I DSP: Recurrence Relations in a Linear Algebra Equation

Destroxia
Messages
204
Reaction score
7
Hello, I've been working through some Digital Signal Processing stuff by myself online, and I saw a system that I wanted to write down as a Linear Algebra Equation. It's a simple delay feedback loop, looks like this:

system.jpg


The (+) is an adder that adds 2 signals together, so the signal from x[n] and z^-1 will be added each time until a full signal is formed for the output y[n].

However, on the first run of the loop, z^-1 is initialized to 0, so it's contribution on the first run is zero. (z^-1 will return always return whatever it was last fed, so on the second run it will have whatever came from x[n] the first time and spit that out)So we can model this system as: $$ y[n]=x[n]+ D\lbrace y[n] \rbrace=x[n]+y[n-1], \space \space \space \space \space \space y[0]=x[0]=1$$
Where: $$\space x[n], \space y[n] \space \in \space \mathbb{C}^{N}$$
and D is a size NxN linear operator matrix which represents z^-1, which acts like so: ## D[{x[n]}]= x[n-1] ##

If you want, you can generalize the system for more delay: $$ y[n]=x[n]+ D_{N}\lbrace y[n] \rbrace=x[n]+y[n-N], \space \space \space \space \space \space y[0]=x[0]=1$$

Now imagine we have an input x[n] of dimension 4 that is simply a pulse:
$$
x =
\left[ {\begin{array}{c}
1\\
0\\
0\\
0\\
\end{array} } \right]
$$

If we push this input into the system, with say a delay of Z^-2, we should get a pulse every 2 loops, so the output would look like so:
$$
y =
\left[ {\begin{array}{c}
1\\
0\\
1\\
0\\
\end{array} } \right]
$$

My issue is, when I try to model this with a linear algebra equation, there is recursion, and I can't seem to get a closed form due to an non-invertible matrix. I was looking at recurrence relations on wikipedia to try to solve my issue, but I guess I just don't understand them yet.

My process for modeling this in Linear Algebra has gone like so:

$$ y = x + Dy $$
$$ y - Dy = x$$
$$ (I - D)y = x $$
$$ I -D =
\left[ {\begin{array}{cccc}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
\end{array} } \right]
-
\left[ {\begin{array}{cccc}
0 & 0 & 0 & 1\\
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
\end{array} } \right]
=
\left[ {\begin{array}{cccc}
1 & 0 & 0 & -1\\
-1 & 1 & 0 & 0\\
0 & -1 & 1 & 0\\
0 & 0 & -1 & 1\\
\end{array} } \right] )

$$

This matrix appears to be uninvertible, but I should be able to somehow create a closed form for this system... I just need to be pointed in the right direction.

(I apologize if this is the wrong board for this, but this is more of a linear algebra question than a signal processing question, as I understand the signal processing part.)
 

Attachments

  • system.jpg
    system.jpg
    7.4 KB · Views: 952
Physics news on Phys.org
I believe I may have solved my own issue:

Instead of treating this as a linear equation right away, something called the "z-transform" can be taken of the equation. And this allows us to easily solve the problem in the frequency domain.

(The Z-transform is the discrete time version of the laplace transform)
 
Thread 'Determine whether ##125## is a unit in ##\mathbb{Z_471}##'
This is the question, I understand the concept, in ##\mathbb{Z_n}## an element is a is a unit if and only if gcd( a,n) =1. My understanding of backwards substitution, ... i have using Euclidean algorithm, ##471 = 3⋅121 + 108## ##121 = 1⋅108 + 13## ##108 =8⋅13+4## ##13=3⋅4+1## ##4=4⋅1+0## using back-substitution, ##1=13-3⋅4## ##=(121-1⋅108)-3(108-8⋅13)## ... ##= 121-(471-3⋅121)-3⋅471+9⋅121+24⋅121-24(471-3⋅121## ##=121-471+3⋅121-3⋅471+9⋅121+24⋅121-24⋅471+72⋅121##...
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
Back
Top