Z Transform From Difference Equation MCQ Quiz in मल्याळम - Objective Question with Answer for Z Transform From Difference Equation - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Taking z- transform of x[z] and y[z] we get

X[z] = 1 + Z^-1   y(z) = 1 + Z^-2

X[z] × y[z] = 1 + Z^-1 + Z^-2 + Z^-3

​$10y\left(n+2\right)+60y\left(n+1\right)+90y\left(n\right)=2x\left(n+1\right)+6x\left(n\right)$

Taking Z transfer of the difference equation we get

$\begin{array}{rl}& 10z^{2}Y\left(z\right)+60zY\left(z\right)+90Y\left(z\right)=2zX\left(z\right)+6X\left(z\right)\\ & \Rightarrow \frac{Y\left(z\right)}{X\left(z\right)}=H\left(z\right)=\frac{2z+6}{10z^2+60z+90}=\frac{2z+6}{10\left(z^2+6z+9\right)}\\ & =\frac{z+3}{5{\left(z+3\right)}^2}=\frac{1}{5\left(z+3\right)}\end{array}$

Taking z- transform we have

${\mathrm{z}}^{-1}\mathrm{Y}\left(\mathrm{z}\right)+\mathrm{y}\left[-1\right]+2\mathrm{Y}\left(\mathrm{z}\right)=\mathrm{X}\left(\mathrm{z}\right)$

For zero input response we put $\mathrm{X}\left(\mathrm{z}\right)=0$

$\mathrm{Y}\left(\mathrm{z}\right)=\frac{-1}{1+\frac{1}{2}{\mathrm{z}}^{-1}}$

$\Rightarrow \mathrm{y}\left[\mathrm{n}\right]=-{\left(-\frac{1}{2}\right)}^{\mathrm{n}}\mathrm{u}\left[\mathrm{n}\right]$

Given,

$\mathrm{y}\left[\mathrm{n}\right]+3\mathrm{y}\left[\mathrm{n}-1\right]=\mathrm{x}\left[\mathrm{n}\right]$

Taking unilateral Z transform we have

$\begin{array}{l}\mathrm{Y}\left(\mathrm{z}\right)+3\left(z^{-1}\mathrm{Y}\left(\mathrm{z}\right)+\mathrm{y}\left[-1\right]\right)=\mathrm{X}\left(\mathrm{z}\right)\\ \Rightarrow Y\left(z\right)=\frac{\alpha }{\left(1+3z^{-1}\right)\left(1-z^{-1}\right)}-\frac{3\beta }{1+3z^{-1}}\\ \left(\alpha u\left[n\right]\leftrightarrow \frac{\alpha }{1-z^{-1}}\right)\end{array}$

$\Rightarrow Y\left(z\right)=\frac{\alpha -3\beta +3\beta z^{-1}}{\left(1+3z^{-1}\right)\left(1-z^{-1}\right)}$   ______ (i)

Now from $\mathrm{y}\left[\mathrm{n}\right]=\left[3{\left(-3\right)}^n+2\right]\mathrm{u}\left[\mathrm{n}\right]$, we have

Taking unilateral z transform, $Y\left(z\right)=\frac{3}{1+3z^{-1}}+\frac{2}{1-z^{-1}}$

$=\frac{5+3z^{-1}}{\left(1+3z^{-1}\right)\left(1-z^{-1}\right)}$  _____(ii)

Comparing (i) and (ii)

We have β = 1

$\begin{array}{l}\alpha --3\beta =5\\ \Rightarrow \alpha =8\end{array}$

Thus, $\alpha +\beta =8+1=9$

Taking z-transform

$Y\left(z\right)-\frac{1}{3}{z}^{-1}Y\left(z\right)=2X\left(z\right)$

$\Rightarrow H\left(z\right)=\frac{Y\left(z\right)}{X\left(z\right)}=\frac{2}{1-\left(\frac{1}{3}\right)Z^{-1}}$

Taking inverse z-transform

$H\left(n\right)=2{\left(\frac{1}{3}\right)}^{n}u\left(n\right)$

it satisfied H(1) = 2