Dynamic Systems:

Linear Transform

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

# 1. Linear Transforms: Fourier and Laplace Transforms¶

## 1.1. Inner Product¶

• For real valued vector
$$\vec{x}\cdot\vec{y} = \langle \vec{x}, \vec{y} \rangle = x^T y =\sum x_iy_i \quad \text{: similarity index}$$

• Orthogonal
$$x^Ty=0 \implies x \perp y$$

• Suppose $\hat{b}_i$ orthogonal basis
• Any $\vec{x}$ can be represented (decomposed) with linear combinations of $\hat{b}_i$

$$\vec{x} = c_1\hat{b}_1 + c_2\hat{b}_2 + \cdots + c_n\hat{b}_n$$

• Question: How to find $c_i$?

\begin{align*} \langle \vec{x}, \hat{b}_i \rangle &= c_1\langle \hat{b}_1, \hat{b}_i \rangle + \cdots + c_i\langle \hat{b}_i, \hat{b}_i \rangle + \cdots + c_n \langle \hat{b}_n, \hat{b}_i \rangle\\ \\ \therefore c_i & = {\langle x, \hat{b}_i \rangle \over \langle \hat{b}_i, \hat{b}_i \rangle} \end{align*}

• Meaning: $c_i$ indicates how much $\hat{b}_i$ information contains in $\vec{x}$

## 1.2. Hermitian Transpose¶

• For complex values
$$\langle \vec{x}, \vec{y} \rangle = y^Hx \quad \text{where } y^H \text{ is complex conjugate transpose of } y$$
• Want to know how much $e^{j\omega t}$ component is contained in $x(t)$.

\begin{align*} \langle x, e^{j\omega t} \rangle &= \int \left(e^{j \omega t} \right)^Hx(t) \,dt \\ &= \int x(t)e^{-j \omega t} \, dt \\ & = X(j\omega) \end{align*}

## 1.3. Fourier Transform¶

$$X(j\omega) = \int x(t)e^{-j \omega t} \, dt$$

## 1.4. Laplace Transform¶

• Want to know how much $e^{-\sigma+j\omega t} = e^{-\sigma} \, e^{j \omega t}$ component is contained in $x(t)$
• Let $s = \sigma+j \omega t$
\begin{align*} \left\langle x, e^{-\sigma + j\omega t} \right\rangle &= \int \left(e^{-\sigma+j \omega t}\right)^H x(t) \, dt \\ & = \int x(t)e^{-\sigma-j\omega t}dt \\ & = \int x(t)e^{-st}dt\\ & = X(s) \end{align*}
• Laplace Transform
$$X(s) = \int x(t)e^{-st}dt$$

# 2. Laplace Transform¶

• Eigenfunction Property of $e^{st}$
• The system output is given by
\begin{align*} y(t) &= H\{x(t)\} \\ &= h(t) * x(t) \\ &= \int^\infty_{-\infty} h(\tau)x(t - \tau)d\tau \end{align*}

• Let $e^{st}$ be a complex exponential with complex frequency $s = \sigma + j\omega$
• When $x(t) = e^{st}$
\begin{align*} y(t) &= \int^\infty_{-\infty} h(\tau) e^{s(t - \tau)}d\tau \\ &= e^{st} \int^\infty_{-\infty} h(\tau) e^{-s\tau}d\tau \end{align*}
• We define the transfer function
$$H(s) = \int^\infty_{-\infty} h(\tau) e^{-s\tau}d\tau$$
• We may write
$$y(t) = H\{e^{st}\} = H(s)e^{st}$$
• If the output signal is a scalar multiple of the input signal, we refer to the signal as an eigenfunction and the multiplier as the eigenvalue

• Complex exponentials are eigenfunctions of LTI systems.
• If $x(t) = e^{st}$ and $h(t)$ is the impulse response then
\begin{align*} y(t) &= (h*x)(t) = \int^\infty_{-\infty} h(\tau)e^{s(t-\tau)}d\tau = e^{st}\int^\infty_{-\infty}h(\tau)e^{-s\tau}d\tau \\ &= H(s)e^{st} \end{align*}

• The eigenvalue associated with $e^{st}$ is $H(s)$.

## 2.2. Laplace Transform: Definition¶

• Laplace transform maps a function of time $t$ to a function of $s$ (complex frequency)
• The Laplace transform is similar to the Fourier transform.

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

Example

• Find the laplace transform of $x(t)$ (where $u(t)$ is unit step function),
$$x(t) = e^{-t}u(t) = \begin{cases} e^{-t} & \text{if } t > 0 \\ 0 & \text{otherwise} \end{cases}$$

\begin{align*} X(s) &= \int^\infty_{-\infty}x(t) e^{-st}dt = \int^\infty_0 e^{-t}e^{-st}dt = \int^\infty_0 e^{-(s+1)t}dt \\ &= {e^{-(s+1)t} \over -(s+1)}\biggr\rvert^\infty_0 = {1 \over s+1} \end{align*}

• Provided $\text{Re}(s+1) > 0$ which implies that
$$\underbrace{\text{Re}(s) > -1}_{\text{Regions of Convergence (ROC)}}$$

Example

\begin{align*} x(t) &= 3e^{-2t}u(t)-2e^{-t}u(t)\\\\ X(s) &= \underbrace{3 \over s+2}_{\text{Re}(s) > -2} - \underbrace{2 \over s+1}_{\text{Re}(s) > -1} = {3(s+1) - 2(s+2) \over (s+2)(s+1)} = \underbrace{s-1 \over s^2 + 3s + 2}_{\text{Re}(s)>-1} \end{align*}

Example

• Laplace transform of the impulse function
\begin{align*} x(t) &= \delta(t)\\ \\ X(s) &= \int^\infty_{-\infty} \delta(t)e^{-st}dt = \int^\infty_{-\infty} \delta(t)e^{-st}\biggr\rvert_{t = 0}dt \\ &= \int^\infty_{-\infty} \delta(t) \cdot 1 dt = 1 \end{align*}

• Sifting property

• Multiplying $f(t)$ by $\delta(t)$ and integrating over $t$ sifts out $f(0)$

## 2.3. Unit Impulse Signal¶

• 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
• Laplace transform of unit-impulse function
$$D(s) = \int^{\infty}_{-\infty}\delta(t)e^{-st}dt = e^{-s0} = 1$$

## 2.4. Unit Step Signal¶

• The indefinite integral of the unit-impulse is the unit-step

$$u(t) = \int^t_{-\infty}\delta(\tau)d\tau = \begin{cases} 1 & t \geq 0 \\ 0 & \text{otherwise} \end{cases}$$

• Laplace transform of unit-step function

$$U(s) = \int^{\infty}_{-\infty}u(t)e^{-st}dt = \int^{\infty}_{0}1\cdot e^{-st}dt = {e^{-st} \over -s}\biggr\rvert^\infty_0 = {1 \over s}, \qquad \text{ Re}\{s\}>0$$

## 2.5. Convolution Property of Laplace Transform¶

$$f \quad \mathop{\longleftrightarrow}^{\mathscr{L}} \quad F, \qquad F(s) = \int^{\infty}_{-\infty}f(u)e^{-su}du$$$$g \quad \mathop{\longleftrightarrow}^{\mathscr{L}} \quad G, \qquad G(s) = \int^\infty_{-\infty}g(v)e^{-sv}dv$$
• Then
$$f * g \quad \mathop\longleftrightarrow^{\mathscr{L}} \quad F\cdot G$$

\begin{align*} F(s)\cdot G(s) &= \int^\infty_{-\infty}f(u)e^{-su}du \cdot \int^\infty_{-\infty}g(v)e^{-sv}dv \\ \\ &= \int^\infty_{-\infty} \int^\infty_{-\infty} f(u) g(v)e^{-s(u+v)} dudv \end{align*}
• Change of variables
\begin{align*} u + v &= t\\\\ u &= u \\ v &= t - u \end{align*}

\begin{align*} \int^\infty_{-\infty} \int^t_{-\infty} f(u)g(t-u)e^{-st}dudt &= \int^\infty_{-\infty} e^{-st}\biggl[\int^t_{-\infty}f(u)g(t-u)du\biggr]dt \\ \\ & = \int^\infty_{-\infty}e^{-st}(f*g)(t) dt = \mathscr{L}\{f*g\} \end{align*}

• Response to general input
\begin{align*} y(t) &= h(t) * x(t) \\\\ Y(s) &= H(s) \cdot X(s) \end{align*}

## 2.6. Laplace Transform of a Derivative¶

• Assume that
\begin{align*} \mathscr{L}\{x(t)\} &= X(s) \\ \\ X(s) &= \int^\infty_{-\infty}x(t)e^{-st}dt \end{align*}
• Find the Laplace transform of $y(t) = \dot{x}(t)$

\begin{align*} Y(s) &= \int^\infty_{-\infty}y(t)e^{-st}dt = \int^\infty_{-\infty}\dot{x}(t)e^{-st}dt \\\\ &= \underbrace{x(t)e^{-st}\biggr\rvert^\infty_{-\infty}}_{\text{must be zero since X(s) converged}} - \int^\infty_{-\infty}x(t)(-se^{-st})dt, \qquad \text{using } \left(\int\dot{v}u = vu - \int v\dot{u}\right) \\ \\ & = s\int^\infty_{-\infty}x(t)e^{-st}dt\\\\ &= sX(s) \end{align*}

• Solving differential equations with Laplace transform

Example

$$\dot{y}(t) + y(t) = \delta(t)$$
• Laplace transform
$$sY(s) + Y(s) = 1$$
• This is a simple algebraic expression
• Laplace transform converts a differential equation to an equivalent algebraic equation
\begin{align*} Y(s) &= {1 \over s+1} \\\\ y(t) &= e^{-t}u(t) \end{align*}

Example

\begin{align*} \ddot{y}(t) + 3\dot{y}(t) + 2y(t) &= \delta(t)\\\\ s^2Y(s) + 3sY(s) + 2Y(s) &= 1 \end{align*}

\begin{align*} Y(s) &= {1 \over s^2 + 3s + 2} = {1 \over (s+1)(s+2)} = {A \over s+1} + {B \over s+2} \\\\ &= {1 \over s+1} - {1 \over s+2}\\\\ y(t) &= e^{-t}u(t) - e^{-2t}u(t) \end{align*}

## 3.7. Laplace Transform of Integral¶

\begin{align*} \int^t_{-\infty}x(\tau)d\tau &= u(t) * x(t) \\ \\ &\downarrow \mathscr{L}\\ \\ Y(s) &= {1 \over s} X(s) \end{align*}

# 3. Transfer Function¶

• Transfer function of an LTI system is defined as the Laplace transform of the impulse response.

\begin{align*} y(t) &= h(t) * x(t)\\\\ Y(s) &= H(s) \cdot X(s)\\\\ \therefore \; H(s) &= {Y(s) \over X(s)} \end{align*}

• Describe input-output behavior of the system.
• Transfer function of differential equation system

\begin{align*} \sum\limits^N_{k=0}a_k{d^k \over dt^k}y(t) &= \sum\limits^M_{k=0}b_k{d^k \over dt^k}x(t) \\ \\ &\Downarrow \mathscr{L} \\ \\ \sum\limits^N_{k=0}a_ks^kY(s) &= \sum\limits^M_{k=0}b_ks^kX(s) \end{align*}

$$H(s) = {Y(s) \over X(s)} = {\sum\limits^M_{k=0}b_ks^k \over \sum\limits^N_{k=0}a_ks^k}$$
• Example: rational transfer function

$$\ddot{y} + 3\dot{y} + 2y = 2\dot{x} - 3x, \qquad \qquad \text{where }\; y: \text{output and} \; x: \text{input}$$

$$H(s) = {Y(s) \over X(s)} = {2s - 3 \over s^2 + 3s + 2} = {2(s - {3 \over 2})\over (s+1)(s+2)}$$

• The poles and zeros of a rational transfer function offer much insight into LTI system characteristics

$$H(s) = {\tilde{b} \prod\limits^M_{k=1}(s-c_k) \over \prod\limits^N_{k=1}(s-d_k)}, \qquad \qquad c_k \; \text{zeros and }\; d_k \; \text{poles}$$

• Determine the frequency response from poles and zeros

• Transfer function by substituting $j\omega$ for $s$
• Evaluating the transfer function along the $j\omega$-axis in the $s$-plane

$$H(j\omega) \quad \Longrightarrow \quad \lvert H(j\omega) \rvert \; \text{and}\; \angle H(j\omega)$$

$\quad\;\Rightarrow$ Graphical evaluation or Bode plot

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