Partial Fraction Expansion

[zordar]

Hello, I am having trouble expanding this function into partial fractions. It is really starting to frustrate me!

I have Y = (s+1)/(s²-1) + 1/[s²(s²-1)]

My professor wrote from the above step = 1/(s-1) + 1/(s²-1) - 1/(s²)

I have no clue how he got that.

Can't you combine the two terms in Y into [(s+1)s² + 1]/[s²(s²-1)] = [s³+s²+1]/[s²(s²-1)] = [s³+s²+1]/[s²(s+1)(s-1)] = A/s²+B/s+C/(s+1)+D(s-1)

This will obviously give me a different answer than the one my Professor got. What is incorrect here??

 

[Gene]

The trick is that the first term is
(s+1)/((s+1)(s-1)) = 1/(s-1)
Then you need only expand the second term. You will get the last two terms of the Prof's answer.

 

[zordar]

Hi,
That explains how he got the first term, the 1/(s-1), but I still don't get why he has only two terms for the second component.

If I have 1/(s-1) + 1/[s²(s²-1)] = 1/(s-1) + 1/[s²(s+1)(s-1)] = [s²(s+1) + 1]/[s²(s+1)(s-1)] after factoring out the denominator and simplifying, shouldn't my partial fraction expansion be A/s² + B/s + C/(s+1) + D/(s-1) ??

After I solve for A, B, C, and D, the B = 0 and cancels out, leaving me with only three terms, but I still get values for A, C, and D that are different than what my Professor got. The terms in his denominator are different as well, which affect my answer when I take the inverse transform.

 

[pka]

MathCad gets something different as well.

 

[Gene]

I'm rushed at the moment, but I'll bet (as much as a quarter) that if you include the missing term E/(s^2-1) your C & D will drop out also. If they don't, come back and I'll look at later.

 

[zordar]

MathCad gets something different as well.

That's exactly what I got! When I take the inverse transform, I get y(t) = -t-1/2e^-t+3/2e^t (using the Table of Laplace transforms). This is what the Prof. got for his final answer: y(t) = e^t+sinh t - t, which are obviously not the same.

Let me post the whole problem as he did it in class (not that long) to see if I am not missing anything:
Solve: y" - y = t, y(0) = 1, y'(0) = 1

First, take the Laplace of both sides:
L[y" - y] = L[t]

s²Y(s)-sy(0)-y'(0)-Y(s) = L[t] = 1/s²

(s²-1)Y(s) = s + 1 + 1/s² which gives us our
"subsidiary equation"

Then we find that the "transfer function " is Q(s) = 1/(s²-1)
Y(s) = (s+1)/(s²-1) + 1/[s²(s²-1)] = 1/(s-1) + 1/(s²-1) - 1/s² which is where I get stuck...

 

[galactus]

You're on the right track.

You have the correct "transfer function":

(s+1+1/s²)/(s²-1). This is equivalent to:

I took the inverse Laplace of both and arrived at

(3e^t/2)-1/(2e^t)-t. Which is equivalent to

e^t+sinh(t)-t

 

[zordar]

You're on the right track.

You have the correct "transfer function":

(s+1+1/s²)/(s²-1). This is equivalent to:

I took the inverse Laplace of both and arrived at

(3e^t/2)-1/(2e^t)-t. Which is equivalent to

e^t+sinh(t)-t

Can you please explain how they are equivalent? Thanks.

 

[galactus]

IOW, 1/(s-1)+1/(s²-1)=(s+2)/(s²-1)=
3/(2(s-1))-1/(2(s+1)).

 

[zordar]

You're on the right track.

You have the correct "transfer function":

(s+1+1/s²)/(s²-1). This is equivalent to:

I took the inverse Laplace of both and arrived at

(3e^t/2)-1/(2e^t)-t. Which is equivalent to

e^t+sinh(t)-t

Thanks for checking, I would not have known if I was doing it correctly.