Signals and Systems Test 1

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Signals and Systems Test 1

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Weekly Quiz Competition

Question 1
1 / -0

The value of the Integral \(\mathop{\int }_{-5}^{1}7{{e}^{{{t}^{2}}}}\cos t\delta \left( t-2 \right)dt\) is _________.

A
0-0

B
04-01

C
07-06

D
09-02

Solution

Concept:
\(\mathop{\int }_{c}^{d}f\left( t \right)\delta \left( t-a \right)dt=\left\{ \begin{matrix} \begin{matrix} f\left( a \right) & c<a<d \\ \end{matrix} \\ \begin{matrix} 0 & ~~~~~~otherwise \\ \end{matrix} \\ \end{matrix} \right.\)
Calculation:
\(f\left( t \right)=7{{e}^{{{t}^{2}}}}\delta \left( t-2 \right)\) is out of the limit of the integral (-5, 1)
Hence \(\mathop{\int }_{-5}^{1}f\left( t \right)\delta \left( t-2 \right)dt=0\)
Question 2
1 / -0

If a signal and its even part is x[n] and x_e[n] respectively, then which out of the following relation must be true

A
\(\mathop \sum \limits_{n = - \infty }^\infty x\left[ n \right] + \mathop \sum \limits_{n = - \infty }^\infty {x_e}\left[ n \right] = 0\)

B
\(\mathop \sum \limits_{n = - \infty }^\infty x\left[ n \right] = 2\mathop \sum \limits_{n = - \infty }^\infty {x_e}\left[ n \right]\)

C
\(\mathop \sum \limits_{n = - \infty }^\infty x\left[ n \right] + 2\mathop \sum \limits_{n = - \infty }^\infty {x_e}\left[ n \right] = 0\)

D
\(\mathop \sum \limits_{n = - \infty }^\infty x\left[ n \right] = \mathop \sum \limits_{n = - \infty }^\infty {x_e}\left[ n \right]\)

Solution

We know that, x[n] = x_e[n] + x₀[n]
x₀[n] is the odd part of the signal
xe[n] is the even part of the signal
We can write:
\( \mathop \sum \limits_{n = - \infty }^\infty x\left[ n \right] = \mathop \sum \limits_{n = - \infty }^\infty {x_e}\left[ n \right] + \mathop \sum \limits_{n = - \infty }^\infty {x_o}\left[ n \right]\) ---(1)
The summation of the odd part can be written as:
\(\\ \mathop \sum \limits_{n = - \infty }^\infty {x_o}\left[ n \right] = \mathop \sum \limits_{n = - \infty }^0 {x_o}\left[ n \right] + \mathop \sum \limits_{n = 0}^\infty {x_o}\left[ n \right]\)
Since x₀(-n) = - x₀(n), the above equation can be written as:
\( = \mathop \sum \limits_{n = - \infty }^0 - {x_o}\left[ n \right] + \mathop \sum \limits_{n = 0}^\infty {x_o}\left[ n \right]\)
\( = - \mathop \sum \limits_{n = 0}^\infty {x_o}\left[ n \right] + \mathop \sum \limits_{n = 0}^\infty {x_o}\left[ n \right] = 0\)
Equation (1) now becomes:
\(\mathop \sum \limits_{n = - \infty }^\infty x\left[ n \right] = \mathop \sum \limits_{n = - \infty }^\infty {x_e}\left[ n \right]\)
Question 3
1 / -0

A discrete time signal is given as \(f\left( n \right) = \sin \left( {\frac{{n\pi }}{3}} \right)\left[ {u\left( n \right) - u\left( {n - 4} \right)} \right]\). The energy of the signal is_____W (Correct up to two decimal places)

A
1.48-1.51

B
1.484-1.511

C
1.487-1.516

D
1.489-1.512

Solution

Signal energy is given by:
\(E = \mathop \sum \limits_{n = - \infty }^\infty {\left[ {f\left( n \right)} \right]^2}\)
\(= \mathop \sum \limits_{n = - \infty }^\infty [\sin \left( {\frac{{n\pi }}{3}} \right).\left[ {u\left( n \right) - u\left( {n - 4} \right)} \right]]^2\)
\(u\left( n \right) - u\left( {n - 4} \right) = \left\{ {\begin{array}{*{20}{c}} 1&{0 \le n \le 3}\\ 0&{else\;where} \end{array}} \right.\)
\(E = \mathop \sum \limits_{n = 0}^3 {\left[ {\sin \frac{{n\pi }}{3}\left( 1 \right)} \right]^2} \)
\(E= \mathop \sum \limits_{n = 0}^3 {\sin ^2}\left( {\frac{{n\pi }}{3}} \right)\)
\(E= {\sin ^2}\left( {\frac{{(0)\pi }}{3}} \right)+{\sin ^2}\left( {\frac{{(1)\pi }}{3}} \right)+{\sin ^2}\left( {\frac{{(2)\pi }}{3}} \right)+{\sin ^2}\left( {\frac{{(3)\pi }}{3}} \right)\)
\(= 0 + {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} + {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} + 0 \)
\(E= \frac{3}{4} + \frac{3}{4} = 1.5\)
Question 4
1 / -0

Consider the signal \(x\left( t \right) = \cos \frac{{2{\rm{\pi }}}}{3}t + 2\sin \frac{{16\pi }}{3}t\)
y(t) = sin π t
The Period of the signal x(t) y(t) is ___________

A
6-6

B
64-61

C
67-66

D
69-62

Solution

let z(t) = x(t) y(t)
\(z\left( t \right) = \cos \frac{{2\pi }}{3}t\sin \pi t + 2\sin \frac{{16\pi }}{3}t\sin \pi t\)
2 sinA cos B = sin (A + B) + sin (A - B)
2 sinA SinB = cos (A - B) – cos (A + B)
\(\frac{1}{2}\left( {Sin\;\left( {\pi + \frac{{2\pi }}{3}} \right)t + \sin \left( {\pi - \frac{{2\pi }}{3}} \right)t} \right) + Cos\;\left( {\frac{{16\pi }}{3} - \pi } \right)t - \cos \left( {\frac{{16\pi }}{3} + \pi } \right)t\)
\(= \frac{1}{2}\sin \frac{{5\pi }}{3}t + \frac{1}{2}\sin \frac{\pi }{3}t + \cos \frac{{13\pi }}{3}t - \cos \left( {\frac{{19\pi }}{3}} \right)t\)
\({T_1} = \frac{{2\pi }}{{5\pi }} \times 3 = \frac{6}{5}\)
\({T_2} = \frac{{2\pi }}{\pi } \times 3 = 6\)
\({T_3} = \frac{{2\pi }}{{13\pi }} \times 3 = \frac{6}{{13}}\)
\({T_4} = \frac{{2\pi }}{{19\pi }} \times 3 = \frac{6}{{19}}\)
\(LCM = \left( {\frac{6}{5},\frac{6}{1},\frac{6}{{13}},\frac{6}{{19}}} \right)\)
\(= \frac{{LCM\;\left( {6,\;6,6,6} \right)}}{{HCF\left( {5,\;1,\;13,19} \right)}} = 6\)