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These two filters have very similar frequency responses, but those responses are not identical except at a single frequency $\omega_0$. I want to find $\omega_0$.

One way I've seen that done is like this:

$$\begin{align} Y_1(z)&=X(z)+2X(z)z^{-1}+X(z)z^{-2}\\ Y_2(z)&=X(z)+X(z)z^{-1} \end{align}$$

if $Y_1(z)=Y_2(z)$

$$ \begin{align} \Longrightarrow 1+2z^{-1}+z^{-2}&=1+z^{-1}\\ \Longrightarrow z^{-1}+z^{-2}&=0\\ \Longrightarrow z\left(z+1\right)&=0\\ \Longrightarrow z=e^{j\omega}=-1\\ \Longrightarrow \omega=\pi\\ \omega_0=\frac{f_s}2 \end{align} $$

But, to be perfectly frank that is quite confusing to me I don't quite know what's going on there, would someone please break it down for me?

Another way I've seen it done is like this:

$$ H_1\left(\omega_0\right)= H_2\left(\omega_0\right)\rightarrow 1+2e^{-j\omega_0}+e^{-j2\omega_0}=1+e^{-j\omega_0}. $$

With algebraic simplification the above equality can be written as:

$$ e^{-j\omega_0}=-e^{-j2\omega_0}. $$ Here's the tricky part,...we multiply both sides of the above equation by $e^{j\omega_0}$ to give us: $$ 1=-e^{-j\omega_0}. $$

I guess they're equivalent, but, in what ways specifically are they identical and in what ways are they divergent?

If someone could help me to see I would be so grateful.


2 Answers 2


Both methods are equivalent. Look at it like this: first write down the frequency responses (as already indicated in your question)

$$H_1(e^{j\omega})=1+2e^{-j\omega}+e^{-2j\omega}\\ H_2(e^{j\omega})=1+e^{-j\omega}$$

Now want to find $\omega_0$ such that


which gives


or, equivalently,


Equation (1) can only be satisfied if the term in parentheses vanishes. So you get


And since $\omega_0=2\pi f_0/f_s$ (where $f_s$ is the sampling frequency) you finally get


  • $\begingroup$ is that e^-jWo = -1 just a characteristic of e? $\endgroup$
    – user8769
    Commented May 29, 2014 at 1:46
  • $\begingroup$ I really don't understand that last part of what you've written, why do we want fo = fs/2 Isn't the answer plus or minus pi? $\endgroup$
    – user8769
    Commented May 29, 2014 at 1:47
  • $\begingroup$ @matthias.anglicus: $e^{-j\omega_0}=-1$ is the condition that Equ. (1) is satisfied, i.e. the frequency $\omega_0$ which satisfies this equ. is the one for which both frequency responses have the same value. The answer is $\omega_0=\pi$, which is equivalent to $f_0=f_s/2$, because $\omega_0=2\pi f_0/f_s$. $\endgroup$
    – Matt L.
    Commented May 29, 2014 at 11:55
  • $\begingroup$ I guess this filter will attenuate anything greater than pi, since a wave that crosses 180 degrees will be eliminated by the way these components are constructed, is that correct? $\endgroup$
    – user8769
    Commented May 29, 2014 at 16:07
  • $\begingroup$ @matthias.anglicus: $\pi$ is the Nyquist frequency. This is the maximum frequency a discrete-time system can deal with. Anything higher than that will result in aliasing. If you're not sure what I'm talking about, please check this article: en.wikipedia.org/wiki/Nyquist_frequency $\endgroup$
    – Matt L.
    Commented May 29, 2014 at 16:24

It's actually pretty easy to show that $$H_2(\omega)=H_1(\omega)\cdot H_1(\omega)=H_1(\omega)^2$$ The second filter is equivalent to filtering twice with the first filter and hence the magnitude of the transfer function of second filters is the square of the first. The only two numbers where the square is the same as the original number are 0 and 1, so the filters have the same magnitude at frequencies where the magnitude is 0 and 1.

Both filters have DC gain (i.e. the magnitude is not 1 at low frequencies). Both have 0 gain at $\omega = \pi$ but the second filter is steeper.

  • $\begingroup$ isn't DC something about the average of all the terms, like the 0th term in the Fourier transform is the DC component, isn't it? What does that mean in this context? Something different I guess. What do you mean 'steeper'? $\endgroup$
    – user8769
    Commented May 29, 2014 at 1:29
  • $\begingroup$ what about -pi, also they have 0 gain there right? What does 0 gain mean? Please forgive my ignorance! $\endgroup$
    – user8769
    Commented May 29, 2014 at 1:31

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