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The Fourier transform of a periodic function periodically sampled -- the discrete Fourier transform in Software Render Data Matrix ECC200 in Software The Fourier transform of a periodic function periodically sampled -- the discrete Fourier transform barcode for C#.net




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The Fourier transform of a periodic function periodically sampled -- the discrete Fourier transform generate, create none none with none projectscreating pdf-417 c# If a function h(t) has per none for none iodicity S, h(t) = h(t + S), and the samples are taken over this period, S = M t, tk = k t, and h k = h(tk ). The samples can be expressed as:. BIRT Reporting Tools h (t) =. (t tk )h(t). The Fourier transform (nor malized to one period) is given by:. M 1 k=0 H( f ) =. (t tk )h(t)e j2 f t dt = M 1 k=0 h(tk )e j2 f tk 5.8 Fourier transforms hk h(t). t tk S = M t Figure 5.37 Waveform sampled at uniform time intervals t. The sampling frequency is f s = 1/ t. From the above equation, w none for none e see that H ( f ) is nonzero only for values of f that are integer multiples of 1/S. De ne the nth harmonic frequency as f n = n/S = n/(M t):. H ( f n ) = Hn = h k e j2 fn k M 1 t h k e j2 nk/M As shown in the preceding, the discrete Fourier transform (DFT) is a special case of the continuous Fourier transform where the waveform is known only at a nite number of discrete points in time. Outside the time interval where the waveform is known, it is assumed to repeat forever, both in the past and into the future. One important application of the DFT is the determination of the sine and cosine components of a periodic waveform.

In many cases, these components are more useful than the shape of the waveform itself. The waveform h(t) is sampled at M time intervals t0 = 0, t1 = t, tk = k t, . .

. , t M 1 = (M 1) t. The sampling frequency f s = 1/ t.

The full sampling window is S = M t (Figure 5.37). Using the notation h k = h(tk ), the DFT of h k is de ned as:.

Hn = h k e j2 nk/M Note: e j = cos + j sin , e j = cos j sin , e0 = e j2 = 1, and e j = 1. The signi cance of the DFT coef cients is that H0 is the Fourier coef cient at frequency 0 (dc component), H1 is the Fourier coef cient at frequency f 1 = f s /M (1 cycle per S), and Hn is the Fourier coef cient at frequency f n = n f s /M (n cycles per S). EXAMPLE 5.17 Perform the DFT on the waveform: h k = A + B sin(2 k P/M), where Pis an integer.

H0 =. k=0 M 1 A = MA [A + B sin(2 k P/M)][cos(2 kn/M) j sin(2 kn/M)], n > 0 Hn = Data analysis and control Due to the orthogonal natu re of the sine and cosine series (see the example that follows), for this h k :. HP = j B sin2 (2 k P/M) =. j BM 2 j BM 2 HM P = j B sin2 (2 k P/M) =. and all the other Fourier coef cients are zero. Thus, we see that the nth Fourier coef cient describes the amplitude of any sine-wave component having n complete cycles per sampling interval S. EXAMPLE 5.

18 Relate the coef cients a j and b j in the general expansion:. hk = am cos(2 mk/M) + bm sin(2 mk/M). to the complex Fourier coe none for none f cients Hn . Firstly, we perform the discrete Fourier transform of h k :. M 1 M 1 Hn = k=0 m=0 am cos(2 mk/M) + bm sin(2 none none mk/M) [cos(2 nk/M) j sin(2 nk/M)]. Since:. [cos(2 m/M) sin(2 nk/M)] = 0, for all m and n k=0 M 1 [cos(2 mk/M) cos(2 nk/M)] = 0, for all m = n k=0 M 1 [sin(2 mk/M) sin(2 nk/M)] = 0, for all m = n M 1 M 1 [cos2 (2 nk/N )] =. k=0 k=0 [sin2 (2 nk/N )] = M/2. we have for n = 0:. H0 =. a0 = Ma0 5.8 Fourier transforms and for 0 < n < M:. Hn = [an cos2 (2 nk/M) jbn s in2 (2 nk/M)]. = (M/2)(an jbn ) an = (2 none for none /M)Re(Hn ) and bn = (2/M)Im(Hn ). We see that the real part of Hn is associated with the cosine terms and the imaginary part of Hn is associated with the sine terms of the expansion. To what do the Fourier coef cients outside the interval from 0 to M/2 correspond From the de nition of the DFT, the Fourier amplitude for M cycles per sampling interval S is the same as 0 cycles per S:. HM = f k e j2 k = f k = H0 Above M samples per S, all the Fourier amplitudes are equal to their lower counterparts:. HM+n = h k e j2 k e j2 kn/M = M 1 k=0 h k e j2 kn/M = Hn Between M/2 and M samples none for none per S, we have the following result:.
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