Can I know what is the sampling rate of ifft() command which is a build in function in matlab?
The sampling frequency or sampling rate, fs, is the average number of samples obtained in one second (samples per second), thus fs = 1/T.
Sampling a signal
To sample a signal in MATLAB, generate a time vector at the appropiate rate, and use this to generate the signal. Plot using the stem function. For example:
% Sample the sinusoid x = sin(2 pi f t), where f = 2 kHz.
% Let x1 be the signal sampled at 10 kHz.
% Let x2 be the signal sampled at 3 kHz.
f = 2000;
T = 1/f;
tmin = 0;
tmax = 5*T; % Plot 5 cycles
dt1 = 1/10000;
dt2 = 1/3000;
t1 = tmin:dt1:tmax;
t2 = tmin:dt2:tmax;
x1 = sin(2*pi*f*t1);
x2 = sin(2*pi*f*t2);
subplot(211)
stem(t1,x1);
subplot(212)
stem(t2,x2);
X = ifft(Y) computes the inverse discrete Fourier transform of Y using a fast Fourier transform algorithm. X is the same size as Y.
X = ifft(Y,n) returns the n-point inverse Fourier transform of Y by padding Y with trailing zeros to length n.
example
X = ifft(Y,n,dim) returns the inverse Fourier transform along the dimension dim. For example, if Y is a matrix, then ifft(Y,n,2) returns the n-point inverse transform of each row.
example
X = ifft(___,symflag) specifies the symmetry of Y. For example, ifft(Y,'symmetric') treats Y as conjugate symmetric.
https://www.mathworks.com/help/signal/ug/discrete-fourier-transform.html
this link may be help you
https://www.mathworks.com/matlabcentral/answers/418739-how-to-use-frequency-as-input-for-ifft
Hi Jason Law
You need not define 'n' in the command X = ifft(Y,n) if you want output X to be the same size as input Y.
If you want output X to be of some specified length, then 'n' can be defined equal to that length.
Hope this helps!
Jason Law
Although I had not followed up on this paper before. But, as mentioned in the methodology of paper:
" If there exists any a priori knowledge about the target range such as upper and/or lower bounds, the range of the time delays over which the objective function is scanned can be limited and the amount of computation necessary to find the global minimum reduced."
Hence it can be ascertained that one can assume any set of time delay t values but at the cost of computation. If you have a priori knowledge about your model (for example, Beatty Standard used in the paper for illustration), then you can assume a range of t values that could be possible with that model which will help to easily find the global minimum of the objective function as search for global minima would be limited to only a certain set of t values. Otherwise, a random assumption of set of t values could lead to more computation in searching the global minima of the objective function.
Kirandeep Kaur
I dont have any priori knowledge about my model that produced me the set of data. I have a set of data in frequency domain with the number of 2 reflections and 101 points of data .
My questions are:
1. I have a doubt on how many t should I assume, should it be number of reflection or number of point of data?
2. Can I have suggestion on random assumption of set of t values ?
I am using the frequency(bandwidth) of the data now for the set of t(1=1/frequency) , it since like almost there but not really.
Thanks
Jason Law
The number of t values to be assumed should depend upon the number of reflections not on the number of data points as reflections are the only cause for time delays.
Yes, you can take a clue from the bandwidth to assume a range of time delays. In fact, one example on Beatty Standard is clearly mentioned in the paper which you can refer to.
Good Luck!
The IFFT simply reverses the FFT; if you have 5k (complex) frequency components, the IFFT will generate 5k (presumably real) time domain points that accurately reproduce the input that would have created the transformation. As noted previously, the sample rate is simply an external bookkeeping arrangement for the application of proper units for interpretation.
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