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LECTURE 1: REVIEW Continuous-Time Systems

Objectives:The purpose of this lecture is to review the following topics as apreparatory for further chapters in this module.

Signal manipulationsClassification of systemsSolving second order linear differential equation with constantcoefficients

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At the end of this lesson, students will be able to:

Manipulate a given signal by scaling, reflection andtime-shifting

Differentiate and classify different types of systems

Solve second order linear differential equations usingclassical method and Laplace transform

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Content:

Signal manipulation Product of 2 signals Signal operation Classification of system

Solution of differential equations

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REVIEW: CONTINUOUS-TIME SYSTEM

1. Signal Manipulation

Usually in the form of x ( t a ) x ( t + a ) shift left a unit x ( t - a ) shift right a unit x ( - t + a ) = x ( -( t a )) reflect & shift right a unit x ( - t - a ) = x ( -( t + a )) reflect & shift left a unit

Example 1.1 :Given the following function x(t),Sketch:

a) x(t + 1) b) x(t 2)

c) x(-t + 2)d) x(-t - 2)

4

x(t )

t -1 0 1 2

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Solution5

x(t + 1)

t -1 0 1 2

x(t - 2)

t -1 0 1 2

x(-( t - 2))

t -1 0 1 2 3

x(-( t + 2))

t -4 -3 -2 -1 0

a) x(t+1) b) x(t-2)

c) x(-t+2) d) x(-t-2)

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2. Product of 2 signals

Example 2.1:Given x 1(t) = 2 u(t) and x 2(t) = 4 u(-t + 2).Find and sketch x 1(t) x2(t).

Solution:x

1(t) x

2(t) = 8 u(t) u(-t + 2)

Graphically:

6

2 u(t)

t 0

2

4 u(-t + 2) = 4 u(-(t - 2))

t -2 0 2

2

4

8 u(t) u(-t + 2)

t 0 2

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3. Signal Operations

a) Time reflection b) Time scalingc) Time shiftingd) Combination of the above operationsIn the form of ,

a) b)

c) d)

time reflection time shiftingtime scaling

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)( bat x

))(()(abt a xbat x ))(()(

abt a xbat x

))(()(ab

t a xbat x ))(()(ab

t a xbat x

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Example 3.1:Given the following function x(t).

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x(t )

t -1 0 1 2

a) Sketch x(3t + 2)b) Sketch x(3t 2)c) Sketch x(-3t+2)d) Sketch x(-3t 2)

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4. Classification Of System

4.1) Continuous OR Discrete-time system4.2) Causal OR non-causal system4.3) Time-variant OR time-invariant system4.4) Linear OR non-linear system

a) Continuous-time system as a function of time t

y(t) = 2x 2(t + 1) +5

b) Discrete-time system as a function of k y(k) = 2x(k 1) + x(k)

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4.1 Continuous OR Discrete-Time System

This may be the simplest classification to understand as the idea of discrete-timeand continuous-time is one of the most fundamental properties to all of signalsand system.

A system where the input and output signals are continuous is a continuoussystem . This signal is defined over a continuous range of time

System where the input and output signals are discrete is a discrete system .Signals at discrete instant of time such as t 1 , t 2 , t 3 , t 4

Discrete-time systems can arise from sampling continuous-time signal whichcan be done uniformly

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4.2 Causal OR Non-causal System

A causal system is one that is nonanticipative ; that is, the output maydepend on current and past inputs, but not future inputs. All "realtime"systems must be causal, since they can not have future inputs available tothem. The output cannot start before the input is applied.

One may think the idea of future inputs does not seem to make muchphysical sense; however, we have only been dealing with time as our

dependent variable so far, which is not always the case. Imagine rather thatwe wanted to do image processing. Then the dependent variable mightrepresent pixels to the left and right (the "future") of the current position onthe image, and we would have a noncausal system.

Figure 1: For a typical system to be causal...

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The output at time t0 , y(t0) , can only depend on the portion of the inputsignal before t0 .

Figure 2

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4.3 Linear OR Nonlinear System

A linear system is any system that obeys the properties of scaling(homogeneity) and superposition (additivity), while a nonlinear system is any system that does not obey at least one of these.

To show that a system H obeys the scaling property is to show that:

H(kf(t)) =kH(f(t))

Figure 3: A block diagram demonstrating the scaling property of linearity

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To demonstrate that a system H obeys the superposition property of linearity is toshow that

H(f1(t) +f2(t)) =H(f1(t)) +H(f2(t))

Figure 4: A block diagram demonstrating the superposition property of linearity

It is possible to check a system for linearity in a single (though larger) step. To do

this, simply combine the first two steps to getH(k1f1(t) +k2f2(t)) =k1H(f1(t)) +k2H(f2(t))

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A linear system response:

Total response = zero-input response + zero-state response

This componentresulted from initialconditions at t=0

This componentresulted from input x(t)

at t >= 0

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4.4 Time Invariant OR Time Variant System

A time invariant system is one that does not depend on when it occurs: theshape of the output does not change with a delay of the input. That is to say thatfor a system H where H(f(t)) =y(t) , H is time invariant if for all T

H(f(t T)) =y(t T)

Figure 5: This block diagram shows what the condition for time invariance. The

output is the same whether the delay is put on the input or the output.

When this property does not hold for a system, then it is said to be time variant ,or time-varying.

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S

Delay T

Delay T

S

X(t)

X(t)

y(t-T)y(t)

y(t-T)x(t-T)

y (t)

t

y (t)

tT

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How to identify time-invariant and time varying systems?

Time-invariant systems:

A system with and input-output relationship described by a linear differentialequation with constant coefficients is time-invariant. Examples:

RLC networks which consist of passive elements are time-invariant. Theselinear systems are known as Linear time-invariant (LTI) systems.

)()(12)(4)( ''' t xt yt yt y

)()()(15)(

6)(

10 '22

t xt xt ydt

t dydt

t yd

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Time-varying systems:

If the coefficients in a linear differential are functions of time, then thesystem is time-varying. Examples:

Networks with active components e.g. transistors are time-varyingsystems.

xt yt dt dy

)32(2

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5. Solution Of Differential Equations

Methods:

Classical method Laplace transform

Impulse response and convolution State space

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5.1 Classical Method: Homogeneous & ParticularSolutions

For any 2 nd order system, it can be represented by thefollowing general equation:

y(t) + ay(t) + by(t) = x(t)

To solve the above equation, y(t) must be obtained;

y(t) = y h(t) + y p(t)

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Case 1: two real roots

Case 2: a real double root

Case 3:complex conjugate roots

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042 ba

042 ba

042 ba

t r t r

Ae Ae y 21

t r t r Bte Ae y 11

21 , r r

11 , r r

qi p)sincos( qt Bqt Ae y pt

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Example 5.1.1 (Homogenous solution)

Solve

Solution:

Step 1: The characteristic equation,

Step 2: The roots,

Step 3: The roots if of case 1, hence,

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02 y y y

022 r r

2,12,1r

t t Be Ae y 2

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Example 5 .1 .2 (Homogenous solution)

Solve

Solution:

Characteristic equation;

The roots;r = -2, -4 j 3

The homogenous solution;

0)(50)(37)(8)( t yt yt yt y

050378 23 r r r

)3sin3cos()( 42 t C t Be Aet y t t h

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Example 5 .1 .3 (Homogenous solution)

Solve

Solution:

The characteristic equation;

The roots; r = -2, -2, -3

The homogenous solution;

0)(12)(16)(7)( t yt yt yt y

012167 23 r r r

t t t h Ce Bte Aet y

322)(

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Particular Solution, y p (t)

In order to solve the particular solution, first findthe general form and then determine themultiplying constant in the general form by matching coefficients. The general form of theparticular solution to some x(t) is listed in thefollowing table.

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x (t ) y p (t )

Constant Constant

tn P ntn + P n-1 t

n-1 + + P 1t + P0

e t Pe t if is not a characteristic root of the differential equation

Ptet

if is a distinct root of the differential equationP n-1 tn-1 e t if is a (n-1) multiple characteristic root of the differential

equation

cos tsin t

P 1cos t + P2sin t

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Example 5 .1 .4 (Homogenous & Particular solution)

Solve y(t) + 6y(t) + 25y(t) = 50

Solution:

y h(t) = e-3 t ( Acos4 t + Bsin4 t )

y p(t) = C y p (t) = 0 y p(t)= 0

Subsituting into the D.E. and compare coefficients we obtainyp(t) = 2

Therefore, y(t) = e-3 t ( Acos4 t + Bsin4 t ) + 2

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Example 5 .1 .5 (Homogenous & Particular solution)

Solve

Solution:

By comparing coefficient, C = 25/82 and D = 225/82

Hence,

t et yt yt y t 3cos50)(24)(10)( 2

t t h Be Aet y

64)(

)3sin3cos()( 2 t Dt C et y t p

)3sin82

2253cos

8225

()( 264 t t e Be Aet y t t t

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Example 5 .1 .6 (Initial values problem)

Solve where y(0)=1, y(0)=0.

Solution

By comparing coefficient, C = 1/9 and D = .

By considering the initial condition, the value of A and B can beevaluated.

t et yt yt y 31)(9)(6)(

t t h Bte Aet y

33)(t

p e Dt C t y32)(

t t t et Bte Aet y 32332

1

9

1)(

t t t

et teet y3233

2

1

9

1

3

8

9

8)(

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Example 5 .1 .7 (Initial values problem)

Solve where y(0)=1, y(0)=0.

Solution

By comparing coefficient, C = 1/4, D = -4/75 and E = 1/25

By considering the initial condition, the value of A and B can be

evaluated.

t et yt yt y t 2sin)(13)(6)( 3

)2sin2cos()( 3 t Bt Aet y t ht E t DCet y t

p2sin2cos)( 3

t t et Bt Aet y t t 2sin251

2cos75

441

)2sin2cos()( 33

t t et t et y t t 2sin251

2cos75

441

)2sin5077

2cos300241

()( 33

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PRACTICE:

Solve the differential equation:

if the initial conditions are y(0) = 2, y(0) = 2 and the inputs

are:

a) 10e -3t b) 5c) 2e -3t +10 cos 3t

33

)()(2)(

3)(

2

2

t xt ydt

t dydt

t yd

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Reference:

B.P. Lathi (2005), Linear Systems & Signals, Oxford University Press(pg. 68-164)

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