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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.

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

Sei2πf= n =sne(i2πfn) 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(i2πf)=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: Sei2π(f+1)=Sei2πf S 2 f 1 S 2 f . Derive this property from the definition of the DTFT.

### Solution

Sei2π(f+1)= n =sne(i2π(f+1)n)= n =e(i2πn)sne(i2πfn)= n =sne(i2πfn)=Sei2πf 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 ei2πn2=e(iπn)=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,

Sei2πf= n =anune(i2πfn)= n =0ae(i2πf)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= α ,|α|<1:11α n 0 α n α α 1 1 1 α
(5)

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

Sei2πf=11ae(i2πf) S 2 f 1 1 a 2 f
(6)

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

|Sei2πf|=11acos2πf2+a2sin22πf S 2 f 1 1 a 2 f 2 a 2 2 f 2
(7)
Sei2πf=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).

## 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.

Sei2πf= n =0N1e(i2πfn) 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.)

Sei2πf=1e(i2πfN)1e(i2πf)=e(iπf(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 Sei2πf=e(iπf(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.

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

1212e(i2πfm)ei2πfnd f ={1  if  m=n0  if  mn 1 2 1 2 f 2 f m 2 f n 1 m n 0 m n
(13)
Therefore, we find that
1212Sei2πfei2πfnd f =1212 m msme(i2πfm)ei2πfnd f = m msm1212e((i2πf))(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
Sei2πf=n=sne(i2πfn) sn=1212Sei2πfei2πfndf 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|Sei2πf|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|Sei2πf|2d f =1212 n nsne(i2πfn) m msn¯ei2πfmd f = n,m n,msnsn¯1212ei2πf(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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