How to get the third value (A), using partial fractions

[Jeviah]

Homework Statement

y(w)= 3/(iw-1)^2(-4+iw)

Homework Equations

N/A

The Attempt at a Solution

3/(iw-1)^2(-4+iw)
= A/iw-1 + B/(iw-1)^2 + C/-4+iw

for B iw = 1
B=3/-4+1 = -1

for C iw = 4
C= 3/(4-1)^2 = 1/3

I know the answer for A should be -1/3 however I am unsure how to obtain this as if the above cancellation method is used then the denominator equals 0.

 

[Charles Link]

You need to get a common denominator in part 3, and then write 3 equations with $A$, $B$, and $C$, so that the coefficients of the $w^2$ terms add to zero, the coefficients of the $w$ terms add to zero, and the constant terms add to $3$. Do you see why?

 

[tnich]

Homework Statement

y(w)= 3/(iw-1)^2(-4+iw)

Homework Equations

N/A

The Attempt at a Solution

3/(iw-1)^2(-4+iw)
= A/iw-1 + B/(iw-1)^2 + C/-4+iw

for B iw = 1
B=3/-4+1 = -1

for C iw = 4
C= 3/(4-1)^2 = 1/3

I know the answer for A should be -1/3 however I am unsure how to obtain this as if the above cancellation method is used then the denominator equals 0.

I can't follow your logic here. Part of the problem is your notation. What does 3/(iw-1)^2(-4+iw) mean? What does C/-4+iw mean? Try using LaTeX. It is easy to use and makes it easy to express what you mean.

 

[Ray Vickson]

Homework Statement

y(w)= 3/(iw-1)^2(-4+iw)

Homework Equations

N/A

The Attempt at a Solution

3/(iw-1)^2(-4+iw)
= A/iw-1 + B/(iw-1)^2 + C/-4+iw

for B iw = 1
B=3/-4+1 = -1

for C iw = 4
C= 3/(4-1)^2 = 1/3

I know the answer for A should be -1/3 however I am unsure how to obtain this as if the above cancellation method is used then the denominator equals 0.

Your expressions are ambiguous and do not say what you probably mean. Your C/-4+iw means $\frac{C}{-4} + iw$. If you mean $\frac{C}{-4 + iw}$ then you need parentheses, like this: C/(-4+iw)---simple, but effective!

 

[vela]

I know the answer for A should be -1/3 however I am unsure how to obtain this as if the above cancellation method is used then the denominator equals 0.

The method you're using is quick, but as you've found out, it runs into some problems.

Starting from
$$\frac{3}{(i\omega-1)^2(-4+i\omega)} = \frac{A}{i\omega-1} + \frac{B}{(i\omega-1)^2} + \frac{C}{-4+i\omega},$$ if you multiply through by the denominator, you end up with
$$3 = A(i\omega - 1)(i\omega - 4) + B(i\omega-4) + C(i\omega-1)^2.$$
There are different approaches you can take from here. Charles suggested one above, and it's well worth trying it out and understanding how it works.

Alternately, you could, as before, set $i\omega$ to convenient values and solve for $B$ and $C$. Once you have those, plug them back into the equation and then choose another value for $i\omega$ and solve for $A$. Remember you can use any value for $i\omega$ since the equality has to hold for all $\omega$.