What is the significance of j2 in this problem?

[LongApple]

What's j2 in this problem?

1. Homework Statement

Homework Equations

The Attempt at a Solution

a. What's j2 in this problem?

b. How did we use the fact that we have a unit impulse?

c. What are our first thoughts and strategies when seeing the problem ?

What is our strategy? It looks like j2 is 4R/2 almost

 

[collinsmark]

What's j2 in this problem?

1. Homework Statement

View attachment 85706

Homework Equations

The Attempt at a Solution

a. What's j2 in this problem?

I think the j2 is merely two times the unit imaginary number. In other words, it's the same as \left( 2 \right) \left( \sqrt{-1} \right) which is the same thing as saying \sqrt{-4}.

b. How did we use the fact that we have a unit impulse?

c. What are our first thoughts and strategies when seeing the problem ?

What is our strategy? It looks like j2 is 4R/2 almost

You'll have to forgive me for not being more skilled with using the operator expressions. But maybe I can add some insight, albeit just a little bit.

If you're wondering where the \frac{Y}{X} = \frac{3+4R}{1+R^2} came from, it might help by rearranging the original difference equation.
y[n] + y[n-2] = 3x[n] +4x[n-1]
Now transform the equation such that each delay corresponds an R. [n] gets no R, [n-1] corresponds a single R, and [n-2] corresponds to RR = R^2. Now solve for \frac{Y}{X}.

The step where \frac{Y}{X} = \frac{3+4R}{1+R^2} = \frac{\frac{3}{2} - j2}{1-jR} + \frac{\frac{3}{2} + j2}{1+jR} is the result obtained from partial fraction decomposition. Note that 1+R^2 doesn't factor using real numbers. That's where imaginary numbers come in.

 

[collinsmark]

By the way, it looks like what's being done here is solving the problem by using what's more commonly called the "z-transform." I've never seen it done with Rs as the notation, which is what threw me. It's more commonly notated with zs. Anyway, if you want to investigate further, do some online research on the "z-transform."

[Edit: Oh, and the approach your coursework is using uses positive exponents of R, R^2, etc., where the z transform uses negative exponents, z^{-1}, z^{-2}, etc. So the approach that your coursework is using is a little different than the standard z-transform (although it's conceptually equivalent; just substitute R^n \Leftrightarrow z^{-n}). So that's something else to keep in mind.]

 

[rude man]

Transwform the finite-difference equation into z transform form:
Y(z) + z^-2Y(z) = αX(z) + βz^-1X(z)
which gives you Y/X (z).
What is the z transform of the given unit impulse function?
Now use partial fractions to decompose the final Y(z) into manageable terms for the purpose of inverting each term.

BTW I haven't done that myself and I suspect their derivation has a math error in it also.

 

[rude man]

EDIT: nope, the given answer for the last 2 terms is correct:

y[n] = 3δ[n] + (3/2 - j2) jⁿ + (3/2 + j2) (-j)ⁿ.

But the omission of the impulse term remains. In fact, given that Y(z) is not a "strictly proper" fraction (numerator and denominator order of z same) automatically requires such a term in the finite-difference equation solution.

I also have never seen R used in lieu of z^-1 but it seems OK. Not sure what the merit of it is; every z transform table I've encountered is in z, not in R = z^-1. To use these tables the substitution R = z^-1 and then multiplying num. & denom. by z would have to be made.