The discrete-time Fourier transform (and the continuous-time transform as well) can be evaluated when we have an analytic expression for the signal. Suppose we just have a signal, such as the speech signal used in the previous chapter, for which there is no formula. How then would you compute the spectrum? For example, how did we compute a spectrogram such as the one shown in the speech signal example? The Discrete Fourier Transform (DFT) allows the computation of spectra from discrete-time data. While in discrete-time we can exactly calculate spectra, for analog signals no similar exact spectrum computation exists. For analog-signal spectra, use must build special devices, which turn out in most cases to consist of A/D converters and discrete-time computations. Certainly discrete-time spectral analysis is more flexible than continuous-time spectral analysis.

The formula for the DTFT is a sum, which conceptually can be easily computed save for two issues.

We thus define the discrete Fourier transform (DFT) to be

Here, S⁢k S k is shorthand for S⁢ei⁢2⁢π⁢kK S 2 k K .

We can compute the spectrum at as many equally spaced frequencies as we like. Note that you can think about this computationally motivated choice as sampling the spectrum; more about this interpretation later. The issue now is how many frequencies are enough to capture how the spectrum changes with frequency. One way of answering this question is determining an inverse discrete Fourier transform formula: given S⁢k S k , k=0…K−1 k 0 … K 1 how do we find s⁢n s n , n=0…N−1 n 0 … N 1 ? Presumably, the formula will be of the form s⁢n=∑ k =0K−1S⁢k⁢ei⁢2⁢π⁢n⁢kK s n k 0 K 1 S k 2 n k K . Substituting the DFT formula in this prototype inverse transform yields

Note that the orthogonality relation we use so often has a different character now. We obtain nonzero value whenever the two indices differ by multiples of KK. We can express this result as K⁢∑lδ⁢m−n−l⁢K K l l δ m n l K . Thus, our formula becomes The integers nn and mm both range over 0…N−1 0 … N 1 . To have an inverse transform, we need the sum to be a single unit sample for m m, n n in this range. If it did not, then s⁢n s n would equal a sum of values, and we would not have a valid transform: Once going into the frequency domain, we could not get back unambiguously! Clearly, the term l=0 l 0 always provides a unit sample (we'll take care of the factor of K K soon). If we evaluate the spectrum at fewer frequencies than the signal's duration, the term corresponding to m=n+K m n K will also appear for some values of m m, n=0…N−1 n 0 … N 1 . This situation means that our prototype transform equals s⁢n+s⁢n+K s n s n K for some values of n n. The only way to eliminate this problem is to require K≥N K N : We must have at least as many frequency samples as the signal's duration. In this way, we can return from the frequency domain we entered via the DFT.

Another way to understand this requirement is to use the theory of linear equations. If we write out the expression for the DFT as a set of linear equations,

s⁢0+s⁢1⁢e(−i)⁢2⁢πK+…+s⁢N−1⁢e(−i)⁢2⁢π⁢(N−1)K=S⁢1 s 0 s 1 2 K … s N 1 2 N 1 K S 1 ⋮ ⋮ s⁢0+s⁢1⁢e(−i)⁢2⁢π⁢(K−1)K+…+s⁢N−1⁢e(−i)⁢2⁢π⁢(N−1)⁢(K−1)K=S⁢K−1 s 0 s 1 2 K 1 K … s N 1 2 N 1 K 1 K S K 1 we have K K equations in N N unknowns if we want to find the signal from its sampled spectrum. This requirement is impossible to fulfill if K<N K N ; we must have K≥N K N . Our orthogonality relation essentially says that if we have a sufficient number of equations (frequency samples), the resulting set of equations can indeed be solved.

By convention, the number of DFT frequency values K K is chosen to equal the signal's duration N N. The discrete Fourier transform pair consists of

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This work is licensed by Don Johnson under a Creative Commons Attribution License (CC-BY 1.0), and is an Open Educational Resource.

Last edited by Richard Baraniuk on Oct 1, 2009 2:04 pm -0500.