Signals and Systems Test 3

Result

Signals and Systems Test 3

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Weekly Quiz Competition

Question 1
1 / -0

Consider a causal LTI system with frequency response \(H\left( {j\omega } \right) = \frac{1}{{j\omega + 2}}\), for a particular input x(t), this system is observed to produce the output y(t) = (e^-2t - e^-4t) u(t). Determine x(t)

A
e^-4t u(t)

B
2e^4t u(t)

C
2e^-4t u(t)

D
3e^-4t u(t)

Solution

\(H\left( {j\omega } \right) = \frac{1}{{j\omega + 2}}\)
Output, \(y\left( t \right) = \left( {{e^{ - 2t}} - {e^{ - 4t}}} \right)u\left( t \right)\)
By applying the Fourier transform,
\(Y\left( \omega \right) = \frac{1}{{j\omega + 2}} - \frac{1}{{j\omega + 4}} = \frac{{j\omega + 4 - j\omega - 2}}{{\left( {j\omega + 2} \right)\left( {j\omega + 4} \right)}}\)
\(= \frac{2}{{\left( {j\omega + 2} \right)\left( {j\omega + 4} \right)}}\)
We know Y(ω) = X(ω) H(ω)
\(X\left( \omega \right) = \frac{{Y\left( \omega \right)}}{{H\left( \omega \right)}} = \frac{{\frac{2}{{\left( {j\omega + 2} \right)\left( {j\omega + 4} \right)}}}}{{1/\left( {j\omega + 2} \right)}}\)
\(= \frac{2}{{j\omega + 4}} = 2{e^{ - 4t}}u\left( t \right)\)
Question 2
1 / -0

What is the Fourier transform of x(t) = e^-2|t| sgn(t)

A
\(\frac{{ - j2\omega }}{{{\omega ^2} + 4}}\)

B
\(\frac{{j2\omega }}{{{\omega ^2} + 4}}\)

C
\(\frac{{ - j\omega }}{{{\omega ^2} + 4}}\)

D
\(\frac{{j2}}{{{\omega ^2} + 4}}\)

Solution

Given that, x(t) = e^-2|t| sgn(t)
\(sgn\left( t \right) = \begin{array}{*{20}{c}}{1,\;t > 0}\\{ - 1,\;t < 0}\end{array}\)
x(t) = e^-2t t > 0
= -e^-2t t < 0
\(x\left( t \right) = {e^{ - 2t}}u\left( t \right) - {e^{ - 2t}}\;u\left( { - t} \right)\)
By applying Fourier transform,
\(X\left( \omega \right) = \frac{1}{{2 + j\omega }} - \frac{1}{{2 - j\omega }}\)
\(X\left( \omega \right) = \frac{{ - j2\omega }}{{{\omega ^2} + 4}}\)
Question 3
1 / -0

The percentage of total energy in signal \(f\left( t \right) = {e^{ - t}}u\left( t \right)\) is contained in frequency band |ω| < 5 rad/sec is ______

A
87-88

B
874-881

C
877-886

D
879-882

Solution

Step 1:
f(t) = e^-t u(t)
\({E_f}\left( t \right) = \mathop \smallint \limits_{ - \infty }^\infty {\left| {f\left( t \right)} \right|^2}dt = \mathop \smallint \limits_0^\infty {\left( {{e^{ - t}}} \right)^2}dt\)
\(= \frac{1}{2} = 0.5\)
\({E_f}\left( \omega \right)\;for\;\left| \omega \right| \le 5 = \frac{1}{{2\pi }}\;\mathop \smallint \limits_{ - 5}^5 \frac{1}{{1 + {\omega ^2}}}d\omega \)
\( = \frac{1}{{2\pi }}\left( {{{\tan }^{ - 1}}\omega } \right)_{ - 5}^5 = \frac{2}{{2\pi }}\left( {{{\tan }^{ - 1}}\omega } \right)_0^5\)
\( = \frac{1}{\pi }\left( {{{\tan }^{ - 1}}5 - {{\tan }^{ - 1}}0} \right)\)
\(= \frac{{1.37}}{\pi } = 0.437\)
Percentage of energy contained \( = \frac{{0.437}}{{0.5}} \times 100\)
= 87.433%