# Partial fraction decomposition of Laplace transform

I have a sinusoidal input beginning at time (t = 0). $$x(t) = e^{j \omega_0 t} \cdot u(t).$$ The Laplace transform of my input is: $$X(s) = \dfrac{1}{s - j \omega_0}.$$ The Laplace transform of my output is: $$Y(s) = H(s) \cdot \dfrac{1}{s - j \omega_0}, where \ H(s) \ is \ assumed \ to \ be \ a \ rational \ transfer \ function$$ The next bit I don't understand. I can apparently decompose the transform of the output as: $$Y(s) = \dfrac{A_1}{s-p_1} + \dfrac{A_2}{s-p_2} + \cdots + \dfrac{A_N}{s-p_N} + H(j \omega_0) \dfrac{1}{s- j \omega_0}$$ Why has the argument of H(s) changed such that the transform of the unit impulse response is now $H(j \omega_0)$? Why is $H(j \omega_0)$ assigned to the fraction with $s- j \omega_0$, shouldn't there it be an arbitrary constant for all the fractions?

Why is $H(jω_0)$ assigned to the fraction with $s−jω_0$, shouldn't there it be an arbitrary constant for all the fractions?

You can assign an arbitrary constant to the fraction with $s−jω_0$. But you will be getting $H(j\omega_0)$ after evaluating it.

Proof

If Y(s) can be decomposed using partial fraction as follows,

$$Y(s) = \dfrac{A_1}{s-p_1} + \dfrac{A_2}{s-p_2} + \cdots + \dfrac{A_N}{s-p_N}$$

then by residue method, $i^{th}$ coefficient ($i<n$), the coefficient of $\dfrac{1}{s-p_i}$ in partial fraction decomposed form can be calculated as:

$$A_i = \left[Y(s)\times(s-p_i)\right]_{s=p_i}$$

So the coefficient of $\dfrac{1}{s-j\omega_0}$ in your problem will bewill be:

$$= \left[H(s) . \frac{1}{s-j\omega_0}\times(s-j\omega_0)\right]_{s=j\omega_0} = H(j\omega_0)$$

PS: See an example using residue method. See this also.

Writing it out in full, for the general case, is tedious and I'm lazy, so just for illustration consider a simple form of $H(s)$, (and let $jw_0 = jw$ to save ink):

Let $H(s) = 1/(s-p)$, then $Y(s) = 1/(s-p)(s-jw) = A/(s-p) + B/(s-jw)$ for partial fraction expansion.

Solving for A and B:

$1 = A(s-jw) + B(s-p)$

let $s=jw$, then $B = 1/(jw-p)$

let $s=p$, then $A = 1/(p-jw)$

Hence

$$Y(s) = \frac{A}{(s-p)} + \frac{1}{(jw-p)(s-jw)}$$

and we see that the coeff. of $1/(s-jw)$ is

$1/(jw-p) = H(jw)$, therefore

$$Y(s) = \frac{A}{(s-p)} + \frac{H(jw)}{(s-jw)}$$

• This is really hard to read. Please use Latex for mathematical formulas. – Matt L. Mar 27 '15 at 7:04
• Apologies for awkward text. Latex is on my to-do list for Easter Vac. – Chu Mar 27 '15 at 8:36
• Please check what I did to your post. As you can see, you can keep it simple and yet it becomes much more readable. – Matt L. Mar 27 '15 at 10:00