2
$\begingroup$

Notation: I will write $a_k$ to denote a sequence/vector of scalar, non-complex elements.

The discrete Fourier transform $\hat a_k$ of a spatial signal $a_k$ is defined as:

$$\hat a_k = \frac{1}{\sqrt n} \sum_{w=0}^{n-1} a_w e^{2\pi j w/n}$$

The inverse transform of the frequency-domain signal $\hat a_k$ only differs in the sign of the exponent:

$$a_k = \frac{1}{\sqrt n} \sum_{w=0}^{n-1} \hat a_w e^{-2\pi j w/n}$$

So obviously, $a_k$ the inverse discrete Fourier transform is just the complex-conjugate of the regular discrete Fourier transform.

Since both operations are linear, they often also are written as matrix expressions:

$$\hat a_k = F a_k, \quad a_k = F^\mathrm H \hat a_k$$

Here $F^\mathrm H$ denotes the Hermitian of $F$, that is its conjugate-transpose. But why is $a_k = F^\mathrm H \hat a_k$ and not just $a_k = \overline F \hat a_k$, i.e. why is here the conjugate-transpose involved and not only the conjugate?

Update: According to Wikipedia, the matrix $F$ is symmetrical, i.e. its conjugate and its conjugate-transpose are the same. Doesn't that mean that $F^\mathrm H = \overline F$?

$\endgroup$
1
$\begingroup$

You are correct. $F^H = \bar{F}$ since $F^T = F$. Thus $F^{-1} = F^H = \bar{F}$, and so:

$$ \hat{\mathbf{a}} = F\mathbf{a} \Leftrightarrow F^H\hat{\mathbf{a}} = \mathbf{a} \Leftrightarrow \bar{F}\hat{\mathbf{a}} = \mathbf{a} $$

where $\mathbf{a} = \begin{bmatrix} a_0 & a_1 & \cdots & a_{n-1} \end{bmatrix}^T$ and similar for $\mathbf{\hat{a}}$

Why is $a_k=F^H\hat{a}_k$ and not just $a_k=\bar{F}\hat{a}_k$, i.e. why is here the conjugate-transpose involved and not only the conjugate?

You're notation is a little off here. You have a matrix operating on a scalar. I believe what you intended is the DFT matrix acting on the vector of $a_k$'s as I've written above.

As you can see from above, they are equivalent. So it is perfectly acceptable to write it either way.


Now, the question of why is $F^{-1}$ typically written as $F^H$ instead of $\bar{F}$, is a different question all together. I can't say for certain, but I would conjecture that this is because this is because $F$ is a type of matrix from a larger class, called unitary matrices. For a unitary matrix, $P$, it is the case that:

$$ P^{-1} = P^H $$

Writing $F^H$ is perhaps more obvious a reminder that the DFT matrix is unitary, and unitary matrices have useful properties which are often used in derivations (such as norm preservation, i.e., Parseval's theorem).

$\endgroup$
  • $\begingroup$ Thanks, but I'm not quite sure how this relates to my question. Could you please add some details on this to your answer? $\endgroup$ – theV0ID Sep 11 '15 at 14:24
  • $\begingroup$ @theV0ID that should clarify... let me know if not $\endgroup$ – Chester Sep 11 '15 at 14:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.