Frequencay Domain Analysis-Digital Signal Processing-Assignment Solution, Exercises of Digital Signal Processing

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10.1. (a) Using the relation ny = eT we find that the index k = 150 in X{k] corresponds to a continuoms time frequency of 2n{150) (1000)(70-*) 2n(1500) rad/s (b) For this part, it is important to realize that the & = 800 index corresponds to a negative continuous- time frequency. Since the DFT is periodic in & with period V, 2x(800 — 1000) 1000(10-*) —2(2000) rad/s Qis0 Qo = 10.2. Using the relation 2k Man. or tk he= ap we find that the equivalent analog spacing between frequencies is 1 AI= NF Thus, in addition to the constraint that NV is a power of 2, there are two conditions which must be met: % > 10,000Hz (to avoid aliasing) wr < SHz (given) These conditions can be expressed in the form 10,000< 2 <5N T The minimal N = 2” that satisfies the relationship is N = 2048 for which 10,000 Hz < z < 10,240 Hz Thus, Finin = 10,000 Hz, and Fimax = 10, 240 Hz. 10.3. (a) The length of a window is L (16,000 aoe) (20 x 10-3 sec) sec = 320 samples (b) The frame rate is the number of frames of data processed per second, or equivalently, the num- ber of DFT computations done per second. Since the window is advanced 40 samples between computations of the DFT, the frame rate is frame rate = (16,000 22%) (spe zcoese?) sec 40 samples = 49 fmes