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I have a sinusoidal input beginning at time (t = 0). \begin{equation} x(t) = e^{j \omega_0 t} \cdot u(t). \end{equation} The Laplace transform of my input is: \begin{equation} X(s) = \dfrac{1}{s - j \omega_0}. \end{equation} The Laplace transform of my output is: \begin{equation} Y(s) = H(s) \cdot \dfrac{1}{s - j \omega_0}, where \ H(s) \ is \ assumed \ to \ be \ a \ rational \ transfer \ function \end{equation} The next bit I don't understand. I can apparently decompose the transform of the output as: \begin{equation} 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} \end{equation} 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?

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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,

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

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.

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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

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

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

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