1
$\begingroup$
  • 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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.