信号与系统英文版教学课件：ch4 The Continuous-Time Fourier Transform

1、1 INTRODUCTIONRepresent continuous-time aperiodic signals as linear combinations of complex exponentials.Fourier transform and inverse Fourier transform. （傅立叶变换） （傅立叶逆变换）Use Fourier methods to analyze and understand signals and LTI systems. 121. Representation of Aperiodic Signals: Continuous-Time F

2、ourier Transform1) Development of the Fourier transform representation of an aperiodic signal (1) Example( Go from Fourier series to Fourier transform )Over one period of the continuous-time periodic square wave:We can regard Tak as samples of an envelope function, specifically -continuous variable-

3、the envelope of Tak-equally spaced samples23 As T increases with T1 fixed, we have two observations: A. The harmonics are packed more and more closer to each other; B. However, the shape (the envelope) of the spectrum remains the same (and it is determined by the pulse shape of the signal).T=4T1T=8T

4、1T=16T13What do we know from the above graph?445(2) Go from periodic to aperiodic Construct a periodic signal from the aperiodic signal x(t); later, let For periodic signal , we have its Fourier serires: For aperiodic signal x(t) : WhenHow to represent the aperiodic signal x(t) ?56As , Use X(j) to d

5、enote this integral, then we have:Since spectrum of x(t) Although X(j) is often abbreviated as “spectrum”, it is different from ak, which is the spectrum of periodic signals. Fourier transform of x(t)X(j) is in fact spectrum-density function(频谱密度函数)67 Thus, we obtainInverse Fourier transform78Fourie

6、r transform pair :orImportant: The aperiodic signals can still be represented as a linear combination of complex exponentials. The magnitude of component with frequence isAn useful relationship:where X(j) is the Fourier transform of x(t), ak is the Fourier coefficients of . x(t) is one period of the

7、 periodic signal 89orExample: periodic square wave:So aperiodic signal x(t): 9102) Convergence of Fourier transformDirichlet conditions: (1) x(t) is absolutely integrable. (2) x(t) have a finite number of maxima and minima within any finite interval. (3) x(t) have a finite number of discontinuity wi

8、thin any finite interval. Furthermore, each of these discontinuities must be finite.If impulse functions are permitted in the transform, some signals which are not absolutely integrable over an infinite interval, can also be considered to have Fourier transforms. This will be convenient in the discu

9、ssion of Fourier methods.1011.Example 4.1 Consider the signal 3) Examples of Continuous-Time FT|X(j)|1/ - -/4-/2/2/4 argX(j)magnitude spectrum of x(t)phase spectrum of x(t)1112The Fourier transform of the signal is Example 4.2Consider the signal x(t)x(t)1t2/X(j)1/ - In this case X(j) is real.1213 Th

10、at is, the unit impulse has a Fourier transform consisting of equal contribution at all frequencies. This spectrum is referred to as white-spectrum. (because the white color has the same spectrum).Example 4.3 Let us determine the Fourier transform of the unit impulse Example 4.4 Consider the rectang

11、ular pulse signal-/T1 /T12T1X(j)1314Example 4.5 Consider the signal x(t) whose Fourier transform isUsing the synthesis equation, we can determine1-W WX(j)W/-/W /Wx(t)tsinc functions:1415The relationship between the time and frequency domains is inverse.For all In this case15162. The Fourier Transfor

12、m for Periodic Signals To obtain the general result for periodic signals, let us first consider the Fourier transform of the complex exponential From the analysis equation,However, this integral does not converge. Consider the Fourier transform pair (t) and 1. 1617This equation says that the Fourier

13、 transform of unit dc is .This equation shows that the Fourier transform of the complex exponential signal is an impulse located at with its area 2.For an arbitrary periodic signal x(t), First representing x(t) with the Fourier series ascalculating the Fourier transform on both sides of this equatio

14、n, 1718 The Fourier transform of a periodic signal with Fourier series coefficients ak can be interpreted as a train of impulses occurring at the harmonically related frequencies and for which the area of the impulse at the kth harmonic frequency k0 is 2 times the kth Fourier series coefficient ak.E

15、xample 4.6 Consider again the periodic square wave. Its Fourier series coefficients are1819 0 0X(j) 2 2 Fourier transform of a symmetric square wave for T = 4T1. then its Fourier transform is1920Example 4.7 Consider and The Fourier series coefficients for x1(t) areThe Fourier series coefficients for

16、 x2(t) are -0 0 0 X1(j)-/j/j -0 0 0 X2(j)2021The Fourier series coefficients for this signal areThus, its Fourier transform isExample 4.8 Consider the impulse train t1 -2T -T 0 T 2Tx(t)X(j)2/T -4/T -2/T 0 2/T 4/T21223. Properties of Continuous-Time Fourier Transform 2) Time Shifting IfthenConsequenc

17、e: a signal which is shifted in time does not have the magnitude of its Fourier transform altered. thenIf1) Linearity22Example 4.9 1 1 0 t -/2 /2 t 0 Example 4.10 f1(t) 1 - 0 t f2(t) 2323243) Conjugation and Conjugate SymmetryIfthen|X(j)| = | X(j)|arg X(j) = arg X( j) even function of If x(t) is rea

18、l, then Re X(j) = Re X(j)Im X(j) = Im X( j) odd function of If x(t) is both real and even, then X(j) will also be real and even. If x(t) is real and odd, then X(j) is purely imaginary and odd.2425Example 4.11 Consider again the Fourier transform evaluation of andFrom the symmetry properties of the F

19、ourier transform, we have25264) Differentiation and Integration IfthenExample 4.12 Determine the Fourier transform of the unit step x(t) = u(t).Consequence: increase the high-frequencies components in a signal. Consequence: decrease the high-frequencies components in a signal. 2627Example 4.13 Calcu

20、late the Fourier transform X(j) for the signal x(t) displayed in the following figure: x(t)t -1 1 1 -1t -1 1 1t -1 1 (-1) ( -1)+27The applied condition of Integration property:suppose then andSo in general:That is the value of original function in the negative infinity is zero!2828Example 4.14 Calcu

21、late the Fourier transform for the signal 0 t 1 - 0 t -1/ - 0 t (2/) 2929Example 4.153030,So 运用积分性质从导函数的频谱求原函数的频谱时,原函数若不含直流分量,则其频谱就不含冲激函数,否则,其频谱等于导函数的频谱除以因子 后再加上直流分量的频谱.3131325) Time and Frequency Scaling Ifthen6) Duality IfthenThis property shows that for any Fourier transform pair there is a dual

22、pair with the time and frequency variables interchanged.A linear scaling in time by a factor of a corresponds to a linear scaling in frequency by a factor of 1/a, and vice versa. Especially, 3233Example: 3334Example 4.16 Let us consider using duality to find the Fourier transform G(j) of the signalW

23、e have known thatFrom duality property, we have3435other important results from duality:A.Differentiation in frequency-domain:B.Integration in frequency-domain:C.Frequency Shifting: Amplitude modulation35A -/2 0 /2 t A A/2 -c 0 c t 3636377) Parsevals RelationParsevals relation says that this total e

24、nergy may be determined either by computing energy per unit time ( ) and integrating over all time or by computing the energy per unit frequency ( ) and integrating over all frequencies. energy-density spectrum （能量密度谱）3738 -1 0 1 x(j)(a) (b) -1 -0.5 0.5 1 x(j)Example 4.17 For each of the Fourier tra

25、nsforms shown in the following figure, evaluate the following time-domain expressions:Use Parsevals relation, we may evaluate E in the frequency domain:3839To evaluate D in the frequency domain, first use the differentiation property: Since Da = 0 Since Xa(j) is even-function, so Ga(j) is odd-functi

26、on, thusand39404. The Convolution Property Consider the convolution integral:The Fourier transform of y(t) is: Interchanging the order of integration and noting that x() does not depend on t, we have 4041The Fourier transform maps the convolution of two signals into the product of their Fourier tran

27、sforms. H(j), the frequency response, is the Fourier transform of the impulse response. It captures the change in complex amplitude of the Fourier transform of the input at each frequency . The frequency response H(j) also can characterize an LTI system, just as its inverse transform, the unit impul

28、se response h(t) .x(t)H1(j)y(t)H2(j)x(t)H1(j) H2(j)y(t)x(t)H2(j)y(t)H1(j)Three equivalent LTI systems. 4142The frequency response cannot be defined for every LTI system. Since essentially all physical or practical signals satisfy the last two conditions in Dirichlet conditions, the condition of abso

29、lutely integrable becomes the determining factor which can guarantee the existence of the Fourier transform H(j) of h(t). That is, only a stable LTI system has a frequency response H(j).If an LTI system is stable , then ,its impulse response integrable; that is In order to use transform techniques t

30、o examine unstable LTI system , we will develop a generalization of the continuous-time Fourier transform ,the Laplace transform. 4243Example 4.18 Consider an integrator that is, an LTI system specified by the equationSince The impulse response for this system is the unit step u(t). the frequency re

31、sponse of the system isUsing the convolution property, we have 4344to the input signalExample 4.19 Consider the response of an LTI system with impulse responsethe Fourier transforms of x(t) and h(t) areTherefore, Expanding Y(j) in a partial-fraction expansion（部分分式展开） where A and B are constants to b

32、e determined. When b a 4445Therefore, When b = a Recognizing this as Consequently,We can use the differentiation in the frequency-domain property. Thus,4546Example 4.20 Determine the response of an ideal low-pass filter to an input signal x(t) that has the form of a sinc function. That is,The impuls

33、e response of the ideal low-pass filter is of a similar form:Therefore, Finally, the inverse Fourier transform of Y(j) is given byThat is, depending on which of i andc is smaller, the output is equal to either x(t) or h(t). 46475. The Multiplication Property modulation property （调制定理）s(t)p(t)r(t)Mul

34、tiplication of one signal by another can be thought of as using one signal to scale or modulate the amplitude of the other, and consequently, the multiplication of two signals is often referred to as amplitude modulation.4748Example 4.21 Let s(t) be a signal whose spectrum S(j) is depicted in Figure

35、 (a). Also, consider the signal (a) Then -1 1S(j)A -0 0 P(j) (b) the spectrum R(j) of r(t) = s(t)p(t) is obtained by an application of the multiplication property:4849The spectrum of r(t) consists of the sum of two shifted and scaled versions of S(j). -0 0 (-0-1) (-0+1) (0-1) (0+1)R(j)= S(j)* P(j)/2

36、 A/2Example 4.22 Let us consider r(t) as obtained in Example 4.21, and let g(t) = r(t)p(t). -0 0R(j) A/2 A/4 A/4 -20 -1 1 20G(j)A/24950Example 4.23 Determine the Fourier transform of the signal Recognize x(t) as the product of two sinc functions: Applying the multiplication property of the Fourier t

37、ransform, we obtainNoting that the Fourier transform of each sinc function is a rectangular pulse, thus X(j) can be obtained by convolving these pulses: -3/2 -1/2 1/2 3/2X(j)50516. Systems Characterized by Linear Constant-Coefficient Differential EquationsConsider the question of determining the fre

38、quency response of an LTI system which is described by a linear constant-coefficient differential equation of the form (assuming that the system is stable) :From the linear property, this becomes5152 from the differentiation property,or equivalently,From the convolution property ( Y(j) = H(j) X(j) )

39、 , H(j) is a ratio of polynomials in (j) . coefficients of the numerator polynomial = coefficients appearing on the right side of the differential equation.coefficients of the denominator polynomial = coefficients appearing on the left side of the differential equation.5253Example 4.24 Consider a st

40、able LTI system that is characterized by the differential equation Determine its impulse response.The frequency response is To determine the corresponding impulse response, we use the method of partial-fraction expansion: Thus, the impulse response is5354Example 4.25 Determine the output of the syst

41、em in Example 4.24, and suppose that the input is In this case, the partial-fraction expansion takes the formwhere A11, A12, and A2 are constants to be determined. 5455So that55567. Frequency-Selective Filters Filtering: a process in which the relative amplitudes of the frequency components in a sig

42、nal are changed or some frequency components are eliminated entirely.Frequency-selective filters（频选滤波器） : systems that are designed to pass some frequencies essentially undistorted and significantly attenuate or eliminate others. Types of frequency-selective filters low-pass filter （低通滤波器） high-pass

43、 filter （高通滤波器） band-pass filter（带通滤波器） band-stop filter （带阻滤波器） 1) The concept and types of filters5657Ideally, the frequency responses of a low-pass filter, a high-pass filter, a band-pass filter and a band-stop filter are illustrated in the following figures, respectively:Passband（通带） H(j) 1 c 0

44、cStopbandStopband（阻带）cutoff frequency（截止频率） 5758 H(j) 1 c 0 c H(j) 1c2 c1 0 c1 c2 H(j) 1c2 c1 0 c1 c2upper cutoff frequency（上截止频率） lower cutoff frequency（下截止频率） 58592)The impulse responses of the ideal low-pass filter and ideal high-pass filter (1) The impulse response of the ideal low-pass filter T

45、he impulse response of the ideal low-pass filter is:c/-/c /ch(t)t5960Thus, the impulse response of the ideal high-pass filter is: The frequency response of the ideal high-pass filter can be represented in terms of the frequency response of the low-pass filter as:h(t) for both low-pass and high pass

46、filters are not causal, so they are ideal.(2) The impulse response of the ideal high-pass filter 6061(1) A Simple RC Low-pass Filtervs(t) + vr(t) R+ C vc(t) +Input: source voltage vs(t); Output: capacitor voltage vc(t) 3) Examples of Continuous-time Filters Described by Differential Equations6162t s

47、(t) 11-1/e= RC = RC h(t) 1/ 1/e 1/RC 0 1/RC|H(j)| 1nonideal low-pass filter 62 A 0 t 6363 let 646465656666A 0 t The rise and fall time characteristics： Rapid change (jump) Gradual change (Smooth)High-frequency component is reduced! A 0 t 676768(2) A Simple RC High-pass FilterInput: source voltage vs

48、(t); Output: resistor voltage vr(t) 1/RC 0 1/RC|H(j)| 1 t =RC s(t) 1 1/enonideal high-pass filter 68%Fourier分析clear% 得到掺杂了均值为0，方差为0.2的高斯白噪声的信号y% 原始 y 信号有三个频率分量：10，100，180 Hzrandn(state,0);Ts=0.002;t=0:Ts:0.4;y=sin(2*pi*10*t)+cos(2*pi*100*t)+2*sin(2*pi*180*t)+0.2*randn(size(t);% 利用FFT求 y 的离散 Fourier

49、系数 YY=fft(y);fs=1/Ts; %采样频率fn=fs/2; % Nyquest频率%得到10阶ButterWorth带通滤波器，截止频率是90110HzB,A=butter(10,90/fn,110/fn);y100=filter(B,A,y); %滤波，得到100Hz的信号subplot(2,1,1)plot(t,y,b-,t,y100,r-) %作图显示噪声信号和滤波后信号axis(0.2,0.3,-4,4), xlabel (t)subplot(2,1,2)f=linspace(0, fn,length(t)/2); %半采样频率中相应的刻度Ya=Ts*abs(Y(1:len

50、gth(t)/2);plot(f,Ya) , xlabel (f) %做出半边频谱69708. Transmission without Distortion (无失真传输)Linear distortion: without new frequency components produced.Input: x(t); Output: y(t) If y(t) = K x(t-t0), we say x(t) is transmitted without distortion.Types of linear distortion: magnitude distortion and phase

51、distortion.7071Examplex(t) are consist of the first harmonic components and the second harmonic components. Physics analyzing： without Distortion the first harmonic components delay :Phase shift: the second harmonic components delay :Phase shift: 7172SupposeThen For being transmitted without phase d

52、istortion, there must beTherefore0Conditions of Transmission without Distortion7273signals process ：sample D/AQuantity & codingprocessAnalog signalsSampling signalsDigital signalsAnalog signals 1) Sample and the spectrum of sampling signalx(t) 0 t 12ssampler sketch mapSampler is a switch p(t) : 0 T

53、t 1 0 T t maths model The interval of each sample pulse is same，the interval is T-uniformity sampleSample frequency ,9. Sampling7374questions：What is the spectrum of xp(t) ？And What relation is it with the spectrum of x(t) ？(2)When we can resume x(t) from xp(t) without distortion ?Suppose from the M

54、ultiplication PropertySinceandSo7475Conclusion: The shape of spectrum of sample signal is determined by , and its magnitude is determined by ,and that, is periodic signal, its period is . But is only relational with k, and it is determined by the shape of p(t). 7576xp(t)x(t)p(t) x(t) 0 t T 1 p(t) 0

55、t T 0 xp(t) x(0) x(T) t Impulse-train sampling 2) Impulse-train sampling（冲激抽样） 7677-M MX(j) 1 P(j) 2/T -2s -s 0 s 2s Xp(j)1/T -M0M s (s M)Xp(j)1/T 0 s (s M)Effect in the frequency domain of sampling Xp(j) is a periodic function of consisting of a superposition of shifted replicas of X(j), scaled by 1/T.7778Sampling Theorem（抽样定理）:Let x(t) be a band-limited signal with X