SE380: Introduction to Feedback Control Systems

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notes on learn

Course is about classical control; most prevalent, assumes linear and time invariant plant


Chapter 1 review

Block Diagrams are the basis of control engineering. Each block represents a function (not like free body diagrams)

Review of Equations:

Chapter 2 Modelling

State-space models are mathematical expressions of systems

One key property of linearity is the superposition principle: the net responses caused by multiple stimuli is the sum of the responses caused by each stimuli individually

Linearization is approximating a nonlinear system, by looking at the system's behaviour near an equilibrium configuration (physical state is at rest). We then calculate the four jacobians, A, B, C and D.

Transfer functions exist only for LTI systems, and are expressed in the frequency domain. They tell us how the system responds at different frequencies (which is only possible because we assume time invarance and so the response does not depend on the current time)

Chapter 3 Linear System Theory


For sketching bode plots, the magnitude plot is on a scale of dB, so \(20 \log|G(s)|\) for transfer function G(s). The phase plot is on a scale of degrees. The x axis is on a logarithmic scale. The sketch is called an asymptotic bode plot. We use straight line approximations to draw the plot.

Chapter 4 Prototypes

Prototype first order system equation lets us compare transfer functions to a parametric form, allowing us to easily derive properties: everything depends on time constant \(\tau\)

Prototype second order system is \(G(s) = \frac{K \omega_n}{s^2 + 2 \zeta \omega_n s + \omega_n^2}\)

Chapter 5 Feedback control theory

analyzing existing controllers

In lab 2, we looked at a second order circuit system.

Pole-zero cancellation is if some pole in either the controller C or plant P gets cancelled by the numerator

Chapter 6 Root-Locus

Want to look at how the poles move in the complex plane as parameters vary (in this case K, the gain of the controller)

m poles tend towards the zeroes, but n- m branches tend towards asymptotes (to infinity since they're poles) that form the following patterns:

Chapter 7 PID control

PID controllers are popular for their simplicity. Two classical PID controllers: error feedback and two degrees-of-freedom

Chapter 8 Frequency

Cauchy's Principle of the argument: we take some contour \(\Gamma_s\) and apply some mapping (rational function). As s travels around our contour, the complex numbers G(s) can be visualized \(\Gamma_g\). by taking the angle relative to the zeroes of G, we can plot the change of phase using just \(\Gamma_g\).

How to read \(K_{gm}\) and \(\phi_{pm}\) from a bode plot

Nyqust criterion is based on the priciple of the arguments: a curve in the complex plane and a complex valued function of a complex variable

Chapter 9 Control design in the frequency domain

design specs in the frequency domain, sometimes by converting time domain specs

Procedure for lag controller design, either given steady state specs (tracking or disturbance rejection) or asked to increase phase margin to bbe greater than or equal to \(\phi_{pm}\) desired (from damping ratio/robustness requirements)

  1. FVT, choose K (as a part of the controller) to meet tracking spec.
  2. Draw frequency response bode plot (KP(jw))
  3. If we don't meet robustness spec, find desired gain crossover frequency
  4. Shift gain so that the gain crossover happens at the desired frequency
  5. Put the zero far away from gc frequency, \(\frac{10}{\alpha T} \leq \omega_{gc}^{des}\)

Lead compensation has the same controller block diagram, with \(\alpha > 1\)

  1. ss spec to pick \(\hat{K}\), and boost \(\hat{K}\) by some guess (like 10dB) to account for magnitude distortion
  2. Draw bode plot of frequency response \(\hat{K} P(j \omega)\)
  3. If we don't meet robustness specs, find how much you need to increase the phase margin by, which determines \(\alpha\)
  4. Use given design equations to find \(\alpha\)


// positive if too close, neg if too far
distance_adjustment = r(t) - y(t)
u(t) = -Kp * distance_adjustment

Better algorithm, we can add another term with an integral that effectively keeps a history of choices to better influence the velocity.

Design Cycle

  1. Study the system to be controlled, decide on sensors and actuators. Sensors change what information is at your disposal, and actuators represent choices you can make in response to the inputs.
  2. Model the resulting system
  3. Simplify model if necessary
  4. Analyze the resulting system
  5. Determine specifications: stability, good steady-state behaviour, robustness, good transient performance
  6. Decide on type of controller
  7. Design the controller
  8. Simulate (usually using Matlab)
  9. Return to step 1 if neecessary
  10. Implement controller

e.g. follower is \(u(t) = -K_p (r(kT)-y(kT)), \quad kT \le t \lt (k+1)T\)


  1. Apply known laws to get a system of differential equations
  2. Linearize the model at an operating point to get a system of linear equations
  3. Take the Laplace transform with zero initial conditions to get a system of linear algebraic equations
  4. Isolate input and output to get a transfer function
  5. Experimentally determine parameter values for the transfer function (e.g. weight of your robot)

Applying known laws to get equations

e.g. Spring

\(q \in \mathbb{R}\) is the position of the mass M

\(\dot{q} := \frac{dq}{dt}, \quad \ddot{q} := \frac{d^2q}{dt^2}\)

Assume that \(q=0\) corresponds to the mass location at which the spring is neither stretched nor compressed.

Newton's 2nd law: \(F = ma\) or \(M\ddot{q} = \sum{ \text{forces acting on } M}\)

Force due to spring: \(F_K(q) = Kq\), assumed to be linear

Force due to damper, possibly nonlinear: \(C(\dot{q})\)

Altogether, we get a second order nonlinear ODE:
\[M\ddot{q} = -Kq - C(\dot{q}) + u\]

Note: if the damper is linear (\(C(\dot{q})=bq\)), then the overall system is linear

e.g. Resistor

Non linear resistor just means it's a function that's non linear with respect to the current

\(V_R(t) = h(i(t)), \quad h : \mathbb{R} \rightarrow \mathbb{R}\) is possibly nonlinear

\(u(t)\): applied voltage; \(y(t)\): voltage across capacitor

Apply Kirchoff's Voltage Law, recall that it states that conservation of energy holds for voltage in a circuit:
\[\begin{align} -u(t) + V_R + y &= 0\\ \\ i(t) &= C\frac{dy}{dt} \quad \text{(capacitor equation)}\\ V_R &= h(i(t)) = h(C\dot(y))\\ \\ -u(t) + h(C\dot{y}) + y &= 0\\ \end{align}\]

Note: if the resister were linear (\(h(i) = Ri\)), the whole system would be linear (see 2.3.4 in notes)

Observe that, no matter what you're trying to model, there are governing laws that define modelling in that domain

State models

e.g. Cart

Newton's second law: \(M\ddot{y} = u - D(\dot{y})\)

We put this model into a standard form by defining two state variables:
\[x_1 := y \text{ (position)}, \quad x_2 := \dot{y} \text{ (velocity)}\]

Together, \(x_1\) and \(x_2\) make up the state of the system. We do some rewriting to have a systematic way to do linearization:

\[\begin{align} \dot{x_1} &= x_2 & \text{(state equation)}\\ \dot{x_2} &= \frac{1}{M}u - \frac{1}{M}D(x_2) & \text{(state equation)}\\ y &= x_1 & \text{(output equation)}\\ \end{align}\]

Together, these equations make up the state-space model. These equations have the general form:

\[\dot{x} = f(x,u) \text{ where } y = h(x)\]

In this example:

\[\begin{align} X &= (x_1, x_2) \in \mathbb{R}^2\\ f(x,u) &= \begin{bmatrix} x_2 \\ \frac{1}{M}u - \frac{1}{M}D(x_2) \end{bmatrix}\\ \end{align}\]

When they're non-linear we can't say much, but in the special case where air resistance is a linear function of \(x_2\) (\(D(x_2)=d x_2\)), then \(f(x,u)\) becomes a linear function of \(x\) and \(u\):

\[f(x,u) = \begin{bmatrix} 0 & 1 \\ 0 & \frac{-d}{M} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix} 0 \\ \frac{1}{M} \end{bmatrix} u \]

Define \(C = \begin{bmatrix}1 & 0\end{bmatrix}\). In the linear case, we get:
\[\begin{align} \dot{x} &= Ax + Bu\\ y &= Cx\\ \end{align}\]

This is a linear, time-invariant (LTI) model.

Expect at least one question like this on the midterm


Generalizing, an important class of systems have models of the form:

\[\begin{align} \dot{x} &= f(x, u), & f: \mathbb{R}^n \times \mathbb{R}^m \rightarrow \mathbb{R}^n\\ y &= h(x, u), & h: \mathbb{R}^n \times \mathbb{R}^m \rightarrow \mathbb{R}^p\\ \end{align}\]

The linear special case is:
\[\begin{align} \dot{x} &= Ax + Bu, & A \in \mathbb{R}^{n \times m}\\ y &= Cx + Du, & C \in \mathbb{R}^{p \times m}\\ \end{align}\]

In this course, we look at single-input, single-output systems: \(m=p=1\).

e.g. more carts

\[\begin{align} m&=2, & u &= (u_1, u_2)\\ p&=2, & y &= (y_1, y_2)\\ m&=4, & x &= (x_1, x_2, x_3, x_4) := (y_1, \dot{y_1}, y_2, \dot{y_2})\\ \end{align}\]

e.g. drones

\[\begin{align} u &= (y_1, u_2, u_3, u_4)= (f_1, f_2, f_3, f_4)\\ p&=3\\ y&=(y_1, y_2, y_3)\\ x &\in \mathbb{R}^{12} = (\text{position}, \text{orientation}, \text{velocity}, \text{angular velocity})\\ \end{align}\]

Determining state

What is the state of a system?

The state vector \(x(t_0)\) encapsulates all of the system's dynamics up to time \(t_0\).

More formally: For any two times \(t_0 \lt t\), knowing \(x(t_0)\) and knowing \(\{u(t) : t_0 \le t \lt t_1\}\), we can compute \(x(t_1)\) and \(y(t_1)\).

e.g. cart with no forces

We know \(M\ddot{y} = 0\) from Newton's laws.

If we try a 1-dimensional state, say \(x := y\), then knowing \(x(t_0)\) without knowing \(\dot{y}\) is not enough information to find the position in the future, \(x(t)\) for \(t \gt t_0\). We have the same problem if we define \(x = \dot{y}\).

Since the governing equation is second order, we need two initial conditions. So, \(x=(y, \dot{y}) \in \mathbb{r}^2\) is a good choice.

In general, it is a good idea to use:

e.g. 2.5.2: Pendulum

Model: \(\ddot{\theta} = \frac{3}{Ml^2} u - 3\frac{g}{l} \sin(\theta)\)

In state space form:
\[\begin{align} x &= (\theta, \dot{\theta})\\ y &= \text{angular position} = \theta\\ \\ \dot{x}&=f(x,u) \quad & & \dot{x_1}=x_2\\ \dot{y}&=h(x,u) \quad & & \dot{x_2}=\frac{3}{Ml^2}u - 3\frac{s}{l} \sin(x_1)\\ & & & y = x_1 \end{align}\]

This is nonlinear due to the sine term.

e.g. 2.4.5 Circuit

\[\begin{align} x_1 &:= \text{voltage across capacitor} = \frac{1}{C} \int_{0}^{t}{y(\tau)d\tau}\\ x_2 &:= \text{current through inductor} = y\\ \\ -u + V_R + V_C + V_L &= 0\\ \Rightarrow -u + Rx_2 + x_1 + L\dot{x_2} &= 0, \\ \\ \dot{x_1} &= \frac{1}{C}x_2(t)\\ \dot{x_2} &= \frac{-1}{L}x_1 - \frac{R}{L}x_2 + \frac{1}{L}u\\ y &= x_2\\ \\ \dot{x} &= \begin{bmatrix} 0 & \frac{1}{C}\\ \frac{-1}{2} & \frac{-R}{L}\end{bmatrix} \begin{bmatrix}x_1\\ x_2\end{bmatrix} + \begin{bmatrix}0 \\ \frac{1}{L}\end{bmatrix} u\\ \dot{x} &= \begin{bmatrix} 0 & \frac{1}{C}\\ \frac{-1}{2} & \frac{-R}{L}\end{bmatrix} \begin{bmatrix}x_1\\ x_2\end{bmatrix} + \begin{bmatrix}0 \\ \frac{1}{L}\end{bmatrix} u\\ y &= \begin{bmatrix}0 & 1\end{bmatrix}\\ \end{align}\]

recall that capacitors hold charge, that converges over time:

recall that inductors impede/release current over time:



This is the process of aproximating a nonlinear state-space model with a linear model.

e.g. 2.5.2

Linearize \(y=x^3\) at the point \(\bar{x}=1\).

Let \(\bar{y} := f(\bar{x}) = 1^3 = 1\)

Taylor series at \(x=\bar{x}\) is:
\[y=\sum_{n=0}^\infty c_n (x-\bar{x})^n, \quad c_n = \frac{1}{n!} \frac{d^n f(x)}{dx^n} \biggr|_{x=\bar{x}}\]

\[\begin{align} f(x) &= f(\bar{x}) + \frac{df(x)}{dx}\biggr|_{x-\bar{x}} (x-\bar{x}) + \text{higher order terms}\\ \text{Keep only the terms $n=0$ and $n=1$:}\\ f(x) &\approx f(\bar{x}) + \frac{df(x)}{dx}\biggr|_{x-\bar{x}} (x-\bar{x})\\ y - \bar{y} &\approx + \frac{df(x)}{dx}\biggr|_{x-\bar{x}} (x-\bar{x})\\ \end{align}\]

If we define the derivations \(\partial y := y - \bar{y}, \partial x := x - \bar{x}\), then \(\partial y = \frac{df}{dx} \bigg|_{x=\bar{x}} \partial x\), i.e. \(\partial y = 3 \partial x\)

e.g. 2.5.3

\[y = \begin{bmatrix}y_1 \\ y_2\end{bmatrix} = f(x) = \begin{bmatrix}x_1 x_2 - 1 \\ x_3^2 - 2x_1 x_3\end{bmatrix} =: \begin{bmatrix}f_1(x) \\ f_2(x)\end{bmatrix}\]

Linearize at \(\bar{x}=(1, -1, 2)\).


Multivariable Taylor series

\[d(x)=f(\bar{x})+\frac{\partial f}{\partial x} \biggr|_{x=\bar{x}} (x-\bar{x}) \text{ + higher order terms}\]

The Jacobian of \(f\) at \(\bar{x}\) is:
\[\begin{align} \frac{\partial f}{\partial x} \biggr|_{x = \bar{x}} &= \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & \frac{\partial f_1}{\partial x_3} \\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} & \frac{\partial f_2}{\partial x_3} \\ \frac{\partial f_3}{\partial x_1} & \frac{\partial f_3}{\partial x_2} & \frac{\partial f_3}{\partial x_3} \\ \end{bmatrix}\\ &= \begin{bmatrix} x_2 & x_1 & 0 \\ -2x_3 & 0 & 2x_3-2x_1 \\ \end{bmatrix}_{x=(1, -1, 2)}\\ &= \begin{bmatrix} -1 & 1 & 0\\ -4 & 0 & 2 \end{bmatrix}\\ &= A \end{align}\]

i.e. \(y-\bar{y} \approx A(x-\bar{x})\)

By extension, near \((x,u)=(\bar{x}, \bar{u})\):
\[f(x, u) \approx f(\bar{x}, \bar{u})+ \frac{\partial f}{\partial x} \biggr|_{(x,u)=(\bar{x}, \bar{u})} (x-\bar{x}) + \frac{\partial f}{\partial u} \biggr|_{(x,u)=(\bar{x}, \bar{u})} (u-\bar{u})\]

Let's apply this to \(\dot{x}=f(x,u), y=h(x,u)\), a dynamic system; non-linear state space model

Definition: A constant pair \((\bar{x}, \bar{u}) \in \mathbb{R}^n \times \mathbb{R}^m\) is an equilibrium configuration of the system \(\dot{x}=f(x,u), y=h(x,u)\) if \(f(\bar{x}, \bar{u})=(0,...,0)\). The constant \(\bar{x}\) is the equilibrium point.

2.5.4 e.g. Find all the equilirium configurations at which the pendulum is upright

\[\begin{align} \dot{x} &= f(x,u)\\ y &= h(x,u)\\ x_1 &= \theta\\ x_2 &= \dot{\theta}\\ f(x,u) &= \begin{bmatrix}x_2\\ \frac{3}{Ml}u-\frac{3g}{l}\sin(x_1)\end{bmatrix}\\ h(x) &= x_1 \end{align}\]

If \(y=\pi\) (upright), then \(\bar{x_1}=\pi\). So we have to solve:
\[ \begin{bmatrix}0\\0\end{bmatrix} = \begin{bmatrix} \bar{x_2}\\ \frac{3\bar{u}}{Ml} - \frac{3g}{l}\sin(\bar{x_1}) \end{bmatrix} \Rightarrow \bar{x_2}=0, \quad \bar{u}=0 \]

Therefore the equilibria are:
\[ \begin{bmatrix}\bar{x_1}\\\bar{x_2}\end{bmatrix} = \begin{bmatrix}\pi + 2\pi k \\ 0\end{bmatrix}, \quad \bar{u}=0 \]

Assume that \(\dot{x}=f(x,u)\) (the nonlinear state-space model) has an equilibrium configuration at \((x,u)=(\bar{x}, \bar{u})\). Forgetting about the higher order terms due to linearization,

\[ f(x,u) \approx \underbrace{f(\bar{x}, \bar{u})}_{=0} + \underbrace{\frac{\partial f}{\partial x} \biggr|_{(x,u)=(\bar{x}, \bar{u})} (x - \bar{x})}_{=: A} + \underbrace{\frac{\partial f}{\partial u} \biggr|_{(x,u)=(\bar{x}, \bar{u})} (u - \bar{u})}_{=:B}\]

Consider deviations from \((\bar{x}, \bar{u})\), where \(||\partial x||, ||\partial u||\) are assumed to be small:
\[\begin{align} \partial x(t) &:= x(t) - \bar{x}\\ \partial u(t) &:= u(t) - \bar{u}\\ \end{align}\]

Then we get linearized state equations:
\[\dot{\partial x} = \dot{x} - 0 = f(x,u) \approx A\partial x + B \partial u\\ \dot{\partial x} = A\partial x + B\partial u\]

Linearized output equation:

\[\partial y = \underbrace{\frac{\partial h}{\partial x} \biggr|_{(x,u)=(\bar{x},\bar{u})} \partial x}_{=:C} + \underbrace{\frac{\partial h}{\partial u} \biggr|_{(x,u)=(\bar{x},\bar{u})} \partial u}_{=:D}\\ \partial y := y-\bar{y}=y-h(\bar{x},\bar{u})\]


Linearizing \(\dot{x}=f(x,u)\) and \(y=h(x,u)\):

  1. Select an equilibrium configuration \((\bar{x}, \bar{u}) \in \mathbb{R}^n \times \mathbb{R}^m\): \(\bar{y}=h(\bar{x},\bar{u}), f(\bar{x},\bar{u})=0\), where all derivatives vanish
  2. Compute Jacobians of \(f,h\) to get \(A,B,C,D\), matrices have to be constants, no x's or u's
  3. Linearization: \(\dot{\partial x} = A \partial + B \partial u, \partial y = C \partial x + D \partial u\)

We do this to simplify our mathematical model, by finding a good linear approximation at some point, allowing us to solve things about non-linear systems

e.g. 2.5.5 (Pendulum)

  1. equalibrium config is \(\bar{x} = (\pi, 0), \bar{u} = 0\)
  2. $ A = \frac{\delta f}{\delta x}, B = (0, \frac{3}{M l^2}), C = (1, 0), D = 0$
  3. Linearize model. \(\dot{\delta x}\)

Note that in practice, our system is still non-linear, so we need to bring it close to equilibrium configuration for our controller to be effective

2.8 Transfer Functions

u(t) -> |LTI system| -> y(t)

Transfer function of the system is the ratio \(\frac{Y(s)}{U(s)}\) where all Laplace transforms are taken with zero initial conditions

example: Recall the mass spring damper

\(M\ddot{q} = u - kq - c(\dot{q})\)

If the damper is nonlinear, then this system doesn't have a transfer function. If \(C \dot{q} = b \dot{q}\), b is a constant, then taking the Laplace transform of the differential equation:

\(\begin{align} s^2 M Q(s) &= U(s) - K Q(s) - s b Q(s) \\ \frac{Q(s)}{U(s)} &= \frac{1}{s^2 M + b s + K} \end{align}\)

Block Diagrams

full table:

Let \(\mathbb{R}(s)\) be the set of all real rational transfer functions.

A complex number \(x \in \mathbb{C}\) is a zero of \(G(s)\) if \(\lim_{s \rightarrow x} |G(s)| = 0\).

2.8.1 Obtaining a TF from a state model

2.8.6 Linearized pendulum

\[\begin{align} \bar{x} &= \begin{bmatrix}\pi\\0\end{bmatrix}, \quad \bar{u}=0\\ \\ \partial \dot{x} &= \begin{bmatrix} 0 & 1 \\ \frac{3g}{l} & 0 \end{bmatrix} \partial x + \begin{bmatrix}0 \\ \frac{3}{Ml^2}\end{bmatrix} \partial u\\ \partial y &= \begin{bmatrix}1 & 0\end{bmatrix} \partial x\\ \\ G(s) &= C(sI-A)^{-1}B + D, \quad (sI-A)^{-1} = \frac{\text{adj}(sI-A) \text{ <-- n x n}}{\det(sI-A) \text{ <--- polynomial}}\\ &= \begin{bmatrix}1 & 0\end{bmatrix}\begin{bmatrix}s & -1 \\ \frac{-3g}{l} & s\end{bmatrix}^{-1}\begin{bmatrix}0 \\ \frac{3}{Ml^2}\end{bmatrix}\\ &= \begin{bmatrix}1 & 0\end{bmatrix}\begin{bmatrix}s & \frac{-3g}{l} \\ 1 & s\end{bmatrix}^T\begin{bmatrix}0 \\ \frac{3}{Ml^2}\end{bmatrix}\\ &= \frac{\frac{s}{Ml^2}}{s^2 - \frac{3g}{l}}\\ \end{align}\]

2.9 Block diagram manipulations

  U(s) -->| G(s) |--> Y(s)      Y(s) = G(s)U(s)

        +->| G(s) |--+
        |  +------+  |
  U(s) -+            +--> Y(s)   Y(s) = (G(s) + H(s))U(s)
        |  +------+  |
        +->| H(s) |--+

                  D(s)                | 1/G |
                    |                 +-----+   
          +------+  v                     |   +------+
  U(s) -->| G(s) |--o--> Y(s)   =  U(s) --o-->| G(s) |   Y(s) = D(s) + G(s)U(s)
          +------+                            +------+

  U(s) --o-->| G(s) |--+--> Y(s)      Y(s) = G(s)U(s)/(1 + G(s)H(s))
         ^-  +------+  |
         |             |
         |   +------+  |
         +---| H(s) |--+

We have the tools to simplify block diagrams now, simplifying series/parallel blocks and rearranging them relative to summing junctions. So some strategies we can use:

  1. mathematically; write equation for Y(s) and rearrange using our laws
  2. re-arrange blocks to reveal common configurations listed above
  3. systematic method below, for complicated diagrams

Systematic method of finding transfer functions

  1. Introduce new variables \(\{v_1, v_2, ... \}\) at the output of every summer
  2. Write expressions for inputs of summers in terms of \(\{u, y, v_1, v_2, ...\}\)
  3. Write equations for each summer and \(y\)
  4. Eliminate \(\{v_1, v_2, ...\}\) from equations


\[\begin{align} y&=G_3G_2v_2\\ v_2 &= H_2y+G_1v_1-H_2G_2v_1\\ v_1&=u-H_1G_2v_2\\ \\ \begin{bmatrix}1 & H_1 G_2 & 0 \\ -G_1 & 1+H_2G_2 & -H_3 \\ 0 & -G_3G_2 & 1\end{bmatrix} \begin{bmatrix}v_1 \\ v_2 \\ y\end{bmatrix} &= \begin{bmatrix}u \\ 0 \\ 0\end{bmatrix}\\ \\ \text{since # of equations = # of variables, we can apply}&\text{ cramer's rule } Ax = b, x_i = \frac{det(A_i)}{det(A)}\\ \text{where }A_i \text{ is the matrix formed by replacing the}&\text{ i-th column of A by the column vector b}\\ \text{By cramer's rule:}\\ Y(s) - \frac{ \det\begin{bmatrix}1 & H_1G_2 & u \\ -G_1 & 1+H_2G_2 & 0 \\ 0 & -G_2G_3 & 0\end{bmatrix} }{ \det\begin{bmatrix}1 & H_1G_2 & 0 \\ -G_1 & 1+H_2G_2 & -H_3 \\ 0 & -G_3G_2 & 1\end{bmatrix} } &= \frac{G_1 G_2 G_3}{1 + H_1 H_2 G_2 - H_3 G_3 G_2 + G_1 H_1 G_2} U(s)\\ \end{align}\]

Chapter 4: First and Second Order Systems

The order of a dynamic system is the order of the highest derivative of its governing differential equation.

First order

\[\tau \dot{y} = ku\]
\[\frac{Y(s)}{U(s)} = \frac{K}{\tau s + 1}\]
\[\begin{align} \dot{x}&=\frac{-x}{\tau} + \frac{K}{\tau} u\\ y &= x \end{align}\]


Step response:
\[y(t) = \mathcal{L}^{-1}\{G(s)U(s)\} = \mathcal{L}^{-1}\left\{\frac{K}{\tau s + 1} \frac{1}{s}\right\} = \mathcal{L}^{-1}\left\{\frac{K}{s} - \frac{K}{s + \frac{1}{\tau}}\right\}\]
\[K(1-e^{\frac{-t}{\tau}}), \quad t \ge 0\]


  1. After \(4\tau\) seconds, \(y(t)\) is within 2% of its steady-state value
  2. For all \(t \gt 0\), \(y(t) \lt y_{ss}\) (no overshoot)
  3. \(y(t)\) increases monotonically (no oscillations)
  4. Decrease in \(\tau \Leftrightarrow\) step response gets faster \(\Leftrightarrow\) bandwidth goes up \(\Leftrightarrow\) poles move to the left

Second order

\[\ddot{y}+2\zeta\omega_n \dot{y} + \omega_n^2 y = K\omega_n^2 u\]
\[ \frac{Y(s)}{U(s)} = \frac{K\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}\]
\[\begin{align} \dot{x} &= \begin{bmatrix}0&1\\-\omega_n^2&-2\zeta\omega_n\end{bmatrix}x + \begin{bmatrix}0\\K\omega_n^2\end{bmatrix}u\\ y &= \begin{bmatrix} 1 & 0\end{bmatrix}x\\ \end{align}\]


\[\begin{align} M \ddot{q} &= u - K_{spring} q - b\dot{q}\\ \frac{Y(s)}{U(s)} &= \frac{\frac{1}{M}}{s^2 + \frac{b}{M}s + \frac{K_{spring}}{M}}\\ \\ \omega_n &= \sqrt{\frac{K_{spring}}{M}}\\ \zeta &= \frac{b}{2 \sqrt{K_{spring}M}}\\ K &= \frac{1}{K_{spring}} \end{align}\]

Pole locations

From the quadratic formula, find the zeroes of the denominator:
\[s = -\zeta \omega_n \pm \omega_n \sqrt{\zeta^2 - 1} = \omega_n\left(-\zeta \pm \sqrt{\zeta^2 - 1}\right)\]

Pole locations are used to categorize the system:

Steady-state gain: \(K\)
Zeroes: none

Step response

Underdamped Systems

Impulse response

\[g(t) = K\frac{\omega_n}{\sqrt{1-\zeta^2}} \underbrace{e^{-\zeta \omega_n t}}_\text{decay rate} \sin\underbrace{\left(\omega_n \sqrt{1-\zeta^2} t\right)}_\text{oscillation rate}, \quad t \ge 0\]

Observe: If we fix \(\zeta \in (0,1)\), then larger bandwidth \(\Leftrightarrow\) faster decay

Step response

\[\begin{align} u(t) &= 1(t)\\ \Rightarrow U(s) &= \frac{1}{s}\\ \\ Y(s) &= G(s)U(s)\\ \Rightarrow y(t) &= \mathcal{L}^{-1}{G \dot U}\\ &= K\left(1 - \frac{1}{\sqrt{1 - \zeta^2}} e^{-\zeta \omega_n t} \sin\left(\omega_n \sqrt{1 - \zeta^2}t + \theta\right)\right), \quad \theta = \arccos \zeta\\ \end{align}\]


General characteristics of step response


e.g. mass-spring damper

\[\frac{Y(s)}{U(s)} = \frac{\frac{1}{M}}{s^2 + \frac{b}{M}s + \frac{K_{spring}}{M}}\]

\[\begin{align} \zeta &= \frac{b}{2\sqrt{MK_{spring}}}\\ \zeta &\ge \frac{-\ln(\%OS_{max})}{\sqrt{\pi^2 + (\ln\%OS_{max})^2}} =:\zeta_{min}\\ \\ \text{To meet specs:}\\ \frac{b}{2\sqrt{MK_{spring}}} \ge 0.6901\\ \end{align}\]

The angle the poles make is \(\pm(\pi - \arccos \zeta)\).
\[ \zeta \ge \zeta_{min} \Leftrightarrow \theta \le \arccos(\zeta_{min})\]
In this example, \(\theta \le 46^{\circ}\)

Therefore the overshoot spec is not met if there are poles in the shaded region.

Settling time

e.g. Mass-spring damper

Find the condition so that \(T_s \le T_s^{max} = 3\)

\[\begin{align} G(s) &= \frac{\frac{1}{M}}{s^2 + \frac{b}{M}s + \frac{K_{spring}}{M}}\\ \frac{4}{\zeta \omega_n} &\le 3 = T_s^{max}\\ \Leftrightarrow \zeta\omega_n &\ge \frac{4}{T_s^{max}}\\ \Leftrightarrow \frac{b}{2M} &\ge \frac{4}{3}\\ \end{align}\]


e.g. mass-spring damper again

Spec: \(T_p \le T_p^{max} = 3 \text{ seconds }\)

\[\begin{align} \omega_n\sqrt{1-\zeta^2} &\le \frac{\pi}{T_p^{max}}\\ \Leftrightarrow \sqrt{\frac{K_{spring}}{M} - \frac{b^2}{4M^2}} &\le \frac{\pi}{3}\\ \end{align}\]

Changes to poles

decrease real part increase imaginary part angle of poles to \(\pm \pi\) increase magnitude
\(\omega_n\) + + no change +
\(\zeta\) + - + no change
%OS - + - no change
\(T_s\) - no change - -
\(T_p\) no change - + -

4 State Response

\[\begin{align} \dot{x}&=Ax\\ x &\in \mathbb{R}^n\\ A &\in \mathbb{R}^{n \times n}\\ x(0) &= x_0 \in \mathbb{R}^n \text{ (initial condition) }\\ \end{align}\]


  1. When \(n=1\) (A is scalar), the solution is \(x(t)=e^{tA}x_0\)
  2. Taylor series expansion of \(e^{At}=1 + At + \frac{(At)^2}{2!} + ...\)

Motivated by 1 and 2, define the matrix exponential:
\[e^A := I + A + \frac{A^2}{2!} + ...\]

e.g. 3.1.1

\[\begin{align} A &= \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}\\ \Rightarrow e^A &= I + 0 + 0 + ...\\ &= \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\\ \end{align}\]

e.g. 3.1.2

\[\begin{align} A &= \begin{bmatrix}1 & 0 \\ 0 & 2\end{bmatrix}\\ \text{For a diagonal matrix:}\\ A^k &= \begin{bmatrix}1^k & 0 \\ 0 & 2^k\end{bmatrix}\\ \Rightarrow e^A &= I + \begin{bmatrix}1 & 0 \\ 0 & 2\end{bmatrix} + \begin{bmatrix}1^2 & 0 \\ 0 & 2^2\end{bmatrix} + ...\\ &= \begin{bmatrix}e^1 & 0 \\ 0 & e^2\end{bmatrix}\\ \end{align}\]

e.g. 3.1.3

\[ A = \begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{bmatrix}\\ \]
Check that \(A^3=0\) (i.e. \(A\) is nilpotent).
\[ e^A = I + A + \frac{A^2}{2} = \begin{bmatrix}1 & 1 & \frac{1}{2} \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}\\ \]

Replace \(A\) with \(tA\) to get a function of time:
\[e^{At} = I + tA + \frac{t^2A^2}{2!} + ...\]

Theorem: The unique solution to \(\dot{x}=Ax, \quad x(0)=x_0\) is \(x(t)=e^{tA}x_0\).

Using Laplace

Take the Laplace transform of \(\dot{x}=Ax\) without assuming \(x(0)=0\):
\[\begin{align} sX(s) - x(0) &= AX(s)\\ X(s) &= (sI-A)^{-1} x(0)\\ \end{align}\]

Conclusion: \(e^{At}\) and \((sI-A)^{-1}\) are Laplace transform pairs.


\[\begin{align} A &= \begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}\\ sI-A &= \begin{bmatrix}s&-1&0\\0&s&-1\\0&0&s\end{bmatrix}\\ (sI-A)^{-1} &= \frac{\text{adj}(sI-A)}{\det(sI-A)}\\ &= \frac{1}{s^3} \begin{bmatrix}s^2&s&1\\0&s^2&s\\0&0&s^2\end{bmatrix}\\ e^{tA} &= \mathcal{L}^{-1} \left\{ (sI-A)^{-1} \right\}\\ &= \begin{bmatrix}1&t&\frac{t^2}{2}\\0&1&t\\0&0&1\end{bmatrix}, \quad t \ge 0\\ \end{align}\]

Total Response

The solution of \(\dot{x}=Ax+Bu\), \(y=Cx+Du\), \(x(0)=x_0\) is:
\[x(t) = \underbrace{e^{At}x_0}_\text{initial state response} + \underbrace{\int_0^t e^{A(t-\tau)} Bu(\tau)d\tau}_\text{forced response}\]

In SISO (single-input-single-output) special case where \(x(0)=0\), we get the familiar result:
\[\begin{align} y(t)&=(g * u)(t) = \int_0^t g(t-\tau)u(\tau)d\tau\\ g(t)&=Ce^{At}B 1(t) + D\delta(t) \end{align}\]
Where \(1(t)\) is the unit step function and \(\delta(t)\) is the unit impulse.

Stability of state-space models

The system \(\dot{x}=Ax\) is asymptotically stable if \(x(t) \rightarrow 0\) for any initial condition.

\(e^{At} \rightarrow 0\) if and only if all the eigenvalues of \(A\) have a negative real part.

e.g. 3.4.2

\[\begin{align} M\ddot{q}&=u-Kq \quad \text{(mass-spring)}\\ x &= \begin{bmatrix}x_1\\x_2\end{bmatrix} := \begin{bmatrix}q\\\dot{q}\end{bmatrix}\\ \dot{x} &= \begin{bmatrix}0&1\\\frac{-k}{M}&0\end{bmatrix}x + \begin{bmatrix}0\\\frac{1}{M}\end{bmatrix}u\\ \\ \text{Using } M=1, k=4:\\ e^{At} &= \begin{bmatrix}\cos 2t&\frac{1}{2}\sin 2t\\-2\sin 2t & \cos 2t\end{bmatrix}\\ \end{align}\]

Since \(e^{At}\) does not approach 0 as \(t\) grows large, the system is not asymptotically stable.

Let's double-check this using the eigenvalues. Solve for \(s\) such that \(\det(sI-A)=0\).
\[\begin{align} A &= \begin{bmatrix}0 & 1 \\ -4 & 0 \end{bmatrix}\\ \det \begin{bmatrix}s & -1 \\ 4 & s\end{bmatrix} &= 0\\ s^2 + 4 &= 0\\ s &= \pm 2j \end{align}\]
The system is therefore not asymptotically stable since it has at least one eigenvalue (in this case, it has two) with a non-negative real part.

If we introduce friction:
\[\begin{align} \ddot{q} &= u-4q-\dot{q} \end{align}\]

Check that it is asymptotically stable (it should be)

Bounded input, bounded output stability

\[ Y(s)=G(s)U(s) \quad \text{or} \quad y(t)=(g*u)(t), g(t)=\mathcal{L}^{-1}\left\{G(s)\right\}\\ \]

A signal \(u(t)\) is bounded if there exists a constant \(b\) such that, for all \(t \ge 0\), \(|u(t)| \le b\).

For example, \(\sin t\) is bounded by \(b=1\).

If \(u\) is bounded, \(||u||_\infty\) denotes the least upper bound. For example, for \(\sin t\), then \(|u(t)| \le 10\) and \(u\) is bounded.

A linear, time-independent system is BIBO stable if every bounded input produces a bounded output. \(||u||_\infty\) is finite \(\Rightarrow ||y||_\infty\) is finite.

e.g. 3.5.1

\(G(s)=\frac{1}{s+2}\). The impulse response is \(g(t)=\mathcal{L}^{-1}\{G(s)\} = e^{-2t}\). Then:

\[\begin{align} y(t) &=(g * u)(t)\\ &= \int_0^t e^{-2\tau} u(t-\tau) d\tau\\ \forall t \ge 0:\\ |y(t)|&=\left|\int_0^t e^{-2\tau} u(t-\tau) d\tau\right|\\ &\le \int_0^t \left|e^{-2\tau} u(t-\tau) \right| d\tau\\ &\le \int_0^t e^{-2\tau} d\tau ||u||_\infty\\ &\le \int_0^\infty e^{-2\tau} d\tau ||u||_\infty\\ &= \frac{1}{2} ||u||_\infty\\ \\ ||y||_\infty &\le \frac{1}{2} ||u||_\infty\\ \therefore \text{ system is BIBO stable. } \end{align}\]

BIBO and poles

Theorem 3.5.4: Assume that \(G(s)\) is rational and strictly proper. Then the following are equivalent:

  1. \(G\) is BIBO stable.
  2. the impulse response \(g(t)=\mathcal{L}^{-1}\{G(s)\}\) is absolutely integrable: \(\int_0^\infty |g(\tau)| d\tau \lt \infty\)
  3. Every pole of \(G\) has a negative real part

For example, \(\frac{1}{s+1}, \frac{1}{(s+3)^2}, \frac{s-1}{s^2+5s+6}\) are all BIBO stable because their poles have a negative real part.

On the other hand, take \(\frac{1}{s}, \frac{1}{s-1}\). These are all BIBO unstable because they have poles which do not have a negative real part. The function \(\frac{1}{s}\) is an integrator, so when you give it a constant function as an input, the output will be a ramp, which is unbounded.

Theorem 3.5.5: If \(G(s)\) is rational and improper (the degree of the numerator is greater than the degree of the denominator), then \(G\) is not BIBO stable.

Stability of state-space models and BIBO stability

\[\begin{align} \dot{x} &= Ax+Bu\\ y *= Cx+Du\\ \Rightarrow Y(s) &= \left(C(SI-A)^{-1}B + D\right)U(s)\\ &= \left(C\frac{\text{adj}(sI-A)}{\det(sI-A)}B + D\right)U(s)\\ \end{align}\]
This is BIBO stable if all poles in \(\Re(s) \lt 0\)
This is asymptotically stable if all eigenvalues of \(A \in \Re(s) \lt 0\)
Eigenvalues of \(A\) = roots of \(\det(sI-A) \supseteq\) poles of \(G(s)=C(sI-A)^{-1}B+D\)

e.g. 3.5.5: mass-spring

\[\begin{align} \dot{x} &= \begin{bmatrix}0&1\\-4&0\end{bmatrix}x + \begin{bmatrix}0\\1\end{bmatrix}u\\ y &= \begin{bmatrix}1 & 0\end{bmatrix} x \end{align}\]

Eigenvalues of \(A\) are \(\pm 2j \Rightarrow\) the system is not asymptotically stable.

\[\begin{align} \frac{Y(s)}{U(s)} &= \begin{bmatrix}1&0\end{bmatrix}\begin{bmatrix}s&1\\4&s\end{bmatrix}^{-1}\begin{bmatrix}0\\1\end{bmatrix}\\ &=\frac{1}{s^2+4}\\ &=G(s) \end{align}\]
The system is not BIBO stable based on its poles.

In this example, \(C\text{adj}(sI-A)B=1\) and \(\det(sI-A)=s^2+4\) are coprime, so eigenvalues of \(A\) are the poles of \(G\).

Steady-state gain

Apply a constant \(b\) as input to a system. When we observe the output, the steady-state gain of a transfer function \(G(s)\) is \(\frac{Y_{ss}}{b}\).

Final Value Theorem (3.6.1): Given \(F(s)=\mathcal{L}\{f(t)\}\), where \(F(s)\) is rational:

  1. If \(F(s)\) has all of its poles in \(\Re(s) \lt 0\), then \(\lim_{t \rightarrow \infty} f(t) = 0\).
  2. If \(sF(s)\) has all poles in \(\Re(s) \lt 0\), then \(\lim_{t \rightarrow \infty} f(t) = \lim_{s \rightarrow 0} sF(s)\)
  3. If \(sF(s)\) has even one pole with \(\Re(s) \ge 0\), then \(f(t)\) does not converge.

For example: \(F(s) = \frac{1}{s^2}, sF(s) = \frac{1}{s}\). \(f(t)=t\), which does not converge.


\(f(t)\) \(\lim_{t \Rightarrow \infty} f(t)\) \(F(s)\) \(\lim_{s \rightarrow 0} sF(s)\) FVT case
\(e^{-t}\) 0 \(\frac{1}{s+1}\) 0 1 or 2
\(1(t)\) 1 \(\frac{1}{s}\) 1 2
\(t\) \(\infty\) \(\frac{1}{s^2}\) \(\infty\) 3
\(te^{-t}\) 0 \(\frac{1}{(s+1)^2}\) 0 1 or 2
\(e^t\) \(\infty\) \(\frac{1}{s-1}\) 0 3
\(\cos{\omega t}\) N/A \(\frac{s}{s^2 + \omega^2}\) 0 3

Theorem 3.6.2: If \(G(s)\) is BIBO stable and we input \(u(t)=b1(t)\), then the steady state gain \(y_{ss}=bG(0)\). This can be proven using the final value theorem.

This is sto say, steady-state gain is always \(\frac{y_{ss}}{b}=G(0)\) for any \(b\).

e.g. Set-point control

\(\dot{x}=-2x+u\), \(y=x\). This gives the transfer function \(Y(s)=\frac{1}{d+2}U(s)\). Given a constant reference \(r(t)=r_o 1(t)\) where \(r_0\) constant, find a control signal \(u\) to make \(y\) go to \(r\).

We want \(\lim_{t \rightarrow \infty} y(t) = r_0\).

Try open loop:

\[\begin{align} y_{ss} &= \lim_{t \rightarrow \infty} y(t)\\ &=^? \lim_{s\rightarrow 0} sC(s)R(s) \\ &= \lim_{s\rightarrow 0} C(s) \frac{1}{s+2} r_0\\ \end{align}\]

If \(C(s)\) is BIBO stable, then \(y_{ss} = \lim_{s \rightarrow 0} C_s \frac{r_0}{s+2} = C(0) \frac{1}{2} r_0\). So, \(y_{ss} = r_0 \Leftrightarrow C(0) = 2 = \frac{1}{P(0)}\).

The simplest choice is \(C(s) = \frac{1}{P(0)} = 2\), a proportional controller.

Frequency response

\(Y(s)=G(s)U(s)\) or \(y(t)=(g*u)(t)\). Assume \(G\) is BIBO stable, and the input signal \(u\) is a sinusoid: \(u(t) = \cos(\omega t)\). The period is \(\frac{2\pi}{\omega}\).

Theorem 3.7.1: Assuming \(G\) is rational and BIBO stable, then if \(u(t)=\cos(\omega t)\), then the steady-state output is \(y(t) = A\cos(\omega t + \phi)\). \(A=|G(j\omega)|\), and \(\phi = \angle G(j\omega)\)


\(\dot{x}=-10x+u\), \(y=x\). Then \(Y(s) = \frac{1}{s+10}U(s) =: G(s)U(s)\). If \(u(t)=2\cos(3t+ \frac{\pi}{6})\), what is the steady-state output?

\(A=|G(3j)|=\left|\frac{1}{3j+10}\right|\approx 0.1\)
\(\phi = \angle G(3j) = \angle \frac{1}{3j+10} = \angle1 - \angle(3j+10) \approx 0.2915\)

Therefore \(y(t)=0.2\cos(3t+\frac{\pi}{6}-0.2915)\).

Definition 3.7.2. If \(G(s) \in \mathbb{R}(s)\) and is BIBO stable, then:

  1. The function \(\mathbb{R} \rightarrow \mathbb{C}, \quad \omega \mapsto G(j\omega)\) is the frequency response
  2. The function \(\mathbb{R} \rightarrow \mathbb{R}, \quad \omega \mapsto |G(j\omega)|\) is the magnitude response
  3. The function \(\mathbb{R} \rightarrow (-\pi, \pi], \quad \omega \mapsto \angle G(j\omega)\) is the phase response

Graphical representations of frequency response

Bode plots

  1. Magnitude plot: \(20\log|G(j\omega)|\) vs \(\log(\omega)\)
  2. Phase plot: \(\angle G(j \omega)\) vs \(\log(\omega)\)

To sketch the Bode plot of any rational transfer function, we only need to know how to sketch four terms:

  1. Pure gain: \(G(s)=K\)
  2. First-order terms: \(G(s) = \tau s \pm 1, \quad \tau \gt 0\)
  3. Zeroes at \(s=0\): \(G(s)=s^n\)
  4. Complex conjugate roots: \(G(s)=s^2 + 2\zeta \omega_n s + \omega_n^2 = \omega_n^2\left(\frac{s^2}{\omega_n^2} + \frac{2\zeta s}{\omega_n} + 1\right)\)

Given a transfer function, we can decompose it into these terms.

Polar plots

\(\Re(G(j\omega))\) vs \(\Im(G(j\omega))\)

e.g. 3.8.5

\[\begin{align} G(s) &= \frac{40s^2(s-2)}{(s+5)(s^2 + 4s+100)}\\ &= \frac{40s^2(2)\left(\frac{s}{2}-1\right)}{5(100)\left(\frac{3}{5}+1\right)\left(\frac{s^2}{10^2}+\frac{4s}{10}+j\right)}\\ &= \frac{40(2)}{5(100)} \cdot \frac{s^2(\frac{s}{2}-1)}{(\frac{s}{5}+1)(\frac{s^2}{10^2}+\frac{4s}{10^2}+1)}\\ \end{align}\]

e.g. 3.8.6

To plot the Bode plot, we need:
\[\begin{align} 20\log|G(j\omega)|&=20\log\left|\frac{80}{500}\right| + 20\log|(j\omega)^2| + 30\log\left|\frac{j\omega}{2}-1\right|\\ &=-20\log\left|\frac{j\omega}{5}+1\right| - 20\log\left|\frac{(j\omega)^2}{10^2} + \frac{4}{10^2}j\omega + 1\right|\\ \\ \angle G(j\omega)&=\angle\frac{800}{500}+\angle(j\omega)^2 + \angle\frac{j\omega}{2}+1-\angle\frac{j\omega}{5}+1-\angle\left(\frac{(j\omega)^2}{10^2}+\frac{4}{10^2}j\omega + 1\right)\\ \end{align}\]

e.g. plot

For \(G(j\omega)=K\):



For \(G(j\omega)=j\tau \omega + 1\) (the transfer function with a zero at \(s=\frac{-1}{\tau}\))



Approximations for sketching:

  1. For \(\omega \lt \frac{1}{\tau}\), \(\Im(G(j\omega)) \approx 0 \Rightarrow \forall \omega \lt \frac{1}{\tau}, \quad 20\log|G(j\omega)| \approx 20\log|1|=0\)
  2. For \(\omega \ge \frac{1}{\tau}\), \(\Re(G) \gt \gt \Im(G) \Rightarrow \omega \ge \frac{1}{\tau}, \quad 20\log|G(j\omega)| \approx 20\log|j \tau \omega|\)
  3. For \(\omega \lt\lt \frac{1}{\tau}\), \(\angle G(j\omega)\approx \angle 0j+1 = 0\) (\(\omega \lt\lt \frac{1}{\tau}\) means \(\omega \le \frac{0.1}{\tau}\))
  4. For \(\omega \gt\gt \frac{1}{\tau}\), \(\angle G(j\omega)\approx \angle j\omega\tau = \frac{\pi}{2}\) (\(\omega \gt\gt \frac{1}{\tau}\) means \(\omega \ge \frac{10}{\tau}\))
  5. Linear interpolation between \(\frac{0.1}{\tau}\) and \(\frac{10}{\tau}\)

Sub-case: \(G(s) = \tau s - 1\) (zero at \(s=\frac{1}{\tau}\))

Polar plot:

e.g. \(G(s) = \frac{100}{s+10} = \frac{100}{10} \cdot \frac{1}{\frac{s}{10}+1} = 10 \frac{1}{\frac{s}{10}+1}\)

Frequency response: \(G(j\omega)=10\frac{1}{\frac{j\omega}{10}+1}\)
Magnitude: \(20\log|G(j\omega)|=\underbrace{20\log10}_{A}-\underbrace{20\log\left|\frac{j\omega}{10}+1\right|}_B\)
Phase: \(\angle G(j\omega) = \underbrace{\angle 10}_A - \underbrace{\angle \frac{j\omega}{10}+1}_B\)

The bandwidth of the above system is the smallest frequency \(\omega_{BW}\) such that:
In dB: \(20\log|G(0)|-20\log|G(j\omega)|=3dB\)

From the Bode plot, \(\omega_{BW}=1.0 \text{ rad/s}\)

e.g. \(G(s)=s^n\)

e.g. Complex conjugate zeroes

\[G(s) = \frac{s^2}{\omega_n^2}+\frac{2\zeta}{\omega_n}s + 1, \quad \zeta \in [0,1), \quad \omega_n \ne 0\]
\[G(j\omega)=\left(1-\frac{\omega^2}{\omega_n^2}\right) + j\cdot 2\zeta \frac{\omega}{\omega_n}\]


For asymptotic Bode plots of complex conjugate roots, approximate \(G(s)\) as two first order terms with roots at \(-\omega_n\). i.e., set \(\zeta=1\):
\[\begin{align} G(s)&=\frac{s^2}{\omega_n^2}+\frac{2\zeta s}{\omega_n} + 1\\ &\approx \frac{s^2}{\omega_n^2}+\frac{2s}{\omega_n}+1\\ &=\left(\frac{s}{\omega_n}+1\right)^2\\ &= (\tau s + 1)^2, \quad \tau = \frac{1}{\omega_n}\\ \end{align}\]



State-space models

\[g(t) = Ce^{At} 1(t) + D \delta(t)\]
\[G(s) = C(sI-A)^{-1}B+D\]