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

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

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

• What have you tried so far? Some effort must be placed before we lend a hand! – Envidia Jan 12 at 20:04
• 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 $||\cdot ||$? – Matt L. Jan 12 at 20:19

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. $$\Re\{H(e^{j\omega})\} = \big|H(e^{j\omega})\big| \cos\big(\arg\{H(e^{j\omega})\}\big) = \Re\{H(e^{-j\omega})\}$$ $$\Im\{H(e^{j\omega})\} = \big|H(e^{j\omega})\big| \sin\big(\arg\{H(e^{j\omega})\}\big) = -\Im\{H(e^{-j\omega})\}$$

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.

• @ Matt L. you are right. I edited it. – DSPinfinity Jan 12 at 21:43

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.

• 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. – robert bristow-johnson Jan 13 at 2:48