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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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Now that we have the underpinnings of digital computation, we need to return to signal processing ideas. The most prominent of which is, of course, the Fourier transform. The Fourier transform of a sequence is defined to be

## Fourier Transform

Sei2πf=n=sne(i2πfn) S 2f n sn 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)=Sei2πf¯ S 2 f S 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 f1 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)snei2πfn=n=snei2πfn=Sei2πf S 2 f1 n sn 2 f1 n n 2 n sn 2 f n n sn 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 12 12 . When the signal is real-valued, we can further simplify our plotting chores by showing the spectrum only over 0 12 0 12 ; 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 at the Nyquist frequency 12Ts 1 2 T s has a sampled waveform that equals

## Sinusoid at Nyquist Freqency 1/2T

cos2π·×12Ts·nTs=cosπn=1n 2 π · 1 2 T s · n T s π n 1 n
(3)

The exponential in the DTFT at frequency 12 1 2 equals e(i2πn)2=e(iπn)=1n 2 π n 2 π n 1 n , meaning that the correspondence between analog and discrete-time frequency is established:

## Analog, Discrete-Time Frequency Relationship

f D = f A Ts f D f A T s
(4)

where 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 Δ Δ decreases, the value of c 0 c 0 , the largest Fourier coefficient, decreases to zero: | c 0 |=AΔT c 0 A Δ T . Thus, to maintain a mathematically viable Sampling Theorem, the amplitude A A must increase as 1Δ 1 Δ , becoming infinitely large as the pulse duration decreases. Practical systems use a small value of Δ Δ , say 0.1·Ts 0.1 · T s and use amplifiers to rescale the signal. Thus, the sampled signal's spectrum becomes periodic with period 1Ts 1 T s . Thus, the Nyquist frequency 12Ts 1 2 T s corresponds to the frequency 12 1 2 .

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

1212e(i2πfm)eiπfndf={1  if  m=n0  if  mn 1 2 1 2 f 2 π f m π f n 1 m n 0 m n
(5)

Therefore, we find that

1212Sei2πfe2πfndf=1212msme(i2πfm)ei2πfndf=msm1212e((i2πf))(mn)df=sn f 1 2 1 2 S 2 π f 2 π f n f 1 2 1 2 m m s m 2 π f m 2 π f n m m s m f 1 2 1 2 2 π f m n s n
(6)

The Fourier transform pairs in discrete-time are

## Fourier Transform Pairs in Discrete Time

Sei2πf=nsne(i2πfn) S 2 π f n n s n 2 π f n
(7)

## Fourier Transform Pairs in Discrete Time

sn=1212Sei2πfei2πfndf s n f 1 2 1 2 S 2 π f 2 π f n
(8)

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|2df n s n 2 f 1 2 1 2 S 2 π f 2
(9)
To show this important property, we simply substitute the Fourier transform expression into the frequency-domain expression for power.
1212|Sei2πf|2df=1212nsne(i2πfn)m s nei2πfmdf=n,msn s m1212ei2πf(mn)df f 1 2 1 2 S 2 π f 2 f 1 2 1 2 n n s n 2 π f n m m s n 2 π f m n,m n,m s n s m f 1 2 1 2 2 π f m n
(10)
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=m n m , giving Parseval's Theorem as a result. We term 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 2

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 Δ Δ . 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 multipled by T.

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