7
$\begingroup$

This question already has an answer here:

INPUT
x[0] = (0.00 + j 0.00)
x[1] = (1.00 + j 0.00)
x[2] = (2.00 + j 0.00)
x[3] = (3.00 + j 0.00)
x[4] = (4.00 + j 0.00)
x[5] = (5.00 + j 0.00)
x[6] = (6.00 + j 0.00)
x[7] = (7.00 + j 0.00)
x[8] = (8.00 + j 0.00)
x[9] = (9.00 + j 0.00)
x[10] = (0.00 + j 0.00)
x[11] = (0.00 + j 0.00)
x[12] = (0.00 + j 0.00)
x[13] = (0.00 + j 0.00)
x[14] = (0.00 + j 0.00)
x[15] = (0.00 + j 0.00)

FFT:
X[0] = (45.00 + j 0.00)
X[1] = (-25.45 + j 16.67)
X[2] = (10.36 + j -3.29)
X[3] = (-9.06 + j -2.33)
X[4] = (4.00 + j 5.00)
X[5] = (-1.28 + j -5.64)
X[6] = (-2.36 + j 4.71)
X[7] = (3.80 + j -2.65)
X[8] = (-5.00 + j 0.00)
X[9] = (3.80 + j 2.65)
X[10] = (-2.36 + j -4.71)
X[11] = (-1.28 + j 5.64)
X[12] = (4.00 + j -5.00)
X[13] = (-9.06 + j 2.33)
X[14] = (10.36 + j 3.29)
X[15] = (-25.45 + j -16.67)

From the above FFT out put I noticed the following:

Re(x[1])=Re(x[15]), Im(x[1])=-Im(x[15])

Re(x[2])=Re(x[14]), Im(x[2])=-Im(x[14])

Re(x[3])=Re(x[13]), Im(x[3])=-Im(x[13])

Re(x[4])=Re(x[12]), Im(x[4])=-Im(x[12])

and so on

Is this a proven result that

 Re(X[n])=Re(x[N-n]), and Im(X[n])=-Im(x[N-n])  for 0<n<N-1, where N is no. of DFT points?

If yes then is there any particular condition under which this is true ? What is the gerenralised result?

If this is a general rule then I can save alot in memory and arithmatics as I am concerned only in magnitudes of the DFT output, and not in phase.

$\endgroup$

marked as duplicate by endolith, sansuiso, Matt L., Peter K. May 4 '13 at 22:12

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

13
$\begingroup$

Yes, this is always true if the input to the DFT is real valued. It's called the "conjugate complex symmetry", because $$ X_{N-n} = {X_n}^* $$ where $X_n$ is the DFT output and $()^*$ denotes the conjugate. It can be proven by inserting the property into the transformation formula of time domain sequence $x_k$: $$ X_n = \sum_{k=0}^{N-1}x_ke^{-j\frac{2\pi k n}{N}}\\ X_{N-n} = \sum_{k=0}^{N-1}x_ke^{-j\frac{2\pi k (N-n)}{N}}\\ =\sum_{k=0}^{N-1}x_k e^{-j 2\pi k}e^{j\frac{2\pi k n}{N}} $$ Using $exp(-j2\pi k) = 1 \:\: \forall \: k$ we find $$ X_{N-n} = \sum_{k=0}^{N-1}x_k e^{j\frac{2\pi k n}{N}} $$ Now we exploit that $x_k$ is real for all $k$ to rewrite the above: $$ X_{N-n} = \sum_{k=0}^{N-1}\left(x_k e^{-j\frac{2\pi k n}{N}}\right)^* = X_n^* $$ It's now straightforward to show that the same is valid for the IDFT. (real spectrum -> complex conjugate symmetric time domain sequence). Moreover, the opposite statement is also true: if the input to the DFT (IDFT) is complex conjugate symmetric its output is real valued.

$\endgroup$
  • $\begingroup$ How you create these boxes of formula ? Is it an images ? like this one XN−n=Xn∗ please suggest. $\endgroup$ – user6363 Jun 1 '17 at 11:25
  • $\begingroup$ You can use Latex markdown in posts. It is rendered by MathJax. $\endgroup$ – Deve Jun 5 '17 at 8:35

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