Laplace techniques: convolution, feedback and frequency response

Lesson 7 introduced the Laplace transform and used it to solve initial value problems with steps and impulses, ending with the idea of a transfer function. This lesson builds the techniques that make the transform the everyday language of control, vibration and signal processing. It completes the partial fraction toolkit for repeated and complex poles, including the resonant case. It develops the convolution theorem, which gives the response to any input from the impulse response. It shows how transfer functions combine in block diagrams and feedback loops, and how the final value theorem predicts steady-state error. Finally, it introduces the frequency response, the Bode plot and a simple test for stability, which together let an engineer judge a system from its transfer function without solving for the time response at all.

Repeated and complex poles

Partial fractions in Lesson 7 handled distinct real poles by the cover-up rule and complex poles by completing the square. A repeated pole needs one term for each power. For a pole at \(s = a\) of multiplicity 2, the expansion contains:

The inverse of the squared term comes from the table entry \(\mathcal{L}\{te^{at}\} = 1/(s - a)^2\). The cover-up rule still gives the coefficient of the highest power; the others come from matching coefficients or substituting convenient values of s.

Repeated complex poles appear in undamped resonance. The table entry needed is:

The convolution theorem

Lesson 7 mentioned that a product in the s-domain corresponds to a convolution in time. Stated fully: if \(F(s) = \mathcal{L}\{f\}\) and \(G(s) = \mathcal{L}\{g\}\), then

Since the output of a system at rest is \(Y = GF\), and \(g(t)\), the inverse of G, is the impulse response, the response to any input is the convolution of the input with the impulse response. This is Duhamel's integral from Lesson 4, now valid for any linear system: measure a system's impulse response once (for example with a hammer test), and its response to any other input can be computed. Convolution is also a quick way to invert some products directly.

Block diagrams and feedback

Transfer functions make it easy to combine components. Two systems in series multiply: \(G_1G_2\). Two in parallel add: \(G_1 + G_2\). The most important combination is the feedback loop, in which a controller C(s) compares the measured output with a set point R(s) and drives the process G(s) with the error. With unity feedback, \(E = R - Y\) and \(Y = CGE\), so:

The denominator \(1 + CG\) sets the closed-loop poles, so feedback changes the speed and stability of a system, not just its gain. The final value theorem (Lesson 7) predicts the steady-state error for a step in the set point.

Frequency response and Bode plots

When a stable linear system is driven by a sinusoid \(\sin\omega t\), the transient dies away and the output becomes a sinusoid at the same frequency, scaled and shifted:

So the frequency response is just the transfer function evaluated at \(s = i\omega\): its magnitude is the gain and its angle is the phase shift. A Bode plot shows both against frequency on logarithmic axes, with the gain in decibels, \(20\log_{10}|G|\). Engineers use Bode plots to design filters, specify sensors and amplifiers, and assess the stability margins of control loops.

For a first-order system \(G = 1/(\tau s + 1)\), the gain is \(1/\sqrt{1 + \omega^2\tau^2}\) and the phase is \(-\arctan\omega\tau\). At the corner frequency \(\omega = 1/\tau\), the gain is \(1/\sqrt{2}\) (−3 dB) and the phase is −45°; above it, the gain falls by 20 dB for every tenfold increase in frequency.

For the second-order system \(G = \omega_n^2/(s^2 + 2\zeta\omega_ns + \omega_n^2)\), the gain is the amplification factor of Lesson 5. A lightly damped system shows a sharp resonant peak of height about \(1/(2\zeta)\) near \(\omega_n\), and the phase swings through −90° at \(\omega_n\) toward −180°.

Stability from the transfer function

A linear system is stable if every pole of its transfer function has a negative real part, so that every mode decays. For a closed loop, the poles are the roots of \(1 + CG = 0\). For low-order polynomials there are simple tests that avoid finding the roots. A second-order polynomial \(s^2 + a_1s + a_0\) has both roots in the left half-plane exactly when \(a_1 > 0\) and \(a_0 > 0\). For a cubic \(s^3 + a_2s^2 + a_1s + a_0\), the Routh-Hurwitz criterion requires all coefficients positive and

import numpy as np

# Process G = 1/(s + 1)^3 with proportional gain K and unity feedback
# Closed-loop characteristic polynomial: s^3 + 3s^2 + 3s + (1 + K)
for K in [2, 6, 8, 10]:
    poles = np.roots([1, 3, 3, 1 + K])
    worst = max(poles.real)
    if worst < -1e-6:
        status = "stable"
    elif worst < 1e-6:
        status = "on the edge"
    else:
        status = "unstable"
    p = sorted(poles, key=lambda z: z.real)
    text = ", ".join(f"{round(z.real, 3) + 0.0:.3f}{z.imag:+.3f}j" for z in p)
    print(f"K = {K:2d}: poles {text}  ({status})")

K =  2: poles -2.260+0.000j, -0.370+1.091j, -0.370-1.091j  (stable)
K =  6: poles -2.817+0.000j, -0.091+1.574j, -0.091-1.574j  (stable)
K =  8: poles -3.000+0.000j, 0.000+1.732j, 0.000-1.732j  (on the edge)
K = 10: poles -3.154+0.000j, 0.077+1.866j, 0.077-1.866j  (unstable)