Can I construct a linear system so that it results in a moving average?

Question

I've just learned about linear systems and impulse response functions. I know that in a linear system consisting of $h_1[k]$ and $h_2[k]$, the impulse response of the system is $h_1[k] \ast h_2[k]$.

In an exercise I was given the following system responses:

\begin{align*} h_1(0) &= a \\ h_1(1) &= b \\ h_2(0) &= c \\ h_2(1) &= d \\ h_1(n) = h_2(n) &= 0 \; \text{otherwise} \\ \end{align*}

The question was whether it would be possible to construct the parameters $a$, $b$, $c$, and $d$ in such a way that the linear system $h_1 \ast h_2$ would result in a moving average of the order 3 being applied to the input.

Well, I started with the convolution $h[k] = h_1[k] \ast h_2[k]$ for $k = 0, 1$:

\begin{equation} h[k] = [ac, bc + ad, bd] \end{equation}

I understand that for a moving average of order 3, the required convolution kernel would look like this:

\begin{equation} [1/3, 1/3, 1/3] \end{equation}

But that means I have to solve the following equation:

\begin{align*} ac &= 1/3 \\ bc + ad &= 1/3 \\ bd &= 1/3 \\ \end{align*}

And that's where I'm stuck. From my guts I'd say it's not possible to get to an answer here, because even if $a = 1$ and $b = 1$, $c$ and $d$ would have to be $1/3$ and $c + d$ would have to be $1/3$ as well. Which isn't solvable.

Answer 1

The function you are trying to solve is equivalent to $\frac{b}{3 \cdot a} + \frac{a}{3 \cdot b}=\frac{1}{3}$

It is also $a^2-ab+b^2=0$

but this function does not have a real number solution.

Answer 2

The gain of the desired moving average filter is not important; you can stick it after you have done the important work, if you like. What you are asking is whether $G(z)H(z) = (a +bz^{-1})(c + dz^{-1})$ can equal $(1 + z^{-1} + z^{-2})$. Unfortunately, the roots of $1+z^{-1}+z^{-2}$ are the two complex cube roots of unity, that is, $$ 1 + z^{-1} + z^{-2} = \left(1+e^{j2\pi/3}z^{-1}\right)\left(1+e^{j4\pi/3}z^{-1}\right)$$and so the only way this is going to work is if we allow the coefficients to be complex numbers.