Discrete Fourier Transform and Handwavingby mathman44 Tags: discrete, fourier, handwaving, transform 

#1
Feb413, 06:15 PM

P: 210

Hi all,
I'm reading the following PDF about the DFT: http://www.analog.com/static/importe...p_book_Ch8.pdf Please see pages 152153. So the inverse DFT (frequency to space, x[i] = ...) is given on page 152. Then it is claimed that the amplitudes for the spacedomain synthesis (inverse DFT) are different than the frequency domain of a signal (page 153). In short, the amplitude for the spacedomain synthesis is the corresponding frequency domain multiplied by 2/N, except for the zero frequency term and the last frequency term, which are multiplied by 1/N. A short digression on 'spectral densities' are made, and I really don't see the connection. Why are the different frequency components weighted differently in the spacedomain synthesis? Any help would be great. Thanks. 



#2
Feb413, 07:26 PM

Sci Advisor
HW Helper
PF Gold
P: 2,928

If we define
$$X(k) = \sum_{n=0}^{N1} x(n) e^{i2\pi k n/N}$$ for ##k=0, 1, \ldots, N1##, and we want to recover ##x(n)## from ##X(k)##, then $$x(n) = \frac{1}{N} \sum_{k=0}^{N1} X(k) e^{i2\pi k n/N}$$ Here the ##X(k)## are complexvalued. We can obtain an equivalent expression in terms of the real and imaginary parts of ##X(k)## by writing ##X(k) = R(k) + iI(k)## and simplifying: $$x(n) = \frac{1}{N} \sum_{k=0}^{N1} R(k) e^{i2\pi k n/N} + i \frac{1}{N} \sum_{k=0}^{N1} I(k) e^{i2\pi k n/N}$$ where ##R(k)## and ##I(k)## are realvalued. Now suppose the original data ##x(n)## is realvalued. Then notice that if we take the complex conjugate of the first equation above, we get $$X^*(k) = \sum_{n=0}^{N1} x(n) e^{i2\pi k n/N}$$ Now replace ##k## by ##Nk##: $$X^*(Nk) = \sum_{n=0}^{N1} x(n) e^{i2\pi (Nk) n/N} = \sum_{n=0}^{N1} x(n) e^{i2\pi k n/N} = X(k)$$ So we have established that ##X^*(Nk) = X(k)##. Now take the real and imaginary parts of this equation. We see that ##R(Nk) = R(k)## and ##I(Nk) = I(k)##. This allows us to rewrite the two sums involving ##R(k)## and ##I(k)##. I will show how this works for ##R(k)##. You can do something similar for ##I(k)##. $$\begin{align} \sum_{k=0}^{N1} R(k) e^{i2\pi k n/N} &= R(0) + \sum_{k=1}^{N/2  1} R(k) e^{i2\pi k n/N} + R(N/2) e^{i\pi n} + \sum_{k=N/2+1}^{N1} R(k) e^{i2\pi k n/N}\\ &= R(0) + \sum_{k=1}^{N/2  1} R(k) e^{i2\pi k n/N} + R(N/2) \cos(\pi n) + \sum_{k=N/2+1}^{N1}R(Nk)e^{i2\pi kn/N} \\ &= R(0) + \sum_{k=1}^{N/2  1} R(k) e^{i2\pi k n/N} + R(N/2) \cos(\pi n) + \sum_{m=1}^{N/21} R(m) e^{i 2\pi mn/N} \\ &= R(0) + \sum_{k=1}^{N/2  1} R(k) (e^{i2\pi k n/N} + e^{i2\pi k n/N}) + R(N/2)\cos(\pi n) \\ &= R(0) + 2 \sum_{k=1}^{N/2  1} R(k)\cos(2\pi k n/N) + R(N/2) \cos(\pi n) \end{align}$$ This shows why there is a factor of 2 on all of the terms from ##k=1, 2, \ldots N/21## but not on the ##k=0## and ##k=N/2## terms. I have used the following facts above: for the ##k=0## term, $$e^{i2\pi k n/N} = e^{i2\pi 0 n/N} = e^{0} = 1$$ For the ##k=N/2## term: $$e^{i2\pi k n/N} = e^{i2\pi (N/2) n/N} = e^{i \pi n} = \cos(\pi n) + i\sin(\pi n) = \cos(\pi n) + 0 = \cos(\pi n)$$ The RHS of this last equation also equals ##(1)^n## but there was no need to use that fact here. 



#3
Feb413, 08:12 PM

P: 210

Whoa... thanks, that's absolutely perfect.



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