# Figuring magnitude and phase response

I've got a linear time-invariant system $$y[n]=\frac{8}{9}y[n-1]+x[n]$$ which I transformed into a transfer function $$Y(z)=\frac{8}{9}Y(z)*z^{-1}+X(z) =>\frac{Y(z)}{X(z)}=\frac{1}{1-\frac{8}{9}*z^{-1}}=\frac{z}{z-\frac{8}{9}}$$ The pole is positive, which would make the system stabile. However, I am having some trouble getting the magnitude/phase responses $$|H(\omega)|$$ and $$\theta(\omega)$$. How should I proceed with the system to figure these out? I've read about using the substitution for $$z$$ as$$Y(z)=\frac{z}{z-\frac{8}{9}}=\frac{e^{iw}}{e^{iw}-\frac{8}{9}}$$ or even continue to transform the exponentials to their $$\cos$$ + $$\sin$$ forms, but i am not sure how i should proceed with them.

As an addition, how would the system change if I would to swap $$x[n]$$ with a sin or a cosine of some value? I assume it would be the same path towards the responses but with different values.

$$|H(z)| = \sqrt{\Re^2 H(z) + \Im^2 H(z)} \\ \theta \big(H(z)\big) = \arctan \frac{\Im H(z)}{\Re H(z)} \\$$

$$\Re z$$ is the real part of z;

$$\Im z$$ is imaginary part.

For a single-pole filter $$y_n = b_0 x_n - a_1 y_{n-1}$$,

$$H(z) = \frac{b_0}{1 + a_1 z^{-1}}; \\ \Re H(z) = b_0 \frac{1 + a_1 \cos \omega}{1 + 2 a_1 \cos \omega + a_1^2}; \\ \Im H(z) = b_0 \frac{a_1 \sin \omega}{1 + 2 a_1 \cos \omega + a_1^2}.$$

With $$b_0 = 1$$ and $$a_1 = -\frac{8}{9}$$, your filter may be unstable.

For a low-pass filter, try $$b_0 = \frac{1}{9}, a_1 = -\frac{8}{9}$$.

• I now got that $$|H(ω)|=\frac{1}{\sqrt{cos(ω)-\frac{8}{9})^2+(sin(ω))^2}}$$ but i still seem to have little to no idea about the phase response. – TootsieRoll Nov 16 '18 at 20:29
• @TootsieRoll: Remember that $e^{i \omega} = cos \omega + i sin \omega$. – igorinov Nov 16 '18 at 22:34
• I see, now i got $$\theta(\omega)=arctan(\frac{sin(\omega)}{(\frac{9}{8})-cos(\omega)})$$ Any tips on the $$x[n]=cos(\frac{\pi*n}{2})$$ ? – TootsieRoll Nov 17 '18 at 0:20
• @TootsieRoll: This is a sine wave with frequency $\frac{f_s }{4}$. $$\omega = \frac{\pi}{2}; \\ z = e^{i \omega} = i$$ – igorinov Nov 17 '18 at 5:29