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Related material

Discrete Fourier Transform Pair

Summary: How to compute spectra using DFT.

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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 Sk S k , k∈0…K−1 k 0 … K 1 , how do we find sn s n , n∈0…K−1 n 0 … K 1 ? Presumably, the formula will be of the form sn≟∑k=0K−1Skⅇ+ⅈ2πnkK ≟ s n k 0 K 1 S k 2 n k K . Substituting the DFT formula in this prototype inverse transform yields

sn≟∑k=0K−1∑m=0N−1smⅇ-ⅈ2πmkKⅇ+ⅈ2πnkK

≟ s n k 0 K 1 m 0 N 1 s m 2 m k K 2 n k K

(1)

Note that the orthogonality relation we use so often has a different character now.

Orthogonality

∑k=0K−1ⅇ-ⅈ 2 π k m Kⅇ+ⅈ 2 π k n K=Kifm∈nn±Kn±2K…0otherwise

k 0 K 1 2 k m K 2 k n K K m n ± n K ± n 2 K … 0

(2)

We obtain nonzero value whenever the two indices differ by multiples of K

. We can express this result as K∑lδm−(n−lK)

K l l δ m n l K

. Thus, our formula becomes

sn≟∑m=0N−1smK∑t=-∞∞δm−n+lK

≟ s n m 0 N 1 s m K t δ m n l K

(3)

The integers n

and m

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

, n

in this range. If it did not, then sn

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

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∧n∈0…N−1

m n 0 … N 1

. This situation means that our prototype transform equals sn+sn+K

s n s n K

for some values of n

. The only way to eliminate this problem is to require K≥N

K N

: We must have more frequency samples than the signal's duration. In this way, we can return from the frequency domain we entered via the DFT.

Exercise 1

When we have fewer frequency samples than the signal's duration, some discrete-time signal values equal the sum of the original signal values. Given the sampling interpretation of the spectrum, characterize this effect a different way.

Solution

This situation amounts to aliasing in the time-domain.

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,

DFT as Linear Equations

s0+s1+…+sN−1=S0

s 0 s 1 … s N 1 S 0

(4)

s0+s1ⅇ-ⅈ2πK+sN−1ⅇ-ⅈ2πN−1K=S0

s 0 s 1 2 K s N 1 2 N 1 K S 0

(5)

⋮

(6)

s0+s1ⅇ-ⅈ2πK−1K+…+sN−1ⅇ-ⅈ2πN−1K−1K=SK−1

s 0 s 1 2 K 1 K … s N 1 2 N 1 K 1 K S K 1

(7)

we have K

equations in 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