31. The ROC of z-transform of the discrete time sequence $$x\left( n \right) = {\left( {{1 \over 3}} \right)^n}u\left( n \right) – {\left( {{1 \over 2}} \right)^n}u\left( { – n – 1} \right)$$ is

[amp_mcq option1=”$${1 \over 3} < \left| z \right| < {1 \over 2}$$" option2="$$\left| z \right| > {1 \over 2}$$” option3=”$$\left| z \right| < {1 \over 3}$$" option4="$$2 < \left| z \right| < 3$$" correct="option3"]

Detailed SolutionThe ROC of z-transform of the discrete time sequence $$x\left( n \right) = {\left( {{1 \over 3}} \right)^n}u\left( n \right) – {\left( {{1 \over 2}} \right)^n}u\left( { – n – 1} \right)$$ is

32. The Fourier series of a real periodic function has only P. Cosine terms if it is even Q. Sine terms if it is even R. Cosine terms if it is odd S. Sine terms if it is odd Which of the above statements are correct?

P and S
P and R
Q and S
Q and R

Detailed SolutionThe Fourier series of a real periodic function has only P. Cosine terms if it is even Q. Sine terms if it is even R. Cosine terms if it is odd S. Sine terms if it is odd Which of the above statements are correct?

33. The region of convergence of z-transform of the sequence $${\left( {{5 \over 6}} \right)^n}u\left( n \right) – {\left( {{6 \over 5}} \right)^n}u\left( { – n – 1} \right)$$ must be

[amp_mcq option1=”$$\left| z \right| < {5 \over 6}$$" option2="$$\left| z \right| > {5 \over 6}$$” option3=”$${5 \over 6} < \left| z \right| < {6 \over 5}$$" option4="$${6 \over 5} < \left| z \right| < \infty $$" correct="option3"]

Detailed SolutionThe region of convergence of z-transform of the sequence $${\left( {{5 \over 6}} \right)^n}u\left( n \right) – {\left( {{6 \over 5}} \right)^n}u\left( { – n – 1} \right)$$ must be

34. The impulse response of an LTI system can be obtained by

Differentiating the unit ramp response
Differentiating the unit step response
Integrating the unit ramp response
Integrating the unit step response

Detailed SolutionThe impulse response of an LTI system can be obtained by

35. If $$F\left( s \right) = L\left[ {f\left( t \right)} \right] = {{2\left( {s + 1} \right)} \over {{s^2} + 4s + 7}}$$ then the initial and final values of f(t) are respectively

0, 2
2, 0
$$0,{2 over 7}$$
$${2 over 7},0$$

Detailed SolutionIf $$F\left( s \right) = L\left[ {f\left( t \right)} \right] = {{2\left( {s + 1} \right)} \over {{s^2} + 4s + 7}}$$ then the initial and final values of f(t) are respectively

36. The inverse Laplace transform of the function $${{s + 5} \over {\left( {s + 1} \right)\left( {s + 3} \right)}}$$ is

2e-t - e-3t
2e-t + e-3t
e-t - 2e-3t
e-t + 2e-3t

Detailed SolutionThe inverse Laplace transform of the function $${{s + 5} \over {\left( {s + 1} \right)\left( {s + 3} \right)}}$$ is

37. The impulse response and the excitation function of a linear time invariant causal system are shown in figure (a) and (b) respectively. The output of the system at t = 2 sec is equal to

0
$$ rac{1}{2}$$
$$ rac{3}{2}$$
1

Detailed SolutionThe impulse response and the excitation function of a linear time invariant causal system are shown in figure (a) and (b) respectively. The output of the system at t = 2 sec is equal to

38. The z-transform of a signal is given by $$C\left( z \right) = {1 \over 4}{{{z^{ – 1}}\left( {1 – {z^{ – 4}}} \right)} \over {{{\left( {1 – {z^{ – 1}}} \right)}^2}}}.$$ Its final value is

$${1 over 4}$$
Zero
1
Infinity

Detailed SolutionThe z-transform of a signal is given by $$C\left( z \right) = {1 \over 4}{{{z^{ – 1}}\left( {1 – {z^{ – 4}}} \right)} \over {{{\left( {1 – {z^{ – 1}}} \right)}^2}}}.$$ Its final value is

40. Increased pulse-width in the flat-top sampling, leads to

Attenuation of high frequencies in reproduction
Attenuation of low frequencies in reproduction
Greater aliasing errors in reproduction
No harmful effects in reproduction

Detailed SolutionIncreased pulse-width in the flat-top sampling, leads to