PROBLEMS 329 conversational level is of order 10 . Very loud talking generates o

Question

PROBLEMS 329 conversational level is of order 10 . Very loud talking generates on the order of 100 and shouting in excess of about 1000 W PROBLEMS 11.2.1. A proportional bandwidth filter has fl/f r-21/n. Find n and r for a filter designed to have a bandwidth of (a) 1/3 octave, (b) 1/2 octave, and (c) 1/12 octave. 11.2.2. The. even-tempered musical scale is designed so that adjacent semitones have 1 /f-21/12 (a) How many semitones are there in an octave? (b) For an octave with lowest note (first semitone) tuned to 440 Hz, determine the frequencies of the remaining semitones in the octave. (c) Calculate the ratio of frequencies of 8th to 1st semitones. How close is that value to 3/2? (d) Repeat for the 6th and 1st semitones and compare with 4/3. /r and 11.2.3. The proportional bandwidth filter has f. /f = r-2im (a) Show that f- f(vr-1/F). ow that the center frequencies of ith and (i + 1)th contiguous bands are related = f:/./r. (b) Show that for the bandwidth w of each band, w- 11.2.4. Show that for a proportional bandwidth filter the upper and lower frequencies of a

Explanation / Answer

Answer:- In terms of octave the bandwidth of filter is given by-

BW = log2(fu / fl). Now for part-

a) BW = 1/3 of Octave => log2(fu / fl) = 1/3 keeping value for the fu / fl i.e 21/n then we can write-

=> log2(21/n) = 1/3, (1/n)*log2(2) = 1/3, thus n = 3. Hence r = 21/3 = 1.26

b) Similarly here in this case n = 2, r = 1.41

c) Doing same and hence n = 12 and r = 1.06

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