web pages bar code Discrete Fourier Transform in .NET Creation pdf417 in .NET Discrete Fourier Transform

Discrete Fourier Transform generate, create none none with none barcode generator We now estimate the Fourier t none for none ransform of a function from a nite number of its sampled points. Suppose that we have N consecutive sampled values hk h(tk ), tk k , k = 0, 1, 2, . .

., N 1 (12.1.

4). Visual Studio Development Language 12. . Fast Fourier Transform h(t). T (a) H( f ). 0 ( b). aliased Fourier transform H( f ) true Fourier transform 1 2 (c). 1 2 . Figure 12.1.1.

The continuous function shown in (a) is nonzero only for a nite interval of time T . It follows that its Fourier transform, whose modulus is shown schematically in (b), is not bandwidth limited but has nite amplitude for all frequencies. If the original function is sampled with a sampling interval , as in (a), then the Fourier transform (c) is de ned only between plus and minus the Nyquist critical frequency.

Power outside that range is folded over or aliased into the range. The effect can be eliminated only by low-pass ltering the original function before sampling..

so that the sampling interval none for none is . To make things simpler, let us also suppose that N is even. If the function h(t) is nonzero only in a nite interval of time, then that whole interval of time is supposed to be contained in the range of the N points given.

Alternatively, if the function h(t) goes on forever, then the sampled points are supposed to be at least typical of what h(t) looks like at all other times. With N numbers of input, we will evidently be able to produce no more than N independent numbers of output. So, instead of trying to estimate the Fourier transform H(f) at all values of f in the range fc to fc , let us seek estimates only at the discrete values fn n , N n= N N , .

. ., 2 2 (12.

1.5). The extreme values of n in (1 2.1.5) correspond exactly to the lower and upper limits of the Nyquist critical frequency range.

If you are really on the ball, you will have noticed that there are N + 1, not N , values of n in (12.1.5); it will turn out that the two extreme values of n are not independent (in fact they are equal), but all the others are.

This reduces the count to N .. 12.1 Fourier Transform of Discretely Sampled Data The remaining step is to appr oximate the integral in (12.0.1) by a discrete sum:.

N 1 N 1 H(fn ) =. h(t)e2 ifn t dt hk e2 ifn tk = hk e2 ikn/N (12.1.6).

Here equations (12.1.4) and ( 12.

1.5) have been used in the nal equality. The nal summation in equation (12.

1.6) is called the discrete Fourier transform of the N points hk . Let us denote it by Hn ,.

Hn hk e2 ikn/N (12.1.7).

The discrete Fourier transfor m maps N complex numbers (the hk s) into N complex numbers (the Hn s). It does not depend on any dimensional parameter, such as the time scale . The relation (12.

1.6) between the discrete Fourier transform of a set of numbers and their continuous Fourier transform when they are viewed as samples of a continuous function sampled at an interval can be rewritten as H(fn ) Hn (12.1.

8). where fn is given by (12.1.5) none none .

Up to now we have taken the view that the index n in (12.1.7) varies from N/2 to N/2 (cf.


You can easily see, however, that (12.1.7) is periodic in n, with period N .

Therefore, H n = HN n n = 1, 2, . . .

. With this conversion in mind, one generally lets the n in Hn vary from 0 to N 1 (one complete period). Then n and k (in hk ) vary exactly over the same range, so the mapping of N numbers into N numbers is manifest.

When this convention is followed, you must remember that zero frequency corresponds to n = 0, positive frequencies 0 < f < fc correspond to values 1 n N/2 1, while negative frequencies fc < f < 0 correspond to N/2 + 1 n N 1. The value n = N/2 corresponds to both f = fc and f = fc . The discrete Fourier transform has symmetry properties almost exactly the same as the continuous Fourier transform.

For example, all the symmetries in the table following equation (12.0.3) hold if we read hk for h(t), Hn for H(f), and HN n for H( f).

(Likewise, even and odd in time refer to whether the values hk at k and N k are identical or the negative of each other.) The formula for the discrete inverse Fourier transform, which recovers the set of hk s exactly from the Hn s is: hk = 1 N.
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