# Partial Fraction Expansion Problem [SOLVED]

Smells great!Houston, TXRegistered User regular
edited November 2007
This is actually a problem from a Circuits test I took last week and just got back today. I did fairly well, but on one problem I had to convert the circuit to the complex frequency domain, perform nodal analysis on the circuit, get an expression for the load current, and then take the inverse Laplace Transform to get the expression for the current in the time domain. No big deal, right? Here's the issue I ran into. I got an expression for the current in the Complex Frequency domain, and in order to take the inverse transform, I had to perform a partial fraction expansion. I've done these a million times, so I generally don't have any issues with them, but for some reason this one has me stumped, and I have no idea why. Here's the basic s-domain expression without any constants included:

1 / [(s)*((s+1)^2)]

To expand, I set this equal to:

[A / s] + [B / ((s+1)^2)] + [C / (s+1)]

Finding the common denominator and then cancelling it from both sides, I get:

1 = A(s+1)^2(s+1) + Bs(s+1) + Cs(s+1)^2

Now, here's where the problem can go a few different ways in terms of how to solve it. Personally, I like the method of grouping in terms of order, factoring out the power term, and then using the resulting equations to solve for A, B, and C, so that's how I'll keep going. So, multipying out the above term completely:

1 = As^3 + 3As^2 + 3As + A +Bs^2 + Bs +Cs^3 +2Cs^2 + Cs

And now I group and factor out s terms according to order:

1 = (s^3)(A+C) + (s^2)(3A+B+2C) + (s)(3A+B+C) + (A)

Now, here's where I spot a problem immediately. There is obviously no s^3, s^2, or s term on the left side of the equation, which means that the groups associated with those terms on the right side of the equation are all equal to zero according to "rules" of partial fraction expansion that I've learned in the past. That's all fine, and I can immediately see that A = 1. Therefore, A+C=0 (from the s^3 term) means that C = -1. Now I have to find B.

3A+B+2C = 0
3A+B+C = 0

Oh crap, when I plug in my values of A and C, I get two different values of B (B=-1 and B=-2). Obviously this should not happen. So I ask you, fellow HA Math junkies - what am I doing wrong here? Is this just a bad way of performing partial fraction expansion, and if so, what should I have done? With a third-order polynomial expression like this, I honestly don't know of a better way to do it.

By the way, the expansion (according to my professor) is supposed to be:

[1 / s] - [1 / ((s+1)^2)] - [1 / (s+1)]

Which is basically what I get using the above method, except for the whole pesky "two values of B" thing. If anyone can spot what I'm doing wrong here, I would greatly appreciate it. I just know it's going to be something really obvious and I'll hate myself afterwards, but I'd still like to know. Thanks in advance.

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## Posts

• Registered User regular
edited November 2007
This is not quite right, you appear to have over zealously multiplied:

1 = (s^3)(A+C) + (s^2)(3A+B+2C) + (s)(3A+B+C) + (A)

It should look like:

1 = (s^2)(A+C) + (s)(2A+B+C) + (A)

That should give you B = -1 and C = -1.

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• Smells great! Houston, TXRegistered User regular
edited November 2007
Holy crap, you're right. I just multiplied each term by the denominator in the other two terms to find the common denominator without thinking, and multiplied in an extra (s+1) in the process. I knew it'd be something simple and obvious.

Thanks, problem solved. This can be locked now.

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