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Sahithyan's S2
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Sahithyan's S2 — Theory of Electricity

Laplace Transform

A mathematical operation that transforms a function of time f(t)f(t), into a function of a complex variable, denoted as F(s)F(s). Mathematically, the Laplace Transform of a function f(t)f(t) is defined as:

F(s)=L{f(t)}=0estf(t)dtF(s) = \mathcal{L}\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) \, \text{d}t

Here:

  • tt represents time
  • s=σ+jωs = \sigma + j\omega, where σ,ωR\sigma, \omega \in \mathbb{R}
  • este^{-st} is the exponential decay factor that weights the function f(t)f(t).

Can only be applied to casual functions. Converts a differential equation (time domain) into a linear complex algebraic equation (frequency domain). After Laplace transformation is applied, the resulting functions are capitalized by convention.

A function of time f(t)f(t) satisfying:

f(t)=0fort<0f(t) = 0 \quad \text{for} \quad t < 0

If L{f(t)}=F(s)\mathcal{L}\{f(t)\} = F(s), then the inverse Laplace Transform is given by:

f(t)=L1{F(s)}=12πjσjσ+jestF(s)dsf(t) = \mathcal{L}^{-1}\{F(s)\} = \frac{1}{2\pi j} \int_{\sigma - j\infty}^{\sigma + j\infty} e^{st} F(s) \, \text{d}s u(t)={0,t<01,t>0u(t) = \begin{cases} 0, & t<0 \\ 1, & t>0 \end{cases}

Laplace transform of the unit step function is:

L{u(t)}=1s\mathcal{L}\{u(t)\} = \frac{1}{s} δ(t)={0,t0,t=0\delta(t) = \begin{cases} 0, & t\neq 0 \\ \infty, & t=0 \end{cases}

Laplace transform of the unit impulse function is:

L{δ(t)}=1\mathcal{L}\{\delta(t)\} = 1

Area under the curve is 11.

δ(t)dt=1\int_{-\infty}^{\infty} \delta(t)\, \text{d}t = 1

And it has a special property:

f(tk)δ(t)dt=f(k)\int_{-\infty}^{\infty} f(t-k)\,\delta(t)\, \text{d}t = f(k) r(t)={0,t<0t,t>0r(t) = \begin{cases} 0, & t<0 \\ t, & t>0 \end{cases}

Laplace transform of the unit ramp function is:

L{r(t)}=1s2\mathcal{L}\{r(t)\} = \frac{1}{s^2}

Both AC and DC voltage sources are transformed according to the Laplace transform table. Resistors are included as is.

An inductor LL is converted to LsLs. To cater for initial current, a voltage source Li(0+)L i(0^+) is added aiding the voltage.

A capacitor CC is converted to 1Cs\frac{1}{Cs}. To cater for initial voltage, a voltage source v(0+)s\frac{v(0^+)}{s} is added opposing the voltage.

Suppose f(t)f(t) has a Laplace transform F(s)F(s) for the below definitions.

aa and bb are constants.

L{af(t)+bg(t)}=aL{f(t)}+bL{g(t)}\mathcal{L}\{a f(t) + b g(t)\} = a \mathcal{L}\{f(t)\} + b \mathcal{L}\{g(t)\} L{f(t)}=sF(s)f(0+)\mathcal{L}\{f'(t)\} = sF(s) - f(0^+) L{f(t)}=s2F(s)sf(0+)f(0+)\mathcal{L}\{f''(t)\} = s^2F(s) - sf(0^+) - f'(0^+) L{f(n)(t)}=snF(s)k=1n1skf(0+)k=1n1f(k)(0+)\mathcal{L}\{f^{(n)}(t)\} = s^nF(s) - \sum_{k=1}^{n-1} {s^{k} f(0^+)} - \sum_{k=1}^{n-1} f^{(k)}(0^+) L{0tf(t)dt}=F(s)s\mathcal{L}\left\{\int_{0}^{t} f(t) \, \text{d}t\right\} = \frac{F(s)}{s} L{f(ta)}=aF(as)\mathcal{L}\left\{f\left(\frac{t}{a}\right)\right\} = aF\left(as\right) L1{F(sa)}=af(at)\mathcal{L}^{-1}\left\{F\left(\frac{s}{a}\right)\right\} = af\left(at\right) L{tnf(t)}=(1)ndnF(s)dsn\mathcal{L}\left\{t^n f(t)\right\} = (-1)^n \frac{\text{d}^n F(s)}{\text{d}s^n} L{f(tT)}=esTF(s)\mathcal{L}\left\{f(t - T)\right\} = e^{-sT}F(s) L1{F(s+a)}=eatf(t)\mathcal{L}^{-1}\{F(s+a)\} = e^{-at}f(t) f(0+)=limt0+f(t)=limssF(s)f(0^+) = \lim_{t \to 0+} f(t) = \lim_{s \to \infty} sF(s) f()=limtf(t)=lims0sF(s)f(\infty) = \lim_{t \to \infty} f(t)= \lim_{s \to 0} sF(s)
Function NameFunctionLaplace Transform
Unit Impulseδ(t)\delta(t)11
Unit Stepu(t)u(t)1s\frac{1}{s}
Polynomialtnt^nn!sn+1\frac{n!}{s^{n+1}}
Exponentialeate^{-at}1(s+a)\frac{1}{(s + a)}
Sine Wavesinωt\sin \omega tω(s2+ω2)\frac{\omega}{(s^2 + \omega^2)}
Cosine Wavecosωt\cos \omega ts(s2+ω2)\frac{s}{(s^2 + \omega^2)}
Damped Sine Waveeatsinωte^{-at} \sin \omega tω(s+a)2+ω2\frac{\omega}{(s + a)^2 + \omega^2}
Damped Cosine Waveeatcosωte^{-at} \cos \omega t(s+a)(s+a)2+ω2\frac{(s + a)}{(s + a)^2 + \omega^2}
Sinh Wavesinhat\sinh ata(s2a2)\frac{a}{(s^2 - a^2)}
Cosh Wavecoshat\cosh ats(s2a2)\frac{s}{(s^2 - a^2)}
Damped Sinh Waveebtsinhate^{-bt} \sinh ata(s+b)2a2\frac{a}{(s + b)^2 - a^2}
Damped Cosh Waveebtcoshate^{-bt} \cosh ats+b(s+b)2a2\frac{s + b}{(s + b)^2 - a^2}
When aba \neq beatebtba\frac{e^{-at} - e^{-bt}}{b - a}1(s+a)(s+b)\frac{1}{(s + a)(s + b)}
When aba \neq ba.eatb.ebtab\frac{a.e^{-at} - b.e^{-bt}}{a - b}s(s+a)(s+b)\frac{s}{(s + a)(s + b)}