6.9 Laplace Transforms & Transfer-Function Methods

Definition and Purpose

The Laplace transform of a function $f(t)$ defined for $t \geq 0$ is

$\mathcal{L}\{f(t)\} = F(s) = \int_{0}^{\infty} e^{-st}f(t)\,dt$

provided the integral converges. The point of the transform is structural: differentiation in the time domain becomes multiplication by $s$ in the transform domain. A differential equation therefore becomes an algebraic equation, which is solved by ordinary algebra and then inverted.

This is why the transform dominates control and vibration analysis — and why it is listed in the same syllabus bullet as differential equations.

Standard Transform Pairs

Note the sign distinction between the circular and hyperbolic pairs — $s^2 + \omega^2$ versus $s^2 - a^2$ — which is a favourite objective-item discriminator.

The Derivative Property

Integrating the definition by parts gives

$\mathcal{L}\{f'(t)\} = sF(s) - f(0)$

$\mathcal{L}\{f''(t)\} = s^2F(s) - sf(0) - f'(0)$

and in general

$\mathcal{L}\{f^{(n)}(t)\} = s^{n}F(s) - s^{n-1}f(0) - \cdots - f^{(n-1)}(0)$

The initial conditions enter the algebra directly. This is the decisive practical advantage over the complementary-function method, where constants must be fitted to boundary conditions at the very end.

Shifting Theorems

First shifting (frequency shift). Multiplying by a decaying exponential shifts the transform:

$\mathcal{L}\{e^{-at}f(t)\} = F(s + a)$

So $\mathcal{L}{e^{-at}\sin\omega t} = \dfrac{\omega}{(s+a)^2 + \omega^2}$ — precisely the transform of a damped oscillation.

Second shifting (time delay). For a function switched on at $t = a$:

$\mathcal{L}\{f(t-a)u(t-a)\} = e^{-as}F(s)$

A pure time delay multiplies the transform by $e^{-as}$, which is how transport lag in a conveyor or a pipeline is modelled.

Further Operational Theorems

Multiplication by $t$:

$\mathcal{L}\{t\,f(t)\} = -\frac{dF(s)}{ds}$

Division by $t$:

$\mathcal{L}\left\{\frac{f(t)}{t}\right\} = \int_{s}^{\infty}F(u)\,du$

Transform of an integral:

$\mathcal{L}\left\{\int_0^t f(\tau)\,d\tau\right\} = \frac{F(s)}{s}$

Convolution theorem. The inverse transform of a product is the convolution of the individual inverses:

$\mathcal{L}^{-1}\{F(s)G(s)\} = \int_0^t f(\tau)g(t-\tau)\,d\tau$

This is the formal statement that a system's output equals its impulse response convolved with the input.

Initial and Final Value Theorems

These two results extract the endpoints of the response without any inversion at all:

$f(0^+) = \lim_{s\to\infty} sF(s), \qquad f(\infty) = \lim_{s\to 0} sF(s)$

The final value theorem is valid only if the system is stable, that is, if all poles of $sF(s)$ lie in the left half-plane. Applying it to an oscillatory or growing response gives a meaningless answer, and objective items exploit exactly that.

Inverse Transforms by Partial Fractions

Inversion in practice means splitting $F(s)$ into standard pairs. For

$F(s) = \frac{s + 5}{(s+1)(s+3)}$

write $\dfrac{A}{s+1} + \dfrac{B}{s+3}$. Multiplying out gives $s + 5 = A(s+3) + B(s+1)$. Setting $s = -1$ gives $4 = 2A$, so $A = 2$; setting $s = -3$ gives $2 = -2B$, so $B = -1$. Hence

$f(t) = 2e^{-t} - e^{-3t}$

Worked Example: Damped Free Vibration

Solve $x'' + 3x' + 2x = 0$ with $x(0) = 1$ and $x'(0) = 0$.

Transforming and inserting the initial conditions:

$\left[s^2X - s(1) - 0\right] + 3\left[sX - 1\right] + 2X = 0$

$X(s^2 + 3s + 2) = s + 3 \quad \Longrightarrow \quad X = \frac{s+3}{(s+1)(s+2)}$

Partial fractions give $X = \dfrac{2}{s+1} - \dfrac{1}{s+2}$, so

$x(t) = 2e^{-t} - e^{-2t}$

Both exponents are negative, confirming a stable, over-damped return to equilibrium — consistent with $c^2 = 9 > 4mk = 8$. Checking the initial value theorem: $\lim_{s\to\infty} sX = \lim s(s+3)/(s^2+3s+2) = 1$, which matches $x(0) = 1$.

Transfer Functions and System Response

For a system at rest initially, the ratio of the transform of the output to the transform of the input is the transfer function:

$G(s) = \frac{Y(s)}{X(s)}$

For the mass-spring-damper driven by a force $F(t)$, transforming $m\ddot x + c\dot x + kx = F$ with zero initial conditions gives

$G(s) = \frac{X(s)}{F(s)} = \frac{1}{ms^2 + cs + k}$

Writing this in the standard second-order form,

$G(s) = \frac{1/k}{\dfrac{s^2}{\omega_n^2} + \dfrac{2\zeta s}{\omega_n} + 1}, \qquad \omega_n = \sqrt{\frac{k}{m}}, \quad \zeta = \frac{c}{2\sqrt{mk}}$

exposes the two parameters that govern every second-order system: the undamped natural frequency and the damping ratio.

Poles and stability

The poles are the roots of the denominator, and they determine the character of the response entirely:

The rule is compact and worth memorising: a system is stable if and only if all poles lie in the left half of the s-plane. The real part sets the rate of decay and the imaginary part sets the frequency of oscillation.

Standard Test Inputs

Because the impulse transform is simply 1, the transform of the impulse response is the transfer function itself — which is why the impulse response characterises a linear system completely.

Worked example. Find the steady-state deflection of the mass-spring-damper under a step force of magnitude $F_0$.

$X(s) = G(s)\frac{F_0}{s} = \frac{F_0}{s\left(ms^2+cs+k\right)}$

Applying the final value theorem, valid here because the system is stable:

$x(\infty) = \lim_{s\to0} s\,X(s) = \frac{F_0}{k}$

The steady-state deflection is simply the static deflection $F_0/k$, entirely independent of the mass and the damping. Those two parameters govern how the system reaches that value — overshoot, oscillation and settling time — but not the final value itself. This separation between transient behaviour and steady-state value is one of the most useful ideas the transform provides, and it carries directly into the vibration and control content of Paper-II.