Frequency-Domain Control Systems

Continuous-Time

We typically assume the plant satisfies a LTI ODE of the form:

$$\sum _{i=0}^n{a}_i\frac{d_{i}y}{d{t}^i}(t)-\sum _{j=0}^m{b}_j\frac{d^{j}u}{d{t}^j}(t)=0$$

where $a_i,b_i\in \mathbb{R}\forall i,j$.

(One-sided) Laplace transform (LT):

$$X(s)=\mathcal{L}(x(t))={\int }_0^{\infty }x(t)e^{-st}dt$$

Taking the LT of the first ODE with zero initial conditions gives:

$$\begin{array}{rl}\sum _{i=0}^n{a}_i{s}^{i}Y(s)& -\sum _{j=0}^m{b}_j{s}^{j}U(s)=0\\ Y(s)& =\underset{P(s)}{\underbrace{\frac{{\sum }_{j=0}^m{b}_j{s}_j}{{\sum }_{i=0}^n{a}_i{s}^i}}}U(s)\\ Y(s)& =P(s)U(s)\end{array}$$

Discrete-Time

In discrete time, we typically assume the plant satisfies an LTI difference equation of the form:

$$\sum _{i=0}^n{a}_{i}y[k+1]-\sum _{j=0}^m{b}_{j}u[k+j]=0$$

where $a_i,b_j\in \mathbb{R}\forall i,j$.

(One-sided) z-transform:

$$X[z]=\mathcal{Z}(x(t))=\sum _{k=0}^{\infty }x[k]\frac{1}{z^k}$$

Take the z-transform of the difference equation with zero initial conditions:

$$\begin{array}{rl}\sum _{i=0}^n{a}_i{z}^{i}Y[z]& -\sum _{j=0}^m{b}_j{z}^{j}U[z]=0\\ Y[z]& =\underset{G[z]}{\underbrace{\frac{{\sum }_{j=0}^m{b}_j{z}^j}{{\sum }_{i=0}^n{a}_i{z}^i}}}U[z]\\ & =G[z]U[z]\end{array}$$