The Laplace Transform converts linear time-invariant (LTI) differential equations into algebraic equations in the complex frequency domain ((s)-domain). Once you solve for a system transfer function in the (s)-domain, you must convert the expression back to the time domain ((t)-domain) using the Inverse Laplace Transform to obtain physical time-domain responses.
In MATLAB, the Symbolic Math Toolbox provides the ilaplace() function to compute exact analytical inverse Laplace transforms.
The Inverse Laplace Transform converts a frequency-domain transfer function (F(s)) into its corresponding time-domain sequence (f(t)):
[f(t) = mathcal{L}^{-1}{F(s)}]This conversion allows control engineers to inspect system transient responses, stability behavior, and time-domain step or impulse signals.
f = ilaplace(F) f = ilaplace(F, s, t)F: Symbolic transfer function in the Laplace domain.s: Laplace variable (frequency domain).t: Time variable (time domain).Find the inverse Laplace transform of:
[F(s) = frac{1}{s^2 + 4s + 3}]syms s t F = 1 / (s^2 + 4*s + 3);
f = ilaplace(F, s, t)Output:
f = (1/2)*exp(-t) - (1/2)*exp(-3*t)The resulting time-domain expression consists of two decaying exponential terms characteristic of a stable second-order system response.
Compute the inverse Laplace transform containing a symbolic parameter a:
syms s a t F = a / (s^2 + a^2);
f = ilaplace(F, s, t)Output:
f = sin(a*t)Find the time-domain response of the open-loop transfer function:
[G(s) = frac{5}{s(s + 2)}]syms s t G = 5 / (s * (s + 2));
g_t = ilaplace(G, s, t)Output:
g_t = (5/2) - (5/2)*exp(-2*t)Visualize the resulting continuous analytical function using fplot():
figure;
f
plot(g_t, [0, 5], 'LineWidth', 1.5, 'Color', 'b');
title('System Step Response g(t)');
xlabel('Time t (seconds)');
ylabel('Amplitude g(t)');
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