Embed Size (px)
Citation preview
• DT Fourier transform enables us to do a lot of things, e.g.
▫ Analyze frequency response of LTI systems
▫ Modulation …
• Why do we need yet another transform?
▫ One view of Laplace Transform is as an extension of the Fourier transform to allow analysis of broader class of signals and systems
▫ In particular, Fourier transform cannot handle large (and important) classes of signals and unstable systems, i.e. when can not meet.
n
nx )(
• How do we analyze such signals/systems? Recall the eigenfunction property of LTI systems:
▫ zn is an eigenfunction of any DT LTI system
▫ We now do not restrict ourselves just to z = ejω
][nx ][nh ][ny
nn zzHnyznx )(][ ][
LTI DTfor ionEigenfunct
convergesit assuming ][)(
n
nznhzH
Topic
10.1 (Bilateral) z Transform
10.2 Properties of RoCs
10.3 Transform Properties
10.4 Inverse z Transform
10.5 Analysis of LTI Systems using the zT
10.6 Unilateral z Transform and Applications
Topic
10.1 (Bilateral) z Transform
10.2 Properties of RoCs
10.3 Transform Properties
10.4 Inverse z Transform
10.5 Analysis of LTI Systems using the zT
10.6 Unilateral z Transform and Applications
7.1.1 The (Bilateral) z Transform
z is a complex variable – Now we explore the full range of z
Basic ideas
Express the complex variable z in polar form as z=rejω
• X(rejω) is the Fourier transform of x[n] multiplied by a real exponential r-n, if it exists.
• For r=1, the z transform reduces to the Fourier transform.
]}[{][)(][ nxZznxzXnxn
n
n
njn
n
njj ernxrenxreX ][][)(
• In general, the z transform of a sequence has associated it with a range of values of z for which X(z) converges, and this range of values is referred to as the region of convergence (ROC)
• If the ROC includes the unit circle, then the Fourier transform also converges
n
nj rnxrez |][|at which ROC
—depends only on r = |z|, just like the ROC in s-plane only depends on Re(s)
Example 1:
sided-right- nuanx n
-n
)( nn znuazX
0
1)(
n
naz
az
z
az
11
1
This form
for PFE and
inverse z-
transform
This form to find
pole and zero locations||||,i.e.,1 If 1 azaz
That is, ROC |z| > |a|, outside a circle
(a – an arbitrary real or complex number)
Example 2:
sided-left-]1[ nuanx n
1
1 )(
n
nn
n
nn
za
znuazX
0
1
1
1 n
n
n
nn zaza
||||.,.,1 If 1 azeiza Same as X(z) in previous example, but different ROC
Key Point (and key difference from FT): Need both X(z) and ROC to uniquely determine x[n].
No such an issue for FT.
7.1.2 Graphical Visualization of the ROC
Example1
sided-right- nuanx n
Example2
sided-left-]1[ nuanx n
7.1.3 Rational Transforms
Many (but by no means all) z transforms of interest to us are rational functions of s (e.g., Example 1 and 2; in general, impulse responses of LTI systems described by LCCDEs), where
X(z) = N(z)/D(z), N(z),D(z) – polynomials in z
Roots of N(z)= zeros of X(z)
Roots of D(z)= poles of X(z)
Any x[n] consisting of a linear combination of complex exponentials for n > 0 and for n < 0 (e.g., as in Example #1 and #2) has a rational z transform.
Except for a scale factor, a complete specification of a rational z transform consists of the pole-zero plot of the transform, together with its RoC
Topic
10.1 (Bilateral) z Transform
10.2 Properties of RoCs
10.3 Transform Properties
10.4 Inverse z Transform
10.5 Analysis of LTI Systems using the zT
10.6 Unilateral z Transform and Applications
• Property 1: The ROC of X(z) consist of a ring in the z-plane centered about the origin
▫ Proof: In order to make the F.T of x[n]r-n converged, it should be absolutely summable, i.e. the RoC only depends on the radius of z
• Property 2: The ROC does not contain any poles. If X(z) is rational, then its ROC is bounded by poles or extends to infinity.
n
nj rnxrez |][|at which ROC
• Property 3: If x[n] is of finite duration, then the ROC is the entire z-plane, except possibly z=0 and/or z=∞
▫ when N1<N2<0, the RoC is the entire z-plane, excluding z=∞:
▫ when N2>N1>0, the RoC is the entire z-plane, excluding z=0
▫ when N1<0,N2>0, the RoC is the entire z-plane, excluding z=0 and z=∞
2
1
][)(N
Nn
nznxzX
z0
z0
z0
• Property 4: If x[n] is a right-sided sequence, and if |z| = r0 is in the ROC, then all finite values of z for which |z| > r0 are also in the ROC
1
1
0
1
nfaster tha converges
Nn
n
n
Nn
rnx
rnx
• Property 5: If x[n] is a left-sided sequence, and if |z| = r0
is in the ROC, then all finite values of z for which 0 < |z| < r0 are also in the ROC.
• If x[n] is two-sided, and if |z| = r0 is in the ROC, then the ROC consists of a ring in the z-plane including the circle |z| = r0.
What types of signals do the following ROC correspond to?
right-sided left-sided two-sided
Example: 0 ,|| bbnx n
1 nubnubnx nn
b
zzb
nub
bzbz
nub
n
n
1 ,
1
11
,1
1
From：
11
1
bzb
zbbzzX
1 ,
1
1
1
1)(
111
Clearly, ROC does not exist if b > 1 ⇒ No z-transform for b|n|.
• If X(z) is rational, then its ROC is bounded by poles or extends to infinity. In addition, no poles of X(z) are contained in the ROC.
• Suppose X(z) is rational, then▫ If x[n] is right-sided, the ROC is to the right of the
outermost pole.
▫ If x[n] is left-sided, the ROC is to the left of the innermost pole.
• If ROC of X(z) includes the unit circle, then FT of x[n] exists.
Topic
10.1 (Bilateral) z Transform
10.2 Properties of RoCs
10.3 Transform Properties
10.4 Inverse z Transform
10.5 Analysis of LTI Systems using the zT
10.6 Unilateral z Transform and Applications
Linearity
222
111
)(][
)(][
RzXnx
RzXnx
)()(][][ 2121 zbXzaXnbxnax
ROC contains the intersection of R1 and R2, which may be empty, also can be larger than the intersection
0 0
0 0
0
1
0
1 1 1 2
0
1
00 1 2
0
1cos [ ] [ ] [ ]
2
1 (cos )1 1 1[ ] , 1
2 1 1 1 (2cos )
(sin )sin [ ] , 1
1 (2cos )
j n j n
j j
n u n e e u n
zz
e z e z z z
zn u n z
z z
Example：
Time Shift
RoC=R, except for the possible addition or deletion of the origin or infinity.
Example： suppose )(][][ zXnunx
To determine the z transform of the following signal:
n
j
jxng0
)(][
)(1
)(
)()()(
]1[][][][]1[0
1
0
zXz
zzG
zzXzGzzG
nxjxjxngngn
j
n
j
ROC of G(z) is that of X(z), excluding z=1But if z=1 is zero(s) of X(z), the RoC of G(z) is identical to that of X(z)
Time Reversal
Time Expansion
for
then
Scaling in the z_Domain
Conjugation
Consequence, if x[n] is real, then
Thus, if X(z) has a pole/zero at z=z0, it must also have a
pole/zero at the complex conjugation point z=z0*
The Convolution Property
ROC contains the intersection of R1 and R2, which may be empty, also can be larger than the intersection
1)(][][
][][][][*][
][][
0
0 0
0
z
zzXjxng
jxjnujxnunx
jxng
n
j
j
n
j
n
j
Example:
Differentiation in the z Domain
RROCdz
zdXznnx
)(][
azazzX )1log()( 1
][)( nxzX
RzXnx ROC)(][
xdx
xd 1)log(
Example： If
Let
Then following:
1
1
1
1
11
1
1
1
( )[ ]
1
1[ ]
1
1( ) [ ]
1
( ) [ 1]1
[ ] ( 1) [ 1]
( )[ ] [ 1] log(1 )
n
n
n
n
n
dX z azz nx n
dz az
a u n z aaz
a u n z aaz
aza a u n
az
nx n a u n
ax n u n az
n
即：thus
Similarly:
1[ 1] log(1 )na
u n azn
The Initial-Value Theorem
Final-value theorem
If x[n] is a casual sequence, i.e. when n<0, x[n]=0, then
)(lim]0[ zXxz
Hints to prove the initial-value theorem:
according to the definition of the Z transform
If x[n] is a casual sequence, i.e. when n<0, x[n]=0, and the
RoC contains the unit circle, then
)]()1[(lim)]()1
[(lim)(11
zXzzXz
zx
zz
Hints to prove the final-value theorem:
Z{x(n+1)-x(n)}=Z{x(n+1)}-Z{x(n)}
=z(X(z)-x(0))-X(z)
and rewrite Z{x(n+1)-x(n)} according to the definition of the Z Transform
Topic
10.1 (Bilateral) z Transform
10.2 Properties of RoCs
10.3 Transform Properties
10.4 Inverse z Transform
10.5 Analysis of LTI Systems using the zT
10.6 Unilateral z Transform and Applications
7.4.1 The Inverse z Transform
ROC },][{)()( jnj rezrnxFreXzX
dereXreXrnx njjjn
2
1 )(2
1)}({F
d
nz
erreXnx njnj
2
)(2
1
dzzj
ddjredzrez jj 11
dzzzX
jnx n 1)(
2
1
for fixed r:
We should choose any value of r in the RoC to make the integration converged
• The symbol O denotes the integration around a counterclockwise circular contour centered at the origin and with radius r.
• The inverse z transform equation states that x[n] can be represented as a weighted integral of complex exponentials
• The formal evaluation of the integral for a general X(z) requires the use of contour integration (围线积分） in complex plane
• There are two alternatives to obtain a sequence from its z transform:
▫ Partial fractional expansion:
▫ Power series expansion:
• Contour Integration Method
m
zz
n
c
n
mzzXsdzzzX
jnx ])([Re)(
2
1][ 11
Where C is the counterclockwise circular contour centered at the origin and within the ROC of X(z),
zm is the pole of X(z)zn-1 at the left-hand side of C.
If zm is the s-th order pole of X(z)zn-1, then
mm zz
ns
ms
s
zz
n zzXzzdz
d
szzXs
})()[({)!1(
1])([Re 1
1
11
Example:
3
1)3(
23
1)2(
2)1(
237
5)(
2
z
z
z
zz
zzX
（1）when |z|>2, x[n] is a causal sequence,
)2)(3
1(
3
5
3
2
3
73
5
)()(2
1
zz
z
zz
z
zzXzG
nn
nLet
As shown in the figure, the poles at the left-hand side of C are
2,3
121 zz
nn
z
n
z
n
z
z
z
z
nx
)2()3
1(
)3
1(
3
5
)2(
3
5
][ 2
3
1
][])2()3
1[(][ nunx nn
1/3 2
C
Then
Thus:
(2) When 1/3<|z|<2, x[n] is a double-side sequence
)2)(3
1(
3
5
)()( 1
zz
z
zzXzG
n
n
when 0n , there is one pole at the left-hand side of C:3
11 z
n
z
n
z
z
nx )3
1(
)2(
3
5
][3
1
1/3 2
C
When n=-1,
)2)(3
1(
3
5
)(
1
zz
z
zG
3
11 z and 02 zThe poles at the left-hand side of C are
2
1
2
53
)2)(3
1(
3
5
2
3
5
][ 0
3
1
1
zz
zzz
z
nx
Both of them are 1st order poles
When n=-2,
)2)(3
1(
3
5
)(
2
zz
z
zG
At the left-hand side of C, besides the 1st order pole3
11 z
, there is a 2nd order pole: 02 z
2
0
3
1
2
)2
1(
4
359)
)2)(3
1(
3
5
()!12(
1
)2(
3
5
][
z
zzz
dz
d
z
z
nx
Similarly, when n=-3, 3)2
1(][ nx
]1[)2(][)3
1(
]1[)2
1(][)
3
1(][
nunu
nununx
nn
nnIn summary,
For n<0, in order to avoid calculating the residual at z=0, we can calculate the
contour integration along the clockwise circular contour C’, which is centered
at the origin and within the ROC of X(z)
m
zz
n
c
n
mzzXs
dzzzXj
nx
])([Re
)(2
1][
1
'
1
zm the pole of X(z)zn-1 at the left-hand side of C’, i.e.
the right-hand side poles of C
As in the previous example, when n<0, there is only one pole at the
right-hand side of C: z=2, then
0)2(
3
13
5
]
)2)(3
1(
3
5
[Re][ 22
n
z
z
zz
z
snx n
z
n
z
n
1/3 2
C
C'
(3) When |z|<1/3, x[n] is a anti-causal sequence
)2)(3
1(
3
5
)()( 1
zz
z
zzXzG
n
n
As shown in the figure , z=0 is the n-th order pole at the left-hand side of C,
whereas the poles at the right-hand side of C are 2,3
121 zz
Then
0)2()3
1(
3
13
5
2
3
5
]
)2)(3
1(
3
5
[Re]
)2)(3
1(
3
5
[Re][
2
3
1
2
3
1
n
z
z
z
z
zz
z
s
zz
z
snx
nn
z
n
z
n
z
n
z
n
]1[])3
1()2[(][ nunx nn
1/3 2
c
• PFE (Partial fractional expansion) Method
Suppose
rkzazazaa
zbzbzbbzX
r
r
k
k
...
...)(
2
210
2
210
M,...2,1 mmzz ，
izz And a s-th order pole at
It has M 1st order poles at
M
0
1 1
B( )
( )
sjm
jm jm i
zA zX z A
z z z z
00 )]([ zzXA
( )[( ) ]
mm m z z
X zA z z
z
izz
s
ijs
j
jz
zXzz
dz
d
js
})(
)[({)!(
1B
s
][!
)1)...(1(
)( 1nu
m
mnnn
z
z mn
m
aznuam
mnnn
az
z n
m
m
][!
))...(2)(1(
)( 1
1
where
Notes:
Example：)4()2(
402)(
3
3
zz
zzzX 4|| z
3
3
3
2
2
211
3
1
1
)2()2()2(4
)2(4)(
z
zC
z
zC
z
zC
z
zA
z
zC
z
zAzX
jj
j
j
4)4(
402)]()
2[(
1)2(
402)]()
4[(
23
33
3
43
2
41
z
zz
zz
zzzX
z
zC
z
zzX
z
zA
B1, B2, B3=1，4，16
31,CASubstituting in X(z)
)4()2(
402
)2(
4
)2()2(4 3
3
3
3
2
2
21
zz
zz
z
z
z
zC
z
zC
z
z
Equating coefficients, one obtains
36 12 CC ，
3
3
2
2
)2(
4
)2(
6
)2(
3
4)(
z
z
z
z
z
z
z
zzX
aznuam
mnnn
az
z n
m
m
][!
))...(2)(1(
)( 1
1
][2)122(
][]2!2
)2)(1(42)1(6234[][
2 nun
nunn
nnx
nn
nnnn
Then
• Power-Series Expansion
▫ X(z) is irrational:
Example：z
arctgzX1
z)(
。。。753
753 xxxxarctgx
0
321
2
7
2
5
2
3
2
1
2
1
2
1
2
1
12
)1(
...7
1
5
1
3
11
...)7
1
5
1
3
1(
)(
n
nn
zn
zzz
zzzzz
arctgzzzX
][12
)1(][ nu
nnx
n
n
nznxzX )()(
▫ X(z) is rational
)(
)()(
zM
zNzX
Performing long division approach to obtain a power series of z for X(z),
the coefficients in this power series are the sequence of x[n].
（1）
（2）
][|| 1 nxRz x is a right-sided sequence, arrange the
numerator and denominator with a order
of the power of z decreasing
][|| 2 nxRz x is a left-sided sequence, arrange the
numerator and denominator with a
order of the power of z increasing
23
7)(
2
zz
zzX 2|| z
23)('
2
zz
zzX
1 2 3
2
1
1
1 2
1 2
3 7
3 2
3 2
3 2
3 9 6
7 6
z z z
z z z
z z
z
z z
z z
1
4321
)12(7
...)1573(7
)('7)(
n
nn z
zzzz
zXzX
][)12(7][ nunx n
Notes: Power-series expansion may come to open/infinity expression
Example:
Topic
10.1 (Bilateral) z Transform
10.2 Properties of RoCs
10.3 Transform Properties
10.4 Inverse z Transform
10.5 Analysis of LTI Systems using the zT
10.6 Unilateral z Transform and Applications
7.5.1 DT System Function
nx nh nhnxny
Y(z) = H(z)X(z) , ROC at least the intersection of the
ROCs of H(z) and X(z),
can be bigger if there is pole/zero
cancellation. e.g.
zzY
zazzX
azaz
zH
all ROC 1)(
,)(
,1
)(
Function System The )(
n
nznhzH
H(z) + ROC tells us everything about system
7.5.2 CAUSALITY
• h[n] right-sided ⇒ ROC is the exterior of a circle possibly including z =
∞:
1
)(
Nn
nznhzH
. include doesbut , circle a outside ROC
at ][ rerm then the,0 If 1
11
not
zzNhNN
0Causal 1 N No zm terms with m>0
=>z=∞
A DT LTI system with system function H(z) is causal ⇔ the ROC of
H(z) is the exterior of a circle including z = ∞
• Causality for Systems with Rational System Functions
NM
azazaza
bzbzbzbzH
N
N
N
N
M
M
M
M
if ,at poles No
)(01
1
1
01
1
1
A DT LTI system with rational system function H(z) is causal
⇔ (a) the ROC is the exterior of a circle outside the outermost pole;
and (b) if we write H(z) as a ratio of polynomials
then
)(
)()(
zD
zNzH
)(degree)(degree zDzN
7.5.3 Stability
• LTI System Stable⇔ ROC of H(z) includes the unit circle |z| = 1
• A causal LTI system with rational system function is stable ⇔ all poles are inside the unit circle, i.e. have magnitudes < 1
nnh
⇒ Frequency Response H(ejω) (DTFT of h[n]) exists.
2
1
2
3]
2
1
2
3[
2
1)( 21
zz
z
z
z
zzH —Example:
Determine h[n], and the causality and stability of the
system
（1） ]][)2
1(][)
2
3[(
2
1][
2
31 nununhz nn ，
Causal, unstable
（2） ]]1[)2
1(]1[)
2
3([
2
1][
2
12 nununhz nn，
Non-causal, unstable
]][)2
1(]1[)
2
3([][
2
3
2
13 nununhz nn ，（3）
Non-causal, stable
× ×3
2
1
2
7.5.4 LTI Systems Described by LCCDEs
Use the time-shift property
M
k
k
N
k
k knxbknya
00
N
k
M
k
k
k
k
k zXzbzYza
0 0
)()(
N
k
k
k
M
k
k
k
za
zbzH
zXzHzY
0
0)(
)()()(
—Rational
ROC: Depends on Boundary Conditions, left-, right-, or two-sided.
For Causal Systems ⇒ ROC is outside the outermost pole
System Function vs Difference Equation
Suppose when the input is ][)2
1(][ nunx n
the zero-state response of the system is
][])2
1(
2
9)
3
1(4)
2
1(
2
3[][ nuny nnn
To determine h[n] and the difference equation of the system
)3
1)(
2
1(
2
)(
)()(
2
zz
zz
zX
zYzH
（1）
3
1
2
2
1
3)(
z
z
z
zzH
2
1z ][])
3
1(2)
2
1(3[][ nunh nn
2
1
2
1)(
z
z
zzX
2
1
)2
1(
3
1)
2
1(
2
2
12
9
3
14
2
12
3
23
z
zzz
zz
z
z
z
z
z
zzY
]1[2][]2[6
1]1[
6
1][
6
1
6
11
21)(
21
1
nxnxnynyny
zz
zzH
—后向差分
（2）
]1[2]2[][6
1]1[
6
1]2[
6
1
6
1
2)(
2
2
nxnxnynyny
zz
zzzH
—前向差分
Example:
• The input-output relationship in time domain for the causal system could be represented by a 2nd order const-coefficient difference equation, if the input is
• The system response (zero state response) is
▫ If the system is with zero-initial state, determine the 2nd order difference equation
▫ If the initial state of the system is y[-1]=1,y[-2]=2, determine the zero-input response of the system with the 2nd order difference equation obtained in step 1
▫ If the initial state of the system is y[-1]=2,y[-2]=4,, the input is , determine the full response of the system y[n]
[ ] [ ]x n u n
][]10532[][ nung nn then
[ ] 3( [ ] [ 5])x n u n u n
(1)
]2[111]1[85][14]2[10]1[7][
1071
1118514
107
1118514
)(
)()(
1
10
53
2)(
1)(
21
21
2
2
nxnxnxnynyny
then
zz
zz
zz
zz
zX
zGzH
z
z
z
z
z
zzG
z
zzX
(2)
1 2
1 21
21 2
[ ] 2 5
1 11
12[ 1] 1 2 5
[ 2] 2 1 1 252
4 25
[ ] 12 2 25 5
n n
zi
n n
zi
y n c c
c ccy
y cc c
y n
设零输入响应为
将 代入
则
2
1 2
14 85 111
( 5)( 2)
2 5
z zH z
z z
系统特征根为 ＝ ， ＝The characteristic values are
Then zero-input response can be
Substituting
Then
in
(3) Since the system could be described by a 2nd order LCCDE, it
is an LTI system, satisfying linear zero-input, and linear zero-state
properties:
[ 1] 2, [ 2] 4
[ ] 2 12 2 25 5
[ ] 3( [ ] [ 5])
[ ] 3( [ ] [ 5])
[ ] [ ] [ ]
n n
zi
zs
zi zs
y y
y n
x n u n u n
y n g n g n
y n y n y n
当 时
（ ）
当 时
when
when
Example:
• For any n, when
• When ,
nnx )2(][ , y[n]=0
1[ ] ( ) [ ]
2
nx n u n1
[ ] [ ] ( ) [ ]4
ny n n a u n
Determine：
(1) the coefficient a,
(2) When for all n, x[n]=1, determine y[n]
(1)
2
1
4
11
)(
)()(
4
1
4
11)(
z
z
z
az
zX
zYzH
z
z
azzY
2
1
2
1)(
z
z
zzX
（2）when x[n]=1
4
1)(1][ 1z ＝zHny n
8
90)2(
0)()2(][
0][,)2(][
2
aH
zHny
nynx
z
n
nwhen
then
Topic
10.1 (Bilateral) z Transform
10.2 Properties of RoCs
10.3 Transform Properties
10.4 Inverse z Transform
10.5 Analysis of LTI Systems using the zT
10.6 Unilateral z Transform and Applications
7.6.1 Unilateral z-Transform (UZT)
• If x[n] = 0 for n < 0, then
• UZT of x[n] = BZT of x[n]u[n]⇒ROC always outside a circle and includes z = ∞
• For causal LTI systems
• The calculation of the inverse UZT is basically the same as that for BZT, with the constraint that the RoC for a UZT must always be the exterior of a circle.
)()( zHz
0
)(
n
nznxz
Note:Differs from BZT
7.6.2 Properties of Unilateral z-Transform
• Many properties are analogous to properties of the BZT e.g
▫ Convolution property (for x1[n<0] = x2[n<0] = 0)
▫ But there are important differences. For example, time-shift
)()(
2121 zzUZ
nxnx
Initial condition
)(1]1[][ 1 zzxznxny Y
100
111)(
n
n
n
n
n
n znxxznxznyzY
z
zmxzx
m
m
0
11
)(][][ zXnunx More general
1
0
])()([][][m
k
km zkxzXznumnx
1
])()([][][mk
km zkxzXznumnx
Proof：
11
1
0
0
1
1
])()([)()(
)()()(
)]()([)(
)()(][][][][
)(][][
mk
km
mk
kmm
mk n
nm
n mk
nm
mk
m
zkxzXzzkxzXz
zkmnkxzXz
zkmnkxzXz
kmnkxmnumnxnumnx
zXzmnumnx
7.6.3 Use of UZTs in Solving Difference
Equations with Initial Conditions
• For an LSI system
Assume x[n]=0 when n<0, the initial condition of the system is {y(-1),y(-2),…,y(-n)}, then
N
k
M
k
kk knxbknya0 0
)()(
)(
)(
)(
])([
)(
)(00
1
0
zB
zXzb
zM
zlyza
zA
zYzaM
k
k
k
N
k kl
lk
k
N
k
k
k
)()(
)(
)(
)()(
)()()()()(
zXzA
zB
zA
zMzY
zXzBzMzYzA
Zero-input response Zero-state response
Example: ][]1[2 nxnyny
11
,]1[
z
nunxy
11
]}1[{
)(12)(z
ny
zYzz
UZ
Y
ZSR
zz
ZIR
zz
)11)(121(121
2)(
Y
UZT of Difference Equation
ZIR — Output purely due to the initial conditions,
ZSR — Output purely due to the input.
β = 0 ⇒ System is initially at rest:
ZSR
α = 0 ⇒ Get response to initial conditions
ZIR
121
1)()(
)(
11
)(
121
1)()()(
zzHz
z
z
z
zzzz
H
XH
XHY
121
2)(
zzY
][)2(2][ nuny n
Example：]2[2][]2[2]1[][ nxnxnynyny
with y[-1]=2，y[-2]=-1/2 x[n]=u[n]
determine ][],[],[ nynyny zszi
)(2
2
2
)1(2)]2(2)1([
)(21
21
21
)1(2)]2(2)1([)(
2
2
2
2
21
2
21
1
zXzz
z
zz
zyzyy
zXzz
z
zz
zyyyzY
)(2)()]2()1()([2)]1()([)( 2121 zXzzXyzyzYzyzYzzY
)()21(])1(2)2(2)1([]21[ 2121 zXzzyyyzYzz
Substituting y[-1]=2,y[-2]=-1/2, and 1
)(
z
zzX In Y(z), one obtains
2 24 2
( 2)( 1) ( 2)( 1) 1
z z z zY z
z z z z z
][]2
3)1(
2
122[
1
2
3
1
2
1
2
2
)1)(1)(2(
2)(
][])1(22[12
2
)1)(2(
4)(
3
2
nuz
z
z
z
z
z
zzz
zzzY
nuz
z
z
z
zz
zzzY
nn
zs
nn
zi
][]2
3)1(
2
1)2(4[][][][ nunynyny nn
zszi
Homework
• BASIC PROBLEMS WITH ANSWER: 10.13
• BASIC PROBLEMS: 10.21, 10.23, 10.34, 10.36, 10.42(a)
• ADVANCED PROBLEMS: 10.46
