Question

The unit sample response of a LTI system and the output of the system is y(n) when unit step u(n) is applied at the input. The value of $\underset{n\to \infty }{lim}y\left(n\right)ish\left(n\right)=3{\left(\frac{1}{2}\right)}^{n}u\left(n\right)-2{\left(\frac{1}{3}\right)}^{n-1}u\left(n\right)$

• A. $\frac{1}{3}$
• B. $-\frac{1}{3}$
• C. 3
• D. -3

Answer 1

The correct option is B $-\frac{1}{3}$

$y\left(n\right)=h\left(n{)}^{\ast }x\right(n)=\sum _{k=-\infty }^{\infty }h(k).x(n-k)$

if x(n) is a unit step,

$\therefore y\left(n\right)=\sum _{k=-\infty }^{\infty }h\left(k\right).u(n-k)=\sum _{k=-\infty }^{n}h\left(k\right)$
$\therefore \underset{n\to -\infty }{lim}y\left(n\right)=\sum _{k=-\infty }^{\infty }h\left(k\right)$

$\therefore \underset{n\to \infty }{lim}y\left(n\right)=3\sum _{k=0}^{\infty }{\left(\frac{1}{2}\right)}^k-2\sum _{k=0}^{\infty }{\left(\frac{1}{3}\right)}^{k-1}=\frac{3}{1-\frac{1}{2}}-\frac{6}{1-\frac{1}{3}}=6-9=-3$