ECE 2221 SIGNALS AND SYSTEMS - International …staff.iium.edu.my/sigit/ece2221/pdf/03...

E C E 2 2 2 1 S I G N A L S A N D S Y S T E M S

ECE 2221 Signals and Systems, Sem 3 2010/2011, Dr. Sigit Jarot

1

Course Objectives

Dr. Sigit PW JarotECE 2221 Signals and Systems 2

To provide an analysis skill of the continuous‐time signals and systems as reflected to their roles in engineering

practice.

To expose students to both the time‐domain and frequency‐domain methods of analyzing signals and

systems.

To illustrate the potential applications of this course as a Pre‐requisite course to communication engineering and principles, digital signal processing and control system.

OBE (Outcome Based Education)Learning Outcomes


Classify, characterize and conduct basic of signals and systems.

Analyze continuous‐time signals and systems in time domain using convolution.

Analyze continuous‐time signals and systems in frequency domain using Laplace transform.

Analyze continuous‐time signals and systems in frequency domain using Fourier series and Fourier transform.

Acquire introductory‐level knowledge of discrete‐time signals and systems, and sampling theory.

Work in group to perform basic simulation of signals and systems analysis.

After completion of this course the students will be able to:

Course Synopsis


Introduction to Signals

Introduction to Systems

Time‐Domain Analysis of Continuous‐Time Systems

Frequency‐Domain System Analysis: the Laplace Transform

MID‐TERM Examination

Signals Analysis using the Fourier Series

Signals Analysis using the Fourier Transform

Introduction to Discrete Time Signals and Systems Analysis

FINAL Examination

Outline


5

Concept of system and system classification Linear time‐invariant continuous‐time systems Linearity and time‐invariance Causality and stability

System Representation Systems represented by differential equations Convolution integral

System interconnection

System Concept


6

System is a mathematical transformation of an input signal (or signals) into an output signal (or signals)

SIGNAL A set of data or information. A function or sequence of values that represents information. A function of one or more variables (e.g. time, frequency, space,..) that conveys information on the nature of a physical phenomenon.

SYSTEM A system is an entity that processes a set of signals (inputs) to yield another sets of signals (outputs)

SystemInputSignal

OutputSignal

Response of a Linear System

A system’s output for t ≥ 0 is the result of 2 independent causes:1. The initial condition at t = 0 zero‐input response2. The input x(t) for t ≥ 0 zero‐state response

Decomposition property:

Total response = zero‐input response + zero‐state response

x(t) y(t) +y0(t) x(t) ys(t)=

t

C tdxC

tRxvty0

0)(1)()0()(

zero‐input response zero‐state response

The importance of Impulse Response h(t)

Zero‐state response assumes that the system is in “rest” state, i.e. all internal system variables are zero.

Deriving and understanding zero‐state response depends on knowing the impulse response h(t) to a system.

Any input x(t) can be broken into many narrow rectangular pulses. Each pulse produces a system response.

Since the system is linear and time invariant, the system response to x(t) is the sum of its responses to all the impulse components.

h(t) is the system response to the rectangular pulse at t=0 as the pulse width approaches zero.

Zero‐state Response (1)

We now consider how to determine the system response y(t) to an input x(t) when system is in zero state.

Define a pulse p(t) of unit height and width Δτ at t=0 .

Input x(t) can be represented as sum of narrow rectangular pulses.

The pulse at t=nΔτ has a height x(t)=x(nΔτ) . This can be expressed as x(nΔτ)p (t-nΔτ). Therefore x(t) is the sum of all [x(nΔτ)/ Δτ] . such

pulses, i.e,

)()(lim

)()(lim)(

0

0

ntpnx

ntpnxtx


The term [x(nΔτ)/ Δτ]p(t-nΔτ) represents a pulse p(t-nΔτ) with height x(nΔτ).

As Δτ 0, height of strip ∞, but area remain x(nΔτ) and

)()()()(

ntnxntpnx

)()(lim)(

0ntnxtx


x(t) y(t)


Therefore,

Knowing h(t), we can determine the response y(t) to any input x(t). Observe the all‐pervasive nature of the system’s characteristic modes, which

determines the impulse response of the system.

LTI Systemh(t)x(t) y(t)

y(t) = x(t) * h(t)

)()(lim)(

0nthnxty

dthxty )()()(

dthxty )()()(

How to determine the unit impulse response h(t) (1)

Given that a system is specified by the following differential equation, determine its unit impulse response h(t).

Remember the general system equation:

It can be shown that the impulse response h(t) is given by:

where u(t) is the unit step function, and yn(t) is a linear combination of the characteristic modes of the system.

)()()()( txDPtyDQ

)()]()([)( tutyDPth n

tN

ttn

Necececty ......)( 2121

dtdxty

dtdy

dtyd

)(232

2

How to determine the unit impulse response h(t)? (2)

The constants ci are determined by the following initial conditions:

Note yn(k)(0) is the kth derivative of yn(t) at t=0.

The above is true if M, the order of P(D) is less the N, the order of Q(D)(which is generally the case for most stable systems).

0)0(....)0()0()0()2(

N

nnnn yyyy

1)0()1(

N

ny

Example

Find the unit impulse response of a systems specified by the equation :

)()()23( 2 tDxtyDD

Exercise 1

Find the unit impulse response of a systems specified by the equation :

)()5()()34( 2 txDtyDD

Exercise

Find the unit impulse response of a systems specified by the equation : )()5()()34( 2 txDtyDD

Exercise 2

Find the unit impulse response of a systems specified by the equation : )()117()()65( 22 txDDtyDD

Exercise Find the unit impulse response of a systems specified by the equation :

)()117()()65( 22 txDDtyDD

The Convolution Integral

The derived integral equation occurs frequently in physical sciences, engineering, and mathematics.

It is given the name: the convolution integral. The convolution integral of two function x1(t) and x2(t) is denoted symbolically as x1(t)* x2(t)

And is defined as

Note that the convolution operator is linear, i.e. it obeys the principle of superposition.

dtxxtxtx )()()()( 2121

Properties of Convolution (1)

Commutative Property (order of operands does not matter):

Associative Property (order of operator does not matter):

Distributive Property:


Shift Property: if

then

also

Impulse Property Convolution of a function x(t) with a unit impulse results in the function x(t).


Width Property: Duration of x1(t)=T1 , and duration of x2(t)=T2 , then duration of x1(t)*x2(t)=T1 + T2

Causality Property:

Convolution Table (1)

Use table to find convolution results easily:

When the input is complex …

What happens if input x(t) is not real, but is complex? If x(t) = xr(t)+ jxi(t) , where xr(t) and xi(t) are the real and imaginary part of x(t) , then

That is, we can consider the convolution on the real and imaginary part separately.

)()()()()()(

)()()()(

tjytytxtjhtxth

tjxtxthty

ir

ir

ir

Convolution usingGraphical Method

ECE 2 2 2 1 S I GNA L S AND SYST EMS

The Convolution Integral

The derived integral equation occurs frequently in physical sciences, engineering, and mathematics.

It is given the name: the convolution integral. The convolution integral of two function x1(t) and x2(t) is denoted symbolically as x1(t)* x2(t)

And is defined as

Note that the convolution operator is linear, i.e. it obeys the principle of superposition.

dtxxtxtx )()()()( 2121

Convolution Integral

The convolution integral of two function x1(t) and x2(t) is denoted symbolically as x1(t)* x2(t)

And is defined as

System output (i.e. zero‐state response) is found by convolvinginput x(t) with system impulse response h(t) .

dtxxtxtx )()()()( 2121

LTI Systemh(t)x(t) y(t)

y(t) = x(t) * h(t)

dthxty )()()(

Convolution Integral

dthxthtxty )()()()()(

The output is the sum/integration of scaled and shifted impulse responses

The definition of the convolution integral in LTI systems :

Demonstration of convolution sum

-4 -2 0 2 4 6 8 10 120

2

4

6input function,x

point number n

-4 -2 0 2 4 6 8 10 120

1

2

3

4system function,h

point number n

convolution sum for n = 0

m

]m0[h]m[x]0[y

For n = 0:

m

]mn[h]m[x]n[y

The value of the convolution sum is the sum of the product terms over m:

0 5 100

5

point number m

inpu

t-fun

ctio

n,x[

m]

0 5 100

2

4

h[n-

m] f

or n

= 0

point number m

0 5 100

0.5

1

prod

uct

point number m0 5 10

0

50Convolution Sum

point number n

= 0

0 5 100

5

point number m

inpu

t-fun

ctio

n,x[

m]

0 5 100

2

4

h[n-

m] f

or n

= 1

point number m

0 5 100

0.5

1

prod

uct


0

50Convolution Sum

point number n


m

]m1[h]m[x]1[y

For n = 1:

m

]mn[h]m[x]n[y


= 0

0 5 100

5

point number m

inpu

t-fun

ctio

n,x[

m]

0 5 100

2

4

h[n-

m] f

or n

= 2

point number m

0 5 100

0.5

1

prod

uct


0

50Convolution Sum

point number n


m

]m2[h]m[x]2[y

For n = 2:

m

]mn[h]m[x]n[y


= 1

0 5 100

5

point number m

inpu

t-fun

ctio

n,x[

m]

0 5 100

2

4

h[n-

m] f

or n

= 3

point number m

0 5 100

1

2

prod

uct


0

50Convolution Sum

point number n


m

]m3[h]m[x]3[y

For n = 3:

m

]mn[h]m[x]n[y


= 4

0 5 100

5

point number m

inpu

t-fun

ctio

n,x[

m]

0 5 100

2

4

h[n-

m] f

or n

= 4

point number m

0 5 100

2

4

prod

uct


0

50Convolution Sum

point number n


m

]m4[h]m[x]4[y

For n = 4:

m

]mn[h]m[x]n[y


= 10

0 5 100

5

point number m

inpu

t-fun

ctio

n,x[

m]

0 5 100

2

4

h[n-

m] f

or n

= 5

point number m

0 5 100

5

prod

uct


0

50Convolution Sum

point number n


m

]m5[h]m[x]5[y

For n = 5:

m

]mn[h]m[x]n[y


= 20

0 5 100

5

point number m

inpu

t-fun

ctio

n,x[

m]

0 5 100

2

4

h[n-

m] f

or n

= 6

point number m

0 5 100

5

prod

uct


0

50Convolution Sum

point number n


m

]m6[h]m[x]6[y

For n = 6:

m

]mn[h]m[x]n[y


= 30

0 5 100

5

point number m

inpu

t-fun

ctio

n,x[

m]

0 5 100

2

4

h[n-

m] f

or n

= 7

point number m

0 5 100

5

10

prod

uct


0

50Convolution Sum

point number n


m

]m7[h]m[x]7[y

For n = 7:

m

]mn[h]m[x]n[y


= 40

0 5 100

5

point number m

inpu

t-fun

ctio

n,x[

m]

0 5 100

2

4

h[n-

m] f

or n

= 8

point number m

0 5 100

10

prod

uct


0

50Convolution Sum

point number n


m

]m8[h]m[x]8[y

For n = 8:

m

]mn[h]m[x]n[y


= 43

0 5 100

5

point number m

inpu

t-fun

ctio

n,x[

m]

0 5 100

2

4

h[n-

m] f

or n

= 9

point number m

0 5 100

10

20

prod

uct


0

50Convolution Sum

point number n


m

]m9[h]m[x]9[y

For n = 9:

m

]mn[h]m[x]n[y


= 38

0 5 100

5

point number m

inpu

t-fun

ctio

n,x[

m]

0 5 100

2

4

h[n-

m] f

or n

= 1

0

point number m

0 5 100

10

20

prod

uct


0

50Convolution Sum

point number n


m

]m10[h]m[x]10[y

For n = 10:

m

]mn[h]m[x]n[y


= 24

0 5 100

5

point number m

inpu

t-fun

ctio

n,x[

m]

0 5 100

2

4

h[n-

m] f

or n

= 1

1

point number m

0 5 100

0.5

1

prod

uct


0

50Convolution Sum

point number n


m

]m11[h]m[x]11[y

For n = 11:

m

]mn[h]m[x]n[y


= 0

Animated Illustrations of Convolution Integral46

http://www.jhu.edu/signals/convolve/

Graphical MethodStep‐by‐step Procedure

dthxthtxty )()()()()(

1. Choose one signal to be x(t), the other to be h(t) ; draw them both in axis

2. ROTATE one of the signal h() about the axis3. SHIFT the rotated signal h(-) to the right by t

4. MULTIPLY x() with the flipped/shifted version of signal h()

5. Calculate the area under this product using correct limit of INTEGRATION

6. Step 4‐5 produces the equation x(t)*h(t) over specified region

7. Repeat step 4‐5 for all regions of interest

123

4

5

ExampleConvolution using Graphical Method

Convolve the following two functions:

Replace t with in f(t) and g(t) Choose to flip and slide g() since it is simpler and symmetric Functions overlap like this:

3

2

f()2

‐2 + t 2 + t

g(t‐)

*2

2

t

f(t)

‐2 2

3

t

g(t)


Convolution can be divided into 5 partsI. t < ‐2

Two functions do not overlap Area under the product of the

functions is zero

II. ‐2 t < 0 Part of g(t) overlaps part of f(t) Area under the product of the

functions is

62

3262

2322

3)2(3222

0

22

0

tttd

tt

3

2

f()2

‐2 + t 2 + t

g(t‐)

3

2

f()2

‐2 + t 2 + t

g(t‐)


III. 0 t < 2 Here, g(t) completely overlaps f(t) Area under the product is just

IV. 2 t < 4 Part of g(t) and f(t) overlap Calculated similarly to ‐2 t < 0

V. t 4 g(t) and f(t) do not overlap Area under their product is zero

6 22

3 232

0

22

0

d

3

2

f()2

‐2 + t 2 + t

g(t‐)

3

2

f()2

‐2 + t 2 + t

g(t‐)


Result of convolution (5 intervals of interest):

4for 0

42for 24 1223

20for 6

02for 623

2for 0

)(*)()(2

2

t

ttt

t

tt

t

tgtfty

t

y(t)

0 2 4‐2

6

No overlap

Partial overlap

Complete overlap

Partial overlap

No overlap

Convolution using graphical method (1)

Determine graphically y(t)=x(t)* h(t) for x(t)=e-t u(t) and h(t)=e-2t u(t) .

Remember: variable of integration is τ, not t .

Convolution using graphical method (2)

Signals and Systems_Simon Haykin & Barry Van Veen 56

Time‐Domain Representations of LTI Systems

2.4 The Convolution Integral1. A continuous‐time signal can be expressed as a weighted superposition of time‐shifted impulses.

-x(t) x( ) (t - )d

(2.10)

The sifting property of the impulse !

2. Impulse response of LTI system H: LTI systemH

Input x(t) Output y(t)

Output:

y t H x t H x t d

-y(t) x( )H{ (t - )}d

(2.10)

Linearity property

3. h(t) = H{ (t)} impulse response of the LTI system HIf the system is also time invariant, then

H{ (t - )} h(t - ) (2.11) A time‐shifted impulse generates a time‐shifted impulse response

output

Fig. 2.9.-y(t) x( )h(t )d

(2.12)



Convolution integral:

x t h t x h t d

2.5 Convolution Integral Evaluation Procedure1. Convolution integral:

-y(t) x( )h(t )d

(2.13)

2. Define the intermediate signal:

tw x h t = independent variable,

t = constant

h (t ) = h ( ( t)) is a reflected and shifted (by t) version of h().

3. Output:

t-

y(t) w ( )d (2.14) The time shift t determines the time at which we evaluate the output of

the system.



Procedure 2.2: Reflect and Shift Convolution Integral Evaluation1. Graph both x() and h(t ) as a function of the independent variable . Toobtain h(t ), reflect h() about = 0 to obtain h( ) and then h( ) shift by t.

2. Begin with the shift t large and negative. That is, shift h( ) to the far left onthe time axis.

3. Write the mathematical representation for the intermediate signal wt ().4. Increase the shift t (i.e., move h(t ) toward the right) until the mathematicalrepresentation for wt () changes. The value of t at which the change occursdefines the end of the current set and the beginning of a new set.

5. Let t be in the new set. Repeat step 3 and 4 until all sets of shifts t and thecorresponding mathematical representations for wt () are identified. Thisusually implies increasing t to a very large positive number.

6. For each sets of shifts t, integrate wt () from = to = to obtain y(t).



Figure 2.10 (p. 117)Input signal and LTI system impulse response for Example 2.6.

Example 2.6 Reflect‐and‐shift Convolution Evaluation

Given 1 3x t u t u t 2h t u t u t and as depicted in Fig. 2‐10, Evaluate the convolution integral y(t) = x(t) h(t).



Figure 2.10 (p. 117)Input signal and LTI system impulse response for Example 2.6.

<Sol.>1. Graph of x() and h(t ): Fig. 2.11 (a).2. Intervals of time shifts: Four intervals

1’st interval: t < 12’nd interval: 1 ≤ t < 3 3’rd interval: 3 ≤ t < 54th interval: 5 ≤ t

3. First interval of time shifts: t < 1

1, 10, otherwiset

tw

Fig. 2.11 (b).

4. Second interval of time shifts: 1 ≤ t < 3wt() = 0



Figure 2.11 (p. 118)Evaluation of the convolution integral for Example 2.6. (a) The input x() depicted above the reflected and time‐shifted impulse response. (b) The product signal wt() for 1 t < 3. (c) The product signal wt() for 3 t < 5. (d) The system output y(t).

t



5. Third interval: 3 ≤ t < 5

1, 2 30, otherwiset

tw

Fig. 2.11 (c).

6. Fourth interval: 5 ≤ t wt() = 0

7. Convolution integral: 1) For t < 1 and t 5: y(t) = 02) For second interval 1 ≤ t < 3, y(t) = t 13) For third interval 3 ≤ t < 5, y(t) = 3 (t 2)

0, 11, 1 3

5 , 3 50, 5

tt t

y tt t

t

Example 2.7 RC Circuit OutputFor the RC circuit in Fig. 2.12, assume that the circuit’s time constant is RC = 1 sec. Ex. 1.21shows that the impulse response of this circuit is h(t) = e t u(t).Use convolution to determine the capacitor voltage, y(t), resulting from an input voltage x(t) = u(t) u(t 2).

Figure 2.12 (p. 119)RC circuit system with the voltage source x(t) as input and the voltage measured across the capacitor y(t),

as output.



<Sol.>1. Graph of x() and h(t ): Fig. 2.13 (a).

2. Intervals of time shifts: Three intervals

1’st interval: t < 02’nd interval: 0 ≤ t < 2 3’rd interval: 2 ≤ t

RC circuit is LTI system, so y(t) = x(t) h(t).

1, 0 20, otherwise

x

and

3. First interval of time shifts: t < 04. Second interval of time shifts: 0 ≤ t < 2

wt() = 0

For t > 0, , 00, otherwise

t

te tw

Fig. 2.13 (b).

5. Third interval: 2 ≤ t

, 0 20, otherwise

t

tew

Fig. 2.13 (c).

,0, otherwise

tt e th t e u t



6. Convolution integral: 1) For t < 0: y(t) = 02) For second interval 0 ≤ t < 2:

3) For third interval 2 ≤ t:

001

t tt t ty t e d e e e

2 2 2

001t t ty t e d e e e e

2

0, 01 , 0 2

1 , 2

t

t

ty t e t

e e t

Fig. 2.13 (d).


Time‐Domain Representations of LTI SystemsFigure 2.13 (p. 120)

Evaluation of the convolution integral for Example 2.7. (a) The input x() superimposed over the reflected and time‐shifted impulse response h(t – ), depicted as a function of . (b) The

product signal wt() for 0 t < 2. (c) The product signal wt() for t 2. (d) The system output y(t).

t



Example 2.8 Another Reflect‐and‐Shift Convolution EvaluationSuppose that the input x(t) and impulse response h(t) of an LTI system are, respectively, given by

1 1 3x t t u t u t and 1 2 2h t u t u t

Find the output of the system.



<Sol.>1. Graph of x() and h(t ): Fig. 2.14 (a).2. Intervals of time shifts: Five intervals

1’st interval: t < 02’nd interval: 0 ≤ t < 2 3’rd interval: 2 ≤ t < 34th interval: 3 ≤ t < 55th interval: t 5

3. First interval of time shifts: t < 0 wt() = 0

4. Second interval of time shifts: 0 ≤ t < 2

1, 1 1

0, otherwiset

tw

Fig. 2.14 (b).

5. Third interval of time shifts: 2 ≤ t < 3

1, 1 3

0, otherwisetw

Fig. 2.14 (c).

6. Fourth interval of time shifts: 3 ≤ t < 5


Time‐Domain Representations of LTI SystemsFigure 2.14 (p. 121) Evaluation of the convolution integral for Example 2.8. (a) The input x() superimposed on the reflected and time‐shifted impulse response h(t – ), depicted as a function of . (b) The product signal wt() for 0 t < 2. (c) The product signal wt() for 2 t < 3. (d) The product signal wt() for 3 t < 5. (e) The product signal wt() for t 5. The system output y(t).

t

Interconnected Systems (1)

Parallel connected system

Cascade system

And commutative property

)()()()()( 21 txthtxthty

)()]()([)( 21 txththty

Interconnected Systems (2)

Integration

Also true for differentiation

Let x(t)=δ(t) and y(t)=h(t)(x(t)is an impulse, and h(t) is the impulse response of the system),

Then g(t), the step response is : dhtg

t

)()(

Total Response

A system’s output for t ≥ 0 is the result of 2 independent causes:1. The initial condition at t = 0 zero‐input response2. The input x(t) for t ≥ 0 zero‐state response

Decomposition property:

Total response = zero‐input response + zero‐state response

x(t) y(t) +y0(t) x(t) ys(t)=

)()(1

thtxecresponsetotalN

k

tk

k

zero‐input component zero‐state component

ECE 2221 SIGNALS AND SYSTEMS - International …staff.iium.edu.my/sigit/ece2221/pdf/03...

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