Is the convolution of two equal signals the same signal?
I have this system:
and i have to find $h_1[n]$ (which i've been struggling for a while), given
$$ h_2[n] = u[n] - u[n-2] $$
and $h[n] = h_1[n] \star h_2[n] \star h_2[n]$ given by:
I can find a solution if $h_2[n] \star h_2[n] = h_2[n]$, but I don't know if that holds.
First, we define $h_3[n] = h_2[n] \star h_2[n] = [1, 2, 1]$. From this result, we know that $h_1[n]$ must have 5 elements (so that $h[n]$ ends up with 7 elements).
Let's define $h_1[n] = [g_1, g_2, g_3, g_4, g_5]$. We can find these as follows, using the definition of discrete convolution.
First, we know that $g_1 h_3[0] = h[0] = 1$, so $g_1 = 1$.
Then, the next value of the convolution is $h[1] = g_2 h_3[0] + g_1 h_3[1] = 5$, from which we deduce that $g_2 = 3$.
Then, $h[2] = g_3 h_3[0] + g_2 h_3[1] + g_1 h_3[2] = 9$, or $g_3 = 2$.
You can stop here, since (by symmetry, in this particular example) $g_4 = g_2$ and $g_5 = g_1$.
In general, the problem can be seen as solving a system of linear equations. In this example, we need to solve the system:
$$\begin{bmatrix} h_3[0] & 0 & 0 & 0 & 0 \\ h_3[1] & h_3[0] & 0 & 0 & 0 \\ h_3[2] & h_3[1] & h_3[0] & 0 & 0 \\ 0 & h_3[2] & h_3[1] & h_3[0] & 0 \\ 0 & 0 & h_3[2] & h_3[1] & h_3[0] \\ 0 & 0 & 0 & h_3[2] & h_3[1] \\ 0 & 0 & 0 & 0 & h_3[2] \end{bmatrix} \begin{bmatrix} g_1 \\ g_2 \\ g_3 \\ g_4 \\ g_5 \end{bmatrix} = \begin{bmatrix} h[0] \\ h[1] \\ h[2] \\ h[3] \\ h[4] \\ h[5] \\ h[6] \end{bmatrix}$$
Thanks to @Hilmar for pointing out a boneheaded mistake in my original answer!
@MBaz solution is neat. Another vision is to use the idea that discrete convolutions turn into polynomial products in the $z$-transform domain: $$[1,1]\ast [1,1] = [1,2,1]$$ is equivalent to $$(1+z^{-1})(1+z^{-1})=(1+2z^{-1}+z^{-2})$$
So separating a long filter into short convolved filters is equivalent to factorizing a polynomial. To ease notations, I'll write $x= z^{-1}$. So you want to factorize $1+5x+9x^2+10x^3+9x^4+5x^5+x^6$. The lazy way is to use software or online polynomial factorization like dcode polynomial-factorization, and you get: $$(x+1)^2(x^2+1)(x^2+3x+1)$$ So you see the that the problem has indeed a solution, given by $(x^2+1)(x^2+3x+1)$. And for the whole problem, the entire system could be written as a cascade of four filters, two identical of length 2 ($[1,1]$, your $h_2$), and two others of length 3 ($[1,0,1]$ and $[1,3,1]$).
You can reach the result by the Euclidean division of polynomials or Polynomial Long division $1+5x+9x^2+10x^3+9x^4+5x^5+x^6$ by $1+2x+x^2$.
Let us be more witty, and (as shwo by @MBaz) find a degree 4 polynomial $D$ that gives $1+5x+9x^2+10x^3+9x^4+5x^5+x^6$ when multiplied by $(1+x)(1+x)$. Since the constant term and the highest-degree term are given in a unique way by the product of constant or highest degree terms, it is easy to see that the corresponding terms in $D$ are both $1$. So you just seek three coefficients such that:
$$(1+x)(1+x)(1+ax+bx^2+cx^3+x^4) = 1+5x+9x^2+10x^3+9x^4+5x^5+x^6$$
With a little more experience, you could tell that $a=c$ (for symmetry reasons), and this is useful to reduce the numbers of unknowns when the polynomials are much bigger. Here, you can resort to analysis. Replace $x$ by three values $x_k$ and equate:
$$(1+ax+bx^2+cx^3+x^4) = \frac{1+5x+9x^2+10x^3+9x^4+5x^5+x^6}{(1+x)(1+x)}$$ and you get a system of three linear equations that you can invert. For instance, with $x_1=1$, you have:
$$(1+a+b+c+1) = \frac{1+5+9+10+9+5+1}{(1+1)(1+1)}$$
or
$$a+b+c = 8$$
