Transfer functions

A transfer function $G (s)$ is the system's recipe card. The same recipe handles every kind of input — steps, kicks, sine waves. Below is a second-order plant with two sliders for its parameters; flip the input picker and watch the same recipe cook three different dishes.

Same plant, three inputs

Plant G(s) = ω_n² / (s² + 2ζω_ns + ω_n²).

ω_n 1.00 ζ 0.20 input ω (sine only) 1.00

input

steady-state (step) —

|G(jω)| at picked ω —

∠G(jω) at picked ω —

poles —

what to try

Step: y settles to 1 (DC gain = 1). The ring around the setpoint is the underdamped response we'll dissect in Lesson 04.
Impulse: a hammer-blow input. The plant's impulse response is the time-domain twin of its transfer function.
Sine: now look at the metrics. The steady-state output amplitude matches $∣ G (j o m e g a) ∣$ exactly, and the phase lag matches $an g l e G (j o m e g a)$ . This is the bridge to Lesson 05 (Bode).
With sine mode on, drag the input-ω slider through ω = ω_n. The output amplitude peaks — resonance.

show the math

For a linear time-invariant plant, the output is the convolution of the impulse response with the input. In the Laplace domain that's a multiplication:

Y (s) = G (s), U (s)

Three special inputs: $U_{e x t s t e p} (s) = 1/ s$ , $U_{e x t im p} (s) = 1$ , and $U_{e x t s in e} (s) = o m e g a / (s^{2} + o m e g a^{2})$ . Each picks out a different aspect of $G$ .