If I arbitrarily label the point between the two delay elements as v(n), I come up with the two equations:

y(n)=v(n)+4v(n)

x(n)=v(n+1) + 3v(n) + 2v(n-1)

Then the task is to eliminate v(n) from those two equations. But I'm completely stuck on that. Any help would be fantastic. I've been pulling my hair out for hours today working on this.

 

Any help would be fantastic. I've been pulling my hair out for hours today working on this.... so I will provide $5 paypal to anyone who can give good answer.

This is a real pain to solve in the time domain. You need to be good with generating summation formulas from the recursion relations and using a summation table. Problems like this are much easier to solve in the frequency domain using Z-transforms. Take the Z-transform of the system components, find the system transfer function and then take the inverse Z-transform of the system transfer function to get back to the system time domain solution.

The technique is similar to using Laplace transforms with continuous time systems. With discrete time, delays have a gain of 1/z in the z-frequency domain and the gain and summation blocks are the same. The transfer function is trivial to calculate now and the inverse transform is easily found using a Z-transform table. Give it a try and post any attempt at the solution, if you can't get it.

By the way, most people don't want money here. The best way to pay back is to help others, when you can. That is much more valuable than a little bit of money.

 

y(n)=v(n)+4v(n)
x(n)=v(n+1) + 3v(n) + 2v(n-1)

By the way, I thought I should point out that these equation are not correct.
It should be as follows.

y(n)=v(n)+4v(n-1)
x(n+1)=v(n+1) + 3v(n) + 2v(n-1)

 

Thank you sir. The reason I posted a reward is because I've been stuck on it for so long and was dieing to know the answer. And without it, I was afraid I would get an incomplete answer, and the thread would languish.

We haven't learned about the z transform, but I do know the Laplace transform and how to apply it to systems like this. I found the transfer function to be

H(z) = Y(z)/X(z) = (z+4)/[(z+2)(z+1)] eqn 1

Y(z)[z^2 + 3z + 2] = X(z)[z+4] eqn 2

y(n+2) + 3y(n+1) +2y(n) = x(n+1) + x(n) eqn 3

I think I got it.

 

y(n+2) + 3y(n+1) +2y(n) = x(n+1) + x(n) eqn 3

I think I got it.

I think you got it too, but there is a typo in your last equation. You dropped the factor of 4 by accident.

y(n+2) + 3 y(n+1) +2 y(n) = x(n+1) + 4 x(n)

 

Yes, transcribed it wrong from my notes.

Thank you again!

I also solved a different problem with the Z transform and in the time-domain. The answers agreed. Love those transforms, much faster.