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Discrete-Time Fourier Transform (DTFT)

Module by: Don Johnson

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
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 .
Problem 1
A special property of the discrete-time Fourier transform is that it is periodic with period one: Sⅇⅈ2πf+1=Sⅇⅈ2πf S 2 f 1 S 2 f . Derive this property from the definition of the DTFT.
[ Click for Solution 1 ]
Solution 1
Sⅇⅈ2πf+1=∑n=-∞∞snⅇ-ⅈ2πf+1n=∑n=-∞∞ⅇ-ⅈ2πnsnⅇ-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)
[ Hide Solution 1 ]
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 -1212 1 2 1 2 . When the signal is real-valued, we can further simplify our plotting chores by showing the spectrum only over 012 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 cos2π12TsnTs=cosπn=-1n 2 1 2 T s n T s n 1 n The exponential in the DTFT at frequency 12 1 2 equals ⅇ-ⅈ2πn2=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=0∞aⅇ-ⅈ2πfn 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=0∞αn=∀α,|α|<1:11-α 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=11-aⅇ-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|=11-acos2πf2+a2sin22πf S 2 f 1 1 a 2 f 2 a 2 2 f 2 (7)
∠Sⅇⅈ2πf=-tan-1asin2πf1-acos2πf 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).
Spectrum of exponential signal
spectrum10.png
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.
Spectra of exponential signals
spectrum11.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 ?
Example 2 
Analogous to the analog pulse signal, let's find the spectrum of the length-NN pulse sequence.
sn=1if0≤n≤N-10otherwise 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=0N-1ⅇ-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 1-αN1-α n n 0 N n 0 1 α n α n 0 1 α N 1 α (11)
for all values of α.
Problem 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.
[ Click for Solution 2 ]
Solution 2
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 which, after manipulation, yields the geometric sum formula.
[ Hide Solution 2 ]
Applying this result yields (Figure 3.)
Sⅇⅈ2πf=1-ⅇ-ⅈ2πfN1-ⅇ-ⅈ2πf=ⅇ-ⅈπfN-1sinπfNsinπf 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-ⅈπfN-1dsincπf 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.
Spectrum of length-ten pulse
spectrum12.png
Figure 3: The spectrum of a length-ten pulse is shown. Can you explain the rather complicated appearance of the phase?
The inverse discrete-time Fourier transform is easily derived from the following relationship:
∫-1212ⅇ-ⅈ2πfmⅇⅈ2πfndf=1ifm=n0ifm≠n 1 2 1 2 f 2 f m 2 f n 1 m n 0 m n (13)
Therefore, we find that
∫-1212Sⅇⅈ2πfⅇⅈ2πfndf=∫-1212∑msmⅇ-ⅈ2πfmⅇⅈ2πfndf=∑msm∫-1212ⅇ-ⅈ2πfm-ndf=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 S 2 f n s n 2 f n (15)
sn=∫-1212Sⅇⅈ2πfⅇⅈ2πfndf s n f 1 2 1 2 S 2 f 2 f n (16)
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 f 2 (17)
To show this important property, we simply substitute the Fourier transform expression into the frequency-domain expression for power.
∫-1212|Sⅇⅈ2πf|2df=∫-1212∑nsnⅇ-ⅈ2πfn∑msn¯ⅇⅈ2πfmdf=∑n,msnsn¯∫-1212ⅇⅈ2πfm-ndf 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 (18)
Using the orthogonality relation, the integral equals δm-n 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 ∑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.
Problem 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.
[ Click for Solution 3 ]
Solution 3
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.
[ Hide Solution 3 ]

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