Dynamic Systems:

Forced Response

By Prof. Seungchul Lee
http://iai.postech.ac.kr/
Industrial AI Lab at POSTECH

Table of Contents

# 1. Linear Time-Invariant (LTI) SystemsΒΆ

The system ${H}$ is a transformation (a rule or formula) that maps an input signal $x$ into a output signal $y$

System examples

\begin{align*} &\text{Identity} \quad &y(t) &= x(t) \quad &\forall t\\ &\text{Scaling} \quad &y(t) &= 2\,x(t) \quad &\forall t\\ &\text{Offset} \quad &y(t) &= x(t)\, + \,2 \quad &\forall t\\ &\text{Square signal} \quad &y(t) &= (x(t))^2 \quad &\forall t\\ &\text{Shift} \quad &y(t) &= x(t + 2) \quad &\forall t\\ &\text{Decimate} \quad &y(t) &= x(2t) \quad &\forall t\\ &\text{Square time} \quad &y(t) &= x(t^2) \quad &\forall t \end{align*}

## 1.1. Linear SystemsΒΆ

A system ${H}$ is linear if it satisfies the following two properties:

1) Scaling

$${H} \{\alpha x\} = \alpha {H}\{x\} \quad \forall \, \alpha \in \mathbb{C}$$

2) Additivity

$$\text{If}\,\, y_1 = {H} \{x_1\} \,\, \text{and}\,\, y_2 = {H} \{ x_2 \} \, \, \text{then } {H} \{x_1 + x_2\} = y_1 + y_2$$

## 1.2. Time-Invariant SystemsΒΆ

A system ${H}$ processing infinite-length signals is time-invariant (shift-invariant) if a time shift of the input signal creates a corresponding time shift in the output signal

## 1.3. Linear Time-Invariant (LTI) SystemsΒΆ

We will only consider Linear Time-Invariant (LTI) systems.

\begin{align*} \text{Identity} &\qquad y(t) = x(t) \quad &\forall t &\qquad \text{Linear} &\qquad \text{Time Invariant}\\ \text{Scaling} &\qquad y(t) = 2\,x(t) \quad &\forall t &\qquad \text{Linear} &\qquad \text{Time Invariant}\\ \text{Offset} &\qquad y(t) = x(t)\, + \,2 \quad &\forall t &\qquad \text{Non Linear} &\qquad \text{Time Invariant}\\ \text{Square signal} &\qquad y(t) = (x(t))^2 \quad &\forall t &\qquad \text{Non Linear} &\qquad \text{Time Invariant}\\ \text{Shift} &\qquad y(t) = x(t + 2) \quad &\forall t &\qquad \text{Linear} &\qquad \text{Time Invariant}\\ \text{Decimate} &\qquad y(t) = x(2t) \quad &\forall t &\qquad \text{Linear} &\qquad \text{Time Variant}\\ \text{Square time} &\qquad y(t) = x(t^2) \quad &\forall t &\qquad \text{Linear} &\qquad \text{Time Variant}\\ \end{align*}

# 2. Response to Non-Zero InputΒΆ

• So far, natural response of zero input with non-zero initial conditions are examined.

\begin{align*} \dot{y} + {1 \over \tau}y &= 0 \\\\ \ddot{y} + 2\zeta\omega_n\dot{y} + \omega^2_n y &=0 \end{align*}

Response to non-zero constant input

• Assume all the systems are stable
• Inhomogeneous ODE
\begin{align*} \dot{y} + {1 \over \tau}y &= q' \\\\ \ddot{y} + 2\zeta\omega_n\dot{y} + \omega^2_n y &=p' \end{align*}

$\quad\;\implies$ Same dynamics, but it reaches different steady state

$\quad\;\implies$ Good enough to sketch

\begin{align*} (\dot{y}-\dot{q}) + {1 \over \tau}(y-q) &= 0 \\\\ (\ddot{y}-\ddot{p}) + 2\zeta\omega_n(\dot{y}-\dot{p}) + \omega^2_n (y-p) &=0 \end{align*}
• Dynamic system response = transient + steady state
• Transient response is present in the short period of time immediately after the system is turned on
• It will die out if the system is stable
• The system response in the long run is determined by its steady state component only
• In steady state, all the transient responses go to zero
$${dy \over dt} \rightarrow 0$$
• Example
\begin{align*} \dot{y} + {1 \over \tau}y &= q, \qquad y(0) = 0 \\\\ \dot{y}(\infty) & = 0 \\ \\ \implies \; y(\infty)& = \tau q \end{align*}

• Example
\begin{align*} \ddot{y} + 2\zeta\omega_n\dot{y} + \omega^2_n y &= p, \qquad y(0) = 0, \; \dot{y}(0) = 0 \\ \\ \ddot{y}(\infty) &= \dot{y}(\infty) = 0 \\ \\ \implies \; y(\infty) &= {p \over \omega^2_n} \end{align*}

• Think about mass-spring-damper system in horizontal setting
$$\ddot{y} + {c \over m}\dot{y} + {k \over m}y = 0, \qquad \text{where} \; y(0) = y_0, \; \dot{y}(0) = 0$$

• Mass-spring-damper system in vertical setting
• $y(0) = 0 \;$ no initial displacement
• $\dot{y}(0) = 0\;$ initially at rest

\begin{align*} m\ddot{y} + c\dot{y} + ky &= mg \\ \ddot{y} + {c \over m}\dot{y} + {k \over m}y &= g, \qquad \text{where} \; y(0) = 0, \; \dot{y}(0) = 0 \end{align*}

• Shift the origin of $y$ axis to the static equilibrium point, then act like a natural response with $y(0) = -{mg \over k}$ and $\dot{y}(0) = 0$ as initial conditions

$$\ddot{y} + 2\zeta\omega_n\dot{y} + \omega^2_n y ={\omega_n^2}p \quad \longrightarrow \quad (\ddot{y}-\ddot{p}) + 2\zeta\omega_n(\dot{y}-\dot{p}) + \omega^2_n (y-p) =0$$

# 3. Time Response to General InputsΒΆ

• We studied output response $y(t)$ when input $x(t)$ is constant
• Goal: output response of $y(t)$ to general input $x(t)$

## 3.1. Step ResponseΒΆ

Start with a step response example

$$\dot{x} + 5x = 1 \quad \text{for} \quad t \geq 0, \qquad x(0) = 0$$

or

$$\dot{x} + 5x = u(t), \qquad x(0) = 0 \qquad \text{where }\; u(t) = \begin{cases} 1 & t >0\\ 0 & \text{otherwise}\end{cases}$$
• Step function $u(t)$

$$u(t) = \begin{cases} 1 & t \geq 0 \\ 0 & \text{otherwise} \end{cases}$$
• The solution is given:
$$x(t) = \frac{1}{5}\left( 1-e^{-5t} \right)$$
InΒ [1]:
t = linspace(0,2,100);
x = 1/5*(1-exp(-5*t));

plot(t,x,t,0.2*ones(size(t)),'k--')
ylim([0,0.25])
xlabel('time')


## 3.2. Impulse ResponseΒΆ

• Impulse response: difficult to image
• The unit-impulse signal acts as a pulse with unit area but zero width

$$\delta(t) = \lim_{\epsilon\rightarrow0}p_\epsilon(t)$$

• The unit-impulse function is represented by an arrow with the number 1, which represents its area

• It has two seemingly contradictory properties :
• It is nonzero only at $t = 0$ and
• Its definite integral ($-\infty$, $\infty$) is 1
• The Dirac delta can be loosely thought of as a function on the real line which is zero everywhere except at the origin, where it is infinite,

$$\delta(t) = \begin{cases}\infty & x = 0 \\ 0 & x \neq 0\end{cases}$$

$\quad \;$ and which is also constrained to satisfy the identity

$$\int_{-\infty}^{\infty}\delta(t)dt = 1$$
• Sifting property
\begin{align*} \int_{-\infty}^{\infty}x(t)\delta(t)dt &= \int_{-\infty}^{0^-}x(t)\delta(t)dt + \int_{0^-}^{0^+}x(t)\delta(t)dt + \int_{0^+}^{\infty}x(t)\delta(t)dt \\&= 0 + x(0)\int_{0^-}^{0^+}\delta(t)dt + 0 \\&= x(0)\\\\ \int_{-\infty}^{\infty}x(\tau)\delta(t-\tau)d\tau &= x(t) \end{align*}
• Question: how to realize initial velocity of $v_0 \neq 0$

• Momentum and impulse in physics I

• Consider an "impulse" which is a sudden increase in momentum $0 \rightarrow mv$ of an object applied at time $0$

• To model this,

$$mv - 0 = \int^{0^+}_{0_-}f(t)dt$$

$\quad\;$ where force $f(t)$ is strongly peaked at time 0

• Actually the details of the shape of the peak are not important, what is important is the area under the curve
$$f(t) = mv\delta(t)$$
• This is the motivation that mathematician and physicist invented the delta Dirac function

\begin{align*} m\ddot{y} + c\dot{y} + ky &= mv_0\delta(t) \\\\ \ddot{y} + {c \over m}\dot{y} + {k \over m}y &= v_0\delta(t) \end{align*}

v_0 \delta(t) \quad \longrightarrow \quad \begin{align*} y(0) &= 0\\ \dot{y}(0) &= v_0 \end{align*} \quad \longrightarrow \quad \text{natural response}

• Impulse response to LTI system

\ddot{y} + 2\zeta \omega_n\dot{y} + \omega^2_n y = \delta(t), \qquad \begin{align*} \delta(t) &: \text{input} \\ y &: \text{output} \end{align*}

• Later, we will discuss why the impulse response is so important to understand an LTI system
• Example: now think about the impulse response
$$\dot{y} + 5y = \delta(t), \qquad y(0) = 0$$

$\quad\;$ The solution is given: (why?)

$$y(t) = h(t) = e^{-5t},\quad t\geq0$$

$\quad\;$ Impulse input can be equivalently changed to zero input with non-zero initial condition (by the impulse and momentum theory)

$$\int_{0^-}^{0^+}\delta(t) \, dt = u(0^+)-u(0^-)=1$$$$\dot{y} + 5y = 0, \qquad {y}(0) = 1$$
InΒ [2]:
t = linspace(0,2,100);
h = exp(-5*t);

plot(t,h,t,zeros(size(t)), 'k--'),
ylim([-0.1,1.1])
xlabel('time')


## 3.3. Step Response AgainΒΆ

\begin{align*} \ddot{x} + 2\zeta\omega_n\dot{x}+\omega^2_nx &= u(t) &\text{or} \\ \\ \ddot{x} + 2\zeta\omega_n\dot{x}+\omega^2_nx &= 1, &t > 0 \end{align*}
• Relationship between impulse response and unit-step response

$\quad\;\implies$ impulse response is the derivative of the step response

## 3.4. Response to a General Input (in Time)ΒΆ

• Response to a "general input" in time
$$\dot{y} + 5 y = x(t), \quad y(0) = 0$$
• The solution is given
$$y(t) = h(t) * x(t) \qquad \text{where} \; h(t) \;\text{is impulse response}$$

• If this is true, we can compute output response to any general input if an impulse response is given

• impulse response = LTI system
• Convolution definition
\begin{align*} y(t) = h(t) * x(t) &= \int^\infty_{-\infty}h(\tau)x(t-\tau)d\tau \\ &= \int^\infty_{-\infty}x(\tau)h(t-\tau)d\tau \end{align*}
• $y(t)$ is the integral of the product of two functions after one is reversed and shifted by $t$
• Time-invariant