Question

Consider a causal LTI system whose input x[n] and output y[n] are related by the difference equation

$y\left[n\right]=0.25y[n-1]+x\left[n\right]$ and

$x\left[n\right]=\delta [n-1],$ then the value of y[4]

• A. $\frac{1}{4}$
• B. $\frac{1}{256}$
• C. $\frac{1}{16}$
• D. $\frac{1}{64}$

Answer 1

The correct option is D $\frac{1}{64}$

Since the system is causal

$y\left[n\right]=0forn<1$

$y\left[1\right]=\frac{1}{4}y\left[0\right]+x\left[1\right]=0+1=1$

$y\left[2\right]=\frac{1}{4}y\left[1\right]+x\left[2\right]=\frac{1}{4}1+0=\frac{1}{4}$

$y\left[3\right]=\frac{1}{4}y\left[2\right]+x\left[3\right]=\frac{1}{4}\times \frac{1}{4}+0=\frac{1}{(4{)}^2}$

$\therefore y\left(n\right)={\left(\frac{1}{4}\right)}^{n-1}$

$\therefore y\left(4\right)=\frac{1}{4^3}=\frac{1}{64}$