Laplace transform in electrical engineering

1. Jul 10, 2011

Bassalisk

Hello,

We recently learned in mathematics the Laplace transform. But we didn't actually use it to solve circuits yet.

I took things at my own hand and tried to figure out whats the point of Laplace transform. I figured out all that frequency and time stuff. But here is the thing.

If you Laplace transform an equation that describes lets say capacitor(in time domain), you get

I=V*wC-I0*C right?

I see resemblance with impedance and other phasor techniques we used to solve circuits. But where is that imaginary part?

From transform up top. Xc=1/wC. But its missing that "-j" part. How did people get to the equation Xc=-j*1/wC?

2. Jul 10, 2011

Feldoh

I don't know what system you're trying to solve but if you just look at the Laplace transform for a single capacitor

You have $i(t) = C \frac{dv}{dt} <==> i(s) = C(s*V(s) - V(0)) ==> V(s) = \frac{i(s)}{sC} + \frac{V(0)}{s}$

Notice that i(s) and V(s) both have an s dependence. s is defined as a+jb so there is indeed a complex component.

3. Jul 10, 2011

Bassalisk

oooh i thought s was omega :D Thank you !

4. Jul 15, 2011

MisterX

If H(s) is a transfer function of a stable system and the t-domain input is $cos(\omega t + \theta)$, then the steady state output is $\left|H(j\omega)\right|cos(\omega t + \theta + \angle H(j\omega))$
This means if $Z(s) = V(s)/I(s)$, we may just multiply the current phasor by $Z(j\omega)$ to get the voltage phasor.

assuming zero initial capacitor voltage
$\frac{V(s)}{I(s)} = \frac{1}{sC}$, with $i(t) = Re\left\{e^{j\omega t + \theta} \right\}$
then
$v(t) = Re\left\{\left(\frac{1}{j\omega C}\right)e^{j\omega t + \theta}\right\}$

understanding why this is:

As H(s) is the transfer function of a stable system, the real part of all the poles of H(s) will be less than zero.

In the partial fraction expansion of the system's response to $cos(\omega t)$, only the terms with purely imaginary poles will be relevant to the steady state, as the other terms (with poles in the left half plane) will correspond to decaying exponentials in the t-domain.

With the Laplace transform of $cos(\omega t + \theta) = cos(\omega t)cos(\theta) - sin(\omega t)sin(\theta)$ being $X(s) = cos(\theta)\frac{s}{(s + j\omega)(s - j\omega)} - sin(\theta)\frac{\omega}{(s + j\omega)(s - j\omega)}$, the s-domain response of the system is

$H(s)X(s) = \frac{k}{s - j\omega} + \frac{k*}{s + j\omega} + N(s)$

where N(s) is the fraction expansion for all the other poles of the response. So

$H(s)X(s)(s - j\omega) = k + (\frac{k*}{s + j\omega} + N(s))(s - j\omega)$

(s - j\omega) in the numerator and denominator of the left hand side cancel after substituting in the expression for X(s).

$H(s)\frac{cos(\theta)s - sin(\theta)\omega}{s + j\omega} = k + (\frac{k*}{s + j\omega} + N(s))(s - j\omega)$

taking the limit of both sides as s goes to $j\omega$

$\lim_{s \to j\omega}H(s)\frac{(cos(\theta)s - sin(\theta)\omega)}{(s + j\omega) } = k + (\frac{k*}{s + j\omega} + N(s))(0)$

so

$k = H(j\omega)\frac{(j\omega cos(\theta) - \omega sin(\theta))}{(2j\omega) } = \frac{H(j\omega)}{2}(cos(\theta) - (-jsin(\theta)) = \frac{e^{j\theta}H(j\omega)}{2}$

the s-domain steady state response is

$\frac{k}{s - j\omega} + \frac{k*}{s + j\omega}$

Which in the t domain becomes

$\left|k\right|e^{\angle k}e^{j\omega t} + \left|k\right|e^{-\angle k}e^{-j\omega t} = 2 \left|k\right|cos(\omega t + \angle k)$

substituting for k:
$\left|k\right| = \frac{\left|H(j\omega)\right|}{2}$
$\angle k = \theta + \angle H(j\omega)$

$2 \left|k\right|cos(\omega t + \angle k) = \left|H(j\omega)\right|cos(\omega t + \theta + \angle H(j\omega))$

Last edited: Jul 15, 2011
5. Jul 16, 2011

Bassalisk

Thank you I will have a look into this in more detail later.

6. Jul 17, 2011

MisterX

correction

should be
$i(t) = Re\left\{e^{j(\omega t + \theta)}\right\}$
then
$v(t) = Re\left\{\left(\frac{1}{j\omega C}\right)e^{j(\omega t + \theta)}\right\}$