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Discrete-Time Fourier Transform (DTFT)

Module by: Don Johnson. E-mail the author

Summary: Discussion of Discrete-time Fourier Transforms. Topics include comparison with analog transforms and discussion of Parseval's theorem.

The Fourier transform of the discrete-time signal sn s n is defined to be

Sej2πf= n =sne(j2πfn) S 2 f n s n 2 f n
(1)
Frequency here has no units. As should be expected, this definition is linear, with the transform of a sum of signals equaling the sum of their transforms. Real-valued signals have conjugate-symmetric spectra: Se(j2πf)=Sej2πf* S 2 f S j 2 f .

Exercise 1

A special property of the discrete-time Fourier transform is that it is periodic with period one: Sej2π(f+1)=Sej2πf S 2 f 1 S 2 f . Derive this property from the definition of the DTFT.

Solution

Sej2π(f+1)= n =sne(j2π(f+1)n)= n =e(j2πn)sne(j2πfn)= n =sne(j2πfn)=Sej2πf S 2 f 1 n s n 2 f 1 n n 2 n s n 2 f n n s n 2 f n S 2 f
(2)

Because of this periodicity, we need only plot the spectrum over one period to understand completely the spectrum's structure; typically, we plot the spectrum over the frequency range 12 12 1 2 1 2 . When the signal is real-valued, we can further simplify our plotting chores by showing the spectrum only over 0 12 0 1 2 ; the spectrum at negative frequencies can be derived from positive-frequency spectral values.

When we obtain the discrete-time signal via sampling an analog signal, the Nyquist frequency corresponds to the discrete-time frequency 12 1 2 . To show this, note that a sinusoid having a frequency equal to the Nyquist frequency 12Ts 1 2 Ts has a sampled waveform that equals cos2π×12TsnTs=cosπn=1n 2 1 2 T s n T s n 1 n The exponential in the DTFT at frequency 12 1 2 equals ej2πn2=e(jπn)=1n 2 n 2 n 1 n , meaning that discrete-time frequency equals analog frequency multiplied by the sampling interval

f D = f A Ts f D f A Ts
(3)
f D f D and f A f A represent discrete-time and analog frequency variables, respectively. The aliasing figure provides another way of deriving this result. As the duration of each pulse in the periodic sampling signal p Ts t p Ts t narrows, the amplitudes of the signal's spectral repetitions, which are governed by the Fourier series coefficients of p Ts t p Ts t , become increasingly equal. Examination of the periodic pulse signal reveals that as ΔΔ decreases, the value of c 0 c 0 , the largest Fourier coefficient, decreases to zero: | c 0 |=AΔTs c 0 A Δ Ts . Thus, to maintain a mathematically viable Sampling Theorem, the amplitude AA must increase as 1Δ 1 Δ , becoming infinitely large as the pulse duration decreases. Practical systems use a small value of ΔΔ, say 0.1· Ts 0.1 · Ts and use amplifiers to rescale the signal. Thus, the sampled signal's spectrum becomes periodic with period 1Ts 1 Ts . Thus, the Nyquist frequency 12Ts 1 2 Ts corresponds to the frequency 12 1 2 .

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,

Sej2πf= n =anune(j2πfn)= n =0ae(j2πf)n S 2 f n a n u n 2 f n n 0 a 2 f n
(4)

This sum is a special case of the geometric series.

n =0αn=11α  ,   |α|<1    n 0 α n α α 1 1 1 α
(5)

Thus, as long as |a|<1 a 1 , we have our Fourier transform.

Sej2πf=11ae(j2πf) S 2 f 1 1 a 2 f
(6)

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

|Sej2πf|=11acos2πf2+a2sin22πf S 2 f 1 1 a 2 f 2 a 2 2 f 2
(7)
Sej2πf=tan-1asin2πf1acos2πf S 2 f a 2 f 1 a 2 f
(8)

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 to 12 1 2 —with increasing aa 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=0.5a 0.5) is shown over the frequency range [-2, 2], clearly demonstrating the periodicity of all discrete-time spectra. The angle has units of degrees.
Spectrum of exponential signal
Spectrum of exponential signal (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=0.5a 0.5 ?
Spectra of exponential signals
Spectra of exponential signals (spectrum11.png)

Example 2

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

sn={1  if  0nN10  otherwise   s n 1 0 n N 1 0
(9)

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

Sej2πf= n =0N1e(j2πfn) S 2 f n 0 N 1 2 f n
(10)

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

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
(11)
for all values of α.

Exercise 2

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 αα.

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 which, after manipulation, yields the geometric sum formula.

Applying this result yields (Figure 3.)

Sej2πf=1e(j2πfN)1e(j2πf)=e(jπf(N1))sinπfNsinπf S 2 f 1 2 f N 1 2 f f N 1 f N f
(12)
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 Sej2πf=e(jπf(N1))dsincπf S 2 f f N 1 dsinc f . The discrete-time pulse's spectrum contains many ripples, the number of which increase with NN, 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?
Spectrum of length-ten pulse
Spectrum of length-ten pulse (spectrum12.png)

The inverse discrete-time Fourier transform is easily derived from the following relationship:

1212e(j2πfm)ej2πfnd f ={1  if  m=n0  if  mn=δmn 1 2 1 2 f 2 f m 2 f n 1 m n 0 m n δ m n
(13)
Therefore, we find that
1212Sej2πfej2πfnd f =1212 m msme(j2πfm)ej2πfnd f = m msm1212e((j2πf))(mn)d f =sn f 1 2 1 2 S 2 f 2 f n f 1 2 1 2 m m s m 2 f m 2 f n m m s m f 1 2 1 2 2 f m n s n
(14)
The Fourier transform pairs in discrete-time are
Sej2πf=n=sne(j2πfn) sn=1212Sej2πfej2πfndf S 2 f n s n 2 f n s n f 1 2 1 2 S 2 f 2 f n
(15)

The properties of the discrete-time Fourier transform mirror those of the analog Fourier transform. The DTFT properties table shows similarities and differences. One important common property is Parseval's Theorem.

n =|sn|2=1212|Sej2πf|2d f n s n 2 f 1 2 1 2 S 2 f 2
(16)
To show this important property, we simply substitute the Fourier transform expression into the frequency-domain expression for power.
1212|Sej2πf|2d f =1212 n nsne(j2πfn) m msn*ej2πfmd f = n,m n,msnsn*1212ej2πf(mn)d f f 1 2 1 2 S 2 f 2 f 1 2 1 2 n n s n 2 f n m m s n 2 f m , n m , n m s n s n f 1 2 1 2 2 f m n
(17)
Using the orthogonality relation, the integral equals δmn δ m n , where δn δ n is the unit sample. Thus, the double sum collapses into a single sum because nonzero values occur only when n=mnm, giving Parseval's Theorem as a result. We term n ns2n n n s n 2 the energy in the discrete-time signal sn s n in spite of the fact that discrete-time signals don't consume (or produce for that matter) energy. This terminology is a carry-over from the analog world.

Exercise 3

Suppose we obtained our discrete-time signal from values of the product st p T s t s t p T s t , where the duration of the component pulses in p T s t p T s t is ΔΔ. How is the discrete-time signal energy related to the total energy contained in st s t ? Assume the signal is bandlimited and that the sampling rate was chosen appropriate to the Sampling Theorem's conditions.

Solution

If the sampling frequency exceeds the Nyquist frequency, the spectrum of the samples equals the analog spectrum, but over the normalized analog frequency fT f T . Thus, the energy in the sampled signal equals the original signal's energy multiplied by TT.

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