DTFT_06 - Discrete-Time Fourier Transform Just as it was...

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Discrete-Time Fourier Transform Just as it was possible to extend Fourier Series analysis of periodic, continuous-time signals and systems, to include aperiodic signals, it is possible to do the same for discrete-time signals. Extending the Fourier Series To Include Aperiodic, Discrete-Time Signals Periodic, discrete-time signals can be expanded in a Fourier series. Consider two different discrete-time signals, one an aperiodic signal, x(n) , and the other its period counterpart, , having period N . The following are examples of such signals. ( ) N xn Now the periodic signal can be expanded in a Fourier series as follows. () 0 0 1 jk n Nk kN jk n nN X e Xx n e N = −Ω = = = Here it can be observed that ( ) ( ) for 0 1 N xn x n n N =≤ It will also be assumed that ( ) ( ) lim N N x n →∞ = That is, the aperiodic signal will be thought of as a periodic signal with an infinite period.
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2 As with time-continuous signals, if the period of the periodic signal is increased two important things happen, As N →∞ 1. The separation between spectral lines (both Amplitude and Phase Spectrums) will crowd closer together. In the limit they merge into an continuum. This is due to the fact that the separation between spectral lines is 0 2 N π Ω= 2. The amplitude of all the spectral lines decreases to zero as N As with continuous signals the second property causes problems in the extension of Fourier series analysis of periodic, discrete-time signals to aperiodic, discrete-time signals. To overcome this problem, modifications of the procedures is required. The source of the problem is the factor 1/ N in the determination of the Fourier coefficients. Fortunately, this problem can be dealt with. Following the same procedures that were applied to extend Fourier analysis of periodic, time-continuous signals to include aperiodic signals, the same can be done to extend it to include aperiodic, discrete- time signals. Toward that end, define a new function () () () 1 0 N j jn N nn Xe xne x ne ∞− Ω− =−∞ = = ∑∑ ± But ( ) 0 jk k X N = and 0 0 1 0 jk N j kn N k xn e N = = Now the spectral lines of the Amplitude and Phase Spectrum of the periodic signal are separated by . Rename and substitute into the above expression. 0 2 N 0 ΩΩ ±
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3 Then () ( ) ( ) 0 0 11 00 1 0 2 1 2 jk jk NN jk n j kn N kk N jk jk n k Xe xn e e N e π ∆Ω −− ∆Ω == ∆Ω ∆Ω = ∆Ω =∆ ∑∑ Next define a function 1 2 j jn QX e e ΩΩ Ω= Then 1 2 j kj k n Qk e ∆Ω ∆Ω ∆Ω = and () ( ) 1 0 N N k Q k = Next observe that Q( ) is periodic with a period of 2 π . 22 1 2 2 j e e Ω+ Ω+ Ω+ = But () () 1 2 0 N jj n j njn j n nn n j xne e ππ Ω+ − Ω+ Ω− −Ω = = = Substituting into the preceding expression 2 2 n j jn j n e e X e e e Q Ω+ Ω+ = = =Ω
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4 Now plot this function over the interval from 0 to 2 π . Then But ( ) Area Under kth Rectangle Qk ∆Ω ∆Ω= and () ( ) 1 0 Total Area Under Rectangles between 0 and N N k xn Q k N = =∆ = For small ∆Ω () ( ) 2 1 0 0 N k Q d π = ∆Ω ∆Ω≅ Ω Ω
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This note was uploaded on 10/24/2009 for the course ECE 108 taught by Professor Li during the Spring '08 term at Lehigh University .

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DTFT_06 - Discrete-Time Fourier Transform Just as it was...

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