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Laplace Transform and Inverse

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

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

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

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

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.

Solve this system of differential equations using `laplace`. First treat the Rj, 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 I1 and Q, respectively. To do this, make a series of substitutions. For the purposes of this example, use the quantities R1 = 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 `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 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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