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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
Dr. Sigit PW JarotECE 2221 Signals and Systems 3
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
Dr. Sigit PW JarotECE 2221 Signals and Systems 4
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
ECE 2221 Signals and Systems, Sem 3 2010/2011, Dr. Sigit Jarot
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
ECE 2221 Signals and Systems, Sem 3 2010/2011, Dr. Sigit Jarot
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
Zero‐state Response (2)
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
Zero‐state Response (3)
Zero‐state Response (4)
x(t) y(t)
Zero‐state Response (5)
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:
Properties of Convolution (2)
Shift Property: if
then
also
Impulse Property Convolution of a function x(t) with a unit impulse results in the function x(t).
Properties of Convolution (3)
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:
Convolution Table (2)
Convolution Table (3)
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
point number m0 5 10
0
50Convolution Sum
point number n
convolution sum for n = 1
m
]m1[h]m[x]1[y
For n = 1:
m
]mn[h]m[x]n[y
The value of the convolution sum is the sum of the product terms over m:
= 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
point number m0 5 10
0
50Convolution Sum
point number n
convolution sum for n = 2
m
]m2[h]m[x]2[y
For n = 2:
m
]mn[h]m[x]n[y
The value of the convolution sum is the sum of the product terms over m:
= 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
point number m0 5 10
0
50Convolution Sum
point number n
convolution sum for n = 3
m
]m3[h]m[x]3[y
For n = 3:
m
]mn[h]m[x]n[y
The value of the convolution sum is the sum of the product terms over m:
= 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
point number m0 5 10
0
50Convolution Sum
point number n
convolution sum for n = 4
m
]m4[h]m[x]4[y
For n = 4:
m
]mn[h]m[x]n[y
The value of the convolution sum is the sum of the product terms over m:
= 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
point number m0 5 10
0
50Convolution Sum
point number n
convolution sum for n = 5
m
]m5[h]m[x]5[y
For n = 5:
m
]mn[h]m[x]n[y
The value of the convolution sum is the sum of the product terms over m:
= 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
point number m0 5 10
0
50Convolution Sum
point number n
convolution sum for n = 6
m
]m6[h]m[x]6[y
For n = 6:
m
]mn[h]m[x]n[y
The value of the convolution sum is the sum of the product terms over m:
= 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
point number m0 5 10
0
50Convolution Sum
point number n
convolution sum for n = 7
m
]m7[h]m[x]7[y
For n = 7:
m
]mn[h]m[x]n[y
The value of the convolution sum is the sum of the product terms over m:
= 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
point number m0 5 10
0
50Convolution Sum
point number n
convolution sum for n = 8
m
]m8[h]m[x]8[y
For n = 8:
m
]mn[h]m[x]n[y
The value of the convolution sum is the sum of the product terms over m:
= 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
point number m0 5 10
0
50Convolution Sum
point number n
convolution sum for n = 9
m
]m9[h]m[x]9[y
For n = 9:
m
]mn[h]m[x]n[y
The value of the convolution sum is the sum of the product terms over m:
= 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
point number m0 5 10
0
50Convolution Sum
point number n
convolution sum for n = 10
m
]m10[h]m[x]10[y
For n = 10:
m
]mn[h]m[x]n[y
The value of the convolution sum is the sum of the product terms over m:
= 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
point number m0 5 10
0
50Convolution Sum
point number n
convolution sum for n = 11
m
]m11[h]m[x]11[y
For n = 11:
m
]mn[h]m[x]n[y
The value of the convolution sum is the sum of the product terms over m:
= 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)
ExampleConvolution using Graphical Method
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‐)
ExampleConvolution using Graphical Method
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‐)
ExampleConvolution using Graphical Method
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)
Signals and Systems_Simon Haykin & Barry Van Veen 57
Time‐Domain Representations of LTI Systems
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.
Signals and Systems_Simon Haykin & Barry Van Veen 58
Time‐Domain Representations of LTI Systems
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).
Signals and Systems_Simon Haykin & Barry Van Veen 59
Time‐Domain Representations of LTI Systems
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).
Signals and Systems_Simon Haykin & Barry Van Veen 60
Time‐Domain Representations of LTI Systems
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
Signals and Systems_Simon Haykin & Barry Van Veen 61
Time‐Domain Representations of LTI Systems
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
Signals and Systems_Simon Haykin & Barry Van Veen 62
Time‐Domain Representations of LTI Systems
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.
Signals and Systems_Simon Haykin & Barry Van Veen 64
Time‐Domain Representations of LTI Systems
<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
Signals and Systems_Simon Haykin & Barry Van Veen 65
Time‐Domain Representations of LTI Systems
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).
Signals and Systems_Simon Haykin & Barry Van Veen 66
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
Signals and Systems_Simon Haykin & Barry Van Veen 67
Time‐Domain Representations of LTI Systems
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.
Signals and Systems_Simon Haykin & Barry Van Veen 68
Time‐Domain Representations of LTI Systems
<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
Signals and Systems_Simon Haykin & Barry Van Veen 69
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