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Determine the frequency response $H(e^{j\omega})$ of a system characterized by $h(n) = (0.9)^n u(n)$

using the definition $$H(e^{j\omega})=\sum_{-\infty}^\infty h(n) e^{-j\omega n}$$ $H(e^{j\omega})=\sum_0^\infty (0.9)^n e^{-j\omega n}$ = $\sum_0^\infty (0.9 e^{-j\omega})^n$ = $$\frac{1}{1-0.9e^{-j\omega}}$$

I understand the above but my question comes when the book tries to get the magnitude of $H(e^{j\omega})$.

$$|H(e^{j\omega})| = \sqrt \frac{1}{(1-0.9 \cos\omega)^2+(0.9 \sin\omega)^2}$$

I know that the use of Euler's formula and the complex conjugate is involved. However, I need to see the math.

Ángel Mario Gallegos
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R SMITH
  • 17

1 Answers1

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Using that $\,e^{j \alpha}=\cos \alpha + j \sin \alpha\,$and $\,|z| = \sqrt{\operatorname{Re}^2(z)+\operatorname{Im}^2(z)}\,$:

$$ \begin{align} \left|\frac{1}{1-0.9e^{-j\omega}}\right| &= \frac{1}{\;|1-0.9e^{-j\omega}|\;} \\ &= \frac{1}{\;|1 - 0.9 \big( \cos(-\omega)+j \sin(-\omega)\big)|\;} \\ &= \frac{1}{\;|\big(1 - 0.9 \cos(\omega)\big) - 0.9\,j \sin(\omega))|\;} \\ &= \frac{1}{\sqrt{\big(1 - 0.9 \cos(\omega)\big)^2+\big(0.9 \sin(\omega)\big)^2}} \end{align} $$

dxiv
  • 76,497
  • Thank you. Couldn't have figured it out myself. Btw shouldn't that be '- 0.9jsin(w)' instead of just '-j*sin(w)' in the third line of the solution? – R SMITH Jun 29 '18 at 00:18
  • @RSMITH Right, thanks, fixed now. – dxiv Jun 29 '18 at 00:20