• Example - Consider the causal stable IIR transfer function $$ H(z)=\frac{K}{1-\alpha z^{-1}}, \quad 0 < \lvert \alpha\rvert 1 $$ where $K$ and $\alpha$ are real constants

  • Its square-magnitude function is given by $$ \lvert H\left(e^{j\omega}\right)\rvert^2 = H(z)H\left(z^{-1}\right)\bigg\vert_{z=e^{j\omega}}=\frac{K^2}{\left(1+\alpha^2\right)-2\alpha\cos\omega} $$

Can anyone please explain how to simplify the equation to get this form? I am not able to understand it.

  • $\begingroup$ replace z with the $e^{j\omega}$ and perform all necessary polar to rectangular conversion $e^{j\omega}=\cos(\omega)+j \sin(\omega)$ and do the necessary algebraic simplifications... $\endgroup$ – Fat32 Mar 11 '17 at 15:00

$$H(z)\cdot H(z^{-1})= \frac{K}{1-\alpha \cdot e^{-j \omega}} \cdot \frac{K}{1-\alpha \cdot e^{+j \omega}}$$

$$= \frac{K^2}{1-\alpha \cdot (e^{-j \omega}+e^{+j \omega})+ \alpha^2} = \frac{K^2}{1 + \alpha^2- 2 \cdot \alpha \cdot Re\{e^{j \omega}\}} = \frac{K^2}{1 + \alpha^2- 2 \cdot \alpha \cdot \cos( \omega)} $$

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It is useful to avoid unnecessary complications: $K$ is real, and the square-magnitude of a ratio is the ratio of the square-magnitude of the numerator under the denominator.

So you only have to look at $$|D(z)|^2 = (1-\alpha z^{-1})(1-\alpha (z^{-1})^{-1}) = 1+\alpha^2-\alpha(z^{-1}+z)\,.$$

For a unit-norm $z$ (or a $z$ on the unit circle), you can write $z=e^{j\omega}$, and $z^{-1}=\overline{z}$, which is the complex conjugate. The simple identity:

$$ z+z^{-1}= z+\overline{z} = 2\mathrm{Real}(z)$$

illustrated below finally yields the denominator since

$$ 2\mathrm{Real}(e^{j\omega}) = 2\cos \omega$$

Fortunately, the final result is the same as the one given by @Hilmar

$$ \frac{K^2}{1+\alpha^2 - 2\alpha\cos \omega}$$

complex and conjugate sum

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