Challenge09_DFT

# Challenge09_DFT - ∆ f That is for a given f s and N ∆...

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Digital Signal Processing Dr. Fred J. Taylor, Professor Lesson Title: DFT Challenge Challenge: [09] Challenge Problem DTMF, or dual tone multi frequency modulation associates two tones with each key on a touch tone telephone pad. When a key is pressed, a column and row are generated. For example, '5' produces tones are 770Hz and 1336Hz (see Table below). . 1 2 3 A 697Hz 4 5 6 B 770Hz 7 8 9 C 852Hz * 0 # D 941Hz 1209Hz 1336Hz 1477Hz 1633Hz You are a notorious gadget person as well as being “cheap.” You find the RC filters, gears, switches, and so forth fascinating. Beside, your “plain old telephone” (POTS) has never failed you for 20 years, and you expect it to serve you another 20 before needing replacement. Nevertheless, in an inspired moment you wonder if you could use the MATLAB fft program running on your laptop (with the license still being in a state of dispute) to do the tone detection. Q: Is it possible to create an FFT DTMF signal detector of length less than or equal to N=1024? Response If an FFT is used, all the center frequencies must be related to each other by a common factor

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Unformatted text preview: ∆ f. That is, for a given f s and N, ∆ f= f s /N, it then follows that: 697=n 1 ∆ f. 770=n 2 ∆ f. and so forth. This, in turn means that the ratio 770/697= n 2 ∆ f/ n 1 ∆ f = n 2 ∆ f/ n 1 ∆ f = n 2 /n 1 = k. But computing the mathematical greatest common divisor shows that except for two frequencies (770 and 852), the DTMF frequencies are pairwise prime and, therefore, cannot be related by some factor k. As a consequence, a DFT can not be used as a replacement technology. >> gcd(697,770) 1 >> gcd(852,770) 2 >> gcd(852,941) 1 DFT - 1 Digital Signal Processing Dr. Fred J. Taylor, Professor >> gcd(1209,941) 1 >> gcd(1209,1336) 1 >> gcd(1477,1336) 1 >> gcd(1477,1633) 1 1 The reason that the DFT is unqualified as a replacement technology is due to the fact that the DFT is a constant bandwidth analysis tool and the POTS system is based on a constant Q (Q=BW/f ) analog technology. DFT - 2...
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Challenge09_DFT - ∆ f That is for a given f s and N ∆...

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