# Symmetry of real and imaginary parts in FFT [duplicate]

INPUT
x = (0.00 + j 0.00)
x = (1.00 + j 0.00)
x = (2.00 + j 0.00)
x = (3.00 + j 0.00)
x = (4.00 + j 0.00)
x = (5.00 + j 0.00)
x = (6.00 + j 0.00)
x = (7.00 + j 0.00)
x = (8.00 + j 0.00)
x = (9.00 + j 0.00)
x = (0.00 + j 0.00)
x = (0.00 + j 0.00)
x = (0.00 + j 0.00)
x = (0.00 + j 0.00)
x = (0.00 + j 0.00)
x = (0.00 + j 0.00)

FFT:
X = (45.00 + j 0.00)
X = (-25.45 + j 16.67)
X = (10.36 + j -3.29)
X = (-9.06 + j -2.33)
X = (4.00 + j 5.00)
X = (-1.28 + j -5.64)
X = (-2.36 + j 4.71)
X = (3.80 + j -2.65)
X = (-5.00 + j 0.00)
X = (3.80 + j 2.65)
X = (-2.36 + j -4.71)
X = (-1.28 + j 5.64)
X = (4.00 + j -5.00)
X = (-9.06 + j 2.33)
X = (10.36 + j 3.29)
X = (-25.45 + j -16.67)


From the above FFT out put I noticed the following:

Re(x)=Re(x), Im(x)=-Im(x)

Re(x)=Re(x), Im(x)=-Im(x)

Re(x)=Re(x), Im(x)=-Im(x)

Re(x)=Re(x), Im(x)=-Im(x)

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.

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

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.