Q21GATE 2007MCQ

Consider the discrete-time system shown in the figure where the impulse response of G(z) is g(0)=0, g(1)=g(2)=1, g(3)=g(4)=...= 0 This system is stable for range of values of K

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📖 Explanation

Given: $g(1)=g(2)=1$ i.e. $g[n]=\delta [n-1]+\delta [n-2]$ Therefore, $G(z)=z^{-1}+z^{-2}$ Therefore overall transfer function of closed loop system,

\begin{aligned} T(z)&=\frac{G(z)}{1-kG(z)} \\ &=\frac{z^{-1}+z^{-2}}{1-k(z^{-1}+z^{-2})} \\ &= \frac{z+1}{z^{2-}k(z+1)} \end{aligned}

So the system will be stable if it's outer most pole will lie inside the unit circle. Location of poles,

\begin{aligned} z=\frac{+k\pm \sqrt{k^{2+}4k}}{2}\\ \frac{k+ \sqrt{k^{2+}4k}}{2} \lt 1\\ k+ \sqrt{k^{2+}4k} \lt 2\\ \sqrt{k^{2+}4k} \lt (2-k)\\ k^{2+}4k \lt (2-k)^2\\ k^{2+}4k \lt 4+k^{2-}4k\\ 8k \lt 4\\ k \lt \frac{1}{2} \end{aligned}

Q23GATE 2006MCQ

A discrete real all pass system has a pole at z = 2 $\angle$ 30 $^{\circ}$ : it, therefore

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Q24GATE 2006MCQ

The discrete-time signal $x[n]\leftrightarrow X(z)=\sum_{n=0}^{\infty }\frac{3^{n}}{2+n}z^{2n},$ where $\leftrightarrow$ denotes a transform-pair relationship, is orthogonal to the signal

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📖 Explanation

$x(n)\rightleftharpoons X(z)=\sum_{n=0}^{\infty }\frac{3^n}{2+n}z^{2n}$ In X(z), power of z are even. Therefore, samples in x(n) are available at even instant of time. By observing all the options. Option (B): $y_2(n)\rightleftharpoons Y_2(z)=\sum_{n=0}^{\infty }(5^n-n)z^{-(2n+1)}$ In $y_2(z)$ , powers of z are odd. Therefore, samples in $y_2(n)$ are avialable only at odd instant of time. Hence, $\sum_{n=0}^{\infty }x(n)\cdot y_2(n)=0$ Thus, $x(n)$ is orthogonal to $y_2(n)$ .

Q25GATE 2005MCQ

If u(t) is the unit step and $\delta \left ( t \right )$ is the unit impulse function, the inverse z-transform of F(z)= $\frac{1}{z+1}$ for $k\gt 0$ is

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📖 Explanation

\begin{aligned} F(z) &=1-\frac{z}{z+1} \\ &= 1-\frac{1}{1+z^{-1}}\\ &= \delta (k)-(-1)^k u(k) \end{aligned}