Signal processing

1. {a(n)} is a real-valued periodic sequence with a period N. x(n) and X(k) form N-point Discrete Fourier Transform (DFT) pairs. The DFT Y(k) of the sequence $$y\left( n \right) = \frac{1}{N}\sum\limits_{r = 0}^{N – 1} {x\left( r \right)} x\left( {n + r} \right)$$ is

$${left| {Xleft( k ight)} ight|^2}$$
$$rac{1}{2}sumlimits_{r = 0}^{N - 1} {Xleft( r ight)X'left( {k + r} ight)} $$
$$rac{1}{2}sumlimits_{r = 0}^{N - 1} {Xleft( r ight)Xleft( {k + r} ight)} $$
0

Detailed Solution{a(n)} is a real-valued periodic sequence

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with a period N. x(n) and X(k) form N-point Discrete
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Fourier Transform (DFT) pairs. The DFT Y(k) of the sequence $$y\left( n \right) = \frac{1}{N}\sum\limits_{r = 0}^{N – 1} {x\left( r \right)} x\left( {n + r} \right)$$ is

2. For a function g(t), it is given that $$\int\limits_{ – \infty }^{ + \infty } {g\left( t \right){e^{ – j\omega t}}dt = \omega {e^{ – 2{\omega ^2}}}} $$ for any real value $$\omega $$ . If $$y\left( t \right) = \int\limits_{ – \infty }^t {g\left( \tau \right)} d\tau ,\,{\rm{then}}\,\int\limits_{ – \infty }^{ + \infty } {y\left( t \right)} dt$$ is. . . . . . . .

0
#NAME?
$$ - {j over 2}$$
$${j over 2}$$

Detailed SolutionFor a function g(t), it is given that $$\int\limits_{ – \infty }^{ + \infty } {g\left( t \right){e^{ – j\omega t}}dt = \omega {e^{ – 2{\omega ^2}}}} $$ for any real value $$\omega $$ . If $$y\left( t \right) = \int\limits_{ – \infty }^t {g\left( \tau \right)}

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d\tau ,\,{\rm{then}}\,\int\limits_{ – \infty }^{ + \infty } {y\left( t \right)} dt$$ is. . . . . . . .

3. The impulse response of a system is h(t) = tu(t). For an input u(t – 1), the output is

$${{{t^2}} over 2}uleft( t ight)$$
$${{tleft( {t - 1} ight)} over 2}uleft( {t - 1} ight)$$
$${{{{left( {t - 1} ight)}^2}} over 2}uleft( {t - 1} ight)$$
$${{{t^2} - 1} over 2}uleft( {t - 1} ight)$$

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is h(t) = tu(t). For an input u(t – 1), the output is" class="read-more button" href="https://exam.pscnotes.com/mcq/the-impulse-response-of-a-system-is-ht-tut-for-an-input-ut-1-the-output-is/#more-51186">Detailed SolutionThe impulse response of a
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system is h(t) = tu(t). For an input u(t – 1), the output is

4. Let Y(s) be the unit-step response of a causal system having a transfer function $$G\left( s \right) = {{3 – s} \over {\left( {s + 1} \right)\left( {s + 3} \right)}}$$ That is, $$Y\left( s \right) = {{G\left( s \right)} \over s}.$$ The forced response of the system is

u(t) - 2e-t u(t) + e-3t u(t)
2u(t)
u(t)
2u(t) - 2e-t u(t) + e-3t u(t)

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42.9-11.4 132.3-11.4 132.3s0 89.4 11.4 132.3c6.3 23.7 24.8 41.5 48.3 47.8C117.2 448 288 448 288 448s170.8 0 213.4-11.5c23.5-6.3 42-24.2 48.3-47.8 11.4-42.9 11.4-132.3 11.4-132.3s0-89.4-11.4-132.3zm-317.5 213.5V175.2l142.7 81.2-142.7 81.2z"/> Subscribe on YouTube
a causal system having a transfer function $$G\left( s \right) = {{3 – s} \over {\left( {s + 1} \right)\left( {s + 3} \right)}}$$ That is, $$Y\left( s \right) = {{G\left( s \right)} \over s}.$$ The forced response of the system is" class="read-more button" href="https://exam.pscnotes.com/mcq/let-ys-be-the-unit-step-response-of-a-causal-system-having-a-transfer-function-gleft-s-right-3-s-over-left-s-1-rightleft-s-3-right-that-is-yleft-s-r/#more-51024">Detailed SolutionLet Y(s) be the unit-step response of a causal system having a transfer function $$G\left( s \right) = {{3 – s} \over {\left( {s + 1} \right)\left( {s + 3} \right)}}$$ That is, $$Y\left( s \right) = {{G\left( s \right)} \over s}.$$ The forced response of the system is

5. The Laplace transform of a continuous-time signal x(t) is $$X\left( s \right) = {{5 – s} \over {{s^2} – s – 2}}.$$ If the Fourier transform of this signal exists, then x(t) is

e2tu(t) - 2e-tu(t)
-e2tu(-t) + 2e-tu(t)
-e2tu(-t) - 2e-tu(t)
e2tu(-t) - 2e-tu(t)

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signal x(t) is $$X\left( s \right) = {{5 – s} \over {{s^2} – s – 2}}.$$ If the Fourier transform of this signal exists, then x(t) is" class="read-more button" href="https://exam.pscnotes.com/mcq/the-laplace-transform-of-a-continuous-time-signal-xt-is-xleft-s-right-5-s-over-s2-s-2-if-the-fourier-transform-of-this-signal-exists-then-xt-is/#more-50925">Detailed SolutionThe Laplace transform of a continuous-time signal x(t) is $$X\left( s \right) = {{5 – s} \over
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{{s^2} – s – 2}}.$$ If the Fourier transform of this signal exists, then x(t) is

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

2, 3
$$rac{1}{2},3$$
$$rac{1}{2},rac{1}{3}$$
$$2,rac{1}{3}$$

Detailed SolutionFor the discrete-time system shown

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288 64 288 64S117.2 64 74.6 75.5c-23.5 6.3-42 24.9-48.3 48.6-11.4 42.9-11.4 132.3-11.4 132.3s0 89.4 11.4 132.3c6.3 23.7 24.8 41.5 48.3 47.8C117.2 448 288 448 288 448s170.8 0 213.4-11.5c23.5-6.3 42-24.2 48.3-47.8 11.4-42.9 11.4-132.3 11.4-132.3s0-89.4-11.4-132.3zm-317.5 213.5V175.2l142.7 81.2-142.7 81.2z"/> Subscribe on YouTube

in the figure, the poles of the system transfer function are located at

7. If the region of convergence of x1[n] + x2[n] is $${1 \over 3} < \left| z \right| < {2 \over 3},$$ then the region of convergence of x1[n] - x2[n] includes

”$${1
+ x2[n] is $${1 \over 3} < \left| z \right| < {2 \over 3},$$ then the region of convergence of x1[n] - x2[n] includes

8. A linear discrete-time system has the characteristics equation, z3 – 0.81 z = 0. The system

Is stable
Is marginally stable
Is unstable
Stability cannot be assessed from the given information

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23.7 24.8 41.5 48.3 47.8C117.2 448 288 448 288 448s170.8 0 213.4-11.5c23.5-6.3 42-24.2 48.3-47.8 11.4-42.9 11.4-132.3 11.4-132.3s0-89.4-11.4-132.3zm-317.5 213.5V175.2l142.7 81.2-142.7 81.2z"/> Subscribe on YouTube 0. The system" class="read-more button" href="https://exam.pscnotes.com/mcq/a-linear-discrete-time-system-has-the-characteristics-equation-z3-0-81-z-0-the-system/#more-50843">Detailed SolutionA linear discrete-time system has the characteristics equation, z3 – 0.81 z = 0. The system

9. A real-valued signal x(t) limited to the frequency band $$\left| f \right| \le {W \over 2}$$ is passed through a linear time invariant system whose frequency response is $$H\left( f \right) = \left\{ {\matrix{ {{e^{ – j4\pi f,}}} & {\left| f \right| \le {W \over 2}} \cr {0,} & {\left| f \right| > {W \over 2}} \cr } } \right.$$ The output of the system is

x(t + 4)
x(t - 4)
x(t + 2)
x(t - 2)

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= \left\{ {\matrix{ {{e^{ – j4\pi f,}}} & {\left| f \right| \le {W \over 2}} \cr {0,} & {\left| f \right| > {W \over 2}} \cr } } \right.$$ The output of the system is" class="read-more button" href="https://exam.pscnotes.com/mcq/a-real-valued-signal-xt-limited-to-the-frequency-band-left-f-right-le-w-over-2-is-passed-through-a-linear-time-invariant-system-whose-frequency-response-is-hleft-f-right/#more-50197">Detailed SolutionA
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real-valued signal x(t) limited to the frequency band $$\left| f \right| \le {W \over 2}$$ is passed through a linear time invariant system whose frequency response is $$H\left( f \right) = \left\{ {\matrix{ {{e^{ – j4\pi f,}}} & {\left| f \right| \le {W \over 2}} \cr {0,} & {\left| f \right| > {W \over 2}} \cr } } \right.$$ The output of the system is

10. The z-transform of a system is $$H\left( z \right) = {z \over {z – 0.2}}.$$ If the ROC is |z| < 0.2, then the impulse response of the system is

”(0.2)nu[n
” option2=”(0.2)nu[- n – 1]” option3=”-(0.2)nu[n]” option4=”-(0.2)nu[- n – 1]” correct=”option3″]

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\over {z – 0.2}}.$$ If the ROC is |z| < 0.2, then the impulse response of the system is" class="read-more button" href="https://exam.pscnotes.com/mcq/the-z-transform-of-a-system-is-hleft-z-right-z-over-z-0-2-if-the-roc-is-z-0-2-then-the-impulse-response-of-the-system-is/#more-50006">Detailed SolutionThe z-transform of a system is $$H\left( z \right) = {z \over {z – 0.2}}.$$ If the ROC is |z| < 0.2, then the impulse response of the system is
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