ECE 352 Systems II
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Transcript of ECE 352 Systems II
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ECE 352 Systems II
Manish K. Gupta, PhD Office: Caldwell Lab 278
Email: guptam @ ece. osu. edu Home Page: http://www.ece.osu.edu/~guptam
TA: Zengshi Chen Email: chen.905 @ osu. edu Office Hours for TA : in CL 391: Tu & Th 1:00-2:30 pm
Home Page: http://www.ece.osu.edu/~chenz/
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Acknowledgements
• Various graphics used here has been taken from public resources instead of redrawing it. Thanks to those who have created it.
• Thanks to Brian L. Evans and Mr. Dogu Arifler
• Thanks to Randy Moses and Bradley Clymer
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ECE 352
Slides edited from: • Prof. Brian L. Evans and Mr. Dogu Arifler
Dept. of Electrical and Computer Engineering The University of Texas at Austin course:
EE 313 Linear Systems and Signals Fall 2003
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tfbtfdtdbtf
dtdbtf
dtdb
tyatydtdaty
dtdaty
dtd
m
m
mm
m
m
n
n
nn
n
011
011
1
1
k
kk
dtdD
dtdD
dtdD 2
22
tfbDbDbDbtyaDaDaDDP
mm
mm
DQ
nn
n
)(
011
1
)(
011
1
Continuous-Time Domain Analysis• Example: Differential systems
– There are n derivatives of y(t) and m derivatives of f(t)– Constants a0, a1, …, an-1 and b0, b1, …, bm
– Linear constant-coefficient differential equation• Using short-hand notation,
above equation becomes
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m
l
ll
n
k
kk
DbbDP
DaaDQ
10
10
)(
)(
Continuous-Time Domain Analysis• Polynomials
Q(D) and P(D)where an = 1:
– Recall n derivatives of y(t) and m derivatives of f(t)– We will see that this differential system behaves as an
(m-n)th-order differentiator of high frequency signals if m > n
– Noise occupies both low and high frequencies– We will see that a differentiator amplifies high
frequencies. To avoid amplification of noise in high frequencies, we assume that m n
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tyDQctfDPctfcDP
tfDPtyDQ
111111
tyDQctfcDP 2222
tyDQctyDQc
tfcDPtfcDPtfctfcDP
2211
22112211
tytyDQtftfDP 2121
Continuous-Time Domain Analysis• Linearity: for any complex constants c1 and c2,
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)(on dependent zero are conditions initial all
)( zero-non toresponse
)( oft independenonly conditions system internal from results
0)(when
response state-zero responseinput -zeroresponse Total
tf
tf
tf
tf
Continuous-Time Domain Analysis• For a linear system,
– The two components are independent of each other– Each component can be computed independently of
the other
T[·] y(t)f(t)
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Continuous-Time Domain Analysis• Zero-input response
– Response when f(t)=0– Results from internal
system conditions only– Independent of f(t)– For most filtering
applications (e.g. your stereo system), we want no zero-input response.
• Zero-state response– Response to non-zero f(t)
when system is relaxed– A system in zero state
cannot generate any response for zero input.
– Zero state corresponds to initial conditions being zero.
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Zero-Input Response• Simplest case
• Solution:
• For arbitrary constant C– Could C be complex?– How is C determined?
0000 tyatydtd tyaty
dtd
000
taeCty 0 0
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dtdDtyDQ where00
00011
1 tyaDaDaD n
nn
tnn
nn
t
t
eCdt
ydtyD
eCdt
ydtyD
eCdt
dytDy
00
22
02
02
00
Zero-Input Response• General case:
• The linear combination of y0(t) and its n successive derivatives are zero.
• Assume that y0(t) = C e t
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Zero-Input Response• Substituting into the differential equation
• y0(t) = C et is a solution provided that Q() = 0.• Factor Q() to obtain n solutions:
Assuming that no two i terms are equal
0011
1
zeronon
t
Q
nn
n
zeronon
eaaaC
0)( 21 nQ
tn
tt neCeCeCty 21210
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Zero-Input Response• Could i be complex? If complex,
• For conjugate symmetric roots, and conjugate symmetric constants,
iii j
termgoscillatin termdampening
sincos tjte
eeee
iit
tjttjt
i
iiiii
termgoscillatin
1
termdampening
111 cos2 CteCeCeC ittt iii
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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 tyD r
trr etCtCCty 1
210
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System Response• Characteristic equation
– Q(D)[y(t)] = 0– The polynomial Q()
• Characteristic of system• Independent of the input
– Q() roots 1, 2, …, n
• Characteristic roots a.k.a. characteristic values, eigenvalues, natural frequencies
• Characteristic modes (or natural modes) are the time-domain responses corresponding to the characteristic roots– Determine zero-input
response– Influence zero-state
response
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RLC Circuit• Component values
L = 1 H, R = 4 , C = 1/40 FRealistic breadboard components?
• Loop equations(D2 + 4 D + 40) [y0(t)] = 0
• Characteristic polynomial 2 + 4 + 40 =
( + 2 - j 6)( + 2 + j 6)• Initial conditions
y(0) = 2 Aý(0) = 16.78 A/s
L R
C
y(t)f(t)
Envelope
y0(t) = 4 e-2t cos(6t - /3) A
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ECE 352
Linear Time-Invariant System
• Any linear time-invariant system (LTI) system, continuous-time or discrete-time, can be uniquely characterized by its– Impulse response: response of system to an impulse– Frequency response: response of system to a
complex exponential e j 2 f for all possible frequencies f
– Transfer function: Laplace transform of impulse response
• Given one of the three, we can find other two provided that they exist
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ECE 352
Impulse response
Impulse response of a system is response of the system to an input that is a unit
impulse (i.e., a Dirac delta functional in continuous time)
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ECE 352
Example Frequency Response
• System response to complex exponential e j for all possible frequencies where = 2 f
• Passes low frequencies, a.k.a. lowpass filter
|H()|
p ss p
passband
stopband stopband
H()
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ECE 352
Kronecker Impulse
• Let [k] be a discrete-time impulse function, a.k.a. the Kronecker delta function:
• Impulse response h[k]: response of a discrete-time LTI system to a discrete impulse function
00
01
k
kk
k
[k] 1
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Transfer Functions
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Zero-State Response• Q(D) y(t) = P(D) f(t)• All initial conditions are 0 in zero-state response
• Laplace transform of differential equation, zero-state component
sYty
sFstfdtdtfD
sYstydtdtyD
kk
kk
rr
rr
input
response state zeroL
LsQsP
sFsYsH
sFsPsYsQ
sFtf
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Transfer Function• H(s) is called the transfer function because it
describes how input is transferred to the output in a transform domain (s-domain in this case)Y(s) = H(s) F(s)y(t) = L-1{H(s) F(s)} = h(t) * f(t) H(s) = L{h(t)}
• The transfer function is the Laplace transform of the impulse response
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Transfer Function• Stability conditions for an LTIC system
– Asymptotically stable if and only if all the poles of H(s) are in left-hand plane (LHP). The poles may be repeated or non-repeated.
– Unstable if and only if either one or both of these conditions hold: (i) at least one pole of H(s) is in right-hand plane (RHP); (ii) repeated poles of H(s) are on the imaginary axis.
– Marginally stable if and only if there are no poles of H(s) in RHP, and some non-repeated poles are on the imaginary axis.
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Ts
Ts
esFsYsH
esFsY
Ttfty
Examples• Laplace transform
• Assume input f(t) & output y(t) are causal
• Ideal delay of T seconds
0
dtetfsF ts
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ssFsYsH
sFsfsFssYdtdfty
)(0
s
sH
ysFs
sY
dftyt
1
010
Examples• Ideal integrator with y(0-) = 0
• Ideal differentiator with f(0-) = 0
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Cascaded Systems• Assume input f(t) and output y(t) are causal
• Integrator first,then differentiator
• Differentiator first,then integrator
• Common transfer functions– A constant (finite impulse response)– A polynomial (finite impulse response)– Ratio of two polynomials (infinite impulse response)
1/s sF(s)/s
tdf
0f(t)
F(s)
f(t)
F(s)
s 1/ss F(s)
dtdf
f(t)
F(s)
f(t)
F(s)
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Frequency-Domain Interpretation
• y(t) = H(s) e s t
for a particular value of s
• Recall definition offrequency response:
sH
sts
ts
ts
dehe
deh
ethty
h(t) y(t)est
h(t) y(t)ej 2f t
fH
fjtfj
tfj
tfj
dehe
deh
ethty
2 2
2
2
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Frequency-Domain Interpretation• s is generalized frequency: s = + j 2 f• We may convert transfer function into
frequency response by if and only if region of convergence of H(s) includes the imaginary axis
• What about h(t) = u(t)?We cannot convert this to a frequency responseHowever, this system has a frequency response
• What about h(t) = (t)?
0Refor 1 s
ssH
1 allfor 1 fHssH
fjs
sHfH2
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Unilateral Laplace Transform• Differentiation in time property
f(t) = u(t)What is f ’(0)? f’(t) = (t).
f ’(0) is undefined.By definition of differentiation
Right-hand limit, h = h = 0, f ’(0+) = 0Left-hand limit, h = - h = 0, f ’(0-) does not exist
t
f(t)
1
h
tfhtftfh
0lim
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Block DiagramsH(s)F(s) Y(s)
H1(s) + H2(s)F(s) Y(s)
H1(s)
F(s) Y(s)H2(s)
=
H1(s)F(s) Y(s)H2(s) H1(s)H2(s)F(s) Y(s)=W(s)
G(s) 1 + G(s)H(s)F(s) Y(s)G(s)F(s) Y(s)
H(s)
- =
E(s)
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Derivations• Cascade
W(s) = H1(s)F(s) Y(s) = H2(s)W(s)
Y(s) = H1(s)H2(s)F(s) Y(s)/F(s) = H1(s)H2(s)
One can switch the order of the cascade of two LTI systems if both LTI systems compute to exact precision
• Parallel CombinationY(s) = H1(s)F(s) + H2(s)F(s)
Y(s)/F(s) = H1(s) + H2(s)
H1(s)F(s) Y(s)H2(s) H2(s)F(s) Y(s)H1(s)
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Derivations• Feedback System
• Combining these two equations
sEsGsY
sYsHsFsE
sF
sHsGsGsY
sFsGsYsHsGsY
sYsHsFsGsY
1
• What happens if H(s) is a constant K?– Choice of K controls all
poles in the transfer function
– This will be a common LTI system in Intro. to Automatic Control Class (required for EE majors)
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Stability
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Stability• Many possible
definitions• Two key issues for
practical systems– System response to zero
input– System response to non-
zero but finite amplitude (bounded) input
• For zero-input response– If a system remains in a particular
state (or condition) indefinitely, then state is an equilibrium state of system
– System’s output due to nonzero initial conditions should approach 0 as t
– System’s output generated by initial conditions is made up of characteristic modes
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Stability• Three cases for zero-input response
– A system is stable if and only if all characteristic modes 0 as t
– A system is unstable if and only if at least one of the characteristic modes grows without bound as t
– A system is marginally stable if and only if the zero-input response remains bounded (e.g. oscillates between lower and upper bounds) as t
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Characteristic Modes• Distinct characteristic roots 1, 2, …, n
0Re if
0Re if
0Re if0
lim
10
λ
λe
λ
e
ecty
tjt
t
n
j
tj
j
– Where = + j in Cartesian form
– Units of are in radians/second
Stable Unstable
Im{}
Re{}
MarginallyStable
Left-handplane (LHP)
Right-handplane (RHP)
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Characteristic Modes• Repeated roots
– For r repeated roots of value .
– For positive k
r
i
tii etcty
1
10
0Reif
0Reif
0Reif0
lim
tk
tet
• Decaying exponential decays faster thantk increases for any value of k– One can see this by using
the Taylor Series approximation for et about t = 0:
...61
211 3322 ttt
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Stability Conditions• An LTIC system is asymptotically stable if and
only if all characteristic roots are in LHP. The roots may be simple (not repeated) or repeated.
• An LTIC system is unstable if and only if either one or both of the following conditions exist:(i) at least one root is in the right-hand plane (RHP)(ii) there are repeated roots on the imaginary axis.
• An LTIC system is marginally stable if and only if there are no roots in the RHP, and there are no repeated roots on imaginary axis.
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dtfhdtfhty
dtfh
tfthty
dττhCty Ctf
tCtftf
and ,
then, i.e.bounded, is )( If
Response to Bounded Inputs• Stable system: a bounded input (in amplitude)
should give a bounded response (in amplitude)• Linear-time-invariant (LTI) system
• Bounded-Input Bounded-Output (BIBO) stable
h(t) y(t)f(t)
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Impact of Characteristic Modes• Zero-input response consists of the system’s
characteristic modes• Stable system characteristic modes decay
exponentially and eventually vanish• If input has the form of a characteristic mode,
then the system will respond strongly• If input is very different from the characteristic
modes, then the response will be weak
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tueeA
tuetuAetfthty
tt
tt
responseweak amplitude small
response strongamplitude large
resonance
tuet
ty
t
Impact of Characteristic Modes• Example: First-order system with characteristic
mode e t
• Three cases
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tueRC
th RCt
1
e-1/RC
1/RC
t
h(t)
System Time Constant• When an input is applied to a system, a certain
amount of time elapses before the system fully responds to that input– Time lag or response time is the system time constant– No single mathematical definition for all cases
• Special case: RC filter– Time constant is = RC
– Instant of time at whichh(t) decays to e-1 0.367 of its maximum value
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t0 tht
h(t)
h(t0) h(t)
ĥ(t)
RCt
dteRCRC
t
h
RCt
h
0
11
System Time Constant• General case:
– Effective duration is th seconds where area under ĥ(t)
– C is an arbitrary constant between 0 and 1– Choose th to satisfy this inequality
• General case appliedto RC time constant:
000
)( )( )(ˆ dtthCthtdtth h
th
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h(t) y(t)u(t)
t
u(t)
1t
h(t)
A
t
y(t)
A th
tr
Step Response• y(t) = h(t) * u(t)
• Here, tr is the rise time of the system• How does the rise time tr relate to the system
time constant of the impulse response?• A system generally does not respond to an input
instantaneously
tr
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Filtering• A system cannot effectively respond to periodic
signals with periods shorter than th
• This is equivalent to a filter that passes frequencies from 0 to 1/th Hz and attenuates frequencies greater than 1/th Hz (lowpass filter)– 1/th is called the cutoff frequency– 1/tr is called the system’s bandwidth (tr = th)
• Bandwidth is the width of the band of positive frequencies that are passed “unchanged” from input to output
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Transmission of Pulses• Transmission of pulses through a system (e.g.
communication channel) increases the pulse duration (a.k.a. spreading or dispersion)
• If the impulse response of the system has duration th and pulse had duration tp seconds, then the output will have duration th + tp
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System Realization
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Passive Circuit Elements• Laplace transforms with
zero-valued initial conditions
• Capacitor
• Inductor
• Resistor
sLsIsVsH
sIsLsVdtdiLtv
sCsIsVsH
sIsC
sV
sVsCsIdtdvCti
1
1
RsIsVsH
sIRsVtiRtv
+
–
v(t)
+
–
v(t)
+
–
v(t)
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First-Order RC Lowpass Filter
x(t) y(t)++
C
R
X(s) Y(s)++
R
sC 1
Time domain
Laplace domainCR
s
CRsXsY
sX
sCR
sCsY
sIsC
sY
sCR
sXsI
1
1
)()(
)(
1
1
)(
)(
1)(
1)()(
i(t)
I(s)
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Passive Circuit Elements• Laplace transforms with
non-zero initial conditions
• Capacitor
sisIsL
iLsIsL
isIsLsV
dtdiLtV
0
0
0
• Inductor
0
1
0
1
0
vCsIsC
svsI
sCsV
vsVsCsI
dtdvCti
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Operational Amplifier• Ideal case: model this nonlinear circuit as
linear and time-invariantInput impedance is extremely high (considered infinite)vx(t) is very small (considered zero)
+_
y(t)+
_
+_vx(t)
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Operational Amplifier Circuit• Assuming that Vx(s) = 0,
• How to realize gain of –1?• How to realize gain of 10?
+_
Y(s)+
_
+_Vx(s)
Zf(s)
Z(s)
+_
F(s)
I(s)
sZ
sZsFsYsH
sFsZsZ
sY
sZsFsI
sZsIsY
f
f
f
H(s)
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Differentiator• A differentiator amplifies high frequencies, e.g.
high-frequency components of noise:H(s) = s where s = + j 2fFrequency response is H(f) = j 2 f | H( f ) |= 2 f |
• Noise has equal amounts of low and high frequencies up to a physical limit
• A differentiator may amplify noise to drown out a signal of interest
• In analog circuit design, one would use integrators instead of differentiators
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Initial and Final Values• Values of f(t) as t 0 and t may be
computed from its Laplace transform F(s) • Initial value theorem
If f(t) and its derivative df/dt have Laplace transforms, then provided that the limit on the right-hand side of the equation exists.
• Final value theoremIf both f(t) and df/dt have Laplace transforms, then
provided that s F(s) has no poles in the RHP or on the imaginary axis.
ssFfs
lim0
ssFtfst 0limlim
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52
32102
sssssY
0
521
32
10lim
523210lim
lim0
2
2
2
ss
ss
sss
ssYy
s
s
s
65
3052
3210lim
limlim
20
0
sss
ssYty
s
st
Final and Initial Values Example• Transfer function
Poles at s = 0, s = -1 j2Zero at s = -3/2