text: # Symmetry of real and imaginary parts in FFT [duplicate]
  
  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.
  
  ## 1 Answer
  
  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.
  
  • How you create these boxes of formula ? Is it an images ? like this one XN−n=Xn∗ please suggest. Commented Jun 1, 2017 at 11:25
  • You can use Latex markdown in posts. It is rendered by MathJax.
  – Deve
  Commented Jun 5, 2017 at 8:35
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