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I have a non-trivial task of computing the Fourier transform of a matrix $\text{mat}[\alpha,\beta, t]$ such that the Fourier transform looks like $\text{matFT}[\alpha,\beta, \omega]$. With $\text{mat}[\alpha,\beta, t]$ given by

 mat[α_, β_, t_] = {
  {(Sqrt[α^2 - 4 β^2] Cosh[1/4 t Sqrt[α^2 - 4 β^2]] - α Sinh[1/4 t Sqrt[α^2 - 4 β^2]])^2/(-4 β^2 + α^2 Cosh[1/2 t Sqrt[α^2 - 4 β^2]] - α Sqrt[α^2 - 4 β^2] Sinh[1/2 t Sqrt[α^2 - 4 β^2]]),
   (I β (-α + α Cosh[1/2 t Sqrt[α^2 - 4 β^2]] - Sqrt[α^2 - 4 β^2] Sinh[1/2 t Sqrt[α^2 - 4 β^2]]))/(4 β^2 - α^2 Cosh[1/2 t Sqrt[α^2 - 4 β^2]] + α Sqrt[α^2 - 4 β^2] Sinh[1/2 t Sqrt[α^2 - 4 β^2]])
  },

  {-((I β (-α + α Cosh[1/2 t Sqrt[α^2 - 4 β^2]] - Sqrt[α^2 - 4 β^2] Sinh[1/2 t Sqrt[α^2 - 4 β^2]]))/(4 β^2 - α^2 Cosh[1/2 t Sqrt[α^2 - 4 β^2]] + α Sqrt[α^2 - 4 β^2] Sinh[1/2 t Sqrt[α^2 - 4 β^2]])),
   -((4 β^2 Sinh[1/4 t Sqrt[α^2 - 4 β^2]]^2)/(4 β^2 - α^2 Cosh[1/2 t Sqrt[α^2 - 4 β^2]] + α Sqrt[α^2 - 4 β^2] Sinh[1/2 t Sqrt[α^2 - 4 β^2]]))
  }
};

How can this be achieved in Mathematica?

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  • $\begingroup$ Are you familiar with the commands FourierMatrix and/or FourierMatrix? $\endgroup$
    – user49048
    Commented May 13, 2021 at 9:28
  • $\begingroup$ @DiSp0sablE_H3r0 isn't FourierMatrix just the 1D Fourier transform though - a matrix of the Fourier basis which you dot with a 1d vector? It's not clear if OP wants multiple 1D or a single 2D Fourier transform. $\endgroup$
    – flinty
    Commented May 13, 2021 at 10:50
  • $\begingroup$ @flinty as you pointed out in the last sentence, it's not clear and this is why I wrote it just as a mere suggestion. $\endgroup$
    – user49048
    Commented May 13, 2021 at 10:51

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