![]() ![]() >Rune Allnor (Rick Lyons) wrote in message news. Then, reconstruct the real-valued time domain signal x(n) as Zero and compute the complex IDFT to find z(n) (complex-valued) in timeĭomain. Set the remaining half-band ( f= [fs/2, fs> ) to ![]() Now, you could half the work and take away some book-keeping drudgeryīy manipulating only the "useful" half-band ( f = [0,fs/2>, fs is Properties of real-valued signals in f domain: X(-f) = conj(X(f)). This assumes that you are careful about the symmetry the imaginary parts of the numbers are of vanishing In that case the output will be real-valued but represented by complex. WAV(real) -> FFT -> manipulate in freq domain -> IFFT -> real >There's a 3rd technique, but it's not nearly as Swap real and imag parts of each x'(n) sample Take forward FFT of X'(m) to yield x'(n). Swap real and imag parts of each X(m) freq-domain or am I too quick in dismissing that now.? WAV(real) -> FFT -> manipulate in freq domain -> IFFT -> realĬaveat doctor - I'm just learning scilab and over thirty years since > result of the IFFT to be a real-valued signal, Although Richard didn't say so, I assumed the >How do I perform an inverse FFT in Scilab? Result of the IFFT to be a real-valued signal, in which case the outerĬonj shouldn't matter. Reply Start a New (Rick Lyons) wrote in message news. > There's a 3rd technique, but it's not nearly asĮngineering is the art of making what you want from things you can get. There's a 3rd technique, but it's not nearly as Take forward FFT of X'(m) to yield x'(n).ģ. Swap real and imag parts of each X(m) freq-domainĢ. >that you need to divide by N, the length of X, to get everything >You would need to scale the result properly. >where x is in time domain and X is in frequency domain. >you can use the same routine to compute the ifft like this: But if you manage to do the forward FFT in scilab, > How do I perform an inverse FFT in Scilab?
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