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TECHNICAL FACULTY,CHRISTIAN-ALBRECHTS-UNIVERSITYOF KIEL
DIGITALSIGNAL PROCESSING AND
SYSTEM THEORY
Solution to Problem 24 (multirate digital signal processing)
(a) h(n) = 0 for |n| > (RL − 1). For a causal system we have a delay by RL − 1samples.
(b) General interpolation condition:
h(0) = 1
h(nL) = 0, n = ±1, ±2, . . .
(c)
y(k) =RL−1∑
k=−(RL−1)
h(k)v(n − k) = h(0)v(n) +RL−1∑
k=1
h(k)(v(n − k) + v(n + k))
This requires only RL − 1 multiplications (assuming h(0) = 1).
(d) Show, that only 2R Multiplications per output sample are required.
y(n) =n+RL−1∑
k=n−(RL−1)
v(k)h(n − k)
Due to part b) it has been shown, that only h(0) = 1, and h(Ln) = 0 for n =+−
1,+− 2, ... This is a general condition related to the interpolation (include (L − 1)
zeros).
⇒If n = mL (m an integer), then we don’t have any multiplications since h(0) = 1and the other non-zero samples of v(n) hit at the zeros h(n). Otherwise theimpulse response spans 2RL − 1 samples of v(n), but only 2R of these are non-zero. Therefore, we have 2R multiplications in total.
Digital Signal Processing and System Theory, Prof. Dr.-Ing. Gerhard Schmidt, www.dss.tf.uni-kiel.de
Advanced Digital Signal Processing, Exercises WS 2012/2013
1
TECHNICAL FACULTY,CHRISTIAN-ALBRECHTS-UNIVERSITYOF KIEL
DIGITALSIGNAL PROCESSING AND
SYSTEM THEORY
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h(0−n)
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Solution to Problem 25 (multirate digital signal processing)
Digital Signal Processing and System Theory, Prof. Dr.-Ing. Gerhard Schmidt, www.dss.tf.uni-kiel.de
Advanced Digital Signal Processing, Exercises WS 2012/2013
1
TECHNICAL FACULTY,CHRISTIAN-ALBRECHTS-UNIVERSITYOF KIEL
DIGITALSIGNAL PROCESSING AND
SYSTEM THEORY
1
Y (z) X(z)
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↑ 3
↑ 3
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↑ 3
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↑ 3 ↓ 2
↓ 2
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↓ 2
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G(z)
G0(z)
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G0(z)
G1(z)
G1(z)
G1(z)
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G1(z)
G00(z)
G01(z)
G02(z)
G10(z)
G12(z)
z−1
z−1
z−1
z−1
z−1
z−1
z−1z−1
z−1
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z−3
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b)
c)
d)
e)
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Digital Signal Processing and System Theory, Prof. Dr.-Ing. Gerhard Schmidt, www.dss.tf.uni-kiel.de
Advanced Digital Signal Processing, Exercises WS 2012/2013
2