The Laplace transform and transfer functions

Solving differential equations directly and evaluating convolution integrals by hand both become laborious as soon as systems grow beyond second order or are connected together. The Laplace transform removes most of that labour. It converts linear differential equations with constant coefficients into algebraic equations in a complex variable s, and it converts convolution into simple multiplication. The ratio of output to input transforms, the transfer function, becomes a compact description of an LTI system, and the roots of its denominator, the poles, reveal stability and the shape of the response at a glance. Transfer functions are the native language of classical control: motor models, sensor models, filters and controllers are all written as ratios of polynomials in s, and the block diagrams of lesson 6 multiply them together. This lesson builds the transform, the key pairs and properties, partial-fraction inversion, poles and zeros, and the final value theorem.

The transform

For a causal signal f(t), zero for t < 0, the (one-sided) Laplace transform is

The transform weighs the signal against decaying complex exponentials \( e^{-st} \) and records how much of each is present. For many signals the integral converges only when the real part of s exceeds some value, which defines the region of convergence; for the causal signals used in engineering we rarely need to worry about it, as long as we remember that the transform describes behaviour for t ≥ 0. Starting the integral at \( 0^- \), just before zero, makes sure that an impulse at t = 0 is included.

Key transform pairs and properties

In practice nobody evaluates the integral every time. A short table of pairs and a few properties cover nearly every problem met in robotics.

The properties are what make the transform powerful:

• Linearity: \( \mathcal{L}\{af + bg\} = aF + bG \).
• Differentiation: \( \mathcal{L}\{\dot{f}\} = sF(s) - f(0^-) \), and \( \mathcal{L}\{\ddot{f}\} = s^2F(s) - sf(0^-) - \dot{f}(0^-) \). Differentiation becomes multiplication by s, with initial conditions appearing as extra terms.
• Integration: \( \mathcal{L}\{\int_0^t f\,d\tau\} = F(s)/s \). Integration becomes division by s, which is why 1/s is the symbol of an integrator in block diagrams.
• Time delay: \( \mathcal{L}\{f(t - T)u(t - T)\} = e^{-sT}F(s) \). A transport delay, such as communication latency, multiplies by \( e^{-sT} \).
• Convolution: \( \mathcal{L}\{f * g\} = F(s)\,G(s) \). The integral of lesson 3 becomes a product.
• Final value theorem: if all poles of sF(s) are in the left half-plane, \( \lim_{t\to\infty} f(t) = \lim_{s\to 0} sF(s) \).

Transfer functions

Take the Laplace transform of an LTI differential equation with all initial conditions zero. For the standard first-order system \( \tau\dot{y} + y = Kx \), the differentiation property gives \( \tau sY(s) + Y(s) = KX(s) \), so

The transfer function G(s) is the ratio of the output transform to the input transform. Because \( Y(s) = G(s)X(s) \) and the output is the input convolved with the impulse response, G(s) is simply the Laplace transform of the impulse response. For the second-order system of lesson 4,

and for a DC motor with the inductance kept, eliminating the current from the electrical and mechanical equations gives the transfer function from voltage to speed:

Position is the integral of speed, so the voltage-to-angle transfer function has an extra factor 1/s. Writing transfer functions takes practice but follows one rule: transform each equation, replace each derivative by s, and solve algebraically for the output over the input.

Poles and zeros

A rational transfer function is a ratio of polynomials, which can be factored:

The roots \( z_i \) of the numerator are the zeros; the roots \( p_i \) of the denominator are the poles, which are exactly the roots of the characteristic equation of lesson 4. Each real pole at \( s = -a \) contributes a term \( e^{-at} \) to the impulse response; each complex pair \( -\sigma \pm j\omega_d \) contributes a decaying oscillation \( e^{-\sigma t}\sin(\omega_d t + \phi) \). The pole locations therefore tell the whole story of the natural response:

• Poles in the left half-plane (negative real part) give decaying terms. If all poles are there, the system is stable.
• Any pole in the right half-plane gives a growing term: instability.
• Poles on the imaginary axis give sustained oscillation (a pair) or a constant or ramp (at the origin); such systems, including integrators, are called marginally stable and are not BIBO-stable.
• Distance from the imaginary axis sets decay rate; distance from the real axis sets oscillation frequency. Poles far to the left die out fast; the poles nearest the imaginary axis, called dominant poles, dominate the response.

Zeros do not affect stability but shape the response: a zero near a pole partly cancels it, and a zero in the right half-plane, which appears for example in some flexible-link arms and in the tilt dynamics of balancing robots, makes the response initially move the wrong way, which limits how fast such a system can be controlled.

Inverting transforms by partial fractions

To get a time response from a transform, factor the denominator and split the rational function into simple terms whose inverses are in the table. For distinct real poles,

This cover-up rule finds each coefficient by cancelling its factor and substituting the pole. Repeated poles and complex pairs need slightly more work, but the idea is the same.

The final and initial value theorems

Often we need only the steady state: where will the joint end up, what steady error remains? The final value theorem answers without inverting anything. For the joint above, \( \lim_{s\to 0} s \cdot \frac{6}{s(s+2)(s+3)} = 1 \). The theorem applies only if the response actually settles, meaning all poles of sY(s) are in the left half-plane; applying it to an unstable system gives a meaningless number. A companion initial value theorem, \( f(0^+) = \lim_{s\to\infty} sF(s) \), gives the starting value.

Working with transfer functions in software

Transfer functions are stored as coefficient lists, highest power first. NumPy can find poles and zeros with np.roots, and the step response can be computed by partial fractions or by simulation. The code below finds the poles of the camera-mount model and checks the partial-fraction step response of the joint example against a direct numerical simulation.

import numpy as np
den = [1, 8, 400]                       # camera mount: 2/(s^2 + 8s + 400)
p = np.roots(den)
wn = np.sqrt(den[2]); zeta = den[1]/(2*wn)
print("poles:", np.round(p, 3), " wn =", wn, " zeta =", zeta)
# joint: G = 6/((s+2)(s+3)); compare partial fractions with simulation
h, T = 1e-4, 4.0
x1 = x2 = 0.0                            # y'' + 5y' + 6y = 6u
for k in range(int(T/h)):
    a = 6.0 - 5.0*x2 - 6.0*x1
    x1, x2 = x1 + h*x2, x2 + h*a
t = T
pf = 1 - 3*np.exp(-2*t) + 2*np.exp(-3*t)
print(f"y(4 s): simulation {x1:.5f}, partial fractions {pf:.5f}")

poles: [-4.+19.596j -4.-19.596j]  wn = 20.0  zeta = 0.2
y(4 s): simulation 0.99901, partial fractions 0.99901

The poles at −4 ± 19.6j confirm ζ = 0.2 and a damped ringing frequency of 19.6 rad/s, and the simulated and analytical step responses agree to five decimal places, confirming the partial-fraction result.