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Let $H(z)$ be the rational system function of an LTI system.

How can we show that $|H(e^{j\omega})|=|H(e^{-j\omega})|$?

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  • $\begingroup$ What have you tried so far? Some effort must be placed before we lend a hand! $\endgroup$ – Envidia Jan 12 at 20:04
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    $\begingroup$ You also need to give some context, because the way you formulated it, that equation is actually wrong. I assume you just mean "magnitude" by $||\cdot ||$? $\endgroup$ – Matt L. Jan 12 at 20:19
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Your statement only holds true for a real sequence $h[n]$.

The frequency response of $h[n]$ equals to

$$H(e^{j\omega}) = \sum_{n=-\infty}^{\infty}h[n] e^{-j\omega n}$$

and if $h(n)$ is real, the conjugate symmetry condition holds

$$H(e^{-j\omega}) = \sum_{n=-\infty}^{\infty}h[n] e^{j\omega n}=\big(H(e^{j\omega})\big)^*$$

So we can derive that $$\big|H(e^{j\omega})\big| = \big|H(e^{-j\omega})\big|$$

and $$\arg\{H(e^{j\omega})\} = -\arg\{H(e^{-j\omega})\}$$ where $\arg\{\cdot\}$ represents phase response.

We can also derive that the real part of the frequency response is even symmetric and the imaginary part is odd symmetric. $$\Re\{H(e^{j\omega})\} = \big|H(e^{j\omega})\big| \cos\big(\arg\{H(e^{j\omega})\}\big) = \Re\{H(e^{-j\omega})\}$$ $$\Im\{H(e^{j\omega})\} = \big|H(e^{j\omega})\big| \sin\big(\arg\{H(e^{j\omega})\}\big) = -\Im\{H(e^{-j\omega})\}$$

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A simple counter-example: take a causal two tap FIR filter $H(z)=h[0]+h[1]z^{-1}$ with $h[1]\neq 0$. Now we have $H(z^{-1})=h[0]+h[1]z$. Take any $z\in\mathbb{C}$ satisfying $|z|\neq 1$ and verify that $|H(z)|\neq |H(z^{-1})|$.

So try to figure out what it is that is actually meant. Could it be that you mean that equality holds for $|z|=1$, i.e., on the unit circle?


Now that the question has been edited, we just have to prove the given equality for $|z|=1$. Note that we need yet another requirement, namely that the system is real-valued, i.e., its impulse response $h[n]$ is real. In that case the frequency response is conjugate symmetrical:

$$H(e^{j\omega})=H^*(e^{-j\omega})\tag{1}$$

and, consequently,

$$|H(e^{j\omega})|=|H^*(e^{-j\omega})|=|H(e^{-j\omega})|\tag{2}$$

must hold.

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  • $\begingroup$ @ Matt L. you are right. I edited it. $\endgroup$ – DSPinfinity Jan 12 at 21:43
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Even after your edit, we can't show what is wrong.

Counterexample:

$$h(\tau)=\delta(\tau)+\frac j{\pi\tau}$$

Convolution with that is an LTI system.

That system happens to be the Hilbert transformator.

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    $\begingroup$ only the imaginary part (which is the stuff that multiplies $j$ in the right-hand term) $\frac{1}{\pi \tau}$ is the impulse response of a Hilbert Transformer. $\endgroup$ – robert bristow-johnson Jan 13 at 2:48

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