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Unformatted text preview: X = n = x [ n ] e j n x [ n ]= 1 2 X e jn d DiscreteTime Fourier Transform (DTFT) X [ k ]= n = N 1 x [ n ] e j2 kn / N k = 0, 1, 2, ... , N 1 x [ n ]= 1 N k = N 1 X [ k ] e j2 kn / N n = 0, 1, 2, ... , N 1 Discrete Fourier Transform (DFT) These overheads were originally developed by Mark Fowler at Binghamton University, State University of New York. We can use the DFT to implement numerical FT processing This enables us to numerically analyze a signal to find out what frequencies it contains!!! A CT signal comes in through a sensor & electronics (e.g., a microphone & amp) The ADC creates samples (taken at an appropriate F s ) ADC DFT Processing (via FFT) X [0] X [1] X [2] X [ N1] Inside Computer memory array x t x [ n ] H/W or S/W on processor x [0] x [1] x [2] x [N1] memory array FFT algorithm computes N DFT values DFT values in memory array (they can be plotted or used to do something neat) N samples are dumped into a memory array If we are doing this DFT processing to see what the original CT signal x ( t ) looks like in the frequency domain we want the DFT values to be representative of the CTFT of x ( t ) Likewise If we are doing this DFT processing to do some processing to extract some information from x ( t ) or to modify it in some way we want the DFT values to be representative of the CTFT of x ( t ) So we need to understand what the DFT values tell us about the CTFT of x ( t ) We need to understand the relations between CTFT, DTFT, and DFT Well mathematically explore the link between DTFT & DFT in two cases: 0 0 x [0] x [1] x [2] ... X [ N 1] 0 N nonzero terms (of course, we could have some of the interior values = 0) 1. For x [ n ] of finite duration : For this case well assume that the signal is zero outside the range that we have captured. So we have all of the meaningful signal data. 2. For x [ n ] of infinite duration or at least of duration longer than what we can get into our DFT Processor inside our computer. So we dont have all the meaningful signal data. What effect does that have? How much data do we need for a given goal? DFT and DTFT: Finite Duration Case If x [ n ] = 0 for n < 0 and n N then the DTFT is: X = n = x [ n ] e j n = n = N 1 x [ n ] e j n we can leave out terms that are zero Now if we take these N samples and compute the DFT (using the FFT, perhaps) we get: X [ k ]= n = N 1 x [ n ] e j2 kn / N k = 0, 1, 2, ... , N 1 Comparing these we see that for the finiteduration signal case: X [ k ]= X k 2 N X 1 k 2 /2 /2 2 3 4 5 6 7 X [ k ] DTFT & DFT : DFT points lie exactly on the finiteduration signals DTFT! Summary of DFT and DTFT for a finite duration...
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 Fall '10
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