Discrete-Time Fourier Transform (DTFT)

Transcription

1 Connexions module: m047 Discrete-Time Fourier Transorm DTFT) Don Johnson This work is produced by The Connexions Project and licensed under the Creative Commons Attribution License Abstract Discussion o Discrete-time Fourier Transorms. Topics include comparison with analog transorms and discussion o Parseval's theorem. The Fourier transorm o the discrete-time signal s n) is dened to be S e iπ ) = n= s n) e iπn)) ) Frequency here has no units. As should be expected, this denition is linear, with the transorm o a sum o signals equaling the sum o their transorms. Real-valued signals have conjugate-symmetric spectra: S e iπ)) = S e jπ ). Exercise Solution on p. 7.) A special property o the discrete-time Fourier transorm is that it is periodic with period one: S e iπ+)) = S e iπ ). Derive this property rom the denition o the DTFT. Because o 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 requency range [ ), ]. When the signal is real-valued, we can urther simpliy our plotting chores by showing the spectrum only over [ 0, ] ; the spectrum at negative requencies can be derived rom positive-requency spectral values. When we obtain the discrete-time signal via sampling an analog signal, the Nyquist requency corresponds to the discrete-time requency. To show this, note that a sinusoid having a requency equal to the Nyquist requency T s has a sampled waveorm that equals cos π ) T s nt s = cos πn) = ) n The exponential in the DTFT at requency iπn equals e ) = e iπn) = ) n, meaning that discrete-time requency equals analog requency multiplied by the sampling interval Version.3: Jul 6, 009 5:5 pm GMT-5 "The Sampling Theorem" < D = A T s )

2 Connexions module: m047 D and A represent discrete-time and analog requency variables, respectively. The aliasing gure provides another way o deriving this result. As the duration o each pulse in the periodic sampling signal p Ts t) narrows, the amplitudes o the signal's spectral repetitions, which are governed by the Fourier series coecients 3 o p Ts t), become increasingly equal. Examination o the periodic pulse signal 4 reveals that as decreases, the value o c 0, the largest Fourier coecient, decreases to zero: c 0 = A T s. Thus, to maintain a mathematically viable Sampling Theorem, the amplitude A must increase as, becoming innitely large as the pulse duration decreases. Practical systems use a small value o, say 0. T s and use ampliers to rescale the signal. Thus, the sampled signal's spectrum becomes periodic with period T s. Thus, the Nyquist requency T s corresponds to the requency. Example Let's compute the discrete-time Fourier transorm o the exponentially decaying sequence s n) = a n u n), where u n) is the unit-step sequence. Simply plugging the signal's expression into the Fourier transorm ormula, This sum is a special case o the geometric series. S e ) iπ = n= a n u n) e iπn)) = ae ) ) iπ) n 3) n=0 α n ) = α, α < : n=0 Thus, as long as a <, we have our Fourier transorm. ) α S e iπ ) = 5) ae iπ) Using Euler's relation, we can express the magnitude and phase o this spectrum. 4) S e iπ ) = acos π)) + a sin π) S e iπ )) )) asin π) = tan acos π) No matter what value o a we choose, the above ormulae clearly demonstrate the periodic nature o the spectra o discrete-time signals. Figure Spectrum o exponential signal) shows indeed that the spectrum is a periodic unction. We need only consider the spectrum between ) and to unambiguously dene it. When a > 0, we have a lowpass spectrumthe spectrum diminishes as requency increases rom 0 to with increasing a leading to a greater low requency content; or a < 0, we have a highpass spectrum Figure Spectra o exponential signals)). 6) 7) "The Sampling Theorem", Figure : aliasing < 3 "Complex Fourier Series", 0) < 4 "Complex Fourier Series", Figure <

3 Connexions module: m047 3 Spectrum o exponential signal Se jπ ) Se jπ ) Figure : The spectrum o the exponential signal a = 0.5) is shown over the requency range [-, ], clearly demonstrating the periodicity o all discrete-time spectra. The angle has units o degrees.

4 Connexions module: m047 4 Spectra o exponential signals Spectral Magnitude db) a = 0.9 a = 0.5 a = Angle degrees) a = 0.5 a = 0.5 a = Figure : The spectra o several exponential signals are shown. What is the apparent relationship between the spectra or a = 0.5 and a = 0.5? Example Analogous to the analog pulse signal, let's nd the spectrum o the length- N pulse sequence. i 0 n N s n) = 0 otherwise The Fourier transorm o this sequence has the orm o a truncated geometric series. 8) S e iπ ) = N n=0 For the so-called nite geometric series, we know that or all values o α. N+n 0 n=n 0 e iπn)) 9) α n αn n0 ) = α α Exercise Solution on p. 7.) Derive this ormula or the nite geometric series sum. The "trick" is to consider the dierence between the series' sum and the sum o the series multiplied by α. Applying this result yields Figure 3 Spectrum o length-ten pulse).) 0) S e iπ ) = e iπn) e iπ) = e iπn )) sinπn) sinπ) )

5 Connexions module: m047 5 sinnx) The ratio o sine unctions has the generic orm o sinx), which is known as the discrete-time sinc unctiondsinc x). Thus, our transorm can be concisely expressed as S e ) iπ = e iπn )) dsinc π). The discrete-time pulse's spectrum contains many ripples, the number o which increase with N, the pulse's duration. Spectrum o length-ten pulse Figure 3: The spectrum o a length-ten pulse is shown. Can you explain the rather complicated appearance o the phase? The inverse discrete-time Fourier transorm is easily derived rom the ollowing relationship: ) e iπm) e iπn i m = n d = 0 i m n Thereore, we nd that = δ m n) ) S e ) iπ e iπn d = m s m) e ) iπm) e iπn) d = m s m) ) ) e iπ))m n) d The Fourier transorm pairs in discrete-time are = s n) S e ) iπ = ) n= s n) e iπn) s n) = ) S e ) iπ e iπn d ) 3) 4)

6 Connexions module: m047 6 The properties o the discrete-time Fourier transorm mirror those o the analog Fourier transorm. The DTFT properties table 5 shows similarities and dierences. One important common property is Parseval's Theorem. s n) ) ) = S e iπ ) ) d 5) ) n= To show this important property, we simply substitute the Fourier transorm expression into the requencydomain expression or power. ) S e iπ ) )) d = ) n s n) e iπn) ) m s n)e ) iπm d = n,m) s n) s n) ) 6) ) eiπm n) d Using the orthogonality relation ), the integral equals δ m n), where δ n) is the unit sample 6. Thus, the double sum collapses into a single sum because nonzero values occur only when n = m, giving Parseval's Theorem as a result. We term n s n) ) the energy in the discrete-time signal s n) in spite o the act that discrete-time signals don't consume or produce or that matter) energy. This terminology is a carry-over rom the analog world. Exercise 3 Solution on p. 7.) Suppose we obtained our discrete-time signal rom values o the product s t) p Ts t), where the duration o the component pulses in p Ts t) is. How is the discrete-time signal energy related to the total energy contained in s t)? Assume the signal is bandlimited and that the sampling rate was chosen appropriate to the Sampling Theorem's conditions. 5 "Discrete-Time Fourier Transorm Properties" < 6 "Discrete-Time Signals and Systems", Figure : Unit sample <

7 Connexions module: m047 7 Solutions to Exercises in this Module Solution to Exercise p. ) S e iπ+)) = ) n= s n) e iπ+)n) = n= e iπn) s n) e iπn)) = ) n= s n) e iπn) = S e ) iπ 7) Solution to Exercise p. 4) N+n 0 α n=n 0 N+n 0 α n ) n=n 0 α n ) = α N+n0 α n0 which, ater manipulation, yields the geometric sum ormula. Solution to Exercise 3 p. 6) I the sampling requency exceeds the Nyquist requency, the spectrum o the samples equals the analog spectrum, but over the normalized analog requency T. Thus, the energy in the sampled signal equals the original signal's energy multiplied by T.