Impulse and frequency response of y[n] + 0.55y[n−1] = x[n]

Given: $y[n] + 0.55\,y[n-1] = x[n]$, initially at rest.

Impulse response

Put $x[n] = \delta[n]$, so $h[n] = \delta[n] - 0.55\,h[n-1]$ with $h[-1] = 0$:

• $h[0] = 1$
• $h[1] = -0.55$
• $h[2] = (-0.55)^2 = 0.3025$, and so on.

So

$$h[n] = (-0.55)^n\,u[n]$$

Sketch (not to scale; the magnitudes shrink by a factor of 0.55 at each step):

 h[n]
 1.0 |  o
     |  |     o     o     o
   0 +--+--+--+--+--+--+--+--> n
     |  0  |  2  |  4  |  6
-0.55|     o     o     o

The samples alternate in sign and decay (since $|-0.55| < 1$): causal and stable, $\sum|h[n]| = 1/0.45 = 2.222$.

Frequency response

DTFT of the difference equation:

$$\begin{aligned} Y(e^{j\omega})\left(1 + 0.55e^{-j\omega}\right) &= X(e^{j\omega}) \\ H(e^{j\omega}) &= \frac{1}{1 + 0.55e^{-j\omega}} = \frac{1}{(1 + 0.55\cos\omega) - j0.55\sin\omega} \end{aligned}$$

(This also follows from $a^n u[n] \leftrightarrow \frac{1}{1 - ae^{-j\omega}}$ with $a = -0.55$.)

Magnitude:

$$|H(e^{j\omega})| = \frac{1}{\sqrt{(1 + 0.55\cos\omega)^2 + (0.55\sin\omega)^2}} = \frac{1}{\sqrt{1.3025 + 1.1\cos\omega}}$$

Phase:

$$\theta(\omega) = \tan^{-1}\frac{0.55\sin\omega}{1 + 0.55\cos\omega}$$

 |H|
2.22|*                   *
    | '.               .'
1.38|   *             *
0.88|     *         *
0.65|        '-*-'
    +--+-------+-------+--> w
     -pi       0       pi

The gain is smallest at $\omega = 0$ ($1/1.55 = 0.645$) and largest at $\omega = \pi$ ($1/0.45 = 2.222$), so this system is a high pass filter: the pole at $z = -0.55$ (near $\omega = \pi$) makes $h[n]$ alternate in sign, so it emphasises rapid sample-to-sample changes. $|H|$ is even and $\theta(\omega)$ is odd, and both are periodic with period $2\pi$.