Question 9.12: Find the inverse transform of F(s) = 20(s + 3)/(s + 1)(s² + ......
Find the inverse transform of
F(s) = \frac{20(s + 3)}{(s + 1)(s^{2} + 2s + 5)}
F(s) has a simple pole at s = −1 and a pair of conjugate complex poles located at the roots of the quadratic factor
(s² + 2s + 5) = (s + 1 − j2)(s + 1 + j2)
The partial fraction expansion of F(s) is
F(s) = \frac{k_{1}}{s + 1} + \frac{k_{2}}{s + 1 – j2} + \frac{k^{*}_{2}}{s + 1 + j2}
The residues at the poles are found from the cover-up algorithm.
k_{1} = \frac{20(s + 3)}{s^{2} + 2s + 5}|_{s = -1} = 10
k_{2} = \frac{20(s + 3)}{(s + 1)(s + 1 + j2)}|_{s = -1 + j2} = −5 − j5 = 5\sqrt{2}e^{+ j5π/4}
We now have all of the data needed to construct the inverse transform.
f(t) = [10e^{−t} + 10\sqrt{2}e^{−t}\cos(2t + 5π/4)]u(t)
In this example we used k2 to obtain the amplitude and phase angle of the damped cosine term. The residue k^{∗}_{2} is not needed, but to illustrate a point we note that its value is
k^{∗}_{2} = (-5 – j5)^{*} = -5 + j5 = 5\sqrt{2}e^{−j5π/4}
If k^{∗}_{2} is used instead, we get the same amplitude for the damped sine but the wrong phase angle. Caution: Remember that Eq. (9–23) uses the residue at the complex pole with a positive imaginary part. In this example, this is the pole at s = −1 + j2, not the pole at s = −1 − j2.
f(t) = [… + 2 |k|e^{−αt} \cos(βt + θ) + …]u(t) (9–23)