Signal processing

[amp_mcq option1=”Stable and causal” option2=”Causal but not stable” option3=”Stable but not causal” option4=”Unstable and noncausal” correct=”option1″]

Detailed SolutionA system is defined by its impulse response h(n) = 2n u(n – 2). The system is

[amp_mcq option1=”40″ option2=”41″ option3=”42″ option4=”82″ correct=”option3″]

Detailed SolutionThe power in the signal $$s\left( t \right) = 8\cos \left( {20\pi t – {\pi \over 2}} \right) + 4\,\sin \left( {15\pi t} \right)$$ is

[amp_mcq option1=”$${{\sin \left( t \right)} \over {\pi t}}$$” option2=”$${{\sin \left( {2t} \right)} \over {2\pi t}}$$” option3=”$${{2\sin \left( t \right)} \over {\pi t}}$$” option4=”$${\left( {{{\sin \left( t \right)} \over {\pi t}}} \right)^2}$$” correct=”option4″]

Detailed SolutionIf the signal $$x\left( t \right) = {{\sin \left( t \right)} \over {\pi t}} * {{\sin \left( t \right)} \over {\pi t}}$$ with $$ * $$ denoting the convolution operation, then x(t) is equal to

[amp_mcq option1=”$${z \over {z – {a^T}}}$$” option2=”$${z \over {z + {a^T}}}$$” option3=”$${Z \over {z – {a^{ – T}}}}$$” option4=”$${z \over {z + {a^{ – T}}}}$$” correct=”option1″]

Detailed SolutionThe z-transform of the time function $$\sum\limits_{k = 0}^\infty {\delta \left( {n – K} \right)} $$ is

[amp_mcq option1=”Only S1 and S2 are true” option2=”Only S2 and S3 are true” option3=”Only S1 and S3 are true” option4=”S1, S2 and S3 are true” correct=”option1″]

Detailed SolutionLet h(t) denote the impulse response of a causal system with transfer function $${1 \over {s + 1}}.$$ Consider the following three statements: S1 : The system is stable. S2 : $${{h\left( {t + 1} \right)} \over {h\left( t \right)}}$$ independent of t for t > 0. S3 : A non-causal system with the same transfer function is stable. For the above system,

[amp_mcq option1=”{0, -2 + 2j, 2, -2 – 2j}” option2=”{2, 2 + 2j, 6, 2 – 2j}” option3=”{6, 1 – 3j, 2, 1 + 3j}” option4=”{6, -1 + 3j, 0, -1 – 3j}” correct=”option3″]

Detailed SolutionThe 4-point discrete Fourier Transform (DFT) of a discrete time sequence {1, 0, 2, 3} is

[amp_mcq option1=”A(s) = s” option2=”$$A\left( s \right) = {1 \over {\left( {1 – \exp \left( { – {T_s}} \right)} \right)}}$$” option3=”$$A\left( s \right) = {1 \over {\left( {1 + \exp \left( { – {T_s}} \right)} \right)}}$$” option4=”A(s) = exp(Ts)” correct=”option4″]

Detailed SolutionThe Laplace transform of a function f(t) u(t), where f(t) is periodic with period T, is A(s) times the Laplace transform of its first period. Then

[amp_mcq option1=”$$\frac{{\left( {1 – 0.6{z^{ – 1}}} \right)}}{{{z^{ – 1}}\left( {1 – 0.4{z^{ – 1}}} \right)}}$$” option2=”$$\frac{{{z^{ – 1}}\left( {1 – 0.6{z^{ – 1}}} \right)}}{{\left( {1 – 0.4{z^{ – 1}}} \right)}}$$” option3=”$$\frac{{{z^{ – 1}}\left( {1 – 0.4{z^{ – 1}}} \right)}}{{\left( {1 – 0.6{z^{ – 1}}} \right)}}$$” option4=”$$\frac{{\left( {1 – 0.4{z^{ – 1}}} \right)}}{{{z^{ – 1}}\left( {1 – 0.6{z^{ – 1}}} \right)}}$$” correct=”option3″]

Detailed SolutionTwo systems H1(z) and H2(z) are connected in cascade as shown below. The overall output y(n) is the same as the input x(n) with a one unit delay. The transfer function of the second system H2(z) is $$x\left( n \right) \to \boxed{{H_1}\left( z \right) = \frac{{\left( {1 – 0.4{z^{ – 1}}} \right)}}{{\left( {1 – 0.6{z^{ – 1}}} \right)}}} \to \boxed{{H_2}\left( z \right)} \to y\left( n \right)$$

[amp_mcq option1=”E-3, F-2, G-4, H-1″ option2=”E-1, F-3, G-2, H-4″ option3=”E-1, F-2, G-3, H-4″ option4=”E-2, F-1, G-4, H-3″ correct=”option3″]

Detailed SolutionMatch the following and choose the correct combination. Group I Group II E. Continuous and aperiodic signal 1. Fourier representation is continuous and aperiodic. F. Continuous and periodic signal 2. Fourier representation is discrete and aperiodic. G. Discrete and aperiodic signal 3. Fourier representation is continuous and periodic. H. Discrete and periodic signal 4. Fourier representation is discrete and periodic.

[amp_mcq option1=”Steady state value of the system output” option2=”Initial value of the system output” option3=”Transient behaviour of the system output” option4=”None of these” correct=”option1″]

Detailed SolutionThe final value theorem is used to find the