{e}^{j10t}.e^{-j\omega t}dt=\underset{-1}{\overset{1}{\int }}{e}^{j\left(10-\omega \right)t}dt$

$X(\omega )={\frac{e^{j\left(10-\omega \right)t}}{j\left(10-\omega \right)}|}_{-1}^1=\frac{2\sin \left(\omega -10\right)}{\left(\omega -10\right)}$

h(n) = 2δ (n – 3)

H(z) = 2z-3

x(z) = z4 + z2 – 2z + 2 – 3z-4

Y(z) = X(z) H(z)

= 2(z + z-1 – 2z-2 + 2z-3 – 3z-7)

By applying inverse z-transform,

y(n) = 2[δ (n + 1) + δ (n - 1) -2δ (n - 2) + 2δ (n - 3) – 3δ (n - 7)]

At n = 4, y (4) = 0

$t_1+t_3=3$

$\frac{t_1+t_3}{2}-\frac{1}{2}=1$

$t_2+t_4=7$

$\frac{t_2+t_4}{2}-\frac{1}{2}=3$

Time scaling will not effect the Fourier series coefficient

x (0.5t) → C_k

$x\left(t-0.5\right)\to e^{-j{\omega }_{0}0.5k}{C}_k$

x (2t) → C_k

∴ Fourier series coefficient of the given signal is

Concept:

Initial value theorem:

The initial value theorem is one of the basic properties of the Laplace transform used to find the response of the system at the initial state (t = 0) in the Laplace domain. Mathematically it is given by

$\mathbf{f}\left(0^{+}\right)=\underset{t\to 0}{lim}\mathbf{f}\left(\mathbf{t}\right)=\underset{s\to \mathrm{\infty }}{lim}sF\left(s\right)$

Where

f(t) is system function

F(s) is Laplace transform of system function f(t)

f(0+) is the initial value of the system

NOTE:

• For the final value theorem to be applicable system should be stable in a steady-state and for that real part of the poles should lie on the left side of the s plane.
• In the given problem exactly the final value theorem is not applied but just X(0+) is calculated.
   

Calculation:

Given that, 

$X(s)=\frac{3s+5}{s^2+10s+21}$

$\mathbf{x}\left(0^{+}\right)=\underset{x\to 0}{lim}\mathbf{x}\left(\mathbf{t}\right)=\underset{s\to \mathrm{\infty }}{lim}sX\left(s\right)$

$\mathbf{x}\left(0^{+}\right)=\underset{x\to 0}{lim}\mathbf{f}\left(\mathbf{t}\right)=\underset{s\to \mathrm{\infty }}{lim}s.\frac{3s+5}{s^2+10s+21}$

$=\underset{s\to \mathrm{\infty }}{lim}\frac{3s^2+5s}{s^2+10s+21}$

= 3

Concept:

The instantaneous power of a signal is calculated as:

$p\left(t\right)=\int s^2\left(t\right)dt$

The average power will be:

$P_{avg}=\frac{1}{T}\underset{0}{\overset{T}{\int }}{s}^2\left(t\right)dt$

Calculation:

The average power of x(t),

$P_x=\frac{1}{T_x}{\int }_{T_x/2}^{T_x/2}{\left|x\left(t\right)\right|}^{2}dt$

$=\frac{1}{4}{\int }_0^2{\left(\frac{A}{2}t\right)}^{2}dt$

$=\frac{A^2}{16}{\int }_0^2{t}^{2}dt$

$=\frac{A^2}{16}{\left(\frac{t^3}{3}\right)}_0^2$

$=\frac{A^2}{6}$

Let, y(t) = x(3t + 6)

Since x(t) is a periodic signal, time scaling or shifting operations do not affect its average power.

So, $P_y=P_x=\frac{A^2}{6}$

Given that f(t) = v(t) = e^-2t u(t)

$F\left(\omega \right)=\frac{1}{2+j\omega }$

${\left|F\left(\omega \right)\right|}^2=\frac{1}{4+{\omega }^2}$

The total energy dissipated by the resistor is

$W=\frac{1}{2\pi }\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\left|F\left(\omega \right)\right|}^{2}d\omega$

$=\frac{1}{2\pi }\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{1}{4+{\omega }^2}d\omega$

$=\frac{1}{2\pi }{\left[\frac{1}{2}{\tan }^{-1}\frac{\omega }{2}\right]}_{-\mathrm{\infty }}^{\mathrm{\infty }}=0.25J$

The energy in the frequency band 0 < ω < 10 is,

$W=\frac{1}{2\pi }\underset{0}{\overset{10}{\int }}{\left|F\left(\omega \right)\right|}^{2}d\omega$

$=\frac{1}{2\pi }\underset{0}{\overset{10}{\int }}\frac{1}{4+{\omega }^2}d\omega$

$=\frac{1}{2\pi }{\left[\frac{1}{2}{\tan }^{-1}\frac{\omega }{2}\right]}_0^{10}=0.109J$

The percentage of the total energy is $=\frac{0.109}{0.25}\times 100=43.6$

Given:

f_m = 5 kHz, 

ω_c = 2πfc = 2000π 

f_c = 1 kHz

Maximum frequency in x(t) is given as:

f_c + f_m = 6 kHz

The correct answer is 9t2 +3t

Concept:

The time scaling property of convolution is given by

$x_1(t)\ast x_2(t)=y(t)$

$x_1(at)\ast x_2(at)=\frac{y(at)}{a}$

Solution:

as we can see are having a scaling factor of a = 3

∴ h(t) = [(3t)² + (3t)]

=[9t² + 3t]