Q6GATE 2017NAT

Consider a causal and stable LTI system with rational transfer function H(z). Whose corresponding impulse response begins at n = 0. Furthermore, $H(1)=\frac{5}{4}$ . The poles of H(z) are $P_{k}=\frac{1}{\sqrt{2}}exp(j\frac{(2k-1)\pi }{4})$ for k = 1,2,3,4. The zeros of H(z) are all at z = 0. Let $g[n]=j^nh[n]$ . The value of g[8] equals ___________.

🚀 Solve in Practice Mode

📖 Explanation

Pole locationof H(z) are given as,

\begin{aligned} z&=\frac{1}{\sqrt{2}}e^{j(2K-1)\frac{\pi}{4}};\;\;K=1,2,3,4\\ z_1&=\frac{1}{\sqrt{2}}e^{j\frac{\pi}{4}} =\frac{1}{2}(1+j)\\ z_2&=\frac{1}{\sqrt{2}}e^{j\frac{3\pi}{4}} =\frac{1}{2}(-1+j)\\ z_3&=\frac{1}{\sqrt{2}}e^{j\frac{5\pi}{4}} =\frac{1}{2}(-1-j)\\ z_4&=\frac{1}{\sqrt{2}}e^{j\frac{7\pi}{4}} =\frac{1}{2}(1-j)\\ \text{Now,}\\ H(z)&=\frac{K\cdot z^4}{(z-z_1)(z-z_2)(z-z_3)(z-z_4)}...(i) \end{aligned}

[As h(n) is causal and it starts from n=0, so numerator will have same order as denominator is having]. by solving equation (i)

\begin{aligned} H(z)&=\frac{K\cdot z^4}{z^{4+}\frac{1}{4}}\;\;...(ii)\\ &\text{Given that,}\\ H(1)&=\frac{5}{4}\\ &\text{From equation (ii),}\\ H(1)&=\frac{K}{1+\frac{1}{4}}\Rightarrow \frac{5}{4}=\frac{K}{\frac{5}{4}}\\ \Rightarrow \; K&=\frac{25}{16}\\ H(z)&=\frac{\frac{25}{16}z^4}{z^{4+}\frac{1}{4}}\\ &\text{By using division rule,}\\ H(z)&=\frac{25}{16}\left[ 1+\frac{1}{4}z^{-4}+\frac{1}{16}z^{-8}+... \right]\$\\ \text{Thus, }h(8)&=\frac{1}{16} \times \frac{25}{16}=0.097\\ &\text{it is given that,}\\ g(n)&=j^nh(n)\\ g(8)&=j^8h(8)=h(8)\\ \Rightarrow \; g(8)&=h(8)=0.097 \end{aligned}

Q13GATE 2014MCQ

A discrete system is represented by the difference equation

\begin{bmatrix} X_{1}(k+1)\\ X_{2}(k+1) \end{bmatrix}=\begin{bmatrix} a & a-1\\ a+1& a \end{bmatrix}\begin{bmatrix} X_{1}(k)\\ X_{2}(k) \end{bmatrix}

It has initial conditions $X_1(0)=1;X_2(0)=0$ . The pole locations of the system for a = 1, are

🚀 Solve in Practice Mode

📖 Explanation

Given that,

\begin{bmatrix} X_1(K+1)\\X_2(K+1) \end{bmatrix}=\begin{bmatrix} a & a-1\\ a+1 & 1 \end{bmatrix}\begin{bmatrix} X_1(K)\\X_2(K) \end{bmatrix}

With initial conditions, $X_1(0)=1,\;\;X_2(0)=0$ For a=1, we can write,

\begin{bmatrix} X_1(K+1)\\X_2(K+1) \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 2 & 1 \end{bmatrix}\begin{bmatrix} X_1(K)\\X_2(K+) \end{bmatrix}

$\Rightarrow \; X_1(K+1)=X_1(K) \;\;...(i)$ $\Rightarrow \; X_2(K+1)=2X_1(K)+X_2(k) \;\;...(ii)$ Applying z-transform on equation (i),

\begin{aligned} zX_1(z)-zX_1(0)&=X_1(z) \\ X_1(z)[z-1]&=zX_1(0) \\ X_1(z) =\frac{z}{z-1}\;\;&[Given, \; X_1(0)=1]...(iii) \\ \text{from equation (ii),}\\ zX_2(z)-zX_2(0)&=2X_1(z)+X_2(z)\\ X_2(z)[z-1]&=2X_1(z)+zX_2(0)\\ [Given,\;\; X_2(0)&=0]\\ \frac{X_2(z)}{X_1(z)}&=\frac{2}{z-1} \end{aligned}

Thus, transfer function, $H(z)=\frac{2}{z-1}$ Therefore, pole location is z=1.

Q14GATE 2014MCQ

An input signal x(t)=2+5sin(100 $\pi$ t) is sampled with a sampling frequency of 400 Hz and applied to the system whose transfer function is represented by $\frac{Y(z)}{X(z)}=\frac{1}{N}(\frac{1-z^{-N}}{1-z^{-1}})$ where, N represents the number of samples per cycle. The output y(n) of the system under steady state is

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} f_s&= 400Hz\\ x(t)&=2+5 \sin (100 \pi t) \\ t\rightarrow xT_s &=\frac{n}{f_s}=\frac{n}{400} \\ x(n)&=2+5 \sin \left ( 100 \pi\frac{n}{400} \right ) \\ &= 2+5 \sin \frac{\pi}{4}n\\ \text{Here, }\omega _0 &=\frac{\pi}{4} \end{aligned}

Therefore, N= time-period of $x(n)=\frac{2 \pi }{\omega _0}=\frac{2 \pi}{\pi/4}=0$

\begin{aligned} \text{Now, }H(z)&=\frac{1}{N}\left[ \frac{1-z^{-N}}{1-z^{-1}} \right]\$ \\ \text{Applying, }z &=e^{j\omega } \\ H(e^{j\omega }) &= \frac{1}{N}\left[ \frac{1-e^{-j\omega N}}{1-e^{-j\omega }} \right]\$ \\ \text{For, } &\text{sinusoidal part of }x(n): \\ H(e^{j\omega _0 }) &= \frac{1}{N}\left[ \frac{1-e^{-j\omega _0 N}}{1-e^{-j\omega _0 }} \right]\$ \\ [\text{Note: }] \omega _0 N&=\frac{\pi}{4} \times 8 =2 \pi] \\ &= \frac{1}{N}\left[ \frac{1-e^{-j2 \pi N}}{1-e^{-j \pi/4}} \right]\$=0 \end{aligned}

Therefore, for sinusoidal part of input, system output is zero. For dc part of input,

\begin{aligned} y(n)&=\text{output}\\ &=H(e^{j0}) \times \text{input dc value}\\ \$[\because \; &H(e^{j\omega })_{\omega =0}=1]\$\\ &=1 \times 2=2 \end{aligned}

Thus, steady state output = 2

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