Question

$\text{The output y[n] of a discrete time LTI system is found to be}2{\left(\frac{1}{3}\right)}^{n}u\left[n\right]\text{when input}x\left[n\right]\text{is}u\left[n\right].\text{The output}y\left[n\right]=[k_1{\left(\frac{1}{2}\right)}^n+k_2{\left(\frac{1}{3}\right)}^n]u\left[n\right]\text{when input is}{\left(\frac{1}{2}\right)}^{n}u\left[n\right].$ .
Find the value of $k_1+k_2$?

• A. 4
• B. 2
• C. 1
• D. 0

Answer 1

The correct option is B 2

$\text{Given,}{x}_1\left[n\right]=u\left[n\right]$
$X_1\left(z\right)=\frac{1}{1-z^{-1}}=\frac{1}{1-z}|z|>1$
$y_2\left[n\right]=2{\left(\frac{1}{3}\right)}^{n}u\left[n\right]$
$\Rightarrow Y_1\left(z\right)=\frac{2z}{z-\frac{1}{3}}$
$x_2\left[n\right]={\left(\frac{1}{2}\right)}^{n}u\left[n\right]$

$Y_1\left(z\right)=\frac{2z}{z-\frac{1}{2}};|z|>\frac{1}{3}$
$x_2\left[n\right]={\left(\frac{1}{2}\right)}^{n}u\left[n\right]$
$X_2\left(z\right)=\frac{z}{z-\frac{1}{2}};|z|>\frac{1}{2}$

$Y_2\left(2\right)=H_1\left(z\right)X_2\left(z\right)$
$=\frac{2z(z-1)}{(z-\frac{1}{2})(z-\frac{1}{3})};|z|>\frac{1}{2}$

$\frac{Y_2\left(z\right)}{z}=\frac{2(z-1)}{(z-\frac{1}{2})(z-\frac{1}{3})};|z|>\frac{1}{2}$
Taking inverse Z-transform,
$y_2\left[n\right]=[-6{\left(\frac{1}{2}\right)}^n+8{\left(\frac{1}{3}\right)}^n]u\left[n\right]$
$\therefore \text{By comparing with given}y\left[n\right],$
$k_1=-6\text{and}{k}_2=8$
$\therefore k_1+k_2=-6+8=2$