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TECHNICAL FACULTY, CHRISTIAN-ALBRECHTS-UNIVERSITY OF KIEL DIGITAL SIGNAL 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 - 1 samples. (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) = 1 and the other non-zero samples of v(n) hit at the zeros h(n). Otherwise the impulse 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

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Page 1: Adsp Ws1213 Solution 13

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

Page 2: Adsp Ws1213 Solution 13

TECHNICAL FACULTY,CHRISTIAN-ALBRECHTS-UNIVERSITYOF KIEL

DIGITALSIGNAL PROCESSING AND

SYSTEM THEORY

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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

Page 3: Adsp Ws1213 Solution 13

TECHNICAL FACULTY,CHRISTIAN-ALBRECHTS-UNIVERSITYOF KIEL

DIGITALSIGNAL PROCESSING AND

SYSTEM THEORY

1

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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