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DTFT Examples

Module by: Don Johnson. E-mail the author

Summary: How to compute discrete-time Fourier transforms for decaying sequences.

Example 1

Let's compute the discrete-time Fourier transform of the exponentially decaying sequence s⁢n=an⁢u⁢n s n a n u n , where u⁢n u n is the unit-step sequence. Simply plugging the signal's expression into the Fourier transform formula,

Fourier Transform Formula

S⁢ei⁢2⁢π⁢f=∑n=−∞∞an⁢u⁢n⁢e−(i⁢2⁢π⁢f⁢n)=∑n=0∞a⁢e−(i⁢2⁢π⁢f)n S 2 f n a n u n 2 f n n 0 a 2 f n
(1)

This sum is a special case of the geometric series.

Geometric Series

∀ α ,|α|<1:∑n=0∞αn=11−α α α 1 n 0 α n 1 1 α
(2)
Thus, as long as |a|<1 a 1 , we have our Fourier transform.
S⁢ei⁢2⁢π⁢f=11−a⁢e−(i⁢2⁢π⁢f) S 2 f 1 1 a 2 f
(3)

Using Euler's relation, we can express the magnitude and phase of this spectrum.

|S⁢ei⁢2⁢π⁢f|=11−a⁢cos2⁢π⁢f2+a2⁢sin22⁢π⁢f S 2 f 1 1 a 2 f 2 a 2 2 f 2
(4)
∠⁢S⁢ei⁢2⁢π⁢f=−arctana⁢sin2⁢π⁢f1−a⁢cos2⁢π⁢f S 2 f a 2 f 1 a 2 f
(5)

No matter what value of aa we choose, the above formulae clearly demonstrate the periodic nature of the spectra of discrete-time signals. Figure 1 shows indeed that the spectrum is a periodic function. We need only consider the spectrum between −12 1 2 and 12 1 2 to unambiguously define it. When a>0 a 0 , we have a lowpass spectrum — the spectrum diminishes as frequency increases from 0 0 to 12 1 2 — with increasing a a leading to a greater low frequency content; for a<0 a 0 , we have a highpass spectrum (Figure 2).

Figure 1: The spectrum of the exponential signal ( a=0.5 a 0.5 ) is shown over the frequency range -2 2 -2 2 , clearly demonstrating the periodicity of all discrete-time spectra. The angle has units of degrees.
Figure 1 (spectrum10.png)
Figure 2: The spectra of several exponential signals are shown. What is the apparent relationship between the spectra for a=0.5 a 0.5 and a=-0.5 a -0.5 ?
Figure 2 (spectrum11.png)

Example 2

Analogous to the analog pulse signal, let's find the spectrum of the length- N N pulse sequence.

s⁢n={1  if  0≤n≤N−10  otherwise   s n 1 0 n N 1 0
(6)

The Fourier transform of this sequence has the form of a truncated geometric series.

S⁢ei⁢2⁢π⁢f=∑n=0N−1e−(i⁢2⁢π⁢f⁢n) S 2 f n 0 N 1 2 f n
(7)

For the so-called finite geometric series, we know that

Finite Geometric Series

∑n= n 0 N+ n 0 −1αn=α n 0 ⁢1−αN1−α n n 0 N n 0 1 α n α n 0 1 α N 1 α
(8)
for all values of α α .

Exercise 1

Derive this formula for the finite geometric series sum. The "trick" is to consider the difference between the series'; sum and the sum of the series multiplied by α α .

Solution

α⁢∑n= n 0 N+ n 0 −1αn−∑n= n 0 N+ n 0 −1αn=αN+ n 0 −α n 0 α n n 0 N n 0 1 α n n n 0 N n 0 1 α n α N n 0 α n 0
(9)
which, after manipulation, yields the geometric sum formula.

Applying this result yields (Figure 3.)

S⁢ei⁢2⁢π⁢f=1−e−(i⁢2⁢π⁢f⁢N)1−e−(i⁢2⁢π⁢f)=e(−(i⁢π⁢f))⁢(N−1)⁢sinπ⁢f⁢Nsinπ⁢f S 2 f 1 2 f N 1 2 f f N 1 f N f
(10)

The ratio of sine functions has the generic form of sinN⁢xsinx N x x , which is known as the discrete-time sinc function, dsinc⁢x dsinc x . Thus, our transform can be concisely expressed as S⁢ei⁢2⁢π⁢f=e(−(i⁢π⁢f))⁢(N−1)⁢dsinc⁢π⁢f S 2 f f N 1 dsinc f . The discrete-time pulse's spectrum contains many ripples, the number of which increase with N N , the pulse's duration.

Figure 3: The spectrum of a length-ten pulse is shown. Can you explain the rather complicated appearance of the phase?
Figure 3 (spectrum12.png)

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