This is the first in a series of posts that are an attempt by me to understand why my Industrial Measurement and Control students need to study the Laplace Transform.

Consider a black box model that takes as an input signal a function $x(t)$ and produces an output signal $y(t)$. For reasons that are as of yet not clear to me, we can take all of the derivatives of $y(t)$ (and $x(t)$) to vanish for $t=0$.

### Definition

The transfer function of a black box model is defined as $\displaystyle H(s)=\frac{Y(s)}{X(s)}$,

where $Y(s)$ and $X(s)$ are the Laplace Transforms of $y(t)$ and $x(t)$.

Note we have the Laplace transform of a function $f(t)$ is a function $F(s)$ defined by $F(s)=\mathcal{L}\{f(t)\}=\int_{0}^\infty f(t)e^{-st}\,dt.$

Poles of $H(s)$ are complex numbers $z$ such that $X(z)=0$. For example, for an input signal $x(t)=e^t$, the transfer function has a pole at $s=1$ as $\displaystyle F(s)=\frac{1}{s-1}$.

If the coefficients of $x(t)$ are real, then in general the poles of the transfer are real or come in conjugate root pairs.

### Derivation of Formula for Transfer Function — Linear ODE Case

Consider a linear input/output system described by the differential equation: $\displaystyle \sum_{i=0}^na_i\frac{d^iy}{dt^i}=\sum_{j=0}^mb_j\frac{d^jx}{dt^j}$,

where we can take $a_n=0$.

If we assume that the input signal is of the form $e^{st}$ then because the system is linear we must also have the output signal as $y(t)=y_0e^{st}$. Inserting these signals into the ODE yields: $\displaystyle y_0\sum_{i=0}^na_is^ie^{st}=\sum_{j=0}^mb_js^je^{st}$,

and so the response of the system can be described by the two polynomials $\displaystyle a(s)=\sum_{i=0}^na_is^i$ (with $a_n=0$)

and $\displaystyle b(s)=\sum_{j=0}^mb_js^j$.

The polynomial $a(s)$ is nothing but the characteristic/auxiliary equation of the associated homogenous system. If it is not everywhere zero then we have $\displaystyle y(t)=y_0e^{st}=\frac{b(s)}{a(s)}e^{st}=G(s)u(t)$.

Therefore, the transfer function is $\displaystyle H(s)=\frac{b(s)}{a(s)}=\frac{1/a(s)}{1/b(s)}$.

### Example: Complex Harmonic Signal

Suppose we have an input signal $\displaystyle x(t)=Xe^{i\omega t}=|X|e^{i\phi_1}e^{i\omega t}$,

and an output signal $\displaystyle y(t)=Ye^{i\omega t}=|Y|e^{i\phi_2}e^{i\omega t}$.

Then we have $\displaystyle X(s)=\frac{|X|e^{i\phi_1}}{s-i\omega}$ and $\displaystyle Y(s)=\frac{|Y|e^{i\phi_2}}{s-i\omega}$ $\displaystyle \Rightarrow H(s)=\frac{|Y|}{|X|}e^{i(\phi_2-\phi_1)}\Rightarrow |H(s)|=\frac{|Y|}{|X|}$.

This $|H(s)|$ is the gain.

### Example: Closed Loop Transfer Function

In this model, the output $y(t)$ is measured and compared with the reference (ideal output) value $r(t)$. The controller then takes the error, $e(t)=\pm (y(t)-r(t))$, and changes the input $x(t)$ of the system under control, $P$, using some controller. This is known as a closed-loop controller or a feedback controller.

If we assume that the controller, system and sensor are linear and time-invariant (their transfer functions $C(s)$, $P(s)$ and $F(s)$ do not depend on time), we can use the Laplace transform to analyse the system. We have $\displaystyle Y(s)=P(s)X(s)$, $\displaystyle X(s)=C(s)E(s)$ and $\displaystyle E(s)=R(s)-F(s)Y(s)$.

Some elementary algebra shows that $\displaystyle Y(s)=\left(\frac{P(s)C(s)}{1+F(s)P(s)C(s)}\right)R(s)=H(s)R(s).$
If $|P(s)C(s)|\gg1$ and $|F(s)|\approx 1$ then we have that $H(s)\approx 1$ and so $Y(s)\approx R(s)$,
and so the output, $y(t)$, tracks the reference value, $r(t)$, quite well.