# In OFDM, does $N N^H$ equal $I$?

I'm still new in OFDM and reading about it, but I have a question.

In OFDM system, when having $N$ data symbol assigned to sub-carriers, $N$ should be orthogonal with each others, which means $NN=0$. That's ok. What's about that matrices $N$ itself?

Is there any property in OFDM saying that $N N^H$ should equal $I$ (where $I$ is the identity matrix)?

• I'm a bit confused by your notation. You say you have $N$ data symbols, so $N$ seems to be a natural number. But then, you say that $N$ should be orthogonal to something, thus implyting that $N$ is a vector. Also, you say that $NN=0$ ($N$ orthogonal to itself), which doesn't make any sense. Please clarify your question so we can help you out. – Tendero Sep 3 '18 at 14:16
• @Tendero I modified the question .. could you please re check and help .. – Fatima_Ali Sep 3 '18 at 15:51

• An orthogonal matrix $\mathbf{A}$ is one whose columns are all unit vectors that are jointly orthogonal to one another, such that $\mathbf{AA^H} = I$.
• Ok .. thank you for that clarification... let me explain the question in another way, suppose that we have F which is $N$ by $N$ DFT, does $F F^H$ have any special results in OFDM? – Fatima_Ali Sep 4 '18 at 6:17