Can I construct a linear system so that it results in a moving average? 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) &amp;= a \\
h_1(1) &amp;= b \\
h_2(0) &amp;= c \\
h_2(1) &amp;= d \\
h_1(n) = h_2(n) &amp;= 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 &amp;= 1/3 \\
bc + ad &amp;= 1/3 \\
bd &amp;= 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.
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.