andpass Sapling (2)
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and-liited Signal = 5 MHz = 5 MHz = 5 MHz = 5 MHz MHz MHz 22.5 20 17.5 17.5 20 22.5 22.5 20 17.5 17.5 20 22.5 irror Lowpass Sapling = 5 MHz = 5 MHz andpass Sapling IF filtering Haronic Sapling Sub-Nyquist Sapling MHz 22.5 20 17.5 17.5 20 22.5 2 andpass Sapling 3
Low-pass Signal Sapling = 5 MHz = 5 MHz 22.5 20 17.5 17.5 20 22.5 MHz 22.5 17.5 17.5 22.5 MHz Nyquist Criterion + 2 andpass Sapling 4
and-pass Signal Sapling = 5 MHz = 5 MHz 22.5 20 17.5 17.5 20 22.5 MHz 22.5 20 17.5 17.5 20 22.5 MHz irror andpass Sapling IF filtering Haronic Sapling Sub-Nyquist Sapling 22.5 20 17.5 17.5 20 22.5 MHz Nyquist Criterion 2 22.5 20 17.5 17.5 20 22.5 MHz 2 andpass Sapling 5
Sapling Frequency (1) Assue there are ultiples of Given an integer = Max condition can be decreased according to the following condition without introducing aliasing probles + = (+1) Min condition Given and-pass Signal is characterized by andwidth Carrier Frequency + /2 + + 1 2 andpass Sapling 6
Sapling Frequency (2) + + 1 + ( + 1) Given and-pass Signal is characterized by andwidth 2 ( + 1) 2 f L Carrier Frequency Noralization by = + /2 f L = /2 Highest frequency Lowest frequency + + /2 /2 + f L f L 2 andpass Sapling 7
Exaple =6 (1) + + 1 When = 6 /2 + in + 7 ax 6 6, ax ax = 6 7, in in = + 7 2 andpass Sapling 8
Exaple =6 (2) = 6 ax = 6 /2 1 2 3 4 5 6 MHz 1 < 6 1 > 6 /2 1 2 3 4 5 6 MHz = + 7 in + = 7 + /2 1 2 3 4 5 6 7 MHz 2 andpass Sapling 9
Exaple =6 (3) = 6 ax = 6 /2 MHz < 6 MHz = + 7 in + = 7 MHz + /2 2 andpass Sapling 10
Miniu Plot (1) + + 1 X-Y Plot + /2 = R X highest signal frequency bandwidth Y,in This plot shows in noralized by, for the given bandpass signal that is characterized by R and the given paraeter + ( + 1) =, in iniu sapling rate bandwidth Y Characterized by andwidth R = X Carrier Frequency = + /2 2 andpass Sapling 11
Miniu Plot (2) = + /2 = 1 R = / = 1 X-Y Plot = + /2 = 2 R = / = 2,in Y This plot shows in noralized by, for the given bandpass signal that is characterized by R and the given paraeter = + /2 = 3 R = / = 3 X Characterized by andwidth R = Carrier Frequency = + /2 2 andpass Sapling 12
Miniu Plot (3) + + 1 g (, R) = 2 (+1) = 2 (+1) R = X + /2 = R = 0 = 1 g (0, R) = 2 R g (1, R) = R slope = 2 slope = 1, in = Y + ( + 1) = 2 ( + 1) = 2 = 3 g (2, R) = 2 R 3 g (3, R) = 1 R 2 slope = 2/3 slope = 1/2 g (, R) = 4 g (4, R) = 2 5 R slope = 2/5 + = 5 g (5, R) = 1 3 R slope = 1/3 = 6 g (6, R) = 2 7 R slope = 2/7 = 7 g (7, R) = 1 4 R slope = 1/ 4 f L f L = 8 g (8, R) = 2 9 R slope = 2/9 2 andpass Sapling 13
Miniu Plot (4) g (, R) = 2 (+1) = 2 (+1) R R = +1 g (, +1) = 2 = 0 g (0, R) = 2 R slope = 2 = 0 R = 1 g (0,1) = 2 = 1 g (1, R) = R slope = 1 = 1 R = 2 g (1, 2) = 2 = 2 g (2, R) = 2 3 R slope = 2/3 = 2 R = 3 g (2, 3) = 2 = 3 g (3, R) = 1 2 R slope = 1/2 = 3 R = 4 g (3, 4) = 2 = 4 g (4, R) = 2 5 R slope = 2/5 = 4 R = 5 g (4,5) = 2 = 5 g (5, R) = 1 3 R slope = 1/3 = 5 R = 6 g (5,6) = 2 = 6 g (6, R) = 2 7 R slope = 2/7 = 6 R = 7 g (6,7) = 2 = 7 g (7, R) = 1 4 R slope = 1/ 4 = 7 R = 8 g (7,8) = 2 = 8 g (8, R) = 2 9 R slope = 2/9 = 8 R = 9 g (8,9) = 2 2 andpass Sapling 14
Miniu Plot (5),in k = 1 ( = 0) k = 2 ( = 1) k = 3 ( = 2) 2 2 3 + + 1 4 slope = 2 + 1 =k + k k 1 3 slope = 1 slope = 2/3 + /2 = R 2 highest signal frequency bandwidth 1 + ( + 1) =, in = g (, R) iniu sapling rate bandwidth 1 2 3 4 5 6 R = 2 andpass Sapling 15
Min, Max Condition on (1) + (+1) /2 2 (+1) 2 f L + + 1 =k ax represents how any are in 2 in ax, ax ax = = 2 f L f L + f L in k, in k represents how any are in 2 + in in in = + k = 2 k + 2 andpass Sapling 16
Min, Max Condition on (2) + + 1 /2 + k k 1 + + 1 =k ax in ax, ax 2 k 2 f L f L + f L in k, in + k = 2 k = 3 k = 4 2 f L 2 3 f L 1 2 2 3 f L = 1 = 2 = 3 2 andpass Sapling 17
Exaple k=1 (=0) k = 1 ( = 0) = + /2 = 1 R = / = 1,in,ax = 2 k = 2 = 2( ) (k 1) = + R [1, 2] (, ) = (2, 4) k = 1 ( = 0) = + /2 = 1.5 R = / = 1.5,in = 2 k = 3 slope = 2,ax = 2( ) (k 1) = + (, ) = (1, 2) k = 1 ( = 0) = + /2 = 2 R = / = 2,in = 2 k = 4,ax = 2( ) (k 1) = + 2 andpass Sapling 18
Exaple k=2 (=1) k = 2 ( = 1) k = 2 ( = 1) = + /2 = 2 = + /2 = 2.5 R = / = 2,in,ax = 2 k = 2 = 2( ) (k 1) = 2 R = / = 2.5,in,ax = 2 k = 2.5 = 2( ) (k 1) = 3 R [2, 3] (3, 4) (, ) = (3, 3) slope = 1 (, ) = (2, 2) k = 2 ( = 1) = + /2 = 3 R = / = 3,in = 2 k = 3,ax = 2( ) (k 1) = 4 2 andpass Sapling 19
Exaple k=3 (=2) k = 3 ( = 2) = + /2 = 3 R = / = 3 R [3, 4],in,ax = 2 k = 2 = 2( ) (k 1) = 2 k = 3 ( = 2) = + /2 = 3.5 R = / = 3.5,in = 2 k = 7 3 (3, 4) (, ) = (4, 8 3 ),ax = 2( ) (k 1) = 3 slope = 2 3 (, ) = (3, 2) k = 3 ( = 2) = + /2 = 4 R = / = 4,in = 2 k = 8 3,ax = 2( ) (k 1) = 4 2 andpass Sapling 20
Min Max Plot (1),in k = 1 ( = 0) k = 2 ( = 1) k = 3 ( = 2) 2 2 3 + + 1 4 + 1 =k + k k 1 3 2 1 1 2 3 4 5 6 R = 2 andpass Sapling 21
Min Max Plot (2),in k = 1 ( = 0) k = 2 ( = 1) k = 3 ( = 2) 2 2 3 + + 1 4 Min + k k 1 Max 3 Max Min 2 k = 3 [3, 4 ] k = 2 y = 2( x 2)+2 y = 2 x 2 y = 1( x 2)+2 y = x [ 2 3, ( )] y = 1( x 3)+2 y = 2 3 (x 3)+2 1 k = 2 k = 2 y = x 1 y = 2 3 x [2, 3 ] [, 2( )] 0 1 2 3 4 5 6 R = 2 andpass Sapling 22
Range of (1) For a given + + 1 Nyquist Criterion 2 = 20 MHz = 5 MHz in ax Optiu Sapling Frequency = 1 2 20 + 5 1 + 1 = 22.5 2 20 5 1 = 35 = 22.5 MHz (10 ) = 2 2 20 + 5 2 + 1 = 15 2 20 5 2 = 17.5 = 17.5 MHz (10 ) = 3 2 20 + 5 3 + 1 = 11.25 2 20 5 3 = 11.67 = 11.25 MHz (10 ) = 4 2 20 + 5 4 + 1 = 9 2 20 5 4 = 8.75 X = 5 2 20 + 5 5 + 1 = 7.5 2 20 5 5 = 7.0 X 2 andpass Sapling 23
2 andpass Sapling 24
References [1] http://en.wikipedia.org/ [2] J.H. McClellan, et al., Signal Processing First, Pearson Prentice Hall, 2003 [3] A graphical interpretation of the DFT and FFT, by Steve Mann [4] R. G. Lyons, Understanding Digital Signal Processing, 1997