---
title: Transfer functions and poles
url: https://doc.liz6.com/en/theory/02-control-theory/05-transfer-functions-and-poles
locale: en
area: theory
tags:
- Control theory
- Theory
date: 2026-09-10
modified: 2026-09-10
description: A system can be represented by differential equations, state updates or input/output transfer functions. Transfer functions simplify the composition of linear dynamic blocks. We use operating-point deviations from Dynamic models and the controller in PID control.
---
# Transfer functions and poles
A system can be represented by differential equations, state updates or input/output transfer functions. Transfer functions simplify the composition of linear dynamic blocks. We use operating-point deviations from [Dynamic models](02-dynamic-models.md) and the controller in [PID control](04-pid-control.md).
## Turn derivatives into algebra
The unilateral Laplace transform is $X(s)=\int_0^\infty x(t)e^{-st}dt$ in its convergence region. A derivative transforms as $\mathcal L\{\dot x\}=sX(s)-x(0)$. The complex variable $s=\sigma+j\omega$ represents exponential growth or decay together with oscillation.
Our thermal deviation equation is $2000\dot x+20x=1000v$. At zero initial state,
$$(2000s+20)X=1000V,\qquad P(s)=\frac XV=\frac{50}{100s+1}.$$
Here $v$ is a duty-ratio deviation and $x$ a temperature deviation. A nonzero $x(0)$ adds $2000x(0)/(2000s+20)$ to the transformed response. The transfer ratio alone is not the complete initial-value solution.
$P(0)=50$ K/duty is the static gain. The pole at $-0.01$ corresponds to $e^{-0.01t}$ decay. This LTI representation does not automatically include saturation or changing operating conditions.
## Where the closed-loop denominator comes from
For unity negative feedback, $E=R-Y$, $U=CE$, and $Y=PU$. Rearranging gives
$$(1+PC)Y=PC R,\qquad \frac YR=\frac{PC}{1+PC}.$$
Clear denominators in $1+PC=0$ to obtain the characteristic equation, while checking internal modes in the realization. Sensor dynamics $H(s)$ change the denominator to $1+PCH$; filtering is part of the loop.
For $C=K_p$,
$$\frac YR=\frac{50K_p}{100s+1+50K_p},\qquad p=-\frac{1+50K_p}{100}.$$
At $K_p=0.04$, the pole is $-0.03$ and time constant 33.33 s, but a unit reference-temperature step produces only a $2/3$ K final increment. Speed and accuracy are separate properties.
## PI poles move with gain
With $C=K_p+K_i/s$, the denominator is
$$100s^2+(1+50K_p)s+50K_i.$$
Fixing $K_p=0.04$, $K_i=0.0002$ gives roots $-0.00382,-0.02618$; $K_i=0.001$ gives $-0.015\pm0.01658j$. Increasing integral action can turn two real modes into oscillatory modes. Positive coefficients ensure stability for this delay-free second-order case, not for arbitrary plants.
**Transfer functions and poles · Experiment**
Thermal PI characteristic equation 100s²+3s+50Ki=0, Kp=0.04. Poles coincide at Ki=0.00045, then separate along real part −0.015. These are pole locations, not physical trajectories.
Check that the roots sum to $-0.03$ and multiply to $0.5K_i$. They coincide when the discriminant vanishes at $K_i=0.00045$. A root locus shows pole locations as a parameter varies, not a physical motion path.
## Zeros and hidden states matter
Compare $G_1(s)=1/(s+1)$ and $G_2(s)=(1-s)/(s+1)$. Both have pole $-1$ and static gain 1. Their unit-step responses are $1-e^{-t}$ and $1-2e^{-t}$. The latter jumps to −1 before tending to 1; it includes a direct path. A right-half-plane zero can produce inverse response and constrain design. Stable poles alone do not determine the transient shape.
Cancellation or an unobserved mode can also hide internal instability. For $\dot x_1=-x_1+u$, $\dot x_2=x_2$, $y=x_1$, the transfer is $1/(s+1)$, yet nonzero $x_2(0)$ grows exponentially. Check the realization as well as the reduced denominator.
## Reproduce the pole calculation
This optional example uses the python-control 0.10.2 API. `feedback` defaults to negative feedback; the result is about the stated linear model.
```python
import control as ct
P = ct.tf([50], [100, 1])
C = ct.tf([0.04, 0.001], [1, 0])
closed_loop = ct.feedback(P * C, 1)
print(ct.poles(closed_loop))
# Approximately: -0.015 +/- 0.016583j
```
## Check your understanding
Why is $Y=PU$ insufficient for a nonzero initial temperature deviation?
Reasoning
The transfer relation assumes zero initial conditions. The derivative transform contributes an additional initial-state term, which must be retained.
Reasoning
No. Zeros, gain, direct feedthrough and initial conditions matter. The two example transfers share poles and static gain but initially move in different directions.