Frequency Domain and a la carte.

Introduction and Notes

Open-loop versus closed-loop

Transient and steady-state response

Stability

• Total response = Natural response + Forced response
  • Natural response (homogeneous solution): evolution of system due to initial conditions
  • Forced response (particular solution): evolution of system due to input

Control objects:

• Stabilize the system
• Produce the desired transient response
• Decrease/eliminate steady-state error
• Make system “robust” to withstand disturbances and variations in parameters
• Achieve optimal performance

Block diagram representation of a system

graph LR
    * -- "r(t)" --> System{{System}} -- "c(t)" --> End{{end}}

System as linear differential equation

Laplace Transform

graph LR
    Diff{{differential equations}} -- "Laplace transform" --> Algebraic{{algebraic equations}} -- "inverse Laplace transform" --> End{{time domain solution}}

$\mathcal{L}\{f(t)\}={\int }_0^{\infty }f(t{)}^{-st}dt=F(s)$

$$\begin{array}{ccc}\text{Item no.}& f(t)& F(s)\\ 1.& \delta (t)& 1\\ 2.& u(t)& \frac{1}{s}\\ 3.& tu(t)& \frac{1}{s^2}\\ 4.& t^{n}u(t)& \frac{n!}{s^{n+1}}\\ 5.& e^{-at}u(t)& \frac{1}{s+a}\\ 6.& \sin (\omega t)u(t)& \frac{\omega }{s^2+{\omega }^2}\\ 7.& \cos (\omega t)u(t)& \frac{s}{s^2+{\omega }^2}\end{array}$$

$\delta (t)=0,t\ne 0,{\int }_0^{\infty }\delta (t)dt=1$

Properties

$$\begin{array}{rl}& f(0-)\text{: initial condition just before 0}\\ & \\ & \text{Linearity:}\mathcal{L}\{k_1{f}_1(t)\pm k_2{f}_2(t)\}=k_1{F}_1(s)\pm k_2{F}_2(s)\\ & \\ & \text{Differentiation:}\\ & \mathcal{L}\left\{\frac{df(t)}{dt}\right\}=sF(s)-f(0^{-})\\ & \mathcal{L}\left\{\frac{d^{2}f(t)}{d{t}^2}\right\}=s^{2}F(s)-sf(0^{-})-f^{\prime }(0^{-})\\ & \\ & \text{Frequency Shifting:}\mathcal{L}\{e^{-at}f(t)\}=F(s+a)\\ & \end{array}$$

Transfer function

$n^{th}$ order linear, time-invariant (LTI) differential equation:

takes Laplace transform from both side

$$\begin{array}{rl}& a_n{s}^{n}C(s)+a_{n-1}{s}^{n-1}C(s)+\cdots +a_{0}C(s)\text{ and init terms for }c(t)\\ & =b_m{s}^{m}R(s)+b_{m-1}{s}^{m-1}R(s)+\cdots +b_{0}R(s)\text{ and init terms for }r(t)\\ & \end{array}$$

assume initial conditions are zero

$$\begin{array}{rl}(a_n{s}^n+a_{n-1}{s}^{n-1}+\cdots +a_0)C(s)& =(b_m{s}^m+b_{m-1}{s}^{m-1}+\cdots +b_0)R(s)\\ & \\ \frac{C(s)}{R(s)}& =G(s)=\frac{b_m{s}^m+b_{m-1}{s}^{m-1}+\cdots +b_0}{a_n{s}^n+a_{n-1}{s}^{n-1}+\cdots +a_0}\\ & \end{array}$$

Transfer function

$$G(s)=\frac{C(s)}{R(s)}$$

Q: $G(s)=\frac{1}{S+2}$. Input: $u(t)$. What is $y(t)$ ?

$$Y(s)=G(s)\cdot u(s)\to Y(s)=\frac{1}{s(s+2)}=\frac{A}{s}+\frac{B}{s+2}=\frac{1}{2\cdot s}-\frac{1}{2\cdot (s+2)}$$

$y(t)=-\frac{1}{2}(1-e^{-2t})u(t)$

Inverse Laplace transform

$${\mathcal{L}}^{-1}\{F(s)\}=\frac{1}{2\pi j}\underset{\omega \to \infty }{\lim }{\int }_{\sigma -j\omega }^{\sigma +j\omega }F(s)e^{st}ds$$

Partial fraction expansion

$$\begin{array}{rl}F(s)& =\frac{N(s)}{D(s)}\\ & \\ N(s)& :m^{th}\text{ order polynomial in }s\\ D(s)& :n^{th}\text{ order polynomial in }s\\ & \end{array}$$

Decomposition of $\frac{N(s)}{D(s)}$

1. Divide if improper: $\frac{N(s)}{D(s)}$ such that $\text{degree of }N(s)\le \text{degree of }D(s)$ such that $\frac{N(s)}{D(s)}=\text{a polynomial }+\frac{N_1(s)}{D(s)}$
2. Factor Denominator: into factor form $$(ps+q{)}^m\text{ and }(a{s}^2+bs+c{)}^n$$
3. Linear Factors: $(ps+q{)}^m$ such that: $$\sum _{j=1}^m\frac{A_j}{(ps+q{)}^j}$$
4. Quadratic Factors: $(a{s}^2+bs+c{)}^n$ such that $$\sum _{j=1}^n\frac{B_{j}s+C_j}{(a{s}^2+bs+c{)}^j}$$
5. Determine Unknown

Stability analysis using Root of $D(s)$

roots of $D(s)$

roots of $D(s)$ as poles

$$G(s)=\frac{N(s)}{D(s)}=\frac{N(s)}{{\prod }_{j=1}^n(s+p_j)}=\sum _{j=1}^n\frac{A_j}{s+p_j}$$

$p_i$ can be imaginary

Solving for $g(t)$ gives

$$g(t)=\sum _{j=1}^n{\mathcal{L}}^{-1}\{\frac{A_j}{(s+p_j)}\}=\sum _{j=1}^n{A}_j{e}^{-p_{j}t}$$

stability analysis

Important

If ${\sigma }_i>0$ then pole is in the left side of imaginary plane, and system is ==stable ==

Complex root

For poles at $s={\sigma }_i\pm j\omega$ we get

$$\frac{\alpha +j\beta }{s+{\sigma }_i+j{\omega }_i}+\frac{\alpha -j\beta }{s+{\sigma }_i-j{\omega }_i}$$

Wants to be on LHP for time-function associated with $s$ plane to be stable

Impedance of Inductor

$$Z(s)=\frac{V(s)}{I(s)}=Ls$$

since the voltage-current relation for an inductor is $v(t)=L\frac{di(t)}{dt}$

Impedance of Capacitor

$$Z(s)=\frac{V(s)}{I(s)}=\frac{1}{Cs}$$

since the voltage-current relation for a capacitor is $v(t)=\frac{1}{C}{\int }_0^{t}i(\tau )d\tau$