Signals and Sys...

If the z-transform of a sequence x[n] = {1, 1, -1, -1} is X[z], then the value of X(1/2) is

It is given that \(H\left( z \right) = \frac{{11}}{{\left( {1 - \frac{1}{3}{z^{ - 1}}} \right)\left( {1 - 4{z^{ - 1}}} \right)}}\)

A sequence u[n] is defined as \(u\left[ n \right] = \left\{ {\begin{array}{*{20}{c}}{1,\;if\;n \ge 0}\\{0,\;if\;n < 0}\end{array}} \right.\) for n = {-∞, …, -1,0,1, …, ∞}. The z-transform of u[n] is U(z). The region of convergence for which the z-transform of u[n] exists is

The z-transform of a signal is given by \(H\left( z \right) = \frac{1}{8}\frac{{{z^{ - 1}}\left( {1 + {z^{ - 5}}} \right)}}{{\left( {1 + {z^{ - 1}}} \right)\left( {1 - {z^{ - 1}}} \right)}}\). It final value is –

The z-transform of a sequence x[n] is given as \(X\left( z \right) = \frac{{3{z^3} + 5{z^2} \pm 5z + 4}}{{{z^2}}}\). If y[n] is the second order backward difference (Δ²x_n), of x[n], then y(z) is given by

What is the signal corresponding to the following z-transform? (Where u[n] is the unit-step signal)

\(X\left( z \right) = \frac{{6{z^2}}}{{\left( {2z - 1} \right)\left( {3z - 1} \right)}},\frac{1}{3} < \left| z \right| < \frac{1}{2}\)

The z transform of the signal \(x\left[ n \right] = n{\left( {\frac{1}{2}} \right)^{\left| n \right|}}\) is

The following is known about a discrete-time LTI system with input x[n] and output y[n]:

1. If x[n] = (-2)ⁿ for all n, then y[n] = 0 for all n.

2. If \(x\left[ n \right] = {\left( {\frac{1}{2}} \right)^n}u\left[ n \right]\) for all n, then y[n] for all n is of the form \(y\left[ n \right] = \delta \left[ n \right] + a{\left( {\frac{1}{4}} \right)^n}u\left[ n \right]\).

The value of constant ‘a’ is

The system represented by the difference equation \(y\left[ n \right] - 2\;y\left[ {n - 2} \right] = x\left[ n \right] - \frac{1}{2}x\left[ {n - 1} \right]\) is