... from the rules. Writing out Rule 7 using the definitions of Fourier transform and convolution [(2 .1) and (2 .15 )] gives 26 Fourier Transforms in Radar and Signal Processing Taking the particular ... begin by defining the notation used. Some of these terms, such as rect and sinc, have been adopted more widely to some extent, but rep 11 14 Fourier Transforms in...
Ngày tải lên: 18/06/2014, 10:05
... Rising edge of width . 52 Fourier Transforms in Radar and Signal Processing Figure 3 .13 Power spectra for rect and exponential impulse responses. 32 Fourier Transforms in Radar and Signal Processing This ... lim → 0 ͩ − 1 − 2 iy − 1 + 2 iy ͪ = 1 iy The Fourier transform of h(x) is now found to be, using P1b, 1 2 ͫ ␦ ( y) + 1 iy ͬ 50...
Ngày tải lên: 18/06/2014, 10:05
Fourier Transforms in Radar And Signal Processing_4 ppt
... 0 (t ∈ޒ) (3. 21) 72 Fourier Transforms in Radar and Signal Processing Figure 4.7 Maximum sampling rate. 2f u /k = 2(k + ␣ )W /k ≤ 2W ′≤2(k − 1 + ␣ )W /(k − 1) = 2f l /(k − 1) (4 .10 ) It is convenient ... sampling it at a rate 2rW, where r is given in (4 .13 ) above. 56 Fourier Transforms in Radar and Signal Processing Using T 0 and ⌬T as given in Figur...
Ngày tải lên: 18/06/2014, 10:05
Fourier Transforms in Radar And Signal Processing_5 doc
... sampling with a finite window width on a high IF may require 82 Fourier Transforms in Radar and Signal Processing above, in particular from the point of view of finding the minimum sampling rate needed ... + ␣ + 1) /(n + 1) , r M = (k + ␣ − 1) /(n − 1) , and r c = (k + ␣ )/n, where n ≤ k (n and k integers and 0 ≤ ␣ < 1) . When n = k, these become r m = 1 + ␣ /(...
Ngày tải lên: 18/06/2014, 10:05
Fourier Transforms in Radar And Signal Processing_6 ppt
... (5.24) = sinc x ͫ sinc y + 1 2 sinc ( y − 1) + 1 2 sinc ( y + 1) ͬ Putting sinc ( y ± 1) = sin ( y ± 1) ( y ± 1) = −sin y ( y ± 1) we have 11 4 Fourier Transforms in Radar and Signal Processing 5.3.3 ... 0.5), x = qFt = r − , and y = (q − 1) x/q ; then 10 4 Fourier Transforms in Radar and Signal Processing Figure 5 .13 Raised cosine rou...
Ngày tải lên: 18/06/2014, 10:05
Fourier Transforms in Radar And Signal Processing_7 docx
... (6 .12 ) 13 4 Fourier Transforms in Radar and Signal Processing snc 0 ( y) = ∑ ∞ n = 0 ( 1) n ( y) 2n (2n + 1) ! (6.23) snc 1 ( y) = ∑ ∞ n = 0 ( 1) n 2n( y) 2n − 1 (2n + 1) ! = ∑ ∞ n = 1 ( 1) n 2n( y) 2n − 1 (2n ... Transforms in Radar and Signal Processing snc r ( y) = 1 r d r dy r sinc ( y) (6 .17 ) then, from (6 .12 ), (6 .16 ) becomes ramp...
Ngày tải lên: 18/06/2014, 10:05
Fourier Transforms in Radar And Signal Processing_8 pdf
... elements, Figure 6 .11 Effect of increasing oversampling rate. 15 5 Equalization bandwidth at baseband is shown in Figure 6 .15 . (This is for a fractional bandwidth of 10 0%.) We first show the gain in linear ... Transforms in Radar and Signal Processing Figure 6 .14 Difference beam response: (a) difference beam with narrowband weights; (b) difference beam with equalization...
Ngày tải lên: 18/06/2014, 10:05
Fourier Transforms in Radar And Signal Processing_9 doc
... raised cosine weighting (solid line). 18 0 Fourier Transforms in Radar and Signal Processing and the weight function, given by the Fourier transform of (7.28), is a(x) = (u 0 /U ) comb 1/ U [sinc (u 0 x) ... are at ± 0 /2 and the corresponding u 0 value is given by 17 0 Fourier Transforms in Radar and Signal Processing where snc 1 is the derivative o...
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Fourier Transforms in Radar And Signal Processing_10 pot
... oversampling and, Wiener-Khinchine relation, 26–27, 11 1 Woodward, P. M., 2–3, 6 510 3 19 8 Fourier Transforms in Radar and Signal Processing Rules and pairs method (continued) Sinc functions, 3, 13 15 , 12 5 amplitudes, ... of, 16 continuous, 16 7, 18 7 properties, 15 discrete, 18 7 scaled, 16 17 phase shift, 16 2 in time-domain, 16 sampled, 16 4 Array bea...
Ngày tải lên: 18/06/2014, 10:05
Fourier Transforms in Radar And Signal Processing_1 ppt
... P2a, and the comb and rep transforms. Figure 2B .1 Contour for integral. 32 Fourier Transforms in Radar and Signal Processing This is in the form of a Fourier series, with period 1/ X = Y, and the coefficients ... lim → 0 ͩ − 1 − 2 iy − 1 + 2 iy ͪ = 1 iy The Fourier transform of h(x) is now found to be, using P1b, 1 2 ͫ ␦ ( y) + 1 iy ͬ 34...
Ngày tải lên: 18/06/2014, 15:20