The Laplace transform of a function *f*(*t*)
is defined as

$$L\left[f\right]\left(s\right)={\displaystyle \underset{0}{\overset{\infty}{\int}}f(t){e}^{-ts}dt},$$

while the inverse Laplace transform (ILT) of *f*(*s*)
is

$${L}^{-1}\left[f\right](t)=\frac{1}{2\pi i}{\displaystyle \underset{c-i\infty}{\overset{c+i\infty}{\int}}f(s){e}^{st}ds},$$

where *c* is a real number selected so that
all singularities of *f*(*s*) are
to the left of the line *s* = *c*. The notation *L*[*f*]
indicates the Laplace transform of *f* at *s*.
Similarly, *L*^{–1}[*f*]
is the ILT of *f* at *t*.

The Laplace transform has many applications including the solution of ordinary differential equations/initial value problems. Consider the resistance-inductor-capacitor (RLC) circuit below.

Let *Rj* and *Ij*, *j* =
1, 2, 3 be resistances (measured in ohms) and currents (amperes),
respectively; *L* be inductance (henrys), and *C* be
capacitance (farads); *E*(*t*) be
the electromotive force, and *Q*(*t*)
be the charge.

By applying Kirchhoff's voltage and current laws, Ohm's Law, and Faraday's Law, you can arrive at the following system of simultaneous ordinary differential equations.

$$\frac{d{I}_{1}}{dt}+\frac{{R}_{2}}{L}\frac{dQ}{dt}=\frac{{R}_{2}-{R}_{1}}{L}{I}_{1},\text{}{I}_{1}(0)={I}_{0}.$$

$$\frac{dQ}{dt}=\frac{1}{{R}_{3}+{R}_{2}}\left(E(t)-\frac{1}{C}Q(t)\right)+\frac{{R}_{2}}{{R}_{3}+{R}_{2}}{I}_{1},\text{}Q(0)={Q}_{0}.$$

Solve this system of differential equations using `laplace`

.
First treat the R* _{j}*, L, and
C as (unknown) real constants and then supply values later on in the
computation.

clear E syms R1 R2 R3 L C real syms I1(t) Q(t) s dI1(t) = diff(I1(t), t); dQ(t) = diff(Q(t),t); E(t) = sin(t); % Voltage eq1(t) = dI1(t) + R2*dQ(t)/L - (R2 - R1)*I1(t)/L; eq2(t) = dQ(t) - (E(t) - Q/C)/(R2 + R3) - R2*I1(t)/(R2 + R3);

At this point, you have constructed the equations in the MATLAB^{®} workspace.
An approach to solving the differential equations is to apply the
Laplace transform, which you will apply to `eq1(t)`

and `eq2(t)`

.
Transforming `eq1(t)`

and `eq2(t)`

L1(t) = laplace(eq1,t,s) L2(t) = laplace(eq2,t,s)

returns

L1(t) = s*laplace(I1(t), t, s) - I1(0) + ((R1 - R2)*laplace(I1(t), t, s))/L - (R2*(Q(0) - s*laplace(Q(t), t, s)))/L L2(t) = s*laplace(Q(t), t, s) - Q(0) - (R2*laplace(I1(t), t, s))/(R2 + R3) - (C/(s^2 + 1) - laplace(Q(t), t, s))/(C*(R2 + R3))

Now you need to solve the system of equations ```
L1 =
0
```

, `L2 = 0`

for `laplace(I1(t),t,s)`

and `laplace(Q(t),t,s)`

,
the Laplace transforms of I_{1} and *Q*,
respectively. To do this, make a series of substitutions. For the
purposes of this example, use the quantities *R**1*
= 4 Ω (ohms), *R2*
= 2 Ω, *R3* = 3 Ω, *C* =
1/4 farads, *L* = 1.6 H (henrys), *I1*(0) = 15 A (amperes), and *Q*(0) = 2 A*sec.
Substituting these values in `L1`

syms LI1 LQ NI1 = subs(L1(t),{R1,R2,R3,L,C,I1(0),Q(0)}, ... {4,2,3,1.6,1/4,15,2})

returns

NI1 = s*laplace(I1(t), t, s) + (5*s*laplace(Q(t), t, s))/4 + (5*laplace(I1(t), t, s))/4 - 35/2

The substitution

NQ = subs(L2,{R1,R2,R3,L,C,I1(0),Q(0)},{4,2,3,1.6,1/4,15,2})

returns

NQ(t) = s*laplace(Q(t), t, s) - 1/(5*(s^2 + 1)) -... (2*laplace(I1(t), t, s))/5 + (4*laplace(Q(t), t, s))/5 - 2

To solve for `laplace(I1(t),t,s)`

and `laplace(Q(t),t,s)`

,
make a final pair of substitutions. First, replace the strings `laplace(I1(t),t,s)`

and `laplace(Q(t),t,s)`

by
the `sym`

objects `LI1`

and `LQ`

,
using

NI1 = subs(NI1,{laplace(I1(t),t,s),laplace(Q(t),t,s)},{LI1,LQ})

to obtain

NI1 = (5*LI1)/4 + LI1*s + (5*LQ*s)/4 - 35/2

Collecting terms

NI1 = collect(NI1,LI1)

gives

NI1 = (s + 5/4)*LI1 + (5*LQ*s)/4 - 35/2

A similar string substitution

NQ = ... subs(NQ,{laplace(I1(t),t,s), laplace(Q(t),t,s)}, {LI1,LQ})

yields

NQ(t) = (4*LQ)/5 - (2*LI1)/5 + LQ*s - 1/(5*(s^2 + 1)) - 2

which, after collecting terms,

NQ = collect(NQ,LQ)

gives

NQ(t) = (s + 4/5)*LQ - (2*LI1)/5 - 1/(5*(s^2 + 1)) - 2

Now, solving for `LI1`

and `LQ`

[LI1, LQ] = solve(NI1, NQ, LI1, LQ)

you obtain

LI1 = (5*(60*s^3 + 56*s^2 + 59*s + 56))/((s^2 + 1)*(20*s^2 + 51*s + 20)) LQ = (40*s^3 + 190*s^2 + 44*s + 195)/((s^2 + 1)*(20*s^2 + 51*s + 20))

To recover `I1`

and `Q`

, compute
the inverse Laplace transform of `LI1`

and `LQ`

.
Inverting `LI1`

I1 = ilaplace(LI1, s, t)

produces

I1 = 15*exp(-(51*t)/40)*(cosh((1001^(1/2)*t)/40) -... (293*1001^(1/2)*sinh((1001^(1/2)*t)/40))/21879) - (5*sin(t))/51

Inverting `LQ`

Q = ilaplace(LQ, s, t)

yields

Q = (4*sin(t))/51 - (5*cos(t))/51 +... (107*exp(-(51*t)/40)*(cosh((1001^(1/2)*t)/40) +... (2039*1001^(1/2)*sinh((1001^(1/2)*t)/40))/15301))/51

Now plot the current `I1(t)`

and charge `Q(t)`

in
two different time domains, 0 ≤ `t`

≤ 10 and 5 ≤ `t`

≤
25. The following statements generate the desired plots.

subplot(2,2,1) fplot(I1,[0,10]) title('Current') ylabel('I1(t)') xlabel('t') grid subplot(2,2,2) fplot(Q,[0,10]) title('Charge') ylabel('Q(t)') xlabel('t') grid subplot(2,2,3) fplot(I1,[5,25]) title('Current') ylabel('I1(t)') xlabel('t') grid text(7,0.25,'Transient') text(16,0.125,'Steady State') subplot(2,2,4) fplot(Q,[5,25]) title('Charge') ylabel('Q(t)') xlabel('t') grid text(7,0.25,'Transient') text(15,0.16,'Steady State')

Note that the circuit's behavior, which appears to be exponential decay in the short term, turns out to be oscillatory in the long term. The apparent discrepancy arises because the circuit's behavior actually has two components: an exponential part that decays rapidly (the "transient" component) and an oscillatory part that persists (the "steady-state" component).

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