When you perform a discrete Fourier transform (DFT), and compute the DFT coefficients, $c_k$ and $X(K)$, what do these values tell us about the signal it came from?
$\begingroup$
$\endgroup$
1
-
1$\begingroup$ "Discrete-Time Signal Processing,2E, A.Oppenheim, CH-10: Fourier Analysis of Signals using the Discrete Fourier Transform" gives you what you need. $\endgroup$– Fat32Jan 11, 2015 at 15:51
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
5
Wikipedia would be a better place to get a full answer, but in short: Discrete Fourier Transform is defined over a finite set of $N$ samples $x_n$ by the formula $$ X_k=\sum\limits_{n=0}^{N-1} x_n e^\frac{-i2\pi kn}{N} $$ The resulting DFT coefficients contains information about the frequency content of the original signal.
-
$\begingroup$ And how do you read those values.. X_k is a complex value, and how does it tell me anything about the frequency content. The only way i see that it will tell me anything about the frequency content is by calculating $\sqrt{|X(k)|}$ and where $f = \frac{k·f_s}{N}$ $\endgroup$– user7304Dec 17, 2013 at 10:08
-
$\begingroup$ In the general case, $X_k$ is indeed complex: $X_k = \left| X_k\right| \cdot e^{j\phi (X_k)}$. The "frequency content" includes information about the phase. But in some cases, you could use only the absolute value. $\endgroup$– HeliosDec 17, 2013 at 12:12
-
$\begingroup$ So what does those values tell me?.. where would i use them.. $\endgroup$– user7304Dec 17, 2013 at 15:48
-
$\begingroup$ For example if you want to filter out some particular frequency component (this may be noise), you can 'edit' the Fourier coefficients and filter out these frequency components and then you can reconstruct your time domain signal via IDFT. Therefore you'll have a denoised signal (this is a very rough explanaion though). $\endgroup$– DenizDec 21, 2013 at 14:08
-
1$\begingroup$ @user, you gotta learn about the concept of a "phasor", not to be confused with the directed energy weapons depicted on Star Trek. $\endgroup$ Jan 20, 2014 at 16:48
$\begingroup$
$\endgroup$
at its core, the DFT operates on a discrete and periodic function (of period $N$) of some dimension (most often time) using the data of one period of that function and transforms it to a discrete and periodic function (with the same period $N$) of the reciprocal dimension (most often frequency).
$$ x[n] \ \triangleq \ \sum_{k=0}^{N-1} \ X[k] \ e^{j 2 \pi nk/N} $$
$$ X[k] \ = \ \frac{1}{N} \sum_{n=0}^{N-1} \ x[n] \ e^{-j 2 \pi nk/N} $$
(i switched, for the sake of cleaner discussion, where the scaling factor $\frac{1}{N}$ goes.)
both functions are fully defined by a finite set of coefficients. $X[k]$ are the Fourier coefficients of the $x[n]$ discrete periodic function. $x[n]$ are the Fourier coefficients of the $X[k]$ discrete periodic function.
they tell you about the magnitude and phase of each sinusoidal component ($e^{j \theta}$ is a sinusoidal function for real $\theta$) of the Fourier series.
answered Feb 19, 2014 at 20:04