81. 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

82. 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?

83. 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

87. 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

88. 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