Linear-phase lowpass filters: frequency responses alignment

Question

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.

Answer 1

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

$$H_1(e^{j\omega_0})=H_2(e^{j\omega_0})$$

which gives

$$1+2e^{-j\omega_0}+e^{-2j\omega_0}=1+e^{-j\omega_0}$$

or, equivalently,

$$e^{-j\omega_0}(1+e^{-j\omega_0})=0\tag{1}$$

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

$$e^{-j\omega_0}=-1\Longrightarrow\omega_0=\pi$$

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

$$f_0=f_s/2$$

Answer 2

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.