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1Signal and System Analysis(MCT 2121)
Department of Mechatronics Engineering
International Islamic University Malaysia
Wahju [email protected]
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2Tell me and I'll forget;
Show me and I may remember;
I do and I'll understand.
Chinese proverbs
bung macht den Meister
(Practice makes perfect) German proverbs
That man can have nothing but what he strives for
Al Quran (An-Najm) 53:39
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3Contents
System response (CT-domain analysis)
Zero input response
Impulse response
Zero state response
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4System Response
For a linear system:LTI y(t)x(t)
Systems total response
Zero-input response
Response when x(t) = 0
Results from internal system conditions only
Independent of x(t)
For most filtering applications (e.g. your stereo system), we
want no zero-input response.
Zero-state response
Response to non-zero x(t)
A system in zero state cannot generate any response for zero
input
Zero state corresponds to initial conditions being zero
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5 Example: Differential systems
There are N derivatives of y(t) and M derivatives of x(t)
Constants a1, , aN and bN-M, , bN
Using short-hand notation,above equation becomes
txbtxdt
dbtx
dt
dbtx
dt
db
tyatydt
daty
dt
daty
dt
d
NNM
M
MNM
M
MN
NNN
N
N
N
11
1
1
11
1
1
k
kk
dt
dD
dt
dD
dt
dD
2
22
txbDbDbDbtyaDaDaDDP
NN
M
MN
M
MN
DQ
NN
NN
)(
1
1
1
)(
1
1
1
Continuous-Time Domain Analysis
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6 Polynomials Q(D) and P(D)
o Normalization: a0 = 1
o N derivatives of y(t)
o M derivatives of x(t)
This differential system behaves as (M-N)th-order differentiator
if M > N
o Noise occupies both low and high frequencies
o Differentiator amplifies high frequencies
o To avoid amplification of noise in high frequencies, we assume
that M N
M
l
l
lN DbDP0
)(
Continuous-Time Domain Analysis
N
k
kN
k DaDQ0
)(
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7 Linearity: for any complex constants c1 and c2
txDPtyDQ
tyDQctxcDP 2222
tyDQctyDQc
txcDPtxcDPtxctxcDP
2211
22112211
tytyDQtxtxDP 2121
Continuous-Time Domain Analysis
tyDQctxDPctxcDP 111111
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8 System response (CT-domain analysis)
Zero input response
Impulse response
Zero state response
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9Zero Input Response: Example 1Determine the zero input response
for the system described by the differential equation:
,)(
)(2)(
3)(
2
2
dt
tdxty
dt
tdy
dt
tyd
given that the initial conditions: ,50,00 yy
5,5
52)0(
0)0(
2)(
)(
,2,1
023023
)(23
,)(,)(,)(
21
02
2
0
1
02
2
0
1
2
21
2
21
21
22
2
2
CC
eCeCy
eCeCy
eCeCty
eCeCty
Ce
dt
tdxCeCeCe
CetyCetyCety
tt
tt
t
ttt
ttt
Let:
Substitute:
Factorised:
Substitute:
Therefore:
To determine C1 and C2, utilise the initial conditions and solve
the simultaneous
equations :
Characteristic
equation
Characteristic polynomial
Characteristic roots
Sol.:
tt eety 255)(
Zero input: no input signal
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10
Notes on System Characteristics
Characteristic equation
Q(D) y(t) = 0
Polynomial Q()
o Characteristic of system
o Independent of the input
Roots 1, 2, , N
Characteristic roots a.k.a. characteristic values, eigenvalues,
natural frequencies
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11
General case:
The linear combination of y0(t) and its Nsuccessive derivatives
are zero
Assume that y0(t) = C e t
dt
dDtyDQ where0 0
0 0111 tyaDaDaD NNNN
tkk
kk
t
t
eCdt
ydtyD
eCdt
ydtyD
eCdt
dytDy
00
2
2
0
2
0
2
00
Zero-Input Response
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12
Zero-Input Response
Substituting into the differential equation
y0(t) = C e t is a solution provided that Q() = 0
Factorize Q() to obtain N solutions:
Assuming that no two i terms are equal
011
1
zeronon
t
Q
NN
NN
zeronon
eaaaC
0)( 21 NQ
tNtt NeCeCeCty
21 210
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13
Zero-Input Response
Could i be complex?
If complex, we can write it in Cartesian form
Exponential solution e t becomes product of two terms
For conjugate symmetric roots, and conjugate symmetric
constants,
iii j
termgoscillatin termdamping
sincos tjteeeee iittjttjt iiiiii
termgoscillatin
1
termdamping
111 cos2 CteCeCeC ittt iii
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14
Zero-Input Response
For repeated roots, the solution changes
Simplest case of a root repeated twice:
With r repeated roots
002
tyD
tetCCty 210
0 0
tyDr
trr etCtCCty 1210
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15
Zero Input Response: Example 2Determine the zero input response
for the RLC system below
given that the initial conditions: ,50,00 vy
)(tv)(tx
H1 3
F2
1)(ty
Solution: First, determine the model of the system:
dt
tdxty
dt
tdy
dt
tyd
dytydt
tdytx
dytvtvtytvdt
tdytv
tvtvtvtx
t
t
CRL
CRL
)()(2
)(3
)(
)(2)(3)(
)(
)(2)()(),(3)(,)(
)(
)()()()(
2
2
0
0
Use this model to find
the systems zero input
response
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16
Zero Input Response: Example 2
5,5
52)0(
5)0(
5)0(3)0(0
)()0(3)0()0(
0)0(
2)(
)(
,2,1
023023
)(23
,)(,)(,)(
21
02
2
0
1
02
2
0
1
2
21
2
21
21
22
2
2
CC
eCeCy
y
y
tvyyx
eCeCy
eCeCty
eCeCty
Ce
dt
tdxCeCeCe
CetyCetyCety
tt
tt
t
ttt
ttt
Let:
Substitute:
Factorised:
Substitute:
Therefore:
To determine C1 and C2, utilise the initial conditions and solve
the simultaneous
equations :
tt eety 255)(
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17
System response (CT-domain analysis)
Zero input response
Impulse response
Zero state response
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18
Impulse Response
Impulse response is the systems response when the input is an
impulse:
system)(
)(
t
tx
)(
)(
th
ty
There are 2 methods in determining the impulse response of a
system.
)()()()(
ttxtyth
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19
)( )(
,032,032
)(,)(
0)(3)(
2
,0
)(5)(3)(
2
2
3
23
tuCeth
CeCe
Cedt
tdhCeth
thdt
tdh
t
tthdt
tdh
t
tt
tt
Impulse Response: Example 3
)()()()(
ttxtyth
Method 1 Determine the impulse response for the system described
by the differential equation:
)(5)(3)(
2 txtydt
tdy
Substitute:
For:
Let:
Therefore:
Solution:
Linear Time Invariant
Causal System
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20
Impulse Response: Example 3
)(2
5)(
2
5
)(5)( 3)()(2
32
),(5)(3)(
2
2
3
23-
23-
23-
tueth
C
ttuCetCetuCe
tthdt
tdh
t
ttt
Substitute into the
equation:
)()( 2
3)( 23
23
tCetuCedt
tdh tt
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21
Impulse Response
Method 2
For a system described by the differential equation:
MM
MM
N
N
N
N
MM
MM
N
N
N
N
bDbDbDbDP
aDaDaDDQ
txDPtyDQ
txbDbDbDbtyaDaDaD
1
1
10
1
1
1
1
1
1
10
1
1
1
1
)(
)(
)()()()(
)()(
or:
where:
)( )()()()( 0 tutyDPtAth n
)(tyn
Its impulse response is given by:
where is subjected to the following initial conditions:
,1)0(,0)0(,0)0(,3
,1)0(,0)0(,2
1)0(,1
nnn
nn
n
yyyN
yyN
yN
,NM But, if its impulse response is given by: )()()( tyDPth
n
where 00 A
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22
Impulse Response: Example 4
Determine the zero impulse response for the system described by
the
differential equation:
)()(
)(6)(
5)(
2
2
txdt
tdxty
dt
tdy
dt
tyd
tt
n
n
n
eCeCty
tutyDth
DDP
tutyDPth
NM
txDtyDD
txdt
tdxty
dt
tdy
dt
tyd
2
2
3
1
21
2
2
2
2
)(
,2,3
065
)( )()1()(
1)(
)( )()()(
2,1
)( 1)(65
)()(
)(6)(
5)(
Solution:
Therefore:
where:
The characteristic equation:
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23
Impulse Response: Example 4
)( 2
)( 23)(
)( )()()( )()(
)( )()1()(
)(
1,1
123)0(
0)0(
23)(
)(
,1)0(,0)0(
2
23
2323
23
21
)0(2
2
)0(3
1
)0(2
2
)0(3
1
2
2
3
1
2
2
3
1
tuee
tueeeeth
tutytytutydt
tdytutyDth
eety
CC
eCeCy
eCeCy
eCeCty
eCeCty
yy
N
tt
tttt
nnnn
n
tt
n
n
n
tt
n
tt
n
nn
Therefore:
Therefore:
Substitute:
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24
System response (CT-domain analysis)
Zero input response
Impulse response
Zero state response
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25
Zero State Response
Zero-state response is the output which arises when all systems
initial conditions are zero.
dtxh
dthx
thtx
responseimpulseinputtyzs
)()(
)()(
)()(
)()(
LTI y(t)x(t)
h(t)
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26
Convolution Integral
Graphical View of Staircase Approximation
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27
Example: (Midterm question Sem 1 2013/2014)
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28
Convolution table
a)
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29
b)
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30
Assignment 2 (Due 31/03/2014)
Q1. Consider the low pass filter shown below. Find its
impulse
response:
y(t)h(t) = ?
x(t) = (t)
Input
Signal
Output
Signal
The differential equation for the above system is:
The impulse response function is:
EquationalDifferentiOrderFistRC
tx
RC
ty
dt
tdy
)()()(
)( 1
)( tueRC
th RCt
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31
Assignment 2
Q2. Find the unit impulse response of an LTIC system specified
by the equation:
(D2 + 6D + 9) y(t) = (2D + 9) x(t)
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32
Assignment 2
Q3.
