Multirate Signal Processing: Theory and Applications

Multirate 1
Multirate signal processing
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Multirate 2
Question.1
Noble identity z- transform for down sampling is given by;
Ck [ n ] =δu ∨n [ n ] = 1
K ∑
k=0
k −1
e
j 2 kπn
K
X k [ n ]= {x [n ] k [ n ]
0 k ≠ n =Ck [ n ] x [ n ]
X k [ z ]=∑
n
X k [ n ] z−n= 1
K ∑
n
∑
k=0
k−1
e
j 2 πkn
K x [ n ] z−n= 1
K ∑
k=0
k−1
∑
n
x [n](e
j 2 πkn
K z)−n
¿ 1
K ∑
k=0
k−1
x (¿ e
− j2 πkn
K z )¿
Thus
X k ( z ) =Y ( zk)
Y ( z )=X k (z
1
k )= 1
K ∑
k =0
k−1
x(e¿¿ − j 2 πkn
K z
1
k )¿
Given;
y [ n ]= ( ↓ 3 ) {h [ n ]∗x [n]} The Z-transform Y (Z) is given as;
Y ( Z ) = ( ↓3 ) { H ( Z )∗X ( Z ) } But H ( Z )= 1−z−1
1−z−3
Substituting, we get;Y ( Z ) = ( ↓3 ) {1−z−1
1−z−3 ∗X ( Z ) } taking inverse z transform
y [ n ]= 1
1−z−1 ∗{ ( ↓ 3 ) (1−z−1∗x [ n ] ) }=h1 [ n ]∗{ ( ↓3 ) (h2 [ n ]∗x [n])}
Hence the process can efficiently be implemented as shown.
Question.2
The discrete time Fourier transform of a signal is given by;
X ( w ) =∑
−∞
∞
x (n)e− jwn =∑
− π
12
π
12
e− jwn
But w=π /12

Multirate 3
y1 ( w ) = 1
6 π ∑
−π / 12
π /12
e− jwn= 1
6 π e− jπ / 12n
| π /12
−π /12
y2 ( w ) = 2
π ∑
−π / 12
π /12
e− jwn= 2
π e− jπ / 12n
| π /12
−π /12
y1 ()
Question.3
For interpolation, the cutoff frequency ¿ 2∗π /4
It takes into interpretation that after the interpolation procedure the signal is made up of (L–1)
zero coefficients, and the decimation progression infers that only one out of every M samples is
needed at the productivity of the converter. To ensure the system is more resourceful, the low-
pass filter in is substituted with a bank of filters organized in parallel. The sampling-rate
π /12−π /12