Transcript ECE 352 Systems II

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/
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
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
Continuous-Time Domain Analysis
• Example: Differential systems
dn
d n 1
d




y
t

a
y
t



a
y t   a0 y t  
n 1
1
n
n 1
dt
dt
dt
dm
dm
d
bm m f t   bm 1 m f t     b1
f t   b0 f t 
dt
dt
dt
– 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
2
k
• Using short-hand notation,
d
d
d
D
D2  2 Dk  k
above equation becomes
dt
dt
dt
n
D
 an1D n1    a1D  a0  yt   bm D m  bm1D m1    b1D  b0  f t 





Q( D)
P( D)
Continuous-Time Domain Analysis
• Polynomials
Q(D) and P(D)
where an = 1:
n
Q( D)  a0   ak D k
k 1
m
P( D)  b0   bl D l
l 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
Continuous-Time Domain Analysis
• Linearity: for any complex constants c1 and c2,
QD  yt   PD f t 
PDc1 f1 t   c1PD  f1 t   c1QD y1 t 
PDc2 f 2 t   c2QD y2 t 
PDc1 f1 t   c2 f 2 t   PDc1 f1 t   PDc2 f 2 t 

c1QD y1 t   c2QD y2 t 
PD f1 t   f 2 t   QD y1 t   y2 t 
Continuous-Time Domain Analysis
• For a linear system,
T otalresponse
f(t)
T[·]
y(t)
zero- input response  zero- stateresponse



when f ( t ) 0
results from internalsystem conditions only
response to non - zero f ( t )
all initialconditions are zero
independent of f ( t )
dependent on f ( t )
– The two components are independent of each other
– Each component can be computed independently of
the other
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.
Zero-Input Response
• Simplest case
d
y0 t   a0 y0 t 
dt
d
y0 t   a0 y0 t   0
dt
• Solution: y0 t   C ea t
0
• For arbitrary constant C
– Could C be complex?
– How is C determined?
Zero-Input Response
QD  y0 t   0
• General case:
D
n
where
D

d
dt
 an1Dn1   a1D  a0 y0 t   0
• The linear combination of y0(t) and its n
successive derivatives are zero.
dy0
t


Dy
t


C

e
• Assume that
0
dt

t
y0(t) = C e
d 2 y0
2
2 t
D y0 t  

C

e
2
dt

d n y0
n
n t
D y0 t  

C

e
n
dt
Zero-Input Response
• Substituting into the differential equation


n
n 1
t
C


a




a


a
e
1
1
0
 n
 0









non  zero
non  zero
Q  
• y0(t) = C e t is a solution provided that Q() = 0.
• Factor Q() to obtain n solutions:
Q( )    1   2   n   0
y0 t   C1e1t  C2e2t   Cnent
Assuming that no two i terms are equal
Zero-Input Response
• Could i be complex? If complex,
i   i  ji
eit  e i  ji t  e it e jit 
 it
e

cosi t   j sin i t 

dampening term
oscillating term
• For conjugate symmetric roots, and conjugate
symmetric constants,
i t
 i t
1
C1e  C e
 2 C1 e it cosi t  C1 



 
dampening term
oscillating term
Zero-Input Response
• For repeated roots, the solution changes.
• Simplest case of a root  repeated twice:
D   2 y0 t   0
y0 t   C1  C2 t  et
• With r repeated roots
D    r y0 t   0


y0 t   C1  C2t   Cr t r 1 et
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
RLC Circuit
L
• Component values
L = 1 H, R = 4 , C = 1/40 F
Realistic breadboard components?
f(t)
R

C
y(t)
• 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 A
ý(0) = 16.78 A/s
y0(t) = 4 e-2t cos(6t - p/3) A
Envelope
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 p 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
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)
ECE 352
Example Frequency Response
• System response to complex exponential e j w for
all possible frequencies w where w = 2 p f
 H(w)
|H(w)|
passband
stopband
ws wp
stopband
wp ws
w
• Passes low frequencies, a.k.a. lowpass filter
w
ECE 352
Kronecker Impulse
• Let d[k] be a discrete-time impulse function, a.k.a.
the Kronecker delta function:

1 k  0
d k   

0 k  0
d[k]
1
k
• Impulse response h[k]: response of a discrete-time
LTI system to a discrete impulse function
Transfer Functions
Zero-State Response
• Q(D) y(t) = P(D) f(t)
• All initial conditions are 0 in zero-state response
dr
r
D y t   r y t   s r Y s 
yt   Y s 
dt
f t   F s 
dk
D f t   k f t   s k F s 
dt
k
• Laplace transform of differential equation, zerostate component
Qs Y s   Ps F s 
Y s  Ps  Lzero stateresponse
H s  


F s  Qs 
Linput
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
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 righthand 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.
Examples
• Laplace transform

F s     f t  es t dt
0
• Assume input f(t) & output
y(t) are causal
• Ideal delay of T seconds
y t   f t  T 
Y s   F s  e  s T
H s  
Y s 
 es T
F s 
Examples
• Ideal integrator with
y(0-) = 0
y t     f  d
t
0
Y s  
 
1
F s   y 0 
s
1
H s  
s
• Ideal differentiator with
f(0-) = 0
df
y t  
dt
Y s   s F s   f 0   s F ( s)
H s  
Y s 
s
F s 
 
Cascaded Systems
• Assume input f(t) and output y(t) are causal
f(t)
• Integrator first,
then differentiator F(s)
• Differentiator first, f(t)
then integrator
F(s)
 f  d
t
1/s
s
0
F(s)/s
df
dt
s F(s)
f(t)
s
F(s)
f(t)
1/s
• Common transfer functions
– A constant (finite impulse response)
– A polynomial (finite impulse response)
– Ratio of two polynomials (infinite impulse response)
F(s)
Frequency-Domain Interpretation
est
h(t)
y(t)
• y(t) = H(s) e s t
for a particular value of s
• Recall definition of
frequency response:
ej 2 p f t
h(t)
y(t)
y t   ht   e s t

  h e s t  d

e
st

s 


h

e
d
H s 
y t   ht   e j 2p

ft
  h e j 2p

e
j 2p f t
f t  
d

j 2p f 


h

e
d
Hf

Frequency-Domain Interpretation
• s is generalized frequency: s =  + j 2 p f
• We may convert transfer function into
frequency response by if and only if region of
convergence of H(s) includes the imaginary axis
H  f   H s 
s  j 2pf
• What about h(t) = u(t)?
H s  
1
s
for Res  0
We cannot convert this to a frequency response
However, this system has a frequency response
• What about h(t) = d(t)?
H s   1 for all s  H  f   1
Unilateral Laplace Transform
• Differentiation in time property
f(t) = u(t)
What is f ’(0)? f’(t) = d(t).
f ’(0) is undefined.
By definition of differentiation
f t   lim
h 0
f(t)
1
f t  h   f t 
h
Right-hand limit, h =   h = 0, f ’(0+) = 0
Left-hand limit, h = -  h = 0, f ’(0-) does not exist
t
Block Diagrams
F(s)
F(s)
H1(s)
W(s)
H(s)
Y(s)
H2(s)
Y(s)
=
F(s)
H1(s)H2(s)

Y(s)
=
F(s)
H1(s) + H2(s)
Y(s)
Y(s)
=
F(s)
G(s)
1 + G(s)H(s)
Y(s)
Y(s)
H1(s)
F(s)
H2(s)
F(s)
-

E(s)
G(s)
H(s)
Derivations
• Cascade
F(s)
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)
H1(s)
H2(s)
H2(s)
H1(s)
Y(s)  F(s)
Y(s)
One can switch the order of the cascade of two LTI
systems if both LTI systems compute to exact precision
• Parallel Combination
Y(s) = H1(s)F(s) + H2(s)F(s)
Y(s)/F(s) = H1(s) + H2(s)
Derivations
• Feedback System
E s   F s   H s  Y  s 
Y s   G s  E s 
• Combining these two
equations
Y s   G s   F s   H s  Y s  
Y s   G s  H s  Y s   G s  F s 
Y s  
G s 
F s 
1  G s  H s 
• 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)
Stability
Stability
• Many possible
definitions
• Two key issues for
practical systems
– System response to zero
input
– System response to nonzero 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
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 zeroinput response remains bounded (e.g. oscillates
between lower and upper bounds) as t  
Characteristic Modes
• Distinct characteristic roots 1, 2, …, n
n
y0 t    c j e
 jt
j 1
0

t
lim e  e j w t
t 


if Reλ  0
if Reλ  0
Im{}
Right-hand
plane (RHP)
if Reλ  0
Stable
– Where  =  + j w
in Cartesian form
– Units of w are in
Left-hand
plane (LHP)
radians/second
Unstable
Re{}
Marginally
Stable
Characteristic Modes
• Repeated roots
r
y0 t    ci t i 1 et
i 1
– For r repeated roots of
value .
0 if Re   0

k t
lim t e   if Re   0
t 

 if Re   0
– For positive k
• Decaying exponential
decays faster than
tk increases for any value
of k
– One can see this by using
the Taylor Series
approximation for et about
t = 0:
1 2 2 1 33
1  t   t   t  ...
2
6
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.
Response to Bounded Inputs
• Stable system: a bounded input (in amplitude)
should give a bounded response (in amplitude)
• Linear-time-invariant (LTI) system
y t 
 ht   f t 
f(t)

 h  f t   d
y t    h  f t   d   h   f t    d
If f (t ) is bounded, i.e. f t   C   t , then

f t     C  , and yt   C  hτ  dτ







h(t)
y(t)
• Bounded-Input Bounded-Output (BIBO) stable
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
Impact of Characteristic Modes
• Example: First-order system with characteristic
mode e  t
y t   ht   f t 
 Ae t u t   e t u t 
A

e t  e  t u t 
 


• Three cases
t et u t 



   resonance


yt   large amplitude    strongresponse


small amplitude    weak response
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
t
1  RC
ht  
e
u t 
RC
h(t)
1/RC
e-1/RC

– Instant of time at which
h(t) decays to e-1  0.367 of its maximum value
t
System Time Constant
• General case:
h(t)
ĥ(t)
h(t0)
h(t)
t0
t
th
– Effective duration is th seconds where area under ĥ(t)

th
0

ˆ
h(t ) dt  th h(t0 )  C  h(t ) dt
0
– C is an arbitrary constant between 0 and 1
– Choose th to satisfy this inequality
• General case applied
to RC time constant:
t
 1

1
th

e RC dt
0 RC
RC
t h  RC

Step Response
• y(t) = h(t) * u(t)
u(t)
h(t)
y(t)
h(t)
u(t)
y(t)
A
1
t
A th
tr
t
tr
• 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
t
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
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
System Realization
Passive Circuit Elements
• Laplace transforms with
zero-valued initial
conditions
• Capacitor
+
dv
v(t)
i t   C
dt
–
I s   C s V s 
1
V s  
I s 
Cs
V s 
1
H s  

I s  C s
• Inductor
di
vt   L
dt
V s   L s I s 
+
v(t)
–
V s 
H s  
Ls
I s 
• Resistor
vt   R it 
V s   R I s 
V s 
H s  
R
I s 
+
v(t)
–
First-Order RC Lowpass Filter
R
+
x(t)
i(t)
C
+
y(t)
Time domain
R
+
X(s)
I(s)
X (s)
1
R
Cs
1
Y (s) 
I (s)
Cs
1
Y ( s)  C s X (s)
1
R
Cs
1
Y (s)
 RC
X (s) s  1
RC
I ( s) 
1
Cs
Laplace domain
+
Y(s)
Passive Circuit Elements
• Laplace transforms with
non-zero initial
conditions
• Capacitor
• Inductor
di
V t   L
dt

 
 L s I s   L i 0 

i 0 


Ls I s 
V s   L s I s   i 0 

i t   C
dv
dt


 
v0 


I s 
I s   C s V s   v 0


1
V s  
Cs
s
1

I s   C v 0 
Cs

 



s 
Operational Amplifier
• Ideal case: model this nonlinear circuit as
linear and time-invariant
Input impedance is extremely high (considered infinite)
vx(t) is very small (considered zero)
+
vx(t) _
_
+
+
y(t)
_
Operational Amplifier Circuit
• Assuming that Vx(s) = 0,
Y s    I s  Z f s 
F s 
Z s 
F(s)
Z f s 
Y s   
F s 
+
_
Z s 
Z f s 
Y s 
H s  

F s 
Z s 
H(s)
I s  
Z(s)
• How to realize gain of –1?
• How to realize gain of 10?
I(s)
Zf(s)
+ _
Vx(s) _ +
+
Y(s)
_
Differentiator
• A differentiator amplifies high frequencies, e.g.
high-frequency components of noise:
H(s) = s where s =  + j 2p f
Frequency response is H(f) = j 2 p f  | H( f ) |= 2 p | 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
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,
sF s  provided that the limit on the rightthen f 0   lim
s 
hand side of the equation exists.
• Final value theorem
If both f(t) and df/dt have Laplace transforms, then
lim f t   lim sF s  provided that s F(s) has no poles in the
t 
s 0
RHP or on the imaginary axis.
Final and Initial Values Example
• Transfer function
Poles at s = 0, s = -1  j2
Zero at s = -3/2
Y s  
102s  3
s s 2  2s  5


lim y t   limsY s 
t 
s 0
 102s  3 
 lim 2
s 0 s  2 s  5 


30

5
6
 
y 0   limsY s 
s 
 102 s  3 
 lim  2
s  s  2 s  5 


 2 3 
  s  s2  

 lim 10 
s 
 1 2  5 

s s 2 
0