Transfer function and magnitude for y[n] = 0.3y[n−1] + x[n]

System: $y[n]=0.3y[n-1]+x[n]$, initially at rest.

Transfer function

Take the z-transform:

$$\begin{aligned} Y(z)&=0.3z^{-1}Y(z)+X(z)\\ H(z)=\frac{Y(z)}{X(z)}&=\frac{1}{1-0.3z^{-1}}=\frac{z}{z-0.3}, \qquad |z|>0.3 \end{aligned}$$

There is a pole at $z=0.3$ (inside the unit circle) and a zero at $z=0$. On the unit circle, $z=e^{j\omega}$:

$$H(e^{j\omega})=\frac{1}{1-0.3e^{-j\omega}}$$

Magnitude

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

The magnitude is even in $\omega$ and periodic with period $2\pi$:

 |H(e^jw)|
 1.43 |            *
      |         *     *
 1.0  |      *           *
      |   *                 *
 0.77 |*                       *
      +--+--------+--------+---> w
       -pi        0        pi

It is a low pass response: maximum $1/0.7$ at $\omega=0$, minimum $1/1.3$ at $\omega=\pm\pi$.

Impulse response

Using $a^n u[n]\leftrightarrow\frac{1}{1-az^{-1}}$ ($|z|>|a|$) with $a=0.3$:

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

Recursion check with $x=\delta[n]$: $h[0]=1,\ h[1]=0.3,\ h[2]=0.09,\ h[3]=0.027,\dots$

Answer: $H(z)=\dfrac{1}{1-0.3z^{-1}}$, $|H(e^{j\omega})|=\dfrac{1}{\sqrt{1.09-0.6\cos\omega}}$, $h[n]=(0.3)^n u[n]$ (causal and stable).