ve216_l4.md

VE216 Lecture 4

Continuous-time system

Multiple Representations of CT Systems

the same as DT Systems

Differential Equations

New methods based on block diagrams and operators, provide new ways to think about system's behaviors.

Block Diagrams

Key difference is the delays in DT are replaced by integrators in CT.

• DT:

  adders, scalers, delays

• CT:

  adders, scalers, integrators

Operator Representation

• use the $A$ operator.
• $A$ in CT signal generates a new signal that equals to the integral of the input signal at all points.
• $Y=AX$ is equal to $y(t) = \int^t_{-\infty}x(\tau)d\tau$ for all the time $t$.

Unit Impulse Signal

Properties

• Nonzero only at $t=0$.
• Integral $(-\infty, +\infty)$ is $1$.

Represented by an arrow with number $1$, representing the area or weight.

Unit Impulse and Unit Step

Unit Step Definition

Indefinite integral of unit impulse is unit step.

$$u(t) = \int^t_{-\infty}\delta(\lambda)d\lambda = \begin{cases}1&t\ge0\0&\text{otherwise} \end{cases}$$

Then we can see the block diagram:

Impulse Response of Acyclic CT System

If the block diagram of CT system has no feedback (acyclic, no circle), then the corresponding expression is imperative.

$Y=(1+A)^2X$

if $x(t) = \delta(t)$, then $y(y) = \delta(t) + 2u(t) + t\cdot u(t)$

CT Feedback Methods

Method 1: Differential Equation Method

$\dot{y}(t) = x(t) + py(t)$ this is a linear, first order equation with constant coefficients.

try $y(t) = C e^{\alpha t}u(t)$, then we get:

$\dot y(t) = \alpha C e^{\alpha t} u(t) + C e^{\alpha t} \delta (t) = \alpha C e^{\alpha t} u(t) + C\delta(t) = \delta(t) + pC e^{\alpha t} u(t) = x(t) + py(t)$

if $y(t) = e^{pt}u(t)$, the equation is satisfied ($\delta(t) = e^{pt}\delta(t)$).

Method 2: Operator Method

$Y=A(X+pY) \leftrightarrow \frac{Y}{X} = \frac{A}{1-pA} = A(1 + pA + p^2A^2 + \cdots)$

Let $X = x(t) = \delta(t)$ then:

$y(t) = A(1+pA + p^2A^2 + \dots)\delta(t) \=(1+pA+p^2A^2 + \dots)u(t) \= (1 + pt + \frac{1}{2}p^2t^2 + \frac{1}{6}p^3t^3+\cdots)u(t) = e^{pt}u(t)$.

Convergent and Divergent Poles

check the CT system:

then we get a $y(t) = e^{pt}u(t)$.

We decide the convergence by this method:

if $Re(p) < 0$ then it is convergent, $Re(p) > 0$ then it is divergent.

Feedback Comparison (CT, DT)

CT Feedback

In CT, each cycle adds new integration.

$y(t) = A(1+pA + p^2A^2 + \dots)\delta(t) \=(1+pA+p^2A^2 + \dots)u(t) \= (1 + pt + \frac{1}{2}p^2t^2 + \frac{1}{6}p^3t^3+\cdots)u(t) = e^{pt}u(t)$

PT Feedback

In DT, each cycle creates another sample in output:

$y[n] = (1+pR+p^2R^2+p^3R^3 +\cdots)\delta[n]\=\delta[n] + p\delta[n-1] + p^2\delta[n-2]+p^3\delta[n-3] + \cdots$

Diagram Comparison

Equation:	$\dot{y}(t) = x(t)+py(t)$	$y[n] = x[n]+py[n-1]$
Block diagram:
Operator:	$\frac A {1-pA}$	$\frac 1 {1-pR}$
Solution	$e^{pt}u(t)$	$p^nu[n]$
Convergence:

Convergent? (Exercise)

• $\frac{1}{1-\frac{1}{4}R^2}$

  $Y = \frac{1}{1-\frac{1}{4}R^2}X\ = k_1\frac{1}{1-\frac{1}{2}R}X+k_2\frac{1}{1+\frac{1}{2}R}X \= k_1 (1 + \frac{1}{2}R + \frac{1}{4}R^2 + \cdots)X + k_2(1 + (-\frac{1}{2}R) + (-\frac{1}{2}R)^2 + \cdots)X$

  Poles of $\pm \frac{1}{2}$ for discrete case, so convergent.

• $\frac{1}{1-\frac{1}{4}A^2}$

  $Y = \frac{1}{1-\frac{1}{4}A^2}X\=k_1\frac{1}{1-\frac{1}{2}A}X + k_2\frac{1}{1+\frac{1}{2}A}X\=k_1(1+\frac{1}{2}A + (\frac{1}{2}A)^2 + \cdots)X + k_2(1+(-\frac{1}{2}A) + (-\frac{1}{2}A)^2 + \cdots)X\ = k_1 e^{1/2t} + k_2 e^{-1/2t}$

  Poles of $\pm \frac{1}{2}$ in continuous case, so divergent ($\frac{1}{2}$).

• $\frac{1}{1+2R+\frac{3}{4}R^2}$

  $Y = \frac{1}{1+2R+\frac{3}{4}R^2}X\=k_1\frac{1}{1+\frac{1}{2}R}X+k_2\frac{1}{1+\frac{3}{2}R}X$

  We have pole for $-\frac{3}{2}$ and $-\frac{1}{2}$ in discrete case, so divergent ($-\frac{3}{2}$).

• $\frac{1}{1+2A+\frac{3}{4}A^2}$

  $Y = \frac{1}{1+2A+\frac{3}{4}A^2}X\=k_1\frac{1}{1+\frac{1}{2}A}X+k_2\frac{1}{1+\frac{3}{2}A}X$

  The poles are $-\frac{1}{2}$ and $-\frac{3}{2}$ in continuous case, so convergent.

Mass-Spring System (Exercise)

$F = K(x(t) - y(t)) = M \dot{\dot{y}}(t)$

$\frac{Y}{X} = \frac{\frac{K}{M}A^2}{1+\frac{K}{M}A^2}$, thus $p = \pm j\sqrt{\frac{K}{M}}$.

$e^{j\omega_0t} = cos\omega_0t+jsin\omega_0t$ and $e^{-j\omega_0t} = cos\omega_0t-jsin\omega_0t$.

$\frac{Y}{X} = \frac{\omega_0}{2j} (\frac{A}{1-j\omega_0A}) - \frac{\omega_0}{2j}(\frac{A}{1+j\omega_0A})$, with $\frac{A}{1\pm j\omega_0A}$ as two modes (check lecture 2 for fundamental modes)

Then check lecture 3 for fundamental mode and complex poles, we get:

$y(t) = \frac{\omega_0}{2j}(e^{j\omega_0t} - e^{-j\omega_0t}) = \omega_0 sin\omega_0t$, $t > 0$.

An alternative (ugly) approach

$\frac{Y}{X} = \frac{\omega_0^2A^2}{1+\omega_0^2A^2} = \omega_0^2A^2\sum^\infty_{l=0}(-\omega_0^2A^2)^l$

then if $x(t) = \delta(t)$:

$y(t) = \sum^\infty_{l=0}\omega_0^2(-\omega_0^2)^lA^{2l+2}\delta(t) = \omega_0^2t-\omega_0^4\frac{t^3}{3!} + \omega_0^6\frac{t^5}{5!} \cdots = \omega_0sin\omega_0t$