H(e^(jω)) and h[n] for y[n] = 0.3y[n−1] + x[n]

Given $y[n]=0.3y[n-1]+x[n]$, i.e. $y[n]-0.3y[n-1]=x[n]$ (system initially at rest).

Frequency response

Take the DTFT of both sides, using the time-shift property $y[n-1]\leftrightarrow e^{-j\omega}Y(e^{j\omega})$:

$$\begin{aligned} Y(e^{j\omega})-0.3e^{-j\omega}Y(e^{j\omega}) &= X(e^{j\omega})\\ H(e^{j\omega})=\frac{Y(e^{j\omega})}{X(e^{j\omega})} &= \frac{1}{1-0.3e^{-j\omega}} \end{aligned}$$

Magnitude and phase (using $e^{-j\omega}=\cos\omega-j\sin\omega$):

$$|H(e^{j\omega})|=\frac{1}{\sqrt{(1-0.3\cos\omega)^2+(0.3\sin\omega)^2}}=\frac{1}{\sqrt{1.09-0.6\cos\omega}}$$ $$\angle H(e^{j\omega})=-\tan^{-1}\!\left(\frac{0.3\sin\omega}{1-0.3\cos\omega}\right)$$

The gain falls from $1/0.7$ at $\omega=0$ to $1/1.3$ at $\omega=\pi$, so it is a low pass system.

Impulse response

Use the standard DTFT pair $a^n u[n]\leftrightarrow\frac{1}{1-ae^{-j\omega}}$ for $|a|<1$. With $a=0.3$:

$$h[n]=(0.3)^n\,u[n]$$

Check by recursion with $x[n]=\delta[n]$, $h[-1]=0$: $h[0]=1$, $h[1]=0.3$, $h[2]=0.09$, $h[3]=0.027,\dots$ which matches $(0.3)^n$.

Answer: $H(e^{j\omega})=\dfrac{1}{1-0.3e^{-j\omega}}$ and $h[n]=(0.3)^n u[n]$. The system is causal and stable ($\sum h=1/0.7\approx1.43$).