# ps5soln - MIT OpenCourseWare http/ocw.mit.edu 2.161 Signal...

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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 2.161 Signal Processing: Continuous and Discrete Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms . MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING 2.161 Signal Processing - Continuous and Discrete Fall Term 2008 Solution of Problem Set 5: The Discrete Fourier Transform Assigned: October 16, 2008 Due: October 23, 2008 Problem 1: Given a signal of duration T = 0 . 128 ms, sampled at a rate of F s = 8 kHz, the number of samples is L = T f s = 1024. If only 256 samples are taken, the frequency spacing in the computed DFT is Δ f = f s /N = 31 . 25 Hz. (a) The number of multiplications required for the direct computation of the DFT is N 2 = 256 2 = 65 , 536. (b) The number of multiplications required for the computation of the DFT using the FFT is N/ 2 log 2 ( N ) = 128 × 8 = 1024. Note that F m = F ∗ ( j 2 πm ) = F ∗ ( j 2 πf s m ). Hence, the m th component of the DFT corresponds to N Δ T N the actual frequency of m f s = m Δ f = 31 . 25 m Hz. N 1 + ∞ j Ω t d Ω f ( t ) = 2 π −∞ F ( j Ω) e N − 1 f n = f ( n Δ t ) = N 1 F m e j mn 2 π N m =0 Comparing above formulas, we can realize that, if there is a ONE-TO-ONE correspondence between discrete values of F m and continuous values of F ( j Ω), then at Ω = 2 πf s m the F ( j Ω) is a delta N function with an amplitude equal to 2 N π F m . This can be verified for example for either a DC signal or a sine signal without spectral leakage . Note that to find the scale factor, we cannot use the relation between F and F ∗ ( F ∗ ( j Ω) = n =+ ∞ n =+ ∞ 1 1 ∗ F ( j (Ω − n 2 πf s )) = F ( j (Ω − n Ω s )) ). That’s because in relating F m to F , we Δ T Δ T n = −∞ n = −∞ assumed that the rest of the sample points are zero. On the other hand, DFT assumes that the...
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ps5soln - MIT OpenCourseWare http/ocw.mit.edu 2.161 Signal...

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