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Using interpolation of a signal to change the frequency.

If the original signal / sample length = 200 I know if I double sample length I get 400 and interpolate the new sample length (400) to the old sample length (200) I will get twice the frequency (hz) of the original signal. But what if I want to add only 4.83Hz to the original signal and not double the frequency how would I figure out how much of the array I need to add at the end? And yes I did try changing (*) to (+) chop_off_amnt=round(length(ya)*freq_decimal_portion) the line chop_off_amnt=round(length(ya)+freq_decimal_portion) and that doesn't work.

Steps / logic:
1) sample length = 200
2) double sample length = 400
3) interpolate double sample (400) to original sample length (200) 
I get twice the frequency (hz) of the original sample length.

My question is:
4) If instead of doubling the frequency I just want to add 4.83hz 
to the original signal how can I figure out how much of the array 
I need to add to the end?  At the moment it will give me 4.83 times the 
original signal I'm looking just to add 4.83hz to the original signal.

I won't know ya or the original frequency (original_freq) they are there as just an example. The audio files that will be imported will be human speech audio files (so they won't be made up of just one frequency but made up of many different frequencies)

test code below

%Matlab Test code below:
Fs = 200 %Sampling frequency
t=linspace(0,1,Fs);

%1a create signal
original_freq=2
ya = .5*sin(2*pi*original_freq*t);

%2a adjust array
freq_wanted=4.83;
freq_rnd_up=ceil(freq_wanted) %get rounded up value
freq_decimal_portion=freq_rnd_up-freq_wanted %minus this amount from array
new_freq_array_tmp=repmat(ya,[1 freq_rnd_up]); %repmat this amount of array
chop_off_amnt=round(length(ya)*freq_decimal_portion) %amount of cells to delete from end
new_freq=new_freq_array_tmp(1:end-chop_off_amnt); %create new array after deleting values from end
%length(new_freq)

%3a Interpolation 
xxo=linspace(0,1,length(new_freq))(:)'; 
xxi=linspace(0,1,length(ya))(:)'; 
yi_t=interp1(xxo,new_freq,xxi,'linear'); %interpolation 
plot(t,yi_t)

This is what the plot looks like now. This is not what I want

This is what I want the plot to look like This what I want

Ps It's a periodic signal and I'm using octave 3.8.1 which is like matlab. And I know I could change the equation but the equation is just an example I will be importing audio files where I won't have an equation and I need to change the frequency by a given amount in Hz. I won't know ya or the original frequency (original_freq) they are there as just an example. The audio files that will be imported will be human speech audio files (so they won't be made up of just one frequency but made up of many different frequencies)

Rick T
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    When you want to increase your signal frequency and keep your sampling interval respectively sampling frequency constant you'd rather have to decrease the number of samples per period. – Matthias W. Jan 19 '16 at 15:33
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    Instead of juggling with samples and interpolation, you should really try and (find a tool to) go the FFT/iFFT way. Think about it: 1Hz -> 5.83Hz, 10000Hz -> 10004.83Hz, ... That's why you cannot just interpolate the signal to `x + c` samples to have all frequencies adjusted by the same number of Hz. – JimmyB Jan 19 '16 at 15:34
  • @Hanno Binder I tried it with fft / ifft but didn't really get much of an answer so I decided to try another route ... see question with code I posted http://stackoverflow.com/questions/34773412/changing-frequency-using-fft-and-ifft-not-using-whole-numbers – Rick T Jan 19 '16 at 16:21

1 Answers1

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% You have to know what the sampling frequency is of your new signal
% otherwise you can't know how to add 4.83Hz because if you want to be all
% relative to Nyquist frequency, that 4.83 should be expressed over the
% Nyquist frequency too.

%If it's a regular audio file it will likely be 44100Hz though, so try like
%this and if you happen to know that the sampling frequency is different,
%just change it here:

F_sampling = 44100; %Hz

%supposing you have 10 seconds, let's create an example variable "audio"
%that contains this 10 seconds long signal sampled ad 44100 Hz:
audio = randn(1,441000);

N = numel(audio); % obbiously in this case is 441000

T = N/F_sampling; %this is the length in SECONDS, you want to keep this constant when resampling

F_add = 4.83; %Hz, frequency to add to original sampling frequency

%let's add it to the old and compute the resampling frequency:
F_sampling_new = F_sampling + F_add; %44100 + 4.83 = 44104.83 Hz

% this is the number of samples the new signal will need to have
N_new = T*F_sampling_new;

%but you want to resample it in the "old space" so you need to create a
%resampling vector that goes from 1 to N(old) but made of N_new samples
% otherwise the resampling function won't know which data interpolate
XX = linspace(1,N,N_new);

% and here it is the signal resampled
audio_resampled = interp1(1:N,audio,XX);

% to answer your question this is the number of samples that i needed to
% add
N_added = N_new - N;

%the problem with this is though that you need some simplification because
% N_new is not an integer: 44104.83 and unless the signal is long enough to
% overcome the decimals you won't be precise.
% So in this case, for example, having 4.83 two decimals, the length of the
% original signal must be exactly 100s so that N_new will be integer
% with 44100Hz*100s=>44104.83Hz*100s = 4410483 samples. So you see why this is not feasable, as the number product between the new sampling frequency and the length in seconds of the original signal must be integer, and it is unlikely to ever happen exactly, so you'd have to get as close as you can
Filippo
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