Revisions to FFT the whole input vs FFT segment

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edited Nov 11, 2015 at 17:55

cpmame

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suppose I have a 4 seconds signal captured at $f_s$ = 1024Hz, which gives 4,096 data array. What is the different between these 2 FFT methods (MATLAB code)

[1] FFT the wholeentire input with 4096-pointpoints FFT in one shot.

N = length(input);
Y = fft(input)/N;

[2] Divide the input array into 8 segmentsegments (5124096/8=512 data point each), and feed each segment to 512-pointpoints FFT, and then combineaverage the output:

N = length(input);
Y = zeros(1,512);
for i = 1:8
  segment = input(1+512*512*(i-0)+1 : 512*i);
  Y = Y + ( abs(fft(segment,512)) / (8*512) );512/8;
end

Question #1: Are these 2 methods equivalent?

Question #2: Is the frequency resolution for both methods equal to $\frac{f_s}{N}$ = 0.25Hz per bin?

suppose I have a 4 seconds signal captured at $f_s$ = 1024Hz, which gives 4,096 data array. What is the different between these 2 FFT methods (MATLAB code)

[1] FFT the whole input with 4096-point FFT in one shot.

N = length(input);
Y = fft(input)/N;

[2] Divide the input array into 8 segment (512 data point each), and feed each segment to 512-point FFT, and then combine the output:

N = length(input);
Y = zeros(1,512);
for i = 1:8
  segment = input(1+512*(i-0) : 512*i);
  Y = Y + ( abs(fft(segment,512)) / (8*512) );
end

Question #1: Are these 2 methods equivalent?

Question #2: Is the frequency resolution for both methods equal to $\frac{f_s}{N}$ = 0.25Hz per bin?

suppose I have a 4 seconds signal captured at $f_s$ = 1024Hz, which gives 4,096 data array. What is the different between these 2 FFT methods (MATLAB code)

[1] FFT the entire input with 4096-points FFT

N = length(input);
Y = fft(input)/N;

[2] Divide the input into 8 segments (4096/8=512 data point each), and feed each segment to 512-points FFT, and then average the output:

N = length(input);
Y = zeros(1,512);
for i = 1:8
  segment = input(512*(i-0)+1 : 512*i);
  Y = Y + abs(fft(segment,512))/512/8;
end

Question #1: Are these 2 methods equivalent?

Question #2: Is the frequency resolution for both methods equal to $\frac{f_s}{N}$ = 0.25Hz per bin?

Rollback to Revision 7

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edited Nov 11, 2015 at 17:52

cpmame

41
1
4

Supposesuppose I have a 4 seconds signal captured at $f_s$ = 1024Hz, which gives a 4,096 data sequencearray. What is the differencedifferent between these 2 FFT methods (in MATLABMATLAB code)?

[1] FFT the entirewhole input sequence with 4096-pointspoint FFT in one shot.

N = length(input);
Y = fft(input)/N;

[2] Divide the input sequencearray into 8 segmentssegment (4096/8=512512 data point each), and feed each segment to 512-pointspoint FFT, and then averagecombine the output:

N = length(input);
Y = zeros(1,512);
for i = 1:8
  segment = input(1+512*(i-0) : 512*i);
  Y = Y + ( abs(fft(segment,512))/512 /8; (8*512) );
end

Question #1: Are boththese 2 methods equivalent?

Question #2: Are the outputs of both methods haveIs the same frequency resolution =for both methods equal to $\frac{f_s}{N}$ = 0.25Hz per bin?

Suppose I have a 4 seconds signal captured at $f_s$ = 1024Hz, which gives a 4,096 data sequence. What is the difference between these 2 FFT methods (in MATLAB code)?

[1] FFT the entire input sequence with 4096-points FFT

N = length(input);
Y = fft(input)/N;

[2] Divide the input sequence into 8 segments (4096/8=512 data point each) and feed each segment to 512-points FFT, and then average the output:

N = length(input);
Y = zeros(1,512);
for i = 1:8
  segment = input(1+512*(i-0) : 512*i);
  Y = Y + abs(fft(segment,512))/512/8;
end

Question #1: Are both methods equivalent?

Question #2: Are the outputs of both methods have the same frequency resolution = $\frac{f_s}{N}$ = 0.25Hz?

suppose I have a 4 seconds signal captured at $f_s$ = 1024Hz, which gives 4,096 data array. What is the different between these 2 FFT methods (MATLAB code)

[1] FFT the whole input with 4096-point FFT in one shot.

N = length(input);
Y = fft(input)/N;

[2] Divide the input array into 8 segment (512 data point each), and feed each segment to 512-point FFT, and then combine the output:

N = length(input);
Y = zeros(1,512);
for i = 1:8
  segment = input(1+512*(i-0) : 512*i);
  Y = Y + ( abs(fft(segment,512)) / (8*512) );
end

Question #1: Are these 2 methods equivalent?

Question #2: Is the frequency resolution for both methods equal to $\frac{f_s}{N}$ = 0.25Hz per bin?

added 15 characters in body

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edited Nov 11, 2015 at 17:45

cpmame

41
1
4

supposeSuppose I have a 4 seconds signal captured at $f_s$ = 1024Hz, which gives a 4,096 data arraysequence. What is the differentdifference between these 2 FFT methods (MATLABin MATLAB code)?

[1] FFT the wholeentire input sequence with 4096-pointpoints FFT in one shot.

N = length(input);
Y = fft(input)/N;

[2] Divide the input arraysequence into 8 segmentsegments (5124096/8=512 data point each), and feed each segment to 512-pointpoints FFT, and then combineaverage the output:

N = length(input);
Y = zeros(1,512);
for i = 1:8
  segment = input(1+512*(i-0) : 512*i);
  Y = Y + ( abs(fft(segment,512)) / (8*512) );512/8;
end

Question #1: Are these 2both methods equivalent?

Question #2: IsAre the frequency resolution foroutputs of both methods equal tohave the same frequency resolution = $\frac{f_s}{N}$ = 0.25Hz per bin?

suppose I have a 4 seconds signal captured at $f_s$ = 1024Hz, which gives 4,096 data array. What is the different between these 2 FFT methods (MATLAB code)

[1] FFT the whole input with 4096-point FFT in one shot.

N = length(input);
Y = fft(input)/N;

[2] Divide the input array into 8 segment (512 data point each), and feed each segment to 512-point FFT, and then combine the output:

N = length(input);
Y = zeros(1,512);
for i = 1:8
  segment = input(1+512*(i-0) : 512*i);
  Y = Y + ( abs(fft(segment,512)) / (8*512) );
end

Question #1: Are these 2 methods equivalent?

Question #2: Is the frequency resolution for both methods equal to $\frac{f_s}{N}$ = 0.25Hz per bin?

Suppose I have a 4 seconds signal captured at $f_s$ = 1024Hz, which gives a 4,096 data sequence. What is the difference between these 2 FFT methods (in MATLAB code)?

[1] FFT the entire input sequence with 4096-points FFT

N = length(input);
Y = fft(input)/N;

[2] Divide the input sequence into 8 segments (4096/8=512 data point each) and feed each segment to 512-points FFT, and then average the output:

N = length(input);
Y = zeros(1,512);
for i = 1:8
  segment = input(1+512*(i-0) : 512*i);
  Y = Y + abs(fft(segment,512))/512/8;
end

Question #1: Are both methods equivalent?

Question #2: Are the outputs of both methods have the same frequency resolution = $\frac{f_s}{N}$ = 0.25Hz?

Rollback to Revision 3

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edited Nov 11, 2015 at 17:27

cpmame

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edited Nov 11, 2015 at 17:27

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edited Nov 11, 2015 at 17:26

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edited Nov 11, 2015 at 17:17

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edited Nov 11, 2015 at 17:11

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edited Dec 27, 2013 at 1:36

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asked Dec 26, 2013 at 8:30

cpmame

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