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DTFT Examples

Module by: Don Johnson. E-mail the author

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

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Example 1

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

Fourier Transform Formula

S2πf=n=-anun-2πfn=n=0a-2πfn 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=0αn=11α α α 1 n 0 α n 1 1 α (2)
Thus, as long as |a|<1 a 1 , we have our Fourier transform.
S2πf=11a-2πf S 2 f 1 1 a 2 f (3)

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

|S2πf|=11acos2πf2+a2sin22πf S 2 f 1 1 a 2 f 2 a 2 2 f 2 (4)
S2πf=-arctanasin2πf1acos2πf 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 -22 -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.

sn=1if0nN10otherwise s n 1 0 n N 1 0 (6)

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

S2πf=n=0N1-2πfn 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 1αN1α 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

αn= n 0 N+ n 0 1αnn= 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.)

S2πf=1-2πfN1-2πf=-πfN1sinπfNsinπf 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 sinNxsinx N x x , which is known as the discrete-time sinc function, dsincx dsinc x . Thus, our transform can be concisely expressed as S2πf=-πfN1dsincπf 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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