# Why do we use conjugate transpose IFFT in OFDM

I have a question regarding the OFDM and using the conjugate transpose of vector (matrix).

As you know, if we have a block of data,let's call it $X_n$, consisting of $N$ data symbols, we should apply for IDFT for that block, and let's say that we get $Y_n$ which is $Y_n = \text{IDFT}(X_n)$.

After that step, the conjugate transpose for the IDFT must be taken. So what's the benefit from taking that? If you have documents which explains that in simple way, it will be appreciated.

• nothing in OFDM says that you need to take the conjugate transpose. That seems to be something specific to whatever literature you're reading at this point (and we don't know what you're reading). – Marcus Müller May 16 '18 at 9:28
• @MarcusMüller thank you for your quick comment, In reality, I've read a paper saying: Let s be the data block consisting of N data symbols with variance σ^2 . Applying IDFT to each block, we get x=FH(s) where F is the N-point DFT matrix and the superscript H represents the conjugate transpose of vector (matrix). ... do you mean that sentence means something else ? could you please explain it again? – Zeyad_Zeyad May 16 '18 at 10:14
• that statement has nothing to do with the question you're asking – the conjugate complex of the DFT Matrix is simply the IDFT matrix. You're transposing the DFT matrix, not the result. – Marcus Müller May 16 '18 at 10:58
• @MarcusMüller OK, so what is the benefit of transposing the DFT matrix ? why should we transpose it? – Zeyad_Zeyad May 16 '18 at 14:40
• Read my comment – conjugate transposed DFT matrix is the IDFT matrix, and you want the IDFT. – Marcus Müller May 16 '18 at 15:30