Impulse response of y[n] − (3/4)y[n−1] + (1/8)y[n−2] = 2x[n]

Given: $y[n] - \frac{3}{4}y[n-1] + \frac{1}{8}y[n-2] = 2x[n]$, system initially at rest.

Step 1: Frequency response

Take the DTFT ($x[n-k] \leftrightarrow e^{-j\omega k}X(e^{j\omega})$):

$$\left(1 - \tfrac{3}{4}e^{-j\omega} + \tfrac{1}{8}e^{-j2\omega}\right)Y(e^{j\omega}) = 2X(e^{j\omega})$$ $$H(e^{j\omega}) = \frac{2}{1 - \frac{3}{4}e^{-j\omega} + \frac{1}{8}e^{-j2\omega}}$$

Step 2: Factorise the denominator

Let $v = e^{-j\omega}$. The roots of $1 - \frac{3}{4}v + \frac{1}{8}v^2$ come from $p^2 - \frac{3}{4}p + \frac{1}{8} = 0$, giving $p = \frac{1}{2}, \frac{1}{4}$. So

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

Step 3: Partial fractions

$$H = \frac{A}{1 - \frac{1}{2}v} + \frac{B}{1 - \frac{1}{4}v}$$ $$\begin{aligned} A &= \left.\frac{2}{1 - \frac{1}{4}v}\right|_{v=2} = \frac{2}{1 - \frac{1}{2}} = 4 \\ B &= \left.\frac{2}{1 - \frac{1}{2}v}\right|_{v=4} = \frac{2}{1 - 2} = -2 \end{aligned}$$ $$H(e^{j\omega}) = \frac{4}{1 - \frac{1}{2}e^{-j\omega}} - \frac{2}{1 - \frac{1}{4}e^{-j\omega}}$$

Step 4: Inverse DTFT

Using $a^n u[n] \leftrightarrow \frac{1}{1 - a e^{-j\omega}}$ ($|a| < 1$):

$$h[n] = \left[4\left(\tfrac{1}{2}\right)^n - 2\left(\tfrac{1}{4}\right)^n\right]u[n]$$

Check by recursion

$h[n] = 2\delta[n] + \frac{3}{4}h[n-1] - \frac{1}{8}h[n-2]$, with $h[-1] = h[-2] = 0$:

Both agree. Both poles ($0.5$, $0.25$) are inside the unit circle, so the system is causal and stable; $h[n]$ decays to zero. Also $H(e^{j0}) = \frac{2}{1 - 0.75 + 0.125} = 5.333 = \sum h[n] = \frac{4}{0.5} - \frac{2}{0.75}$.

Answer: $h[n] = \left[4(0.5)^n - 2(0.25)^n\right]u[n]$