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1、Chapter 3 Discrete-Time Fourier TransformThe DefinitionThe TheoremsDigital Processing of Continuous-Time Signals1How to Represent the Discrete-Time Signal? Time-domain (a weighted linear combination of delayed unit sample sequences.) Transform-domain a. frequency domain b. Z domain (a sequence in te
2、rms of complex exponential sequences of the form and .2Several Forms of FTFT-continuous in time,continuous in frequencyx(t)t3Result:Continuous function in time domain will cause non-periodicity in frequency domain,whereas the non-periodicity in time domain will cause Continuous function in frequency
3、 domain.4FS-continuous in time,discrete in frequencyx(t) T0Where:5Result:Continuous function in time domain will cause non-periodicity in frequency domain,whereas the periodicity in time domain will cause discrete frequency spectrum. 63.1 The Continuous-Time Fourier TransformIts an useful tool to re
4、present a continuous-time signal in frequency-domain.It briefly called CTFT.The relationship between its time and frequency is: continuous and non-periodic in time-domain, continuous and non-periodic in frequency-domain. 73.1.1 The DefinitionThe definition of CTFT is: The CTFT often is referred to a
5、s the Fourier spectrum. The I-CTFT(inverse CTFT) is:8We denote the CTFT pair of above two equations as: The CTFT is a complex function of in the range . It can be expressed in polar form as93.1.2 Energy Density SpectrumBased on Parsevals relation: The total energy of a finite-energy continuous-time
6、complex signal is given by:So we get energy density spectrum of is:103.1.3 Band-limited Continuous-Time SignalsAn ideal band-limited signal has a spectrum as: In practice, for a band-limited signal outside the specified frequency range, the signal energy is arbitrarily small. The bandwidth of a band
7、-limited signal.Lowpass, highpass, bandpass.113.2 The Discrete-Time Fourier TransformIts an useful tool to represent a discrete-time signal in frequency-domain.It briefly called DTFT.The relationship between its time and frequency is: discrete and non-periodic in time-domain, continuous and periodic
8、 in frequency-domain.123.2.1 DefinitionIts inverse transform process: The discrete-time Fourier transform of a sequence is defined by :13The convergence condition of above series is: Result:Discrete time will cause periodicity in frequency,and the non-periodicity in time domain will cause continuous
9、 function in frequency domain.14So: X(ej)is a periodic function of , and the DTFT definition equation is the Fourier series expansion of the periodic function X(ej) , xn are the coefficients of the Fourier series.153.2.2 Basic PropertiesIn general, X(ej ) is a complex function of the real variable a
10、nd can be written as: X(ej ) = Xre(ej ) + j Xim(ej )Where,Xre(ej ) and Xim(ej ) are, respectively, the real and imaginary parts of X(ej ) , and are real functions of .X(ej ) can alternately be expressed as: X(ej ) = | X(ej ) |ej ( )where ( ) = argX(ej ) 16| X(ej ) | is called the magnitude function.
11、 ( ) is called the phase function.Both quantities are real functions of .In many applications, the DTFT is called the Fourier spectrum.Likewise, | X(ej ) | and ( ) are called the magnitude and phase spectra.For a real sequence xn, | X(ej ) | and Xre(ej ) are even functions of , whereas, ( ) and Xim(
12、ej ) are odd functions of .17Note: X(ej ) = | X(ej ) |ej ( )+2 k = | X(ej ) |ej ( )for any integer k.The phase function ( ) cannot be uniquely specified for any DTFT.Unless otherwise stated, we shall assume that the phase function ( ) is restricted to the following range of values: - ( ) called the
13、principal value.18E.g. Compute DTFT of RN(n) Sa(.)linear phaseSolution：DTFTMagnitude：Phase：19)(wX0p2p2-pp-N=8Nw2021The DTFTs of some sequences exhibit discontinuities of 2 in their phase responses.An alternate type of phase function that is a continuous function of is often used.It is derived from t
14、he original phase function by removing the discontinuities of 2. .22The DTFT X(ej ) of a sequence xn is a continuous function of It is also a periodic function of with a period 2 :Please look at the examples(P95) of DTFT computation.233.2.3 Symmetry Relations:xn is a complex sequence24xn is a real s
15、equencePlease look at P98 example3.725For a purely imaginary sequenceSo, for a purely imaginary sequence, Xre(ej ) and Xim(ej ) are , respectively, odd and even function of . It can also be shown that |X(ej )| and ( ) are, respectively even and odd function of . 263.2.4 Convergence ConditionIf we de
16、note:For uniform convergence of X(ej ), that is: Now, if xn is an absolutely summable sequence, i.e., if:27Thenfor all values of Thus, the absolute summability of xn is a sufficient condition for the existence of the DTFT X(ej ).28Sincean absolutely summable sequence has always a finite energy.Howev
17、er, a finite-energy sequence is not necessarily absolutely summable.Please look at book about Examples.29Example 3.8 Consider the DTFTShown below30does not converge uniformly for all values of , but converge in the mean-square sense.So,Examining 31K=10K=20K=30K=40Gibbs phenomenon (discussed in Secti
18、on 10.2.3)32The type of sequences that a DTFT representation is possible using the Dirac Delta function () nneither absolutely summable nor square summablen,cos(0n+) The DTFT of xn is Dirac Delta function 33w Dirac Delta function ( ) is the limiting form of a unit area pulse function p ( ) as goes t
19、o zero satisfyingThe sampling property of the Dirac delta functionThe DTFT of xn is Dirac Delta function 34Example 3.9 Consider the complex exponential sequencewhere ( ) is an impulse function of andIts DTFT is given by35is a periodic function of w with a period 2 and is called a periodic impulse tr
20、ain. To verify that X(ej ) given above is indeed the DTFT of xn=ej 0n we compute the inverse DTFT of X(ej ) The function36Where we have used the sampling property of the Dirac delta function37Commonly used DTFT pairs38p is a positive integer.In practice, the value of p used is typically 1 or 2 or .T
21、he strength of a DTFT is given by its norm. of X(ej) is defied by 3.2.5 Strength of a DTFT39Peak absolute value is the mean absolute value of X(ej) is root-mean-squared(rms) value of X(ej) 3.2.5 Strength of a DTFT40Classroom ExcisesAn LTI discrete-time system is given by H (ej )=1+ 2cos + 3cos2 + 4c
22、os3 . . Determine the impulse response hn of the system. 413.3 DTFT Theorems1.Linearity 2.Time-Reversal3. Frequency-Shifting 4. Differentiation in Frequency5. Convolution 6. Modulation7. Parsevals Relation 4243Application of Above PropertiesCompute DTFT of sequences conveniently:Eg: please look at P
23、106,example3.10,3.11,3.12, 3.13,3.14Compute linear convolution:IDTFT443.4 Energy Density Spectrum of a Discrete-Time SequenceUse Parsevals relation:The quantityis called the energy density spectrum of gn.453.5 Band-Limited Discrete-Time SignalsAn ideal band-limited is defined as: In practice, we can
24、 make the signal energy be very small outside the specified frequency range to get band-limited signal. A classification of the band-limited DTS:Lowpass, highpass,bandpass(narrow-band),46Band-Limited Discrete-Time SignalsExample: Consider the sequence xn=(0.5)nnIts DTFT is given below on the left al
25、ong with its magnitude spectrum shown below on the right47Band-Limited Discrete-Time SignalsIt can be shown that 80% of the energy of this lowpass signal is contained in the frequency range 0|0.5081Hence, we can define the 80% bandwidth to be 0.508148Band-Limited Discrete-Time SignalsExample: Comput
26、e the energy of the sequence hLPn=sincn/n, -n2fh.Several concepts:(P120 Figure3.15)Over-sampling ,Under-sampling , critical-sampling71Sampling Theorem-(Nyquist Theorem)Over-sampling - The sampling frequency is higher than the Nyquist rate.Under-sampling - The sampling frequency is lower than the Nyq
27、uist rate.Critical sampling - The sampling frequency is equal to the Nyquist rate.Note: A pure sinusoid may not be recoverable from its critically sampled version.72Sampling Theorem-(Nyquist Theorem)In digital telephony, a 3.4 kHz signal bandwidth is acceptable for telephone conversation.Here, a sam
28、pling rate of 8 kHz, which is greater than twice the signal bandwidth, is used.73Sampling Theorem-(Nyquist Theorem)In high-quality analog music signal processing, a bandwidth of 20 kHz has been determined to preserve the fidelity.Hence, in compact disc (CD) music systems, a sampling rate of 44.1 kHz
29、, which is slightly higher than twice the signal bandwidth, is used.74ExampleConsider the three continuous-time sinusoidal signals:Their corresponding CTFTs are:75ExampleThese three transforms are plotted below:76ExampleThese continuous-time signals sampled at a rate of T = 0.1 sec, i.e., with a sam
30、pling frequency fT =20 rad/sec.The sampling process generates the continuous-time impulse trains, g1p(t), g2p(t) , and g3p(t).Their corresponding CTFTs are given by: 77ExamplePlots of the 3 CTFTs are shown below:78The Relationship Between G(ejw) and Gp(j)Compare the equation(1) and (2),we can get:79
31、Supplement: The practical samplingIn practical condition,p(t) is a periodic rectangle impulse with -width .If the and T are fixed, then when k varies, the magnitude of Ck varies with:80Supplement: The practical samplingSo we can see, in practical sampling,use Ck to substitute Ak, but Ck is varying.0
32、0813.8.2 Recovery of the Analog SignalIf T2 m , ga(t) can be recovered exactly from gp(t) .Passing it through an ideal lowpass filter Hr(j ) with a gain T and a cutoff frequency c greater than m and less than T - m as shown below:82Recovery of the Analog SignalSo we use a ideal lowpass filter to get
33、 Ga(j) from Gp(j):And its time-domain representation is:83Recovery of the Analog SignalThe spectra of the filter and pertinent signals are shown below:84Recovery of the Analog SignalFor simplicity, assuming c=s/2=/T, we get:nT(n-1)T(n+1)T(n-2)T(n+2)Tt10t3T2TT4T85Recovery of the Analog SignalThe idea
34、l bandlimited interpolation process is illustrated below:863.8.3 Implications of the Sampling ProcessConsider again the three continuous-time signals: g1(t)=cos(6 t), g2(t)=cos(14 t) , and g3(t)=cos(26 t)When they are sampled by a sampling rate of f=10Hz.Please look at P126. 8788解：解：8990913.9 Sampli
35、ng of Bandpass SignalsFor a bandpass signal, there is no need to sample it based on LP sampling theorem, because there is a lot of interval in its spectrum space.Let =H-L define the bandwidth of the bandpass signal. Assume: H=M() we choose the sampling frequency T to satisfy the condition: 92Samplin
36、g of Bandpass SignalsWe choose the sampling frequency T to satisfy the condition T = 2() = 2 H/Mwhich is smaller than 2 H , the Nyquist rate.Substitute the above expression in:93Sampling of Bandpass SignalsThis leads to:As before, Gp(j ) consists of a sum of Ga(j ) and replicas of Gp(j ) shifted by
37、integer multiples of twice the bandwidth and scaled by 1/T.94Sampling of Bandpass SignalsFigure below illustrate the idea behind:0095Bandpass Sampling Theorem(1)If ga(t) is bandlimited with its frequency band in (fL,fH) .(2)If using sampling rate:Where, n is a maximum integer which meets fs=2(fH-fL)
38、=2B (n = 0,1,2)Then, the sampled signal ga(nTs) can be used to represent original signal ga(t).96Note:Before band-sampling, the signal in only one frequency-band can be permitted.For lowest sampling rate, that is fs=2B ,the signal center frequency f0 should be:97The frequency relationship of band-sa
39、mplingThe band-sampling result is moving the signal in the frequency band of (nB, (n+1)B)(n=0,1,2,) to the frequency band of (0, B). When n is odd number, the relationship between them is reversal; while n is even, the relationship is opposite.98HomeworkRead the textbook from p89 to 131Problems: 3.16, 3.18(a)(c)(e), 3.23, 3.26, 3.33, 3.48, 3.61, 3.62 ,3.63, 3.65, 3.66M3.2,M3.3 99
