Changing the frequency of an array does not work always
2 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
farzad
am 27 Nov. 2020
Kommentiert: farzad
am 28 Nov. 2020
Hi All
I have a time array , from which I first drop some elements, then change its frequency based on the input I give. the array is X in the code. what happens is that for some specific values , like reduction = 2 and freq = 1000 Hz if the original frequency of X was 512 Hz, the length of redt2 falls one element shorter than the original X array. how does this happen and what is the solution.
X is any time signal, 1D array with any length and frequency
reduction =2
X =X (1:reduction:numel(X));
redt1=[X];
rf= redt1(2)-redt1(1);
tend=redt1(numel(redt1));
freq= input(frequency)
dt= 1/freq
redt2= 0:dt:tend/rf/freq;
8 Kommentare
Akzeptierte Antwort
Image Analyst
am 28 Nov. 2020
In
freq= input(frequency)
what is frequency? Some string you defined to ask the user for the frequency? Please show how you defined it.
reduction = 2
X_Original = 1 : 1000;
whos X
samplingFreqOriginal = (X_Original(end) - X_Original(1)) / numel(X_Original)
X_Reduced = X_Original(1:reduction:numel(X_Original));
whos X
samplingFreqReduced = (X_Reduced(end) - X_Reduced(1)) / numel(X_Reduced) % Will be 2 times samplingFreqOriginal
% Not sure what anything after here is intended to do.
redt1 = X_Reduced;
rf = redt1(2) - redt1(1)
tend = redt1(numel(redt1))
frequency = 'Enter the frequency : ';
freq = input(frequency)
dt = 1 / freq
redt2 = 0:dt:tend/rf/freq;
9 Kommentare
Image Analyst
am 28 Nov. 2020
Then simply do
% Sample values:
startingValue = 1
deltaX = 3
numElements = 9
% Create array:
x = startingValue : deltaX : (startingValue + (numElements - 1) * deltaX)
fprintf('x is %d elements long.\n', length(x));
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Signal Processing Toolbox finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!