Signal processing

101. Consider a six-point decimation-in-time Fast Fourier Transform (FFT) algorithm, for which the signal-flow graph corresponding to X[I] is shown in the figure. Let $${W_6} = \exp \left( { – \frac{{j2\pi }}{6}} \right).$$ In the figure, what should be the values of the coefficients a1, a2, a3 in terms of W6 so that X[I] is obtained correctly?

$${a_1} = 1,{a_2} = W_6^2,{a_3} = {W_6}$$
$${a_1} = - 1,{a_2} = W_6^2,{a_3} = {W_6}$$
$${a_1} = - 1,{a_2} = {W_6},{a_3} = W_6^2$$
$${a_1} = 1,{a_2} = {W_6},{a_3} = W_6^2$$

Detailed SolutionConsider a six-point decimation-in-time Fast Fourier Transform (FFT) algorithm, for which the signal-flow graph corresponding to X[I] is shown in the figure. Let $${W_6} = \exp \left( { – \frac{{j2\pi }}{6}} \right).$$ In the figure, what

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should be the values of the coefficients a1, a2, a3 in terms of W6 so that X[I] is obtained correctly?

102. An FIR system is described by the system function $$H\left( z \right) = 1 + \frac{7}{2}{z^{ – 1}} + \frac{3}{z}{z^{ – 2}}$$ The system is

Maximum phase
Minimum phase
Mixed phase
Zero phase

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z \right) = 1 + \frac{7}{2}{z^{ – 1}} + \frac{3}{z}{z^{ – 2}}$$ The system is" class="read-more button" href="https://exam.pscnotes.com/mcq/an-fir-system-is-described-by-the-system-function-hleft-z-right-1-frac72z-1-frac3zz-2-the-system-is/#more-44534">Detailed SolutionAn FIR system is described by the system function $$H\left( z \right) = 1 + \frac{7}{2}{z^{ – 1}} + \frac{3}{z}{z^{ – 2}}$$ The system is

103. Given that $$L\left[ {f\left( t \right)} \right] = {{s + 2} \over {{s^2} + 1}},L\left[ {g\left( t \right)} \right] = {{{s^2} + 1} \over {\left( {s + 3} \right)\left( {s + 2} \right)}},$$ $$h\left( t \right) = \int\limits_0^t {f\left( \tau \right)} g\left( {t – \tau } \right)d\tau $$ L[h(t)] is

$${{{s^2} + 1} over {s + 3}}$$
$${1 over {s + 3}}$$
$${{{s^2} + 1} over {left( {s + 3} ight)left( {s + 2} ight)}} + {{s + 2} over {{s^2} + 1}}$$
None of the above

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\right] = {{s + 2} \over {{s^2} + 1}},L\left[ {g\left( t \right)} \right] = {{{s^2} + 1} \over {\left( {s + 3} \right)\left( {s + 2} \right)}},$$ $$h\left( t \right) = \int\limits_0^t {f\left( \tau \right)} g\left( {t – \tau } \right)d\tau $$ L[h(t)] is" class="read-more button" href="https://exam.pscnotes.com/mcq/given-that-lleft-fleft-t-right-right-s-2-over-s2-1lleft-gleft-t-right-right-s2-1-over-left-s-3-rightleft-s-2-right/#more-44474">Detailed SolutionGiven that $$L\left[ {f\left( t \right)} \right] = {{s + 2} \over {{s^2} + 1}},L\left[ {g\left( t \right)} \right] = {{{s^2} + 1} \over {\left( {s + 3} \right)\left( {s + 2} \right)}},$$ $$h\left( t \right) = \int\limits_0^t {f\left( \tau \right)} g\left( {t – \tau } \right)d\tau $$ L[h(t)] is

104. A 1.0 kHz signal is flat-top sampled at the rate of 1800 samples/sec and the samples are applied to an ideal rectangular LPF with cut-off frequency of 1100 Hz, then the output of the filter contains

Only 800 Hz component
800 Hz and 900 Hz components
800 Hz and 1000 Hz components
800 Hz, 900 Hz and 1000 Hz components

Detailed SolutionA 1.0 kHz signal is flat-top sampled at the rate of

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1800 samples/sec and the samples are
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applied to an ideal rectangular LPF with cut-off frequency of 1100 Hz, then the output of the filter contains

105. The input x(t) and the output y(t) of a continuous- time system are related as $$y\left( t \right) = \int\limits_{t – T}^t {x\left( u \right)du} $$ The system is

Linear and time-variant
Linear and time-invariant
Non-linear and time-variant
Non-linear and time-invariant

Detailed SolutionThe input x(t) and the output y(t) of a

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continuous- time system are related as $$y\left( t \right) = \int\limits_{t – T}^t {x\left( u \right)du} $$ The system is

106. The input to a channel is a bandpass signal. It is obtained by linearly modulating a sinusoidal carrier with a single-tone signal. The output of the channel due to this input is given by $$y\left( t \right) = \left( {{1 \over {100}}} \right)\cos \left( {100t – {{10}^{ – 6}}} \right)\cos \left( {{{10}^6}t – 1.56} \right)$$ The group delay (tg) and the phase delay (tp) in seconds, of the channel are

tg = 10-6, tp = 1.56
tg = 1.56, tp = 10-6
tg = 10-8, tp = 1.56 × 10-6
tg = 108, tp = 1.56

Detailed SolutionThe input to a channel is a bandpass signal. It is obtained by linearly modulating a sinusoidal carrier with a single-tone signal. The output of the channel due to this input is given by $$y\left( t \right) = \left( {{1 \over {100}}} \right)\cos \left( {100t – {{10}^{

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– 6}}} \right)\cos \left( {{{10}^6}t – 1.56} \right)$$ The group delay (tg) and the phase delay (tp) in seconds, of the channel are

107. The z-transform F(z) of the function f(nT) = anT is

$${z over {z - {a^T}}}$$
$${z over {z + {a^T}}}$$
$${z over {z - {a^{ - T}}}}$$
$${z over {z + {a^{ - T}}}}$$

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z-transform F(z) of the function f(nT) = anT is" class="read-more button" href="https://exam.pscnotes.com/mcq/the-z-transform-fz-of-the-function-fnt-ant-is/#more-43673">Detailed SolutionThe z-transform F(z) of the function f(nT) = anT is

108. The response of an initially relaxed linear constant parameter network to a unit impulse applied at t = 0 is 4e-2tu(t). The response of this network to a unit step function will be

”2[1
u(t)” option2=”4[e-t – e-2t]u(t)” option3=”sin2t” option4=”(1 – 4e-4t)u(t)” correct=”option1″]

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class="youtube-subscribe-button"> Subscribe on YouTube unit impulse applied at t = 0 is 4e-2tu(t). The response of this network to a unit step function will be" class="read-more button" href="https://exam.pscnotes.com/mcq/the-response-of-an-initially-relaxed-linear-constant-parameter-network-to-a-unit-impulse-applied-at-t-0-is-4e-2tut-the-response-of-this-network-to-a-unit-step-function-will-be/#more-43661">Detailed SolutionThe response of an initially relaxed linear constant parameter network to a unit impulse applied at t = 0 is 4e-2tu(t). The response of this network to a unit step function will be

109. A network consisting of a finite number of linear resistor (R), inducer (L), and capacitor (C) elements, connected all in series or all in parallel, is excited with a source of the form $$\sum\limits_{k = 1}^3 {{a_x}\,\cos \left( {k{\omega _0}t} \right),{\rm{were}}\,{a_k} \ne 0,} \,{\omega _0} \ne 0.$$ The source has nonzero impedance. Which one of the following is a possible form of the output measured across a resistor in the network?

$$sumlimits_{k = 1}^3 {{b_x},cos left( {k{omega _0}t + {phi _k}} ight),{ m{were}},{b_k} e {a_k},} , orall K$$
$$sumlimits_{k = 1}^3 {{b_x},cos left( {k{omega _0}t + {phi _k}} ight),{ m{were}},{b_k} e 0,} , orall K$$
$$sumlimits_{k = 1}^3 {{a_x},cos left( {k{omega _0}t + {phi _k}} ight)} $$
$$sumlimits_{k = 1}^2 {{a_x},cos left( {k{omega _0}t + {phi _k}} ight)} $$

Detailed SolutionA network consisting of a finite number of linear resistor

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(R), inducer (L), and capacitor (C) elements, connected all in series or all in parallel, is excited with a source of the form $$\sum\limits_{k = 1}^3 {{a_x}\,\cos \left( {k{\omega _0}t} \right),{\rm{were}}\,{a_k} \ne 0,} \,{\omega _0} \ne 0.$$ The
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source has nonzero impedance. Which one of the following is a possible form of the output measured across a resistor in the network?

110. The Fourier Transform of a function x(t) is X(f). The Fourier transform of $${{dx\left( t \right)} \over {dt}}$$ will be

$${{dxleft( t ight)} over {dt}}$$
$$j2pi fXleft( f ight)$$
$$jfXleft( f ight)$$
$${{Xleft( f ight)} over {jf}}$$

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t \right)} \over {dt}}$$ will be" class="read-more button" href="https://exam.pscnotes.com/mcq/the-fourier-transform-of-a-function-xt-is-xf-the-fourier-transform-of-dxleft-t-right-over-dt-will-be/#more-43305">Detailed SolutionThe Fourier Transform of a function x(t) is X(f). The Fourier transform of $${{dx\left( t \right)} \over {dt}}$$ will be

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