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 $R$s 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.