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- Stable;
- Non-Causal – output depends on future input;
- Linear – T(ax[n] + by[n]) = aT(x[n]) + bT(y[n]);
- Time Variant;
- Not Memoryless;
P2.
As [ 2 ] 2 [ 1 ] 8
[ 1 ]
y [ n ]− yn − + yn − = xn − , the Fourier transform is
jw jw jw jw jw jw jw Y e Y e e Y e e X e e
− − − − + ( ) = 2 ( ) 8
2
When [ ]= [ ]⇒ ( )= 1
jw
xn δ n X e , thus
− − − −
−
jw jw jw jw
jw jw
e e e e
e Y e 1 ( 1 / 4 )
2
[ ]
y [ n ] 8 un
n n
P2.
As h[n] = [1 1 1 1 –2 –2] for n from 0 to 5 and x[n] = u[n-4], the system response is:
y[n] = x[n] * h[n] = [0 0 0 0 1 2 3 4 2 0 …..]; The sketch is shown as follows:
HW#2: P2.5; P2.18; P2.29 (a), (c), (e)
P2.
(a) The roots for polynomial 1 5 6 0
1 2 − + =
− − z z are 2 and 3, so the homogeneous
response for the system is:
n n y [ n ]= A 1 2 + A 23
(b) As y [ n ]− 5 y [ n − 1 ]+ 6 y [ n − 2 ]= 2 x [ n − 1 ] and x [ n ]= δ[ n ], the impulse response of
the system is:
[ ] 2 ( 3 2 ) [ ]
2
hn u n
e e e e
e H e
n n
jw jw jw jw
jw jw
− − − −
−
(c) As y [ n ]− 5 y [ n − 1 ]+ 6 y [ n − 2 ]= 2 x [ n − 1 ]and x [ n ]= u [ n ], the step response of the
system is:
[ ] ( 3 2 1 ) [ ]
1 2
1 2 1 1 1 1
1
yn u n
z z z z z z z
z Y z H z X z
n n ⇒ = − +
− − − − − −
−
P2.
(a) h [ n ] ( 1 / 2 ) u [ n ]
n
Causal, the output of the system does not depend on future input;
(b) h [ n ]= ( 1 / 2 ) u [ n − 1 ]
n
Causal, the output of the system does not depend on future input;
(c)
n h [ n ]=( 1 / 2 )
Non-Causal, the output of the system depends on future input;
(d) h [ n ]= u [ n + 2 ]− u [ n − 2 ]
Non-Causal, the output of the system depends on future input;
(e) h [ n ]= ( 1 / 3 ) u [ n ]+ 3 u [− n − 1 ]
n n
Non-Causal, the output of the system depends on future input;
P2.
As x[n] = [1 1 1 1 1 1/2] for n from –1 to 4,
(a) x[n-2] = [1 1 1 1 1 1/2] for n from 1 to 6; The sketch is:
x [ n ] cos( n ) u [ n ] ( 1 ) u [ n ]
n = π = −
) [ ]
[ ] ( un
j h n
n
) [ ] ( 1 ) (
[ ] [ ]* [ ] [ ][ ] (
1
0
j
j
j uk un k
j yn hn xn hk xn k
n n
k
n
k
k nk n k
k
∞
= −∞ =
−
∞
=−∞
∑ ∑ ∑
Since 1 / 2
lim
1
j j
j
n
n (^) +
→∞
The steady state response to the excitation x [ n ] ( 1 ) u [ n ]
n = − is
cos( )
j
n
j
n
π
P2.
Given a periodic impulse train (^) ∑
∞
=−∞
k
x [ n ] δ [ n kN ], we can write its Fourier transform as
∑
∞
=−∞
k
j k N N
X e ( 2 / )
(Refer to Signal and Systems , 2 nd edition by A.V. Oppenheim and A.S. Willsky, Page 371 for its proof)
In problem, 2.41, N =16, so its Fourier transform is
∑
∞
=−∞
k
j X e ( 2 k / 16 ) 16
Let ( )
j ω Y e denotes the output of the system, then
j ω j ω j ω Y e = X e He (3)
If | ω |< 3 π/ 8 ,
3 ( )
j ω j ω H e e
−
[ ( ) ( / 8 ) ( / 8 )]
−
∞
=−∞
− ∑
j j
k
j e k e e (4)
If | ω |≥ 3 π/ 8 , ( )= 0
j ω H e , thus ( )= 0
j ω Y e , (5)
So [ ( ) ( / 8 ) ( / 8 )] 16
3 / 8 3 / 8
j j − j Y e e e (6)
Take the inverse Fourier transform, we can get
cos( 8
( 1 2 cos 16
[ ]
3 / 8 / 8 3 / 8 / 8 ( 3 ) / 8 ( 3 ) / 8
− − − − −
n n
y n e e e e e e
j n j n n n
π π π π π π
Note: Take a look at (3), ( )
j ω H e is band limited, ( )
j ω X e is infinite pulse train. If we
multiply them together, we can only consider those pulses falling into the band
( − 3 π/ 8 , 3 π / 8 ). The rest pulses are cancelled due to the multiplication with 0. There are
three pulses falling into the band ( ) 16
− , so we get
P3.
(a)
1 1 1 2 1 1 12 1 1 2 1 3
2
− − − − − − − −
z
D
z
C
z
B
z
A
z z z
X z
X (z)’s poles are z=-1/2, 2, 3, if it is stable, the ROC is | z |∈( 1 / 2 , 2 )
1 / (^211)
1 2
− −
=− − −
− z z z z
A X z z
1 2 1
2
1 =−
− −
=
− z z
z z
C X z z
1 2 1
3
1
− −
=
− z z
z z
D X z z
Also, Let 0
1
− z at both sides,
X z A B C D z
− (^) =
(^10)
Thus,
1225
B = 1 − A − C − D =
1 1 1 2 1 1 12 1 1 3
− − − − − − − −
z z z z z z z
X z
Since the ROC is | z |∈( 1 / 2 , 2 ),
3 [ 1 ]
2 [ 1 ]
) [ ]
) [ 1 ]
x [ n ]= n + − un + + − un + u − n − − u − n −
n n n n
Note: To get the inverse Z-Transform of second-order term or multiple order term, we
can use the differentiation property
dz
dX z nx n z
[ ]↔ − (Refer to page122 of textbook for
its proof)
E.g. right side sequence 1 1
[ ] [ ]
− −
az
x n a un
n (ROC: | z |>| a |)
1 2
1 1
]
[ ] [
−
− −
az
az
dz
az
d
na n z
n , So 1 2
1 1
( 1 )
[ ]
−
− −
−
az
z na n
n
(c)
1
2
3
[ ] ( )
− −
z
z z z
z z xn X z
[ 1 ]
[ ]
[ ] 1 [ ] 2 [ ]
[ 1 ]
2 [ ]
[ ]
[ ]
1 [ ]
1 1
1 4 3
3 2
1 2 1
1
1 3 3
2 2
1 1
1
1
1 1
1
^ −
− − − − − −
−
− − − −
−
−
− −
−
u n a
un a
xn x n x n
un a
z x n a
z a
z a
z z a
az
un a
un a a
z x n a
z a
z z a a a
z a
az
z a z a
az X z
n n
n
n n
L
L
- Fourier Transform exists when the ROC including unit circle, which means
a < 1.
P3.
(a) As the ROC of X (z) is |z| > 3/4, and the ROC of Y (z) is |z| > 2/3, the ROC of H (z)
should be |z| > 2/3 ;
(b) As the ROC of X (z) is |z| < 1/3, and the ROC of Y (z) is 1/6 < |z| < 1/3, the ROC of
H (z) should be |z| > 1/6 ;
P4.
As x (^) c ( t )= sin[ 2 π( 100 t )]and T = 1/400 sec,
[ ] (^)
[ ] ( ) sin 2 ( 100 ) sin
n x n xc nT nT
P4.
As x (^) c ( t )= cos[ 4000 π t ]and
[ ] cos
n π
x n ,
(a) Let 12 , 000
x [ n ]= xc ( nT )⇒ T =
(b) T is not unique, for example, 12 , 000
T =
HW#5: P4.5; P4.7; P5.2; P5.
P4.
(a) From Nyquist Sampling theorem, to avoid aliasing in the C/D converter, the sampling
frequency Hz Hz T
m s
s
4 2 2 * 5000 10
Ω = ≥ Ω = = , so Ts s
4 10
− ≤
(b) Hz
f s
cutoff cutoff^10625 2
4 Ω = Ω = =
(c) Hz kHz
f s
cutoff cutoff^21012501.^25 2
4 Ω = Ω = × = =
Note: The relation between digital frequency f and analog frequency Ω is
f
s
where Ω (^) s is the sampling frequency, f is in radians.
P4.
(a)
xc ( t )= sc ( t )+ α s c ( t − τ d )
j d Xc j Sc j e
τ
−Ω Ω = Ω +
Consider sampling, x [ n ]= xc ( nT ), in frequency domain ( refer to Eq4.19 in textbook, P147),
∞
=−∞
Ω = Ω− k
c
jT
T
k X j T
Xe
/
∞
=−∞
−
∞
=−∞
−Ω
∞
=−∞
Ω
k
c
j T
k
c
j
k
c
j jT
T
k
T
e S j T
T
k e S j T T
k X j T
Xe Xe
d
d
ω π α
π α
π
ωτ
ω τ
(b)
j j d T H e e
/ ( ) 1
ω ωτ α
− = +
(c)
τ π
τ π α π
π
α ω π
ω π
α ω π
ω π
π
π
π
π
ω ω τ
π
π
ωτ ω
π
π
ω ω
sin( ) sin[( / ) ]
[ ]
/ ( / )
n T
n T
n
n
hn He e d e e d e d e d
d
d
j jn j d T jn jn j n d T
− −
− −
−
−
−
if τ (^) d = T , [ ] [ 1 ]
( 1 )
sin( ) sin[( 1 ) ] [ ] = + − −
= + n n n
n
n
n hn δ αδ π
π α π
π
if τ (^) d = T / 2 ,
π
π δ α π
π α π
π
sin[( 1 / 2 ) ] [ ] ( 1 )
sin( ) sin[( 1 / 2 ) ] [ ] −
n
n n n
n
n
n hn
P5.
[ ] [ 1 ] [ ]
y [ n − 1 ]− yn + yn + = xn
1 z Y z − Y z + zY z = X z
−
P5.
(a)
As the poles of H(z) are 0.9, -0.9, ROC includes the unit circle, the system is stable.
(b)
1
()
1 1
1 1
()
1 1
1 1 1
1 1
1 1 1
1
H zH z
z z
z z
j z j z
z z z
j z j z
z z z H z
ap
H z Hapz
− −
− −
− −
− − −
− −
− − −
P5.
Generalized Linear Phase – GLP;
Linear Phase – LP;
(a) As
jw h n n n n H jw we
− [ ]= 2 δ [ ]+δ[ − 1 ]+ 2 δ[ − 2 ]⇒ ( )=( 1 + 4 cos ) ,
A ( jw )= 1 + 4 cos w and α = 1 , β= 0 , it is a GLP, but not LP as A ( jw )is not always
nonnegative for all w;
(b) It is not a GLP or LP as it is not a symmetric filter;
(c) As
jw h n n n n H jw we
− [ ]= δ [ ]+ 3 δ[ − 1 ]+ δ[ − 2 ]⇒ ( )=( 3 + 2 cos ) ,
A ( jw )= 3 + 2 cos w and α = 1 , β= 0 , it is a GLP and also LP as A ( jw )is always
nonnegative for all w;
(d) As
( / 2 ) [ ] [ ] [ 1 ] ( ) 2 cos( / 2 )
jw h n n n H jw w e
− =δ +δ − ⇒ = ,
A ( jw )= 2 cos( w / 2 ) and α = 1 / 2 ,β= 0 , it is a GLP, but not LP as A ( jw )is not always
nonnegative for all w;
(e) As
( / 2 ) [ ] [ ] [ 2 ] ( ) 2 sin
π δ δ
− − = − − ⇒ =
jw h n n n H jw we ,
A ( jw )= 2 sin w and α = 1 , β= π/ 2 , it is a GLP, but not LP as A ( jw )is not always
nonnegative for all w;
HW#7: P6.7; P6.8; P6.11; P6.25; P7.
P6.
The difference equation is: [ ]
4
[ 2 ] [ 2 ]
y [ n ]− yn − = xn − − xn
Z-Transform: ( )
4
2 2 Y z − Y zz = X zz − X z
− −
Transfer function:
2
2
−
−
z
z
X z
Y z H z
P6.
y [ n ]− 2 y [ n − 2 ]= 3 x [ n − 1 ]+ x [ n − 2 ]
P6.
1
1 2 3
1
1 1 1
−
− − −
−
− − −
z
z z z
z
z z z H z
(a)
− 1 z
− 1 z
x [ n ] y [ n ]
− 1 z
− 1 z
y [ n ]
− 1 z 8
Cutoff: ω c = 0. 3 π
The peak approximate error 20 log 10 δ<− 26. 02 dB
Among the windows in Table7.1 (Page 471), Hanning, Hamming, Blackman can be used
(b)
Hanning: M
0. 1 = , M = 80 , L = M + 1 = 81
Hamming: M
0. 1 = , M = 80 , L = M + 1 = 81
Blackman: M
0. 1 = , M = 120 , L = M + 1 = 121
Note that the estimation is not accurate. We can use MATLAB to find the minimum of
filter order to meet the requirements.
