DSP: Recurrence Relations in a Linear Algebra Equation

Click For Summary
SUMMARY

The discussion centers on modeling a delay feedback loop in Digital Signal Processing (DSP) using Linear Algebra equations. The user presents a system defined by the equation $$ y[n]=x[n]+ D\lbrace y[n] \rbrace=x[n]+y[n-1] $$, where D represents a delay operator. The user encounters difficulties in obtaining a closed form due to an uninvertible matrix when attempting to express the system in matrix form. The solution proposed involves utilizing the Z-transform, which simplifies the analysis by transitioning to the frequency domain.

PREREQUISITES
  • Understanding of Digital Signal Processing concepts
  • Familiarity with Linear Algebra, particularly matrix operations
  • Knowledge of recurrence relations and their applications
  • Basic understanding of the Z-transform and its significance in DSP
NEXT STEPS
  • Study the properties and applications of the Z-transform in DSP
  • Explore methods for solving linear recurrence relations
  • Learn about the implications of non-invertible matrices in linear systems
  • Investigate the relationship between time-domain and frequency-domain representations in signal processing
USEFUL FOR

Students and professionals in Digital Signal Processing, Linear Algebra enthusiasts, and anyone interested in solving linear systems involving delay operators and recurrence relations.

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: 980
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)
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 23 ·
Replies
23
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K