Solution:

In this example, we determine the impulse response for a stable linear time-invariant system for which the input $x[n]$ and output $y[n]$ satisfy the linear constant-coefficient difference equation,

$$ y[n]-\frac{1}{2} y[n-1]=x[n]-\frac{1}{4} x[n-1] $$

However, this example offers a hint of the utility of transform methods in the analysis of linear systems. To find the impulse response, we set $x[n]=\delta[n]$; with $h[n]$ denoting the impulse response, Eq. becomes,

$$ h[n]-\frac{1}{2} h[n-1]=\delta[n]-\frac{1}{4} \delta[n-1] ....(1) $$

Applying the Fourier transform to both sides of Eq., we obtain,

$ \begin{gathered}\\ H\left(e^{j \omega}\right)-\frac{1}{2} e^{-j \omega} H\left(e^{j \omega}\right)=1-\frac{1}{4} e^{-j \omega},...(2) \\\\ H\left(e^{j \omega}\right)=\frac{1-\frac{1}{4} e^{-j \omega}}{1-\frac{1}{2} e^{-j \omega}} ...(3).\\ \end{gathered}\\ $

To obtain $h[n]$, we want to determine the inverse Fourier transform of $H\left(e^{j \omega}\right)$. Toward this end, we rewrite Eq. (3) as,

$$ H\left(e^{j \omega}\right)=\frac{1}{1-\frac{1}{2} e^{-j \omega}}-\frac{\frac{1}{4} e^{-j \omega}}{1-\frac{1}{2} e^{-j \omega}}...(4) . $$

Combining this transformation, we obtain,

$$ -\left(\frac{1}{4}\right)\left(\frac{1}{2}\right)^{n-1} u[n-1] \stackrel{\mathcal{F}}{\longleftrightarrow}-\frac{\frac{1}{4} e^{-j \omega}}{1-\frac{1}{2} e^{-j \omega}}...(5) . $$ Based on property , then, $$ h[n]=\left(\frac{1}{2}\right)^n u[n]-\left(\frac{1}{4}\right)\left(\frac{1}{2}\right)^{n-1} u[n-1] ...(6). $$