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I am a beginner to study about the filter notion and property

Being a real digital filter, (here "real filter" I means that its impulse response is real-valued)

this formula is established. But I have no idea how to prove it

$$ |H(\pi +w )|=|H(\pi -w)|$$

What should be $H(w)$ or $|H(w)| $ for generalized proof procedure?

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1 Answer 1

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HINT:

From the definition of the DTFT

$$H(\omega)=\sum_{n=-\infty}^{\infty}h[n]e^{-jn\omega}\tag{1}$$

derive the following facts:

  1. $H(\omega)=H(\omega+2\pi)$
  2. $H(\omega)=H^*(-\omega)$ for real-valued $h[n]$

where $*$ means complex conjugation. Combine these two results to prove the given equation.

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  • $\begingroup$ Looking at OP's question he wants to show the relation between the "magnitudes" of $H(e^{j\omega})$ and for that purpose you can also add a 3rd hint as: $H(e^{j\omega}) = |H(e^{j\omega})| e^{j\phi(\omega)}$, from which he can translate relations of 1 and 2 into the relations of magnitudes. $\endgroup$
    – Fat32
    Mar 23, 2016 at 18:15
  • $\begingroup$ @Fat32 and Matt L. For all your help :), I can reach the conclusion, I really appreciate for hints and comments !!! In conclusion, $H(\pi+w) = H^{*}(\pi -w) $ and magnitude used $\endgroup$ Mar 24, 2016 at 2:54

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