ECE 352 Systems II
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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
an1D n1 a1D a0 yt bm D m bm1D m1 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,
QD yt PD f t
PDc1 f1 t c1PD f1 t c1QD y1 t
PDc2 f 2 t c2QD y2 t
PDc1 f1 t c2 f 2 t PDc1 f1 t PDc2 f 2 t
c1QD y1 t c2QD y2 t
PD f1 t f 2 t QD 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 ea t
0
• For arbitrary constant C
– Could C be complex?
– How is C determined?
Zero-Input Response
QD y0 t 0
• General case:
D
n
where
D
d
dt
an1Dn1 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 C1e1t C2e2t Cnent
Assuming that no two i terms are equal
Zero-Input Response
• Could i be complex? If complex,
i i ji
eit e i ji t e it e jit
it
e
cosi 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 cosi 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 et
• With r repeated roots
D r y0 t 0
y0 t C1 C2t Cr t r 1 et
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
yt 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
Qs Y s Ps F s
Y s Ps Lzero stateresponse
H s
F s Qs
Linput
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 es 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
es 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 ht e s t
h e s t d
e
st
s
h
e
d
H s
y t ht e j 2p
ft
h e j 2p
e
j 2p f t
f t
d
j 2p f
h
e
d
Hf
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 Res 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 Reλ 0
if Reλ 0
Im{}
Right-hand
plane (RHP)
if Reλ 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 et
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 et 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
ht 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 yt C hτ 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 ht f t
Ae t u t e t u t
A
e t e t u t
• Three cases
t et u t
resonance
yt 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
ht
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
vt L
dt
V s L s I s
+
v(t)
–
V s
H s
Ls
I s
• Resistor
vt R it
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
v0
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
102s 3
s s 2 2s 5
lim y t limsY s
t
s 0
102s 3
lim 2
s 0 s 2 s 5
30
5
6
y 0 limsY s
s
102 s 3
lim 2
s s 2 s 5
2 3
s s2
lim 10
s
1 2 5
s s 2
0