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
Z-transforms
Z-transforms
• For discrete-time systems, z-transforms play
the same role of Laplace transforms do in
continuous-time systems
Bilateral Forward z-transform
H [ z] 


k  
hk  z k
Bilateral Inverse z-transform
h[k ] 
1
2  j R
H [ z ] z k 1dz
• As with the Laplace transform, we compute
forward and inverse z-transforms by use of
transforms pairs and properties
Region of Convergence
• Region of the complex zplane for which forward
z-transform converges
• Four possibilities (z=0 is
a special case and may
or may not be included)
Im{z}
Entire
plane
Im{z}
Disk
Re{z}
Re{z}
Im{z}
Complement
of a disk
Im{z}
Re{z}
Intersection
of a disk and
complement
of a disk
Re{z}
Z-transform Pairs
• h[k] = d[k]
H [ z] 

 d k  z
• h[k] = ak u[k]
k
0
  d k  z
k
1
H [ z] 
k  
k 0
Region
of convergence:
entire zplane
H [ z] 
1
z   d k  1 z  z
d k of1 convergence:
Region
entire zk  
k
k 1
plane
h[n-1]  z-1 H(z)
k
 a uk  z
k


k
k  
• h[k] = d[k-1]


1
k
a
  a k z k    
k 0
k 0  z 
1
a

if
1
a
z
1
z
Region of convergence: |z| >
|a| which is the
complement of a disk
Stability
• Rule #1: For a causal sequence, poles are inside
the unit circle (applies to z-transform functions
that are ratios of two polynomials)
• Rule #2: More generally, unit circle is included
in region of convergence. (In continuous-time,
the imaginary axis would be in the region of
convergence of the Laplace transform.)
a uk 
k
Z
1
1  a z 1
for
z a
– This is stable if |a| < 1 by rule #1.
– It is stable if |z| > 1 > |a| by rule #2.
Inverse z-transform
c  j
1
f k  
F z z k 1dz

2j c  j
• Yuk! Using the definition requires a contour
integration in the complex z-plane.
• Fortunately, we tend to be interested in only a
few basic signals (pulse, step, etc.)
– Virtually all of the signals we’ll see can be built up
from these basic signals.
– For these common signals, the z-transform pairs have
been tabulated (see Tables)
Example
z 2  2z 1
X [ z] 
3
1
z2  z 
2
2
1  2 z 1  z 2
X [ z] 
3
1
1  z 1  z  2
2
2
1  2 z 1  z 2
X [ z] 
 1 1 
1
1  z  1  z
 2 

X [ z ]  B0 

A1
A2

1 1 1  z 1
1 z
2
• Ratio of polynomial zdomain functions
• Divide through by the
highest power of z
• Factor denominator
into first-order factors
• Use partial fraction
decomposition to get
first-order terms
Example (con’t)
2
1  2 3 1
z  z  1 z  2  2 z 1  1
2
2
z  2  3z 1  2
• Find B0 by polynomial
division
5 z 1  1
 1  5 z 1
X [ z]  2 
 1 1 
1
1  z  1  z
 2 

1  2 z 1  z 2
A1 
1  z 1
1  2 z 1  z  2
A2 
1
1  z 1
2
z 1  2
1 4  4

 9
1 2

z 1 1

1 2 1
8
1
2
• Express in terms of B0
• Solve for A1 and A2
Example (con’t)
• Express X[z] in terms of
B0, A1, and A2
X z   2 
9
8

1
1
1  z 1 1  z
2
• Use table to obtain
inverse z-transform
k
1
xk   2 d k   9   uk   8 uk 
 2
• With the unilateral ztransform, or the
bilateral z-transform
with region of
convergence, the inverse
z-transform is unique.
Z-transform Properties
• Linearity
a1 f1k   a2 f 2 k   a1F1z   a2 F2 z 
• Shifting
f k  m uk  m  z m F z 
 m

f k  m uk   z F z   z   f  k z m 
 k 1

m
m
Z-transform Properties
f1 k  f 2 k  

 f m f k  m
m  
1
2
 

Z  f1 k  f 2 k   Z   f1 m f 2 k  m
m  


 

    f1 m f 2 k  m z  k
k    m  



 f m  f k  mz
m  




m  
1
k  

k
• Convolution definition
• Take z-transform
• Z-transform definition
• Interchange summation
2
f1 m  f 2 r z r  m 
• r=k-m
k  

 

 m 
   f1 mz   f 2 r z  r 
 m  
 k  
•
 F1 z F2 z 
Z-transform definition
Example
g k   k uk   uk  6
 k uk   k uk  6
 k uk   k  6  uk  6  6 uk  6
Gz  

z
1
z
1 z 


6
 6

2
2
6
z
z
z

1
z  1
z  1 

z
1
6


z  12 z 5 z  12 z 5 z  1
z  1 6
z5
z
1
 5


z  z  12 z 5  z  12  z  1 z 5  z  1
z6  6z  5
 5
2
z  z  1
Difference Equations
Linear Difference Equations
• Discrete-time
LTI systems
can be
characterized
by difference
equations
f[k]
+

y[k]
+
Unit
Delay
+
1/2
y[k-1]
Unit
Delay
1/8
y[k-2]
y[k] = (1/2) y[k-1] + (1/8) y[k-2] + f[k]
• Taking z-transform of the difference equation
gives description of the system in the z-domain
Advances and Delays
• Sometimes differential equations will be
presented as unit advances rather than delays
y[k+2] – 5 y[k+1] + 6 y[k] = 3 f[k+1] + 5 f[k]
• One can make a substitution that reindexes the
equation so that it is in terms of delays
Substitute k with k-2 to yield
y[k] – 5 y[k-1] + 6 y[k-2] = 3 f[k-1] + 5 f[k-2]
• Before taking the z-transform, recognize that
we work with time k  0 so u[k] is often implied
y[k-1] = y[k-1] u[k]  y[k-1] u[k-1]
Example
• System described by a difference equation
y[k] – 5 y[k-1] + 6 y[k-2] = 3 f[k-1] + 5 f[k-2]
y[-1] = 11/6, y[-2] = 37/36
f[k] = 2-k u[k]
1
1
1
  1

1
  1

Y z   5 Y z   y 1  6 2 Y z   y 1  y 2  3 F z   f  1  5 2 F z   f  1  f  2
z
z
z
 z

z
 z

11 
3
5
 5 6 


1   2 Y z    3   
z z 
z  z  0.5 z z  0.5


Y z  
26  z  7  z  18  z 

 
 

15  z  0.5  3  z  2  5  z  3 
7 k 18 k 
 26
k
yk    0.5  2   3 uk 
3
5
 15

Transfer Functions
• Previous example describes output in time
domain for specific input and initial conditions
• It is not a general solution, which motivates us
to look at system transfer functions.
• In order to derive the transfer function, one
must separate
– “Zero state” response of the system to a given input
with zero initial conditions
– “Zero input” response to initial conditions only
Transfer Functions
• Consider the zero-state response
– No initial conditions: y[-k] = 0 for all k > 0
– Only causal inputs: f[-k] = 0 for all k > 0
• Write general nth order difference equation
yk   an 1 yk  1    a0 yk  n  bn f k   bn 1 f k  1    b0 f k  n
yk  m uk  
1  a
n 1
1
Y z   0
m
z
f k  m uk  


1
F z   0
m
z

z 1    a1 z1 n  a0 z  n Y z   bn  bn 1 z 1    b1 z1 n  b0 z  n F z 
Y z  Z zero stat eresponse bn  bn 1 z 1    b1 z1 n  b0 z  n
H z  


F z 
Z input
1  an 1 z 1    a1 z1 n  a0 z  n
Y z   H z F z 
Stability
• Knowing H[z], we can compute the output given
any input
F[z]
H[z]
Y[z]
• Since H[z] is a ratio of two polynomials, the roots
of the denominator polynomial (called poles)
control where H[z] may blow up
• H[z] can be represented as a series
– Series converges when poles lie inside (not on) unit circle
– Corresponds to magnitudes of all poles being less than 1
– System is said to be stable
Relation between h[k] and H[z]
• Either can be used to describe the system
– Having one is equivalent to having the other since they
are a z-transform pair
– By definition, the impulse response, h[k], is
y[k] = h[k] when f[k] = d[k]
Z{h[k]} = H[z] Z{d[k]}  H[z] = H[z] · 1
h[k]  H[z]
• Since discrete-time signals can be built up from
unit impulses, knowing the impulse response
completely characterizes the LTI system
Complex Exponentials
• Complex exponentials have
special property when they
are input into LTI systems.
• Output will be same complex
exponential weighted by H[z]
yk   hk  f k   hk  z k


 hmz
k m
m  
z

k
 h[m]z
m
m  
 z k H z 
• When we specialize the z-domain to frequency
domain, the magnitude of H[z] will control
which frequencies are attenuated or passed.
Z and Laplace Transforms
Z and Laplace Transforms
• Are complex-valued functions of a complex
frequency variable
Laplace: s =  + j 2  f
Z:
z = e jW
• Transform difference/differential equations
into algebraic equations that are easier to solve
Z and Laplace Transforms
• No unique mapping from Z to Laplace domain
or vice-versa
– Mapping one complex domain to another is not unique
• One possible mapping is impulse invariance.
– The impulse response of a discrete-time LTI system is a
sampled version of a continuous-time LTI system.
Z
f[k]
H[z]
Laplace
y[k]
~
f t 
H[esT]
~y t 
Z and Laplace Transforms

~
f t    f k d t  kT 
Z
k 0

f[k]
~
y t    yk d t  kT 
H[z]
y[k]
k 0
~
~
Y s   H s  F s 

 
 yk  e
s T k
k 0

 H s  f k  e  k T s
k 0
Let z  e s T :

 yk  z
k o
k
Laplace

 H [ z ]  f k  z
Y [ z]  H [ z] F[ z]
k 0
k
~
f t 
H[esT]
~y t 
Impulse Invariance Mapping
• Impulse invariance mapping is z = e s T
Im{s}
Im{z}
1
-1
1
Re{s}
1
-1
s = -1  j  z = 0.198  j 0.31 (T = 1)
s = 1  j  z = 1.469  j 2.287 (T = 1)
Laplace Domain
Z Domain
Left-hand plane
Inside unit circle
Imaginary axis
Unit circle
Right-hand plane
Outside unit circle
Re{z}
Sampling Theorem
Sampling
• Many signals originate as continuous-time
signals, e.g. conventional music or voice.
• By sampling a continuous-time signal at
isolated, equally-spaced points in time, we
obtain a sequence of numbers
sk   sk Ts 
s(t)
Ts
k  {…, -2, -1, 0, 1, 2,…}
Ts is the sampling period
ssampled t  
t

 skT d t  k T 
k  
s
s[ k ]
s
Sampled analog waveform
Shannon Sampling Theorem
• A continuous-time signal x(t) with frequencies
no higher than fmax can be reconstructed from
its samples x[k] = x(k Ts) if the samples are
taken at a rate fs which is greater than 2 fmax.
– Nyquist rate = 2 fmax
– Nyquist frequency = fs/2.
• What happens if fs = 2fmax?
• Consider a sinusoid sin(2  fmax t)
– Use a sampling period of Ts = 1/fs = 1/2fmax.
– Sketch: sinusoid with zeros at t = 0, 1/2fmax, 1/fmax, …
Sampling Theorem Assumptions
• The continuous-time signal has no frequency
content above the frequency fmax
• The sampling time is exactly the same between
any two samples
• The sequence of numbers obtained by sampling
is represented in exact precision
• The conversion of the sequence of numbers to
continuous-time is ideal
Why 44.1 kHz for Audio CDs?
• Sound is audible in 20 Hz to 20 kHz range:
fmax = 20 kHz and the Nyquist rate 2 fmax = 40 kHz
• What is the extra 10% of the bandwidth used?
Rolloff from passband to stopband in the magnitude
response of the anti-aliasing filter
• Okay, 44 kHz makes sense. Why 44.1 kHz?
At the time the choice was made, only recorders capable of
storing such high rates were VCRs.
NTSC: 490 lines/frame, 3 samples/line, 30 frames/s =
44100 samples/s
PAL: 588 lines/frame, 3 samples/line, 25 frames/s = 44100
samples/s
Sampling
• As sampling rate increases, sampled waveform
looks more and more like the original
• Many applications (e.g. communication
systems) care more about frequency content in
the waveform and not its shape
• Zero crossings: frequency content of a sinusoid
– Distance between two zero crossings: one half period.
– With the sampling theorem satisfied, sampled sinusoid
crosses zero at the right times even though its
waveform shape may be difficult to recognize
Aliasing
• Analog sinusoid
x(t) = A cos(2f0t + f)
• Sample at Ts = 1/fs
x[k] = x(Ts k) =
A cos(2 f0 Ts k + f)
• Keeping the sampling
period same, sample
y(t) = A cos(2(f0 + lfs)t + f)
where l is an integer
y[k] = y(Ts k)
= A cos(2(f0 + lfs)Tsk + f)
= A cos(2f0Tsk + 2 lfsTsk + f)
= A cos(2f0Tsk + 2 l k + f)
= A cos(2f0Tsk + f)
= x[k]
Here, fsTs = 1
Since l is an integer,
cos(x + 2l) = cos(x)
• y[k] indistinguishable from
x[k]
Aliasing
• Since l is any integer, an infinite number of
sinusoids will give same sequence of samples
• The frequencies f0 + l fs for l  0 are called
aliases of frequency f0 with respect fs to because
all of the aliased frequencies appear to be the
same as f0 when sampled by fs
Generalized Sampling Theorem
• Sampling rate must be greater than twice the
bandwidth
– Bandwidth is defined as non-zero extent of spectrum of
continuous-time signal in positive frequencies
– For lowpass signal with maximum frequency fmax,
bandwidth is fmax
– For a bandpass signal with frequency content on the
interval [f1, f2], bandwidth is f2 - f1
Difference Equations and Stability
Example: Second-Order Equation
• y[k+2] - 0.6 y[k+1] - 0.16 y[k] = 5 f[k+2] with
y[-1] = 0 and y[-2] = 6.25 and f[k] = 4-k u[k]
• Zero-input response
Characteristic polynomial g2 - 0.6 g - 0.16 = (g + 0.2) (g - 0.8)
Characteristic equation
(g + 0.2) (g - 0.8) = 0
Characteristic roots
g1 = -0.2 and g2 = 0.8
Solution
y0[k] = C1 (-0.2)k + C2 (0.8)k
• Zero-state response
Example: Impulse Response
• h[k+2] - 0.6 h[k+1] - 0.16 h[k] = 5 d[k+2]
with h[-1] = h[-2] = 0 because of causality
• In general, from Lathi (3.41),
h[k] = (b0/a0) d[k] + y0[k] u[k]
• Since a0 = -0.16 and b0 = 0,
h[k] = y0[k] u[k] = [C1 (-0.2)k + C2 (0.8)k] u[k]
• Lathi (3.41) is similar to Lathi (2.41):
y[k ]  m[k ] u[k ]
Lathi (3.41) balances
impulsive events at origin
Example: Impulse Response
• Need two values of h[k] to solve for C1 and C2
h[0] - 0.6 h[-1] - 0.16 h[-2] = 5 d[0]  h[0] = 5
h[1] - 0.6 h[0] - 0.16 h[-1] = 5 d[1]  h[1] = 3
• Solving for C1 and C2
h[0] = C1 + C2 = 5
h[1] = -0.2 C1 + 0.8 C2 = 3
Unique solution  C1 = 1, C2 = 4
• h[k] = [(-0.2)k + 4 (0.8)k] u[k]
Example: Solution
• Zero-state response solution (Lathi, Ex. 3.13)
ys[k] = h[k] * f[k] = {[(-0.2)k + 4(0.8)k] u[k]} * (4-k u[k])
ys[k] = [-1.26 (4)-k + 0.444 (-0.2)k + 5.81 (0.8)k] u[k]
• Total response: y[k] = y0[k] + ys[k]
y[k] = [C1(-0.2)k + C2(0.8)k] +
[-1.26 (4)-k + 0.444 (-0.2)k + 5.81 (0.8)k] u[k]
• With y[-1] = 0 and y[-2] = 6.25
y[-1] = C1 (-5) + C2(1.25) = 0
y[-2] = C1(25) + C2(25/16) = 6.25
Solution: C1 = 0.2, C2 = 0.8
Repeated Roots
• For r repeated roots of Q(g) = 0
y0[k] = (C1 + C2 k + … + Cr kr-1) gk
• Similar to the continuous-time case
Continuous
Time
Discrete
Time
et u(t )
g k u[k ]
t met u(t )
k mg k u[k ]
Case
non-repeated
roots
repeated
roots
Stability for an LTID System
• Asymptotically stable if and
only if all characteristic roots
are inside unit circle.
• Unstable if and only if one or
both of these conditions exist:
– At least one root outside unit circle
– Repeated roots on unit circle
Im g
Marginally Stable
Unstable
g
|g|
b
-1
Re g
1
Stable
Lathi, Fig. 3.16
• Marginally stable if and only if no roots are
outside unit circle and no repeated roots are on
unit circle (see Figs. 3.17 and 3.18 in Lathi)
Stability in Both Domains
Im 
Im g
Marginally
Stable
Marginally
Stable
Unstable
-1
1
Re 
Re g
Stable
Stable
Discrete-Time Systems
Unstable
Continuous-Time Systems
Marginally stable: non-repeated characteristic roots on the unit
circle (discrete-time systems) or imaginary axis (continuoustime systems)
Frequency Response of
Discrete-Time Systems
Frequency Response
• For continuous-time systems the response to
sinusoids are
• For discrete-time systems in z-domain
• For discrete-time systems in discrete-time
frequency
Response to Sampled Sinusoids
• Start with a continuous-time sinusoid
cos t 
• Sample it every T seconds (substitute t = k T)
• We show discrete-time sinusoid with
• Resulting in
W  T
• Discrete-time frequency is equal to continuoustime frequency multiplied by sampling period
Example
• Calculate the frequency response of the system
given as a difference equation as
• Assuming zero initial conditions we can take
the z-transform of this difference equation
• Since z  e jW
Example
• Group real and imaginary parts
• The absolute value (magnitude response) is
   1  0.8 cosW  j0.8 sin W
He
1
jW

1
1  0.8 cosW2  0.8 sin W2
1

1.64  1.6 cosW
Example
• The angle (phase response) is
where 0 comes from the angle of the nominator and the
term after – comes from the denominator of H e jW 
• Reminder: Given a complex number a + j b the
absolute value and angle is given as
Example
• We can calculate the output of this system for a
sinusoid at any frequency by substituting W
with the frequency of the input sinusoid.
H e jW 
 
H e jW
5

2
3

2
3
W
53.13
W
Discrete-time Frequency Response
• As in previous example, frequency response of
a discrete-time system is periodic with 2
– Why? Frequency response is function of the complex
exponential which is periodic with 2 : e j W  e j W  2  m 
• Absolute value of discrete-time frequency
response is even and angle is odd symmetric.
– Discrete-time sinusoid is symmetric around 
cosW  2m k  cosWk  2m k  cosWk
cos W x k  cosW x k  cosk   sinW x k sink 
 cosW x k  cosk 
 cos Wx   cos Wx 
Aliasing and Sampling Rate
• Continuous-time sinusoid can have a frequency
from 0 to infinity
• By sampling a continuous-time sinusoid,
sample
cos t   cos k T   cosW k 
t  kT
• Discrete-time frequency W unique from 0 to 
0  T    0   

 f s  0  2f  f s  0  f  f s / 2
T
– We only can represent frequencies up to half of the
sampling frequency.
– Higher frequencies exist would be “wrapped” to some
other frequency in the range.
Effect of Poles and Zeros of H[z]
• The z-transform of a difference equation can be
written in a general form as
H z   bn
z  z1 z  z2 z  zm 
z  g 1 z  g 2 z  g m 
• We can think of complex number as a vector in
the complex plane.
Im
– Since z and zi are both complex
numbers the difference is again
a complex number thus a vector
in the complex plane.
zi
z  zi
z
Re
Effect of Poles and Zeros of H[z]
• Each difference term in H[z] may be represented
as a complex number in polar form
– Magnitude is the distance of
the pole/zero to the chosen
point (frequency) on unit circle.
– Angle is the angle of vector
with the horizontal axis.
H e jW   bn
jf1
jf2
jfm
r1e r2e  rme
d1e j1 d 2e j2 d me jm
Im
g1
z2
o
1
x
r1 f
2
g2 x
r1r2  rm j f1 f2 fm 1 2 m 
 bn
e
d1d 2 d m
2
d1
d2
r2
f
z1 1
o
Re
Effect of Poles/Zeros (Lathi)
H
H
T
x

T
o
-
-/2
H
H
H
H
T
o

x

x
o
T
x
x
x

-
-
H
H