Signal processing

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

Stable and causal
Causal but not stable
Stable but not causal
Unstable and noncausal

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

22. The 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

40
41
42
82

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

23. If 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

$${{sin left( t ight)} over {pi t}}$$
$${{sin left( {2t} ight)} over {2pi t}}$$
$${{2sin left( t ight)} over {pi t}}$$
$${left( {{{sin left( t ight)} over {pi t}}} ight)^2}$$

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

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

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

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

25. Let 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,

Only S1 and S2 are true
Only S2 and S3 are true
Only S1 and S3 are true
S1, S2 and S3 are true

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,

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

{0, -2 + 2j, 2, -2 - 2j}
{2, 2 + 2j, 6, 2 - 2j}
{6, 1 - 3j, 2, 1 + 3j}
{6, -1 + 3j, 0, -1 - 3j}

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

27. The 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

A(s) = s
$$Aleft( s ight) = {1 over {left( {1 - exp left( { - {T_s}} ight)} ight)}}$$
$$Aleft( s ight) = {1 over {left( {1 + exp left( { - {T_s}} ight)} ight)}}$$
A(s) = exp(Ts)

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

28. Two 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 $$x\left( n \right) \to \boxed{{H_1}\left( z \right) = \frac{{\left( {1 – 0.4{z^{ – 1}}} \right)}}{{\left( {1 – 0.6{z^{ – 1}}} \right)}}} \to \boxed{{H_2}\left( z \right)} \to y\left( n \right)$$

$$ rac{{left( {1 - 0.6{z^{ - 1}}} ight)}}{{{z^{ - 1}}left( {1 - 0.4{z^{ - 1}}} ight)}}$$
$$ rac{{{z^{ - 1}}left( {1 - 0.6{z^{ - 1}}} ight)}}{{left( {1 - 0.4{z^{ - 1}}} ight)}}$$
$$ rac{{{z^{ - 1}}left( {1 - 0.4{z^{ - 1}}} ight)}}{{left( {1 - 0.6{z^{ - 1}}} ight)}}$$
$$ rac{{left( {1 - 0.4{z^{ - 1}}} ight)}}{{{z^{ - 1}}left( {1 - 0.6{z^{ - 1}}} ight)}}$$

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 $$x\left( n \right) \to \boxed{{H_1}\left( z \right) = \frac{{\left( {1 – 0.4{z^{ – 1}}} \right)}}{{\left( {1 – 0.6{z^{ – 1}}} \right)}}} \to \boxed{{H_2}\left( z \right)} \to y\left( n \right)$$

29. Match 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.

E-3, F-2, G-4, H-1
E-1, F-3, G-2, H-4
E-1, F-2, G-3, H-4
E-2, F-1, G-4, H-3

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.

30. The final value theorem is used to find the

Steady state value of the system output
Initial value of the system output
Transient behaviour of the system output
None of these

Detailed SolutionThe final value theorem is used to find the

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