Challenge12_FFT#2

# Challenge12_FFT#2 - x[k T Σ DC … … T Σ X[n n th...

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Digital Signal Processing Dr. Fred J. Taylor, Professor Lesson Title: More FFTs Challenge: [12] Challenge Problem Goertzel’s Algorithm was presented in the 1 st order DC filters and 2 nd order filters of the form: [ ] ( 29 2 1 1 / 2 cos 2 1 1 - - - - + - - = z z N k z W m H mk N k π which was called a " resonator " (see Figure 1). A bank of the filters is then used to perform a DFT of the signal at the specific frequencies of the implemented resonators. Figure 1: Goertzel DFT It is noted that the poles of the filter are on the unit circle. For example, for N=32 and k=4, the resonator’s poles are located at: d=[1,-2*cos(2*pi*4/32),1]; d 1 -1.414 1 root(d) 0.707-0.707j 0.707+0.707j That is, the poles are located at z=0.707 ± j0.707. Q. Should you be concerned about the filter having poles on the unit circle?

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Unformatted text preview: x[k] T Σ DC … … T Σ X[n] n th harmonic T Σ 2cos(2 π n/N-1-W N n Σ … … FFT#2 - 1 Digital Signal Processing Dr. Fred J. Taylor, Professor Response For k=2, for example, the poles are at the location of the zeros of W122 which implies pole-zero calculation. However, even in floating point the production of W122 contains some error meaning that exact pole-zero is unrealistic. The pole maybe conditionally unstable. If a tone arrives at the 4th harmonic, your dead. It is for this reason that the poles of the resonator are force to be “slightly” interior to the unit circle. FFT#2 - 2...
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