Electronic – Partial fraction decomposition of Laplace transform

laplace transform

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?

Best Answer

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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