Transfer Functions

Wikipedia · Transfer function · CC BY-SA 4.0

The Laplace domain

A linear time-invariant (LTI) system is described by a differential equation. The Laplace transform replaces derivatives with powers of the complex variable s. Differentiation becomes multiplication: if y(t) has transform Y(s), then y'(t) has transform sY(s). This turns calculus into algebra.

Poles and zeros

The transfer function is a ratio of polynomials in s. The roots of the numerator are zeros: frequencies where the output vanishes. The roots of the denominator are poles: frequencies where the output blows up. Poles in the left half-plane decay. Poles in the right half-plane grow. Poles on the imaginary axis oscillate forever.

Step response

Apply a unit step input (off to on at t=0) and watch the output. The step response reveals rise time, overshoot, settling time, and steady-state error. These four numbers summarize the system's behavior for most practical purposes.

First-order and second-order systems

A first-order system has one pole: H(s) = 1/(s + a). It rises exponentially with time constant 1/a. No overshoot. A second-order system has two poles: H(s) = wn^2/(s^2 + 2*zeta*wn*s + wn^2). The damping ratio zeta controls whether the response is overdamped (zeta greater than 1), critically damped (zeta = 1), or underdamped (zeta less than 1, the oscillating case).

; Simulate a first-order step response: y(t) = 1 - e^(-t/tau)
; Time constant tau = 1/a determines how fast it responds

(define tau 2.0)
(define dt 0.5)

(define (exp-approx x n)
  ; Taylor series approximation of e^x
  (if (= n 0) 1.0
    (+ 1.0
       (let loop ((k 1) (term 1.0) (sum 0.0))
         (if (> k n) sum
           (let ((next-term (/ (* term x) k)))
             (loop (+ k 1) next-term (+ sum next-term))))))))

(define (step-response t)
  (- 1.0 (exp-approx (- (/ t tau)) 15)))

(display "First-order step response (tau=2):") (newline)
(let loop ((t 0.0))
  (if (> t 10.0) 'done
    (begin
      (display "  t=") (display t)
      (display "  y=") (display (step-response t))
      (newline)
      (loop (+ t dt)))))

import math

# First-order step response: y(t) = 1 - e^(-t/tau)
tau = 2.0
dt = 0.5

print("First-order step response (tau=2):")
t = 0.0
while t <= 10.0:
    y = 1.0 - math.exp(-t / tau)
    bar = "#" * int(y * 40)
    print(f"  t={t:4.1f}  y={y:.3f}  {bar}")
    t += dt