# ECE 486 Control Systems

Lecture 3

## Impulse Response

Recall from last time, we introduced state-space model, a modern approach with which we can describe control systems with a unified framework.

\begin{align*} \dot{x} &= Ax + Bu \\ y &= Cx, \end{align*}where

- \(x(t) \in \RR^n\) is the
*state*(vector) at time \(t\); - \(u(t) \in \RR^m\) is the
*input*(vector) at time \(t\); - \(y(t) \in \RR^p\) is the
*output*(vector) at time \(t\); - \(A \in \RR^{n \times n}\) is the
*dynamics matrix*; - \(B \in \RR^{n \times m}\) is the
*control matrix*; - \(C \in \RR^{p \times n}\) is the
*sensor matrix*.

A natural question arises

**Question**: Given a dynamical system, how do we determine the output \(y\) for a
given input \(u\)?

**Note**: We will only consider **single-input**, **single-output** (SISO) systems,
i.e., \(u(t),y(t) \in \RR\) for all times \(t\) of interest. In the above
notation for state-space model, \(m=p=1\) means the system is SISO.

### Definition of Impulse Response

Recall from ECE 210, a *unit impulse* or *Dirac’s $δ$-function* is
a function of time which satisfies,

- \(\delta(t) = 0\) for all \(t \neq 0\) and
- \(\displaystyle \int^a_{-a}\delta(t) \d t = 1\) for all \(a > 0\).

Figure 1: Unit impulse or Dirac’s $δ$-function

We can think of \(\delta(t)\) as a limit of impulses of unit area as shown in Figure 2. As \(\varepsilon \to 0\), the impulse gets taller \(\frac{1}{\varepsilon} \to + \infty\), but the area under its graph remains at \(1\).

Figure 2: Step signal with width \(\varepsilon\), height \(\frac{1}{\varepsilon}\)

### System Output as Convolution Integral

Now consider the input \(u(t) = \delta(t-\tau)\) (unit impulse applied at time
\(t=\tau\)), given a **linear** and **time-invariant** (LTI) system with zero
initial conditions (IC) as in Figure 3, the system output \(y(t) =
h(t-\tau)\). This function \(h\) is called the *impulse response* of the system.

Figure 3: Linear system with zero initial conditions \(x(0) = 0\).

Given system impulse response \(h\), we can find the system’s response to other
(arbitrary) inputs. Indeed, recall the *sifting property* of the
$δ$-function: for any function \(f\) which is “well-behaved” at \(t=\tau\),
we have

Equation \eqref{d3_eq1} says the $δ$-function “sifts out” the value of
\(f\) at \(t = \tau\). Therefore, any **reasonably regular** function can be
represented as an integral of impulses.

To compute the system’s response to other (arbitrary) inputs by a given \(h\), we can write this input signal \(u\) in integral form by the above sifting property,

\begin{align*} u(t) = \int^\infty_{-\infty}u(\tau)\delta(t-\tau)\d\tau. \end{align*}
By *superposition principle*, the response of a linear system to a sum (or
integral) of inputs is the sum (or integral) of the individual responses to
these inputs, i.e.,

The integral in Equation \eqref{d3_eq2} that defines \(y(t)\) is a
*convolution* of \(u\) and \(h\).

**Superposition Principle**: For all linear systems, the net response caused
by two or more stimuli is the sum of the responses that would have been
caused by each stimulus individually.

Thus, for an LTI system with **zero initial conditions**, its output is the
convolution of the input with the system impulse response,

Unfortunately, the formula in the definition does not provide a convenient
way of computing the output \(y\) for a given input \(u\). A more practical is to
use *Laplace Transform*.

## Laplace Transforms and the Transfer Function

### Transfer Function

Recall the *two-sided* Laplace transform of a function \(f(t)\) is the integral
transform

where \(s \in \CC\) is a complex variable.

We can use the table below to establish the quick correspondence between time domain and frequency domain signals.

time domain |
frequency domain |

\(u(t)\) | \(U(s)\) |

\(h(t)\) | \(H(s)\) |

\(y(t)\) | \(Y(s)\) |

And convolution in time domain \(y(t) = h(t) \star u(t)\) is *equivalent* to
multiplication in frequency domain \(Y(s) = H(s)U(s)\) through Laplace
transform.

In particular, the Laplace transform of the impulse response
\[H(s) = \int^\infty_{-\infty}h(\tau)e^{-s\tau}\d\tau\]
is called *transfer function* of the system.

When we consider **causal** systems, whose output at time \(t\) is not affected by
inputs at future times \(t' > t\), we agree that \(h(t) =
0\) for \(t < 0\), where \(h(t)\) is the response at time \(t\) to a unit impulse at
time \(0\).

We will take all other possible inputs (not just impulses) to be \(0\) for \(t <
0\), and work with *one-sided* Laplace transforms

### Computing Transfer Functions

Here we want to answer another question—Given \(u(t)\), we can find its
transform \(U(s)\) by using tables of Laplace transforms or `MATLAB`

, but how
about \(h(t)\) or \(H(s)\)? Recall the LTI system in Figure 3, in this
specific case, we can use the following formulae,

More on these formulae will be discussed much later this semester.

But at the moment let’s look at a quick example with a single input and a single output.

**Example 1**: Consider the system dynamics

Compute system transfer function \(H(s)\).

**Solution**: Given \(u(t) = e^{st}, t \ge 0\), we think of \(s\) as some fixed
number. Then

Substitute

\begin{align*} \dot{y}(t) &= \frac{\d}{\d t}\left(H(s)e^{st}\right) = sH(s)e^{st} \end{align*}into the governing differential equation \(\dot{y}(t) = -ay(t) + u(t)\),

\begin{align*}\require{cancel} sH(s)\cancel{e^{st}} &= -a H(s)\cancel{e^{st}} + \cancel{e^{st}}, \qquad \forall~ s, t > 0 \\ sH(s) &= -a H(s) + 1 \tag{3} \label{d3_eq3}. \end{align*}Solve for \(H(s)\) in Equation \eqref{d3_eq3}, we have

\begin{align*} H(s) &= \frac{1}{s+a}, \tag{4} \label{d3_eq4}\\ \implies y(t) &= \frac{e^{st}}{s+a}. \end{align*}If we look up the right hand side of the transfer function in Equation \eqref{d3_eq4} from tables, then the impulse response \(h(t)\) by taking the inverse Laplace transform is

\begin{align*} h(t) &= \begin{cases} e^{-at}, & t \ge 0 \\ 0, & t < 0 \end{cases} \end{align*}This example can be illustrated using the chart,

Figure 4: Linear system with zero initial conditions \(x(0) = 0\).

### Frequency Response

We already know how to find \(y\) for a given \(u\) with a known \(h\). But given an input \(u(t) = e^{st}\), do we follow Example 1 and repeat for all \(s\) of interest in order to compute the ratio \[ H(s) = \frac{y(t)}{u(t)}? \]

That is to say, compute \(H(s) = \dfrac{y(t)}{u(t)}\), repeat for as many values of \(s\) as necessary according to Figure 5.

Figure 5: Compute the transfer function = compute the ratio \(H(s) = \frac{y(t)}{u(t)}\)?

One problem of this approach is that \(e^{st}\) either blows up very quickly if \(s > 0\), or it decays to \(0\) very quickly if \(s < 0\).

Instead, our input signals shall be **sustained**, **bounded** signals. That’s why
when we introduced Laplace transform in Equation \eqref{d3_lt}, \(s\) is allowed
to be a complex number \(s \in \CC\).

**Basics of Complex Numbers**: We can visualize a complex number \(s = a+jb \in
\CC\) with an **Argand diagram**,

We can also use its **polar form**,

Euler’s formula: \(e^{j\varphi} = \cos \varphi + j \sin \varphi\).

**Do you know**? Euler’s identity, which is a special case of the formula,
relates the five most important mathematical numbers, i.e., \(0\), \(1\), \(e\),
\(i\) or \(j\), and \(\pi\) by Euler’s equation \(e^{i \pi} + 1 = 0\).

Consider the LTI system with zero initial conditions, given a sinusoidal input \(u(t) = A\cos(\omega t)\) with amplitude \(A\), frequency \(\omega\) in rad/s, we can compute the system output \(y(t)\).

Recall the complex cosine function in \(z \in \CC\) is defined as \[ \cos (z) \triangleq \frac{e^{iz} + e^{-iz}}2. \]

It also makes sense based on Euler’s formula above. Therefore, by linearity, the response \(y(t)\) is \[ y(t) = \frac{A}{2} \Big( H(j\omega)e^{j\omega t} + H(-j\omega)e^{-j\omega t}\Big), \] where

\begin{align*} H(j\omega) &= \int^\infty_0 h(\tau)e^{-j\omega\tau}\d\tau, \\ H(-j\omega) &= \int^\infty_0 \underbrace{h(\tau)e^{j\omega\tau}}_{\text{complex}\atop\text{conjugate}}\d\tau \\ &= \overline{H(j\omega)}. \end{align*}Note that \(h(\tau) \in \RR\) is a real valued function and \(H(j\omega) \in \CC\). Denote

\begin{align*} H(j\omega) &= M(\omega)e^{j\varphi(\omega)} \\ H(-j\omega) &= M(\omega)e^{-j\varphi(\omega)}, \end{align*}since both are complex numbers thus having magnitudes and phases.

If we rewrite \(y(t)\),

\begin{align*} y(t) &= \frac{A}{2} M(\omega)\Big( e^{j(\omega t + \varphi(\omega))} + e^{-j(\omega t + \varphi(\omega))}\Big) \\ &= A M(\omega) \cos\big(\omega t + \varphi(\omega) \big) \hspace{3cm} \text{(steady state)} \\ \tag{5} \label{d3_eq5} &= A\underbrace{M(\omega)}_{\text{amplitude}\atop\text{magnification}} \cos\big(\omega t + \underbrace{\varphi(\omega)}_{\text{phase}\atop\text{shift}}\big). \end{align*}Equation \eqref{d3_eq5} says the (steady-state) system response to a cosine signal with amplitude \(A\) and frequency \(\omega\) is still a cosine signal but with amplitude scaled \(AM(\omega)\); same frequency \(\omega\), but with phase shift \(\varphi(\omega)\).

We shall learn in the next lecture that system response to general signals
(not necessarily sinusoidal signals) is always given by \(Y(s) =
H(s)U(s)\). For system with **nonzero** initial conditions, the system response
is described by

if the system is *stable*.