Transfer function of y[n] = 0.4y[n−1] + x[n]

Given: $y[n] = 0.4\,y[n-1] + x[n]$, i.e. $y[n] - 0.4\,y[n-1] = x[n]$, initially at rest.

Transfer function (frequency response)

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

$$\begin{aligned} Y(e^{j\omega})\left(1 - 0.4e^{-j\omega}\right) &= X(e^{j\omega}) \\ H(e^{j\omega}) &= \frac{Y(e^{j\omega})}{X(e^{j\omega})} = \frac{1}{1 - 0.4e^{-j\omega}} \end{aligned}$$

(In the z-domain, $H(z) = \frac{1}{1 - 0.4z^{-1}} = \frac{z}{z - 0.4}$, a pole at $z = 0.4$ inside the unit circle.)

With $e^{-j\omega} = \cos\omega - j\sin\omega$:

$$|H(e^{j\omega})| = \frac{1}{\sqrt{(1 - 0.4\cos\omega)^2 + (0.4\sin\omega)^2}} = \frac{1}{\sqrt{1.16 - 0.8\cos\omega}}$$ $$\theta(\omega) = -\tan^{-1}\frac{0.4\sin\omega}{1 - 0.4\cos\omega}$$

 |H|
1.67|          *
    |        .' '.
1.30|      *       *
0.93|    *           *
0.71|  *               *
    +--+-------+-------+--> w
     -pi       0       pi

The magnitude is even and periodic ($2\pi$), maximum $1/0.6 = 1.667$ at $\omega = 0$ and minimum $1/1.4 = 0.714$ at $\omega = \pi$: a low pass system.

Impulse response

Using $a^n u[n] \leftrightarrow \frac{1}{1 - ae^{-j\omega}}$ with $a = 0.4$:

$$h[n] = (0.4)^n\,u[n]$$

(Recursion check: $h[0] = 1$, $h[1] = 0.4h[0] = 0.4$, $h[2] = 0.16$.)

 h[n]
 1.0 |  o
     |  |
 0.4 |  |  o
0.16 |  |  |  o
     |  |  |  |  o  .
   0 +--+--+--+--+--+--> n
        0  1  2  3  4

The system is causal, stable ($\sum h[n] = 1/0.6 = 1.667$) and IIR.