# Tag Info

## Hot answers tagged ofdm

2

Because when a quantity can be complex, and even when it is just real, the absolute squared difference $|f-g|^2$ can be expressed in both domain (complex and real) as: $$|f-g|^2 = (f-g)^H(f-g)$$ and of course this is correct as well for reals. This setting is often related to Hilbert spaces.

1

I would prove it like that (in Matlab) F=dftmtx(4); dot(F(:,1),F(:,2)) ans = 0

1

The other answer points out that the DFT is a matrix multiply. The matrix $\mathbf{D}$ is like this:  \mathbf{D}= \begin{bmatrix} 1 & 1 & 1 & ... & 1 \\ 1 & \omega & \omega^2 & ... & \omega^{N-1} \\ 1 & \omega^2 & \omega^4 & ... & \omega^{2(N-1)} \\ ... & ... & ... & ... & ... \\ 1 & \...

1

Orthogonality is defined as "the inner product of two vectors equals zero". Now, in OFDM, the transmit vector for a single subcarrier is exactly one row vector $\mathbf D_k$ of the DFT Matrix $\mathbf D$, multiplied by the complex value of a symbol $c_k$, i.e. $c_K \mathbf D_K$. Two different subcarriers $k, l, k\ne l$ hence have the inner product \$...

Only top voted, non community-wiki answers of a minimum length are eligible