Post on 17-May-2018
1
2. The Z transform
For a complex variable jz x jy re Ω= + = the Z transform Z of a discrete sequence x[n] is:
[ ]{ }( ) [ ] ( )n
nZ x n z x n z X z
∞−
=−∞
= =∑
The Z transform Z is computed on the unit circle is the discrete-time Fourier transform:
[ ]{ }( ) [ ] [ ]{ }( )j j n
nZ x n e x n e F x n
∞Ω − Ω
=−∞
= = Ω∑
For a linear time-invariant (LTI) system, if
[ ] [ ] ( )n nk k k k k
k kx n c z y n c z H z= ⇒ =∑ ∑
The set of values of z for which the series X(z) is convergent is named region of convergence. ROC Properties
1. The ROC of a bilateral z transform can not contain any pole 2. for right sided signals (including causal signals), the ROC extends outward from
the outermost pole, 3. for left sided signals (including anticausal signals), the ROC extends inward from
the innermost pole 4. for infinite duration signals, ROC is a ring that doesn’t include poles – bounded
on the interior and exterior by a pole. 5. for finite duration signals, the ROC is the entire z -plane, except possibly z=0 or
z=∞ . 6. a stable system’s transfer function ( )H z has the poles inside the unit circle. ROC
is outside the unit circle. Problems 1. a) Proof that an even discrete-time sequence x[n]=x[-n] the following equality holds: X(z)=X(1/z). b) Proof that the poles (zeros) of the Z transform are in pairs z0, 1/z0. c) Proof that the sequence [ ] nx n a= is even. Sketch the signal for a=3/4 and |n|<4. d) Find the Z transform of the signal from c) and its region of convergence. e) Find the poles and the zeros of this transform. 2. Determine the Z transform and the ROC for the following discrete-time sequences. Sketch the pole/zero constellation. Do they have discrete time Fourier transform or not? a) [ ]nδ ,
b) [ ]0.5n nσ ,
c) [ ]0.5n nσ − ,
2
d) 0.5 n , e) [ ] [ ]{ }0.5 2n nσ σ− − ,
f) ( ) [ ]0sin n nσΩ ,
g) ( ) [ ]0sinna n nσΩ . 3. Suppose X(z) is the Z transform of the discrete-time sequence x[n]; compute the Z transform for the following signals, as a function of X(z): a) [ ]{ } [ ] [ ]1x n x n x nΔ = − −
b) [ ] ( ) [ ]1 1 nx n x n= −
c) [ ]2
, even2
0, odd
nx nx n
n
⎧ ⎡ ⎤ −⎪ ⎢ ⎥= ⎣ ⎦⎨⎪ −⎩
d) [ ] [ ]3 2x n x n= . 4. Find the discrete-time signals having the following Z transforms: a) ( )( )1 35 1 1z z− −− +
b) ( )21
1
1 az−− for the two possible ROC’s.
c) 21 1
21 11 12 4
z z− −⎛ ⎞⎛ ⎞− +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
, for 1 14 2
z< <
d) 2
2 0.5 0.6z
z z− +
e) 1 2 3
1 2
11 11 36 3
5 116 6
z z z
z z
− − −
− −
+ + +
+ +
f) ( )
( )( )z e e
z e z e
α β
α β
− −
− −
−
− − ; α, β real and positive.
g) ( )1log 1 z−+ , for z a> 5. Consider the causal discrete time system, with null initial conditions represented below.
3
a) Write the finite difference equation b) Find the impulse response h[n] c) Find the transfer function H(z) d) Find the frequency response of the system e) Compute the magnitude and argument of the frequency response computed at d) f) Find the response of the system for the input signal:
1. cos10 nπ ; 2. [ ]12
n
nσ⎛ ⎞⎜ ⎟⎝ ⎠
; 3. ( )2sin 10 nπ ; 4. ( ) ( )sin 10 cos 5n nπ π ; 5. [ ]nσ
g) Implement the system using canonical form II. 6. Consider the causal discrete-time system represented below:
a) Find its transfer function H(z). Sketch the pole/zero constellation and its region of convergence b) What is the value of k for which the poles of the systems are inside the unit circle?
c) Find y[n] if k=1 and [ ] 12
n
x n ⎛ ⎞= ⎜ ⎟⎝ ⎠
7. A discrete-time linear time-invariant system is described by the finite difference equation:
y[n]-2y[n-1]+y[n-2]=x[n]
a) Find its transfer function H(z). Sketch the pole/zero constellation and its region of convergence b) Find its impulse response h[n] c) Analyze the stability of the system.
+ +
D
+ +
D x[n] y[n]
1/3 1/2 1/2
x[n] + +
z-1
y[n]
-k/5 -k/6
4
Solutions. Solution for problem 1 a) [ ] [ ]x n x n= −
[ ] [ ] [ ] [ ]( ) ( )1 1 1mz n m
n m mx n x n z m n x m z x m z X z X
z
∞ ∞ ∞ −− − −
=−∞ =−∞ =−∞
⎛ ⎞− ←⎯→ − = − = = = ⎜ ⎟⎝ ⎠
∑ ∑ ∑
[ ] ( )x n X z↔
But [ ] [ ] ( ) 1 . . .x n x n X z X q e dz
⎛ ⎞= − ⇒ = ⎜ ⎟⎝ ⎠
b) Suppose we have 0z a zero and zp a pole for ( )X z ⇒
( ) ( ) ( )( ) ( )
( )
( )
0 00 00
1 11 1 111
1 1 1 1 11 ppp p
p
z Pz P zz Pz z P z z zzz z zX z Xz zz z Q z z Q zz Q z Qz z z z z
⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ −− − ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− ⎝ ⎠⎛ ⎞ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠= ⇒ = == =⎜ ⎟ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− ⎝ ⎠ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ −⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠
=> 0
1z
- zero and 1
pz is a pole for 1X
z⎛ ⎞⎜ ⎟⎝ ⎠
.
But ( )1X X zz
⎛ ⎞ =⎜ ⎟⎝ ⎠
, which means that 0
1z
is a zero and 1
pz is a pole for ( )X z .
- the zeros of ( )X z are grouped in pairs of the form 00
1,zz
.
- the poles of ( )X z are grouped în pairs of the form 1,pp
zz
c) [ ] [ ] [ ];n n nx n a x n a a x n−= − = = = [ ]x n⇒ is an even function
n
-3 -2 -1 0 1 2 3
1
3/4
9/16
27/64
x[n]
5
d) [ ] [ ] [ ] [ ] [ ], 0 11 1, 0
nnn n n
n
a nu n a n a n a n n
aa nσ σ σ σ−
−
⎧ ⎫≥ ⎛ ⎞= = + − − = + − −⎨ ⎬ ⎜ ⎟< ⎝ ⎠⎩ ⎭
But : [ ] 1
11
n
z aa n
azσ −>
↔−
and [ ] 1
111
n
z bb n
bzσ −<
− − − ↔−
For 1ba
= ⇒ [ ] [ ]11 11
1 1 1 11 11 11 1
n n
zz aa
n na az z
a a
σ σ− −<<
−⎛ ⎞ ⎛ ⎞− − − ⇔ − − ↔⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠− −
↔
For 34
a = the region of convergence is 3 44 3
z< < and:
[ ] ( )1
1 1 1 1
73 1 1 12
3 4 3 44 1 1 1 14 3 4 3
n zu n U z
z z z z
−
− − − −
−⎛ ⎞= ↔ − = =⎜ ⎟ ⎛ ⎞⎛ ⎞⎝ ⎠ − − − −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
The poles of ( )U z are 1
34pz = and
2
43pz = . We see that
1
2
1p
p
zz
= (1)
The zeros for ( )U z are 10 0z = , and ( )
20lim 0z
U z z→∞
= ⇒ = ∞ . We notice that
2
1
00
1zz
= (2)
From the relations (1) and (2) ⇒ the property from b) is fulfilled. Solution problem 2
a) [ ] [ ] [ ]0 1n
nn n zδ δ δ
∞−
=−∞
↔ = =∑ { }ROC z z= ∈ .
The Z transform doesn’t have poles nor zeros. The sequence has discrete time Fourier transform.
b) [ ] 10,5
10,51 0,5 0,5
n
z
znz z
σ −>↔ =
− −
The sequence has discrete time Fourier transform.
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Verify the Z transform using Matlab: >> syms n >> f=(0.5)^n >> ztrans(f) Plot the PZC: >> % Set up vector for zeros >> z = [0]; >> % Set up vector for poles >> p = [0.5]; >> figure(1); >> zplane(z,p); >> title('Pole/Zero Plot');
c) [ ]0
0 00,5 0,5 0,5
0,5
mm nn n n m m
n m m
zn z zσ∞ ∞=−
−
=−∞ = =
⎛ ⎞− ↔ = = ⎜ ⎟
⎝ ⎠∑ ∑ ∑
The geometric series is convergent if 1 0,50,5z z ROC< ⇔ < ←
[ ] 1 0,50,50,51
0,5
n n z zσ − ↔ =
−− . The discrete-time sequence doesn’t have discrete time
Fourier transform.
0.5 1
Im(z)
Re(z)
7
d) [ ] [ ] [ ] [ ]0,5 , 00,5 0,5 0,5 1 0,5 2 1
0,5 , 0
nn n n n n
n
nn n n n
nσ σ σ σ−
−
⎧ ⎫≥= = + − − = + − −⎨ ⎬
<⎩ ⎭
[ ] [ ]1 10,5 2
1 10,5 ;2 11 0,5 1 2
n n
z zn n
z zσ σ− −> <
↔ − − ↔−− −
=> { }0,5 2ROC z z= ∈ < <
( )( ) ( )( )1
1 1 1 1
1 1 1,5 1,50,51 0,5 1 2 0,5 21 0,5 1 2
n z zz z z zz z
−
− − − −
−↔ − = =
− − − −− −
The sequence has discrete time Fourier transform.
e) [ ] [ ]{ } [ ] [ ]{ } ( )1 0,5 10,5 2 0,5 1 0,5 0,5
zn n n n z
zσ σ δ δ − +
− − = + − ↔ + =
{ } { }0ROC z= ∈ − . The sequence has discrete time Fourier transform.
0.5 2
Im(z)
Re(z)1
Re(z)
Im(z)
10.5
8
f)0 0
0sin2
j n j ne enj
Ω − Ω−Ω =
[ ] ( )0
0 0
0
21
00 0 0
1 1 1sin2 2
n njj n j n n
jn n n
SS
en n e e zj j z e z
σΩ∞ ∞ ∞
Ω − Ω −Ω
= = =
⎡ ⎤⎢ ⎥⎛ ⎞ ⎛ ⎞⎢ ⎥Ω ↔ − = −⎜ ⎟ ⎜ ⎟
⎝ ⎠⎢ ⎥⎝ ⎠⎢ ⎥⎣ ⎦
∑ ∑ ∑
The geometric series 1S is convergent if 0 11 1 1
je zz z
Ω
< ⇔ < ⇔ >
The geometric series 2S is convergent if 0
1 11 1 1j ze z zΩ < ⇔ < ⇔ >
{ }1ROC z z= ∈ > . The discrete-time sequence doesn’t have discrete time Fourier
transform.
[ ]0 0 0
0
0 0
0 0
0 1 1
1 1
01 1 2 1 20
1 1 1 1 1 1sin 12 2 1 111
sin2 1 1 2cos
j j j
j
j j
j j
n nej j e z e z
e zze e z z
j e z e z z z z
σ Ω Ω − Ω− −
Ω
Ω − Ω − −
Ω − Ω− − − − −
⎡ ⎤⎢ ⎥ ⎡ ⎤Ω ↔ − = −⎢ ⎥ ⎢ ⎥− −⎣ ⎦⎢ ⎥−−⎢ ⎥⎣ ⎦
−= = Ω
− − + − Ω +
Which means: [ ]1
0 0 01 2 20 0
sin sin sin1 2cos 2cos 1
z zn nz z z z
σ−
− −Ω ↔ Ω = Ω− Ω + − Ω +
0
1 2
202
0 , 0 0
4cos 42cos 1 0 cos sin
2j
p pz z z j e± ΩΩ −− Ω + = ⇒ = = Ω ± Ω =
Verify the Z transform using Matlab: >> syms n omega0 >> f=sin(omega0*n) >> ztrans(f) Plot the PZC for 0 / 4Ω = π : >> % Set up vector for zeros >> z = [0]; >> % Set up vector for poles >> p = [cos(pi/4)+j*sin(pi/4); cos(pi/4)-j*sin(pi/4)]; >> figure(1); >> zplane(z,p); >> title('Pole/Zero Plot');
9
g) ( )0 00sin
2
nj n j nn aa n e e
jΩ − ΩΩ = −
[ ]0 0
0 0
1 2
00 0 0 0
1 1sin2 2
n nj jj n j nn n n n n
n n n n
S S
ae aea n n a e z a e zj j z z
σΩ − Ω∞ ∞ ∞ ∞
Ω − Ω− −
= = = =
⎛ ⎞⎜ ⎟⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎜ ⎟Ω ↔ − = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎜ ⎟⎝ ⎠
∑ ∑ ∑ ∑
The geometric series 1S is convergent if 0
1 1j aae z az z
Ω
< ⇔ < ⇔ >
The geometric series 2S is convergent if 0
1 1j aae z a
z z
− Ω
< ⇔ < ⇔ >
{ }ROC z z a= ∈ >
[ ]0
0
1 2
00 2 2
0 0
2 2 20 0
, 0 0
sin1 1 1sin2 2 cos11
2 cos 4 cos 4cos sin
2
nj
jp p
aza n n ae jaej z a z azz
a a az a ja ae
σ Ω
± Ω
⎛ ⎞⎜ ⎟ Ω
Ω ↔ − =⎜ ⎟− Ω − Ω +⎜ ⎟−−⎜ ⎟⎝ ⎠
Ω ± Ω −= = Ω ± Ω =
Verify the Z transform using Matlab: >> syms n omega0 a >> f=a^n*sin(omega0*n) >> ztrans(f)
1
Im(z)
Re(z)
10
Plot the PZC for 0.5a = and 0 / 4Ω = π : >> % Set up vector for zeros >> z = [0]; >> % Set up vector for poles >> p1 = 0.5*(cos(pi/4)+j*sin(pi/4)); >> p2 = 0.5*(cos(pi/4)-j*sin(pi/4)); >> p = [p1; p2]; >> figure(1); >> zplane(z,p); >> title('Pole/Zero Plot');
The sequence has discrete time Fourier transform if 1a < . Solution problem 3. a) [ ] ( )x n X z↔ ; [ ] 11 ( )x n z X z−− ↔
[ ]{ } [ ] [ ] ( ) ( )11 1x n x n x n z X z−Δ = − − ↔ −
b) [ ] ( ) [ ] ( ) [ ] [ ]( )1 1 1 ( )n n nnx n x n x n z x n z X z−= − ↔ − = − = −∑ ∑
c) [ ]2
, even20, odd
nx nx n
n
⎧ ⎡ ⎤ −⎪ ⎢ ⎥= ⎣ ⎦⎨⎪ −⎩
( ) [ ] [ ] [ ] [ ]2 (2 1) 2 22 2 2 22 2 1 ( )n p p p
n p p pX z x n z x p z x p z x p z X z− − − + −= = + + = =∑ ∑ ∑ ∑
d) [ ] [ ] ( ) [ ]3 32 2 nx n x n X z x n z−= ↔ =∑
a
Im(z)
Re(z)1
11
( ) [ ] [ ] [ ]2 (2 1)2 2 1n p p
p pX z x n z x p z x p z− − − += = + +∑ ∑ ∑
( ) [ ]( ) [ ]( ) [ ]( )
( ) [ ] [ ]
2 (2 1)
2 (2 1)
2 2 1
2 2 1
n p p
p p
p p
p p
X z x n z x p z x p z
X z x p z x p z
− − − +
− − +
− = − = − + + −
− = − +
∑ ∑ ∑
∑ ∑
=> ( ) [ ] 21 ( ) ( ) 22
p
pX z X z x p z−+ − =∑ => [ ]
1 12 21 ( ) ( ) 2
2p
pX u X u x p u−⎛ ⎞
+ − =⎜ ⎟⎝ ⎠
∑
( )1 12 2
31 ( ) ( )2
X z X z X z⎛ ⎞
= + −⎜ ⎟⎝ ⎠
Solution problem 4. a) ( ) ( )( )1 3 1 3 45 1 1 5 5 5 5X z z z z z z− − − − −= − + = − + −
[ ] [ ] [ ] [ ][ ] 5 5 1 5 3 5 4x n n n n nδ δ δ δ= − − + − − − Verify using Matlab that this is true: >> syms z >> X=5*(1-1/z)*(1+z^(-3)) >> iztrans(X) Please notice that the result looks like this: 5*charfcn[0](n)-5*charfcn[4](n)+5*charfcn[3](n)-5*charfcn[1](n) The function charfcn[A](x) is the equivalent of the unit impulse in discrete time (shifted with A), [ ]Anδ − : mhelp charfcn - The charfcn function is the characteristic function of the ''set'' A. It is defined to be / 1 if x is "in" A charfcn[A](x) = < 0 if x is not "in" A \ 'charfcn[A](x)' otherwise
b) ( )( )21
1 ;1
X zaz−
=−
12
( ) ( )( )
( )( )
( )( ) ( )
( ) ( ) ( )
11 21
21 12
2 2 21
21 12
1Let 1
1
1
dX z d z z a zX zaz dz dz z a z a
dX z dX za az azdz dzz a z a az
dX z dX zzX z za dz a dz
−
−
− −⎛ ⎞= ⇒ = =⎜ ⎟− −⎝ ⎠ −
⇒ − = ⇒ − = =− − −
⎡ ⎤⇒ = − = −⎢ ⎥
⎣ ⎦
But: ( ) [ ][ ]1 1
,11 ,1
n
n n
a n z aX z
a n z aazσ
σ−
⎧ >⎪= ↔ ⎨− − − <− ⎪⎩
1st case: [ ] [ ]1
nz a x n a nσ> ⇒ =
( ) [ ] ( ) ( ) [ ]
[ ] ( ) [ ] [ ] ( ) [ ]
1 11 1
1
1 1 1
1 1 1 1 1n n
dX z dX zzz nx n z n x ndz a dz a
x n n a n x n n a na
σ σ+
⎛ ⎞− ↔ ⇒ + − ↔ + +⎜ ⎟
⎝ ⎠
⇒ = + + ⇔ = + +
Verify the signal’s expression using Matlab: >> syms a z >> X=1/((1-a/z)^2) >> iztrans(X) 2nd case
[ ] [ ] [ ] ( ) [ ]
[ ] ( ) [ ]
11
11 1 2
1 2
n n
n
z a x n a n x n n a na
x n n a n
σ σ
σ
+⎡ ⎤< ⇒ = − − − ⇒ = + − − −⎣ ⎦
= − + − −
c) ( ) 1 22 2
1 11 1 1
21 11 1 11 11 1 14 22 4 4
A A BX zz zz z z− −− − −
= = + +⎛ ⎞⎛ ⎞ ⎛ ⎞+ −− + +⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠
( ) 11
12 22
1
1 2 812 911
4
zzB z X z
z−
−
−==
−
⎛ ⎞= − = =⎜ ⎟⎝ ⎠ ⎛ ⎞+⎜ ⎟
⎝ ⎠
( ) 1 1
21
2 4 41
1 2 21 14 312
z zA z X zz
− −−
=− =−−
⎛ ⎞= + = =⎜ ⎟⎝ ⎠ −
;
13
( ) 1 1
1 11
21 1
14 4
2 22 1
1 1 14 441
1 11 14 4
2 1 1 42 1 41 4 2 2 912
z z
z zz
d dz X z A zdz dz
d zA z z A Adz z
− −
− −−
− −=− =−
− −− −
=− =−=−−
⎡ ⎤⎛ ⎞ ⎡ ⎤+ = +⎢ ⎥⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥ ⎛ ⎞⇔ = − ⇔ − − = − ⇔ =⎢ ⎥ ⎜ ⎟
⎝ ⎠⎢ ⎥−⎣ ⎦
( ) 21 11
4 1 2 1 8 11 19 3 911 114 24
X zz zz− −−
= + +⎛ ⎞+ −+⎜ ⎟⎝ ⎠
[ ] [ ] ( ) [ ] [ ]4 1 2 1 8 1 1 11 1 1 ; ROC:9 4 3 4 9 2 4 2
n n n
x n n n n n z⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + + − + − − − < <⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
σ σ σ
[ ] [ ] ( ) [ ] [ ]4 1 2 1 8 1 11 1 ; ROC:9 4 3 4 9 2 2
n n n
x n n n n n z⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + + − + + >⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
σ σ σ
Verify the signal’s expression using Matlab: >> syms a z >> X=2/(1-1/2/z)/(1+1/4/z)^2 >> iztrans(X)
d) ( ) 1 2
1010 5 6
X zz z− −=
− +
1 2
1,
5 25 240 5 21512 12p p
jz− ± − ±= =
1 2
1 1p pz z m− −= =
( )1 2
1 1 1 1 ;p p
A BX zz z z z− − − −= +
− −
( ) ( ) ( ) ( )1 1
1 1
1 1 1 11 1 1 1;p p
p pA z z X z B z z X z
z z z z− − − −
− − − −= − = −= =
( ) 1 2
1 2
1 1
1 1
1 111
p p
p p
A Bz z
X zz zz z
− −
− −
− −
= +−
−
Case I z m> [ ] [ ] [ ]1 1 2 2
1 1 1 1
1 1n
p p p p
A Bx n n nz z z z
σ σ− − − −
⎛ ⎞ ⎛ ⎞= − −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
Case II z m< [ ] [ ] [ ]1 1 2 2
1 1 1 1
1 11 1n n
p p p p
A Bx n n nz z z z
σ σ− − − −
⎛ ⎞ ⎛ ⎞= − − + − −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
14
e)
( )
( )( )( )
( ) ( )( )
( ) ( ) ( ) ( )
1
3 2 11
12 1
2 1
11
1 11 1
1 1
11 1
1
1 1 1
1 11 11 1
1 11 3 13 6 2 11 5 5 61
6 6162 1 2 1
1 12 3 1 12 3
2 12 3
2 3
2 2; 32 3
X z
z z z zX z zz zz z
zzz zz z z z
A Bzz z
zX zz z
A z X z B z X zz z
− − −−
−− −
− −
−−
− −− −
− −
−− −
−
− −
− −− −
+ + += = + +
+ ++ +
= + + = + +⎛ ⎞⎛ ⎞+ + + +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
= + + ++ +
=+ +
= + = − = += − = −
3=
( ) 1 1
1 1 1 1
2 1 3 1 1 12 1 2 11 1 1 12 31 1 1 12 3 2 3
X z z zz z z z
− −
− − − −= + − + = + − +
+ + + +
Case I 12
z > ;
[ ] [ ] [ ] [ ] [ ]1 12 12 3
n n
x n n n n n⎛ ⎞ ⎛ ⎞= − + − − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
δ δ σ σ
Verify the signal’s expression using Matlab: >> syms a z >> X=(1/3/z^3+11/6/z^2+3/z+1)/(1/6/z^2+5/6/z+1) >> iztrans(X)
Case II 1 13 2
z< < ;
[ ] [ ] [ ] [ ] [ ]1 12 1 12 3
n n
x n n n n n⎛ ⎞ ⎛ ⎞= − + + − − − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
δ δ σ σ
Case III 13
z < ;
[ ] [ ] [ ] [ ] [ ]1 12 1 1 12 3
n n
x n n n n n⎛ ⎞ ⎛ ⎞= − + + − − − − − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
δ δ σ σ
15
f) ( )
( )( )z e e
z e z e
α β
α β
− −
− −
−
− − ; α, β real and positive.
( ) ( )( )( )
( )( )( )
( ) ( ) ( ) ( )
( ) ( ) ( )
( )
1
1 11 1
11
1 11
11
1 11
1 11 1
1 11 1
11 11 1
11
z e e z e e A BX ze z e zz e z e e z e z
z e e e e eA e z X z
z e z ee z e e
z e e eB e z X zz e z ee z e
X ze z
− − − − −
− − − −− − − − − −
− − − − −− −
− −− − −
− − − −− −
− − −− − −
−
− −= = = +
− −− − − −
− −= − = = =
= =− −
− −= − = = = −
= =− −
=−
α β α β
α βα β α β
α β α α βα
α αβ β α
α β β αβ
β βα β α
α 1 1
11 e z− − −−− β
We consider α β< 1st case [ ] [ ] [ ]; n nz e e x n e n e nα β α βσ σ− − − −> > = −
2nd case [ ] [ ] [ ]; 1n ne z e x n e n e nα β α βσ σ− − − −> > = − − − −
3rd case [ ] [ ] [ ]; 1 1n nz e e x n e n e nβ α α βσ σ− − − −< < − − − + − − g) ( ) ( )1log 1 ;X z z z a−= + > ; ( ) ( )1 log 1z u X u u− = ⇒ = + In origin, this can be written in a Taylor series.
( ) ( ) ( ) ( )
( ) ( )
( ) ( )
( ) ( ) ( ) [ ] ( )
2
2 32 3
12 3
2
1 1
2
1 1log 1 log 1 log 1 ...0 0 01! 2!
1 1 1 1 1 20 ...0 0 01!1 2! 3!1 1
1 1 12 ... 1 12! 3! !
1 11 1 1! !
kk
k
k nk
k
X u u u u u uu u u
u u uu u uu u u
u u u k uk
nX z k z x n nk n
σ
∞−
=
∞− −−
=
′ ′′⎡ ⎤ ⎡ ⎤= + + + + + + =⎣ ⎦ ⎣ ⎦= = =
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= + + − + + =⎜ ⎟ ⎜ ⎟= = =+ + +⎝ ⎠ ⎝ ⎠
= − + + = − − ⇒
−⇒ = − − ⇒ = − −
∑
∑ [ ]1
Solution problem 5 a) A first order system is described by finite difference equation: [ ] [ ] [ ] [ ]0 1 0 11 1a y n a y n b x n b x n+ − = + − and the transfer function:
16
( )1
0 11
0 1
b b zH za a z
−
−
+=
+
The canonical form II of implementation of this system is:
The system is obtained by the interconnection in cascade of two first order systems.
For the first system we have: 0 1 11
1 1 11; 1; ;3 2
b a ba= = − = =
The transfer function of the first system is: ( )1
11
112113
zH z
z
−
−
+=
−
For the second system we have: 0 1 11
1 11; 1; ; 12
b a ba= = − = =
The transfer function of the second system is: ( )1
21
1112
zH zz
−
−
+=
−
The transfer function for the system obtained by the interconnection in cascade of two first order systems is:
( ) ( ) ( )1 2 1 2
1 21 21 2
3 1 3 11 12 2 2 2
5 11 1 1 116 63 2 6
z z z zH z H z H z
z zz z
− − − −
− −− −
+ + + += ⋅ = =
⎛ ⎞ − +− + +⎜ ⎟⎝ ⎠
The coefficients of the transfer function are: 0 1 0 1 23 5 11; ; 1; ;2 6 6
b b a a a= = = = − =
The finite difference equation is:
[ ] [ ] [ ] [ ] [ ] [ ]5 1 3 11 2 1 26 6 2 2
y n y n y n x n x n x n− − + − = + − + −
x[n] + +
D
y[n]
b1 -a1
1/a0 b0
17
b) 1 2 2 11,2
35 1 5 11 0 5 6 0;26 6 2
z z z z z− − − − ⎧±− + = ⇔ − + = = = ⎨
⎩
( )
( )
2 1 1
2 1 2 1
1
1 2 1 1 1
3 9 6 24 123 ;5 6 5 6
24 125 6 2 3
z z zH zz z z z
z A BH zz z z z
− − −
− − − −
−
− − − −
+ + −= = +
− + − +−
= = +− + − −
( ) ( )1
11 1 11
24 12 24 2 122 362 23 2 3
zA z H zz zz
−−
− −−
− ⋅ −= − = = = −
= =− −
( ) ( )1
11 1 11
24 12 24 3 123 603 32 1
zB z H zz zz
−−
− −−
− ⋅ −= − = = =
= =−
( ) 1 11 1
1 1 1 136 60 18 201 12 3 1 12 3
H zz z z z− −
− −= − + = −
− − − −
But:
[ ]
[ ]
1
1
1 11 2121 11 313
ncausal
ncausal
nz
nz
−
−
⎛ ⎞↔ ⎜ ⎟⎝ ⎠−
⎛ ⎞↔ ⎜ ⎟⎝ ⎠−
σ
σ
=> So we have: [ ] [ ] [ ] [ ]1 13 18 202 3
n n
h n n n n⎛ ⎞ ⎛ ⎞= + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
δ σ σ
Verify the impulse response’s expression using Matlab: >> syms z >> H=(3/z^2+9/z+6)/(1/z^2-5/z+6) >> iztrans(H)
c) The transfer function has already been computed: ( )2 1
2 1
3 9 65 6
z zH zz z
− −
− −
+ +=
− +
d) ( )2
2
3 9 65 6
j jj
j j
e ez e He e
− Ω − ΩΩ
− Ω − Ω
+ += ⇒ Ω =
− +
e)
( ) ( ) ( )( )
( )( )
3 cos 2 sin 2 9 cos sin 6cos 2 sin 2 5 cos sin 6
3cos 2 9cos 6 3sin 2 9sincos 2 5cos 6 sin 2 5sin
j jH
j j
jj
Ω− Ω + Ω− Ω +Ω = =
Ω− Ω− Ω− Ω +
Ω+ Ω+ − Ω+ ΩΩ− Ω+ − Ω− Ω
18
( ) ( ) ( )( ) ( )
( )( )
2 2
2
2
2
3cos 2 9cos 6 3sin 2 9sin
cos 2 5cos 6 sin 2 5sin
36 2cos 1 108cos 126126 54cos 36cos 2 108cos62 10cos 12cos 2 60cos 12 2cos 1 70cos 62
HΩ+ Ω+ + Ω+ Ω
Ω = =Ω− Ω+ + Ω− Ω
Ω− + Ω++ Ω+ Ω+ Ω= =
− Ω+ Ω− Ω Ω− − Ω+
( )2
2
72cos 108cos 9024cos 70cos 50
H Ω+ Ω+Ω =
Ω− Ω+
f)1˚) [ ] [ ] ( ) ( ){ }( )cos10 10 cos 10 arg 10x n n y n H n Hπ π π π= ⇒ = +
( ) ( ) ( )10 0 5 2 0 4,74H H Hπ π= + ⋅ = = ; ( ){ } ( ){ }arg 10 arg 0 0H Hπ = = [ ] ( )4,74cos 10y n nπ=
2˚) [ ] [ ] ( ) ( ) ( ) ( )1
1 112 12
n
x n n X z Y z X z H zz
σ−
⎛ ⎞= ⇒ = ⇒ = ⇒⎜ ⎟⎝ ⎠ −
( )1
1
21 1 1 1 11
111 121 1 1 1 111 1 1 1 112 3 2 2 32
z z A B CY zz z z z zz
−−
− − − − −−
+ +⇒ = ⋅ ⋅ = + +
⎛ ⎞− − − − −−⎜ ⎟⎝ ⎠
( )( )
1 1
1 1
123 3
1
1 51 1 41 2 21 4013 11 42
z z
z zC z Y z
z− −
− −
−
= =−
⎛ ⎞+ +⎜ ⎟⎛ ⎞ ⎝ ⎠= − = = =⎜ ⎟⎝ ⎠ ⎛ ⎞−⎜ ⎟
⎝ ⎠
( )( )
1 1
1 12
1
2 21
11 11 621 181 12 1
3 3z z
z zB z Y z
z− −
− −
−
= =−
⎛ ⎞+ +⎜ ⎟⎛ ⎞ ⎝ ⎠= − = = =⎜ ⎟⎝ ⎠ −
( )1 1 1
1
1 2 221
2 2 21 2
2
2 22
3 1 3 111 2 2 2 21 1 12 13 3
7 12 6 150 150
13
z z z
z
z z z zd d dA z Y zdz dz dzz z z
zA
z z
− − −
−
− −
−
= = =−
=
⎛ ⎞ ⎛ ⎞+ + + +⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎛ ⎞= − = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠ ⎜ ⎟ ⎜ ⎟⎝ ⎠ − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
+= = ⇒ =⎛ ⎞−⎜ ⎟⎝ ⎠
19
We have: ( )
[ ] [ ]
21 1
1150 12 403
150 181 11 12 2
nnn
n
Y zz z
σσ
− −
⎛ ⎞⎜ ⎟ ⎛ ⎞⎝ ⎠ ⎜ ⎟
⎝ ⎠
= +⎛ ⎞− −⎜ ⎟⎝ ⎠
and 1
2 21 1
118 2361 11 12 2
z z
z z
−
− −
⋅=
⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
and [ ]1
21
112211
2
nzn n
zσ
−
−
⎛ ⎞↔ ⎜ ⎟⎝ ⎠⎛ ⎞−⎜ ⎟
⎝ ⎠
, ⇒ ( ) [ ]1 1
21
11236 36 1 1211
2
nzz n n
zσ
− +
−
⎛ ⎞↔ + +⎜ ⎟⎝ ⎠⎛ ⎞−⎜ ⎟
⎝ ⎠
Therefore:
[ ] [ ] ( ) [ ] [ ]11 1 1150 36 1 1 40
2 2 3
n n n
y n n n n nσ σ σ+
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
3˚) [ ] ( ) ( ) [ ]1 cos 20sin 10 0, 0,
2n
x n n n y n nπ
π−
= = = ∀ ∈ ⇒ = ∀ ∈
4˚) ( ) ( ) [ ] ( )sin 10 cos 5 0, 0,n n n y n nπ π = ∀ ∈ ⇒ = ∀ ∈
5˚) [ ] [ ] ( ) 1
11
x n n X zz
σ −= ⇒ ⇒−
( ) ( ) ( )
( ) ( )( )
1 1
11
1 11 1 1 1
1 1
1
1 11 1
111 121 1 1 11 11 1 1 13 2 3 2
1 31 1 22 2 9;2 11 11 13 23 2
z z
z z A B CY z X z H zz zz z z z
z zA z Y z
z z− −
−−
− −− − − −
− −
−
= =− −
+ +⇒ = ⋅ = ⋅ ⋅ = + +
− −− − − −
⎛ ⎞+ + ⋅⎜ ⎟⎝ ⎠= − = = =⎛ ⎞ ⎡ ⎤ ⋅− −⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦
( )( )
( ) ( )1 1
1 1
1
3 31 1
1 51 1 41 2 21 101 13 1 1 22 2
z z
z zB z Y z
z z− −
− −
−
= =− −
⎛ ⎞+ + ⋅⎜ ⎟⎛ ⎞ ⎝ ⎠= − = = =⎜ ⎟ ⎛ ⎞ ⎛ ⎞⎝ ⎠ − − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
( )( )
( ) ( )1 1
1 1
1
2 21 1
11 11 2 321 18
1 12 1 1 13 3
z z
z zC z Y z
z z− −
− −
−
= =− −
⎛ ⎞+ +⎜ ⎟ ⋅⎛ ⎞ ⎝ ⎠= − = = = −⎜ ⎟ ⎛ ⎞ ⎛ ⎞⎝ ⎠ − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
20
( ) [ ] [ ] [ ] [ ]11 1
9 10 18 1 19 10 181 11 3 21 13 2
n n
Y z y n n n nz z z
σ σ σ−− −
⎛ ⎞ ⎛ ⎞= + + ↔ = + +⎜ ⎟ ⎜ ⎟− ⎝ ⎠ ⎝ ⎠− −
g) ( )1 2
0 1 2 0 1 21 2
3 11 5 1 3 12 2 1; ; ; 1; ;5 1 6 6 2 216 6
z zH z a a a b b b
z z
− −
− −
+ += ⇒ = = − = = = =
− +
Solution problem 6
a) 0 1 10
1 1; 1; ;5 6k kb a b
a−
= = − = − = .
( )1
10 1
0110 1
16 6 ;
6 515 5
p
k kz zb b z k kH z z zk ka a z z z
−−
−−
− −+= = = ⇒ = = −
+ + +
b) 1 1 5 55k k− < − < ⇒ − < <
c) [ ] [ ] [ ]11 1 1 12 2 2
n n n
x n n nσ σ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = + − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
[ ]12
1
1 112 12
zn
nz
σ>
−
⎛ ⎞ ↔⎜ ⎟⎝ ⎠ −
and [ ]12
1
1 11 12 12
zn
nz
σ<
−
−⎛ ⎞ − − ↔⎜ ⎟⎝ ⎠ −
The ROC of the Z transform for the sequence [ ]1x n is the void set.
x[n] + +
D
y[n]
3/2 5/6
D
+ +
-1/6 1/2
21
[ ] [ ] [ ] [ ]
1 , 021 1 2 1
2 21 2 , 02
n
n nn
nn
nx n x n n n
n
σ σ−
⎧ ⎛ ⎞ ≥⎪ ⎜ ⎟⎪ ⎝ ⎠⎛ ⎞ ⎛ ⎞= = ⇒ = + − −⎨⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎛ ⎞⎪ = <⎜ ⎟⎪⎝ ⎠⎩
[ ]12
1
1 112 12
zn
nz
σ>
−
⎛ ⎞ ↔⎜ ⎟⎝ ⎠ −
and [ ]2
1
12 11 2
zn n
zσ
<
−
−− − ↔
−
The ROC of the Z transform for the sequence [ ]x n is 1 22
z z⎧ ⎫∈ < <⎨ ⎬
⎩ ⎭
( )
( ) ( ) ( ) ( )
1 1
11
11 1
1 1 ;1 1 2
11 1 161 1 1 21 15 2
k
X zz z
zY z X z H z Y z
zz z
− −
−=
−− −
= −− −
⎛ ⎞− ⎜ ⎟= ⋅ ⇒ = −⎜ ⎟−⎜ ⎟+ −
⎝ ⎠
( )
( )
( )( )2 2
1 1
1 1 1 1
1 11 16 6
1 1 11 1 1 1 25 2 5
Y z Y z
z zY z
z z z z
− −
− − − −
− −= −⎛ ⎞⎛ ⎞ ⎛ ⎞+ − + −⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠
( )11 1
;1 11 15 2
A BY zz z− −
= ++ −
( )1 1
1
11 5 51
111 1161 15 2112
z z
zA z Y z
z− −
−
−
=− =−−
−⎛ ⎞⇒ = + = =⎜ ⎟⎝ ⎠ −
( )1
1
11 21
111 1061 12 2115
z
zB z Y z
z−
−
−
=−
−⎛ ⎞= − = =⎜ ⎟⎝ ⎠ +
[ ] [ ] [ ]111 1 10 121 5 21 2
n n
y n n nσ σ⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
( ) ( )1 1
1
12 21 15 51
111 1611 1 2 5 1 2 615
z z
zC DY z C z Y zz zz
− −
−
−− −=− =−−
−⎛ ⎞= + ⇒ = + = =⎜ ⎟− −⎝ ⎠+
( ) ( )1 1
1
12 1 11
2 2
11 561 2 1 615
z z
zD z Y z
z− −
−
−
−= =
−= − = =
+
22
( ) [ ] [ ] [ ]2 211
1 1 5 1 1 1 5 216 6 1 2 6 5 615
nnY z y n n n
zz−
−
⎛ ⎞= ⋅ + ⋅ ↔ = − +⎜ ⎟− ⎝ ⎠+σ σ
[ ] [ ] [ ] [ ] [ ] [ ] [ ]1 211 1 10 1 1 1 5 221 5 21 2 6 5 6
n n nny n y n y n n n n nσ σ σ σ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = − + − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
[ ] [ ] [ ] [ ]45 1 10 1 5 2126 5 21 2 6
n nny n n n nσ σ σ⎛ ⎞⎛ ⎞ ⎛ ⎞= − + −⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠ ⎝ ⎠
Solution problem 7
a) 0 1 2 01; 2; 1; 1a a a b= = − = = => ( )( ) 1 2
1 121 2 1
1 1 11 2 1
p pH z z zz z z
− −− − −
= = ⇒ = =− + −
There are two regions of convergence: 1z < or 1z > .
b) ( ) [ ]1
1 1
1 z
H z nz
σ>
−= ↔ ; ( ){ }( ) ( ) ( )
2
1 2 2 21
1 11 1 1 1
d d z z z zH zdz dz z z z z
−
−
− −⎛ ⎞= = = − =⎜ ⎟−⎝ ⎠ − − −
( )( ){ } ( ) [ ]2
121
1 1 11
dz H z n ndzz
σ−
⇒ = − ↔ + +−
The system is strictly stable. If [ ] [ ]0 0x n y n= ⇒ =
If [ ]x n is the simplest discrete-time signal that isn’t null, [ ] [ ]x n nδ= then
[ ] [ ] ( ) [ ]1 1y n h n n nσ= = + + , which means the systems has oscillations.
n
n
x[n]
y[n] …-1 0 1 2…
…-1 0 1 2…