信号与系统复试真题库信号与系统ch3

1、Ch3 Fourier representations of signals and Linear Time-Invariant Systems（信号与LTI系统的傅里叶表示）郝晓莉陈后金北京交通大学电子信息工程学院Ch3.1 IntroductionDiscrete-Time Periodic Signals: Discrete-Time FourierSeries离散周期信号的傅里叶级数Discrete-TimeNon-Periodic Signals: Discrete-Time FourierTransform离散非周期信号的傅里叶变换Continuou

2、s-Time Periodic Signals: Continuous-TimeFourier Series连续周期信号的傅里叶级数Continuous-Time Non-Periodic Signals: Continuous-TimeFourier Transform连续非周期信号的傅里叶变换Properties of Fourier Representations傅里叶表示的性质Frequency-Domain Analysis of LTI systemLTI系统的频域分析Ch3.2 Complex Sinusoids and Frequency Response of LTI Sys

3、tem（复正弦信号和LTI系统的频率响应）复正弦信号输入到 LTI 系统，输出仍为同频率的复正弦信号，其幅度乘以系统的频率响应。H ( jw)e jwte jwty(t) = ejwtejw (t -t ) h(t )dt* h(t) =-we- jwth(t )dt= e jwt H ( jw)=jte-其中称为系统的频率响应。H ( jw) = h(t)e- jwt dt-h(t)Complex Sinusoids and FrequencyResponse of LTI SystemH ( jw) e j argH ( jw )h(t)e- jwt dt=w) =H ( j-Defini

4、tions：H ( jw) e j (wt +argH ( jw )y(t) = e jwt H ( jw)=H ( jw) 改变了输入信号的幅度。系统通过通过 argH ( jw)改变了输入信号的相位。H ( jw)is theMagnitude response(幅度响应).argH ( jw)is the Phase response(相位响应).Example: A RC circuit system with the impulse response- t eRCu(t)1h(t) =RCFind the frequency response of this system.y(t)x

5、(t)C已知连续时间LTI系统的单位冲激响应，求系统的频率响应 Y(jw)。RExample: A RC circuit system with the impulse response- t eRCu(t)1h(t) =RCFind the frequency response of this system.solution：H ( jw) = - 1 t1- jwt dt =eRC u(t)e- jwt dth(t)eRC-( jw + 1 )t-( jw + 1 )t-111dt = 1eRCeRCRCRC0jw +10RC= 1 -1(0 -1)= RCjw + 1RCjw + 1RC

6、RCExample:The frequency response ofthe RC circuit.1/ RCjw +1/ RCH ( jw) e j argH ( jw )H ( jw) =Magnitude response.1/ RCH ( jw) =w 2+ (1/ RC) 2Phase response.argH ( jw)= -arctan(wRC)Example: The frequency response ofthe RC circuit.(a) Magnitude response.(b) Phase response.argH ( jw)= -arctan(wRC)1/

7、RCjw +1/ RCH ( jw) =Ch3.3 Fourier Representations for Four Classes of Signals（ 4种信号的傅里叶表示）Periodic Signals: Fourier Series Representationsxn is a Discrete-time signal with fundamental period N,xn = Ake jkW0nkx(t) is a Continuous-time signal with fundamental period T,x(t) = Ake jkw0tkFourier Represen

8、tations for Four Classes of SignalsNon-Periodic Signals: Fourier Series Representationsx(t) is a Continuous-time signal, 12pX (e jw ) e jwt dwx(t) =-xn is a Discrete-time signal,p X (e jW ) e jWndW-p12pxn =Ch3.4 离散周期信号的傅里叶级数（Discrete-Time Fourier Series: DTFS)1、离散周期信号展开为傅里叶级数N -1xn = X k e jkW0nk =0

9、N周期2pW0=基频NkWk次谐频k次谐波k次谐波系数0e jkW0nX k xn为周期信号， Xk为xn的DTFS系数。DTFS pairs（ DTFS对）N -1xn = X k e jkW0n0 n N -1k =0N -1xne- jkW0n1n=0X k =0 k N -1NDenote the DTFS relationship asxnDTFS X k DTFS系数Xk为xn 的频域表示，称为频谱。Example3.2 Determine the DTFS coefficients ofp8n + p )xn = cos(3using the method of inspecti

10、on.xn = cos(p n + p )solution:83pp3pp3j (n+- j (n+)+ e= e882j pj p n- j p3 e- j p n1212=+e3 e8e8xn = cos(p n + p )Example3.283j pj p n- j p3 e- j p n1212xn =+solution:e3 e8e81j p3 ,- j p3 ,k = 1e21=j arg X k X k eX k = k = -1e20,- 8 k 7 and k 1Hkisthe spectrum(频谱).Hk is theMagnitude spectrum(幅度谱).ar

11、gHk is the Phase response(相位谱).j p3 ,- j p3 ,1k = 1eExample3.2xn = cos(21p8p3X k = ek = -1n +)20,- 8 k 7 and k 1Magnitude Spectrum.1/ 2Xkk- 16- 8816Phase Spectrum.argXkp/ 3k-16- 8816-p/ 3因此，该周期序列的幅度谱和相位谱图为：X (k)1/ 2k- 16-8816argX (k )p / 3k-16-8816- p / 3Example3.2 DTFS representations of a square w

12、ave. 计算周期方波序列的DTFS。xn1n0- N- MNMxnExample3.2.1n- M0N- NMsin 2M +1 kW0 solution:1M 21e- jkW0n n=- M=X k =WkNNsin022M +1 ,k = 0， N ,2N ,LN 2M +1= kW0 sin 12k 0， N ,2N ,L,WkNsin02N = 50:(a) M = 4.(b) M = 12Ch3.5 连续周期信号的傅里叶级数（Continuous-Time Fourier Series: CTFS) 连续周期信号展开为傅里叶级数(Fourier Series)x(t) = X (

13、kwk =-)e jkw0t0其中，x(t)为周期信号，X(k)为傅立叶展开式的系数：1- k T x(t)e- jkw0t dtX (kw) =0Where,T周期2pw0=基频Tkw0k次谐频k次谐波jkw te0Example 3.9 ：Time-domain signal.Magnitude spectra and phase spectra for Example 3.9.Ex3.10 Computation of CTFS coefficients by inspection.x(t) =1-cos(pt) + 2sin(2pt) + cos(3pt)Determine the D

14、TFS coefficients of using the method of inspection.solution: x(t) = 1 - cos(pt) + 2 sin( 2pt) + cos(3pt)e jw0t + e- jw0te j 2w0t- e- j 2w0te j3w0t+ e- j3w0t= 1 -+ 2+22 j2k = 0k = 1 k = 2 k = 3others1,- 1 ,Therefore,2) = m j,X (kw01 ,20,因此，该周期信号的幅度谱和相位谱图为：X (k)11/ 2k-88argX (k)p / 2k-88- pEx3.13 FS r

15、epresentations of a square wave.计算周期方波的CTFS。x(t)t- T- T0TT00solution:2T0k = 0， 2mp ,L,sin (k 2pT0 / T ) = 1T= 2T0TT / 2w t=- jkXke（）02T0T0 Tk 2pT / TT-T / 2sin k 2pk 0， 2mp ,L,0T1(a) T0 /T = 1/4 .(b) T0 /T = 1/16.Ch3.6 离散时间非周期信号的傅里叶变换（Discrete-Time Fourier Transform: DTFT) 1、离散非周期信号分解为一系列复指数信号的叠加：p1

16、- n x(n) = X (e 2pjWjWndW) e-p其中，x(n)为离散时间非周期信号，其傅立叶变换(DTFT)为：k =-X (e jW ) =x(n)e- jWnDTFT of An Exponential Sequencexn = a nuna 0solution:X ( jw ) = x(t)e- jw t dt-a- jw t=teedt011a + jwF ( jw )=+ w 2a2wf (w) = - tan()-1aSpectrum of An Real Decaying Exponential1F ( jw )=+ w 2a2waf (w) = - tan(-1)(

17、a) Real time-domain exponential signal.(b) Magnitude spectrum.(c) Phase spectrum.FT of A Rectangular Pulse (非周期方波序列的DTFT)1/ 2x(t)e- jw t dtw tX ( jw) =- j=edt-1/ 2-1- jw(e- jw / 2- e jw / 2 )= Sa(w / 2)X ( jw)x(t)1- 2pwt2p单位宽度矩形波及频谱1- 0.50.5Inverse FT of A Rectangular SpectrumExample 3.26.(a) Rectan

18、gular spectrum in the frequency domain.(b) Inverse FT in the time domain.FT of The Unit Impulsed- jwd (t)e- jwt dt = 1(t) =tFf (t)edt-d (t)X ( jw)1(1)wt00单位冲激信号及其频谱Inverse FT of An Impulse SpectrumX ( jw) =e- jwt dt=F1-X ( jw)(2p )1wt00直流信号及其频谱函数FT of The Unit Stepu(t) = 1 u(t) + u(-t) + 1 u(t) - u(

19、-t)22Fu(t) = pd (w) +1jwX ( jw)u(t)1(p )wt00阶跃信号及其频谱Signum Function（符号函数）Fsgn( t) = lim Fsgn( t)e-s t o 00st- jwe-ste- jwt dt=(-dt +t1)ee-00e(s - jw )te-(s + jw )t= -s - jw- s +jwt =-t =0-1o - jw1o + jw2jw=+=-1t 0Signum Function（符号函数）X(jw)=|X(jw)|ejf(w)|X(jw)|f(jw) 幅度频谱相位频谱f (w )p / 2X ( jw)w0- p /

20、2w0符号函数的幅度频谱和相位频谱Ch3.8 傅里叶分析的性质(properties of Fourier Represents)(DTFS, FS, DTFT, FT:以FT、其次以DTFT为主)1. 周期性2. 线性3. 对称性 4.时域卷积5.时域微分6.频域微分8. 时移9. 频移10. 部分分式法求傅里叶反变换11. 时域相乘12. 展缩13. 帕斯瓦尔能量定理14. 时间带宽积15. 对称互易7.积分特性Ch3.8周期性（Periodicity）周期性（Periodicity）连 续 离 散 非周期周期Ch3.9 Linearity and Symmetry(线性与对称特性)FT:

21、FS:DTFT:DTFS:其中a和b均为常数。ax n + bx nDTFS aXk + bXk 1212ax n + bx nDTFT aX(e jW ) + bX(e jW )1212ax (t) + bx(t) FS aX(k ) + bX(k )1212ax (t) + bx(t) FT aX( jw) + bX( jw)1212Representation of a step function as the sum of a constant and a signum function.u(t) = 1/ 2 +1/ 2sgn(t)1jwFu(t) = pd(w) +Represen

22、tation of the periodic signal z(t) as a weighted sum of periodic square waves:(a) z(t).(b) x(t).(c) y(t).z(t) =(3/2)x(t) + 1/2y(t).Z(k) = (3/2)X(k) + 1/2Y(k).Symmetry Properties: Real and Imaginary SignalsX(jw)为复数，可以表示为X ( jw) e jj (w ) = ReX ( jw)+ jImX ( jw)X ( jw) =当x(t)为实函数时，有ReX(jw) = ReX(-jw)

23、实部偶对称，ImX(jw) =-ImX(jw) 虚部奇对称X(jw) = |X(-jw)| 幅频偶对称j (w) = - j (-w) 相频奇对称thenx *(t) FT X *(- jw)ifx(t) FT X ( jw)Ch3.10 Convolution Property(时域卷积)Convolution of Non-periodical Signals非周期信号的时域卷积FT:DTFT:y(n) = x(n)* h(n) DTFT X (e jW ) H (e jW )y(t) = x(t) * h(t) FT X ( jw) H ( jw)证明：-x (t )x(t -t )dt

24、 e- jwt dtFx (t) * x(t) =1212-x(t -t )e- jwt dtdtx (t )12x (t ) X( jw)e- jwt dt12= X1 ( jw) X 2 ( jw)Example 3.32 Finding Inverse FTs By Means of TheConvolution Property.求如图所示信号x(t)的频谱。solution：(a) 矩形脉冲 z(t).(b)z(t) 与其本身卷积得到 x(t).x(t) = z(t)* z(t)z(t) 2ATSa (wT ) = 2Sa(w)X ( jw) = 4Sa 2 (w)Qy(t) = x

25、(t) * h(t) FT X ( jw) H ( jw)Convolution of Periodical Signals周期信号的时域卷积FS:(周期卷积)DTFS:(周期卷积)DTFS ; W = 2py(n) = xn zn0N NX k ZkFS; w = 2py(t) = x(t) z(t) 0T TX k ZkCh3.11 Differential and Integration Property(时域微分和积分特性)Ch3.11.1 Differentiation in Time(时域微分)x(t) X ( jw)ifthenndx(t)( jw) X ( jw)ndt nx(

26、t) = 1 2pX ( jw)e jwt dw证明：-dx(t)dt=( jw)X ( jw)e- jwt dw-dx(t) ( jw) X ( jw) dtExample 3.37 Verifying the Differentiation Property. Find thee-at u(t)dFT of. 利用时域微分特性求傅里叶变换。dtsolution:1jw + ae-atu(t) FT 故利用时域微分特性可得: d e-at u(t)FT jwjw + adtCh3.11.2 Differentiation in Frequency（频域微分）ifx(t) FT X ( jw)

27、xnDTFT X (e jW )jWdX (e)-jn xnDTFTthendWX ( jw)x(t)e- jwt dt=证明：-w)dX ( j ddw- jww =(- jt)x(t)e- jwt dt=tx(t)eddw- jt x(t) FT dX ( jw)dwProblem 3.25. Use the frequency-differentiation property to find the FT of te-atu(t).试利用频域微分特性求信号te-atu(t)的傅立叶变换。solution：1jw + ae-atu(t) FT ddw1jw + ate-atu(t) FT

28、j()1( jw + a )2=- jt x(t) FT dX ( jw)dwCh3.11.3 Integration Property(积分特性)x(t) FT X ( jw)if1X ( jw) + X ( j0)d (w) jwx(t )dt FT tthen1-X ( jw) jw若信号不存在直流分量，则- j 0tX ( j0) =dt =x(t)dt = 0x(t)e-Ch3.12.1 Time-Shift Property (时移特性)ifthen证明：-)e- jwt dt = e- jwt0)e- jw (t -t0 )dtFx(t - t) =x(t - tx(t - t0

29、00X ( jw) e jarg X ( jw )-wt0 = X ( jw) e- jwt0 =信号在时域中的时移，对应频谱函数在频域中产 生的附加相移，幅度频谱保持不变。FT- jwtx(t - t0 ) X ( jw) e0x(t) FT X ( jw)Time-Shift Property (时移特性)x(n) DTFT X (e jW )ifthenx(n - n ) DTFT X (ejW ) e- jWn0 0Application of the time-shift property.Ex: Find the spectrum Y(jw) of signale-2t e-5(t

30、-t )dt , (t -2)ty(t) =-2te-2t e-5(t -t ) dty(t) =solution：-2= e-2t u(t + 2) * e-5t u(t)利用Fourier变换的卷积特性可得Y ( jw) = Fe-2t u(t + 2)Fe-5t u(t)= e4 e j2we j2w +41jw + 5=jw + 2(jw) 2 + 7 jw + 10Ch3.12.2 Frequency-Shift Property（频移特性）x(t) FT X ( jw)x(n) FT X (e- jW )证明：由傅里叶变换定义有w tx(t)ejw0t e- jwt dtx(t)e

31、- j(w -w0 )t dtFx(t) e =j0-= X j(w - w0 )=-x(n) e jW0n DTFT X (e j(W-W0 ) )x(t) ejw0t FT X j(w -w )0Frequency-Shift Property（频移特性）1FT 2pd (w)+ e- jw0t FT p d (w + w12) + d (w - wjw tcos(w t) =e0000- e- jw0t FT jp d (w + w 1 2 j) - d (w - w)w tsin( w t) =je0000e jw0t FT 2pd(w - w)0Ch3.13 Finding Inve

32、rse Fourier Transform by Using Partial-Fraction Expansions（部分分式法求傅里叶反变换） Ch3.13.1 Inverse Fourier Transform（由X(jw)求x(t)）b( jw)M( jw)M -1 +L+ b ( jw) + b+ bX ( jw) =M -1M10a( jw) N( jw) N -1 +L+ a ( jw) + a+ aN -1N10+ B( jw)= c+ c ( jw) + c( jw)2 +L+ c( jw)M - NM - N012A( jw) 当MN时存在 真分式d ( M - N ) (t

33、) + F -1 B( jw) x(t) = c d (t) + c d(t) + c d(t) +L+ c A( jw) M - N012ifthenk= ( jw + l ) B( jw)i = 1,2,L, NWhere,jw =-liA( jw)iiEx：X ( jw) = jw + 1, find x(t)( jw)2 + 5 jw + 6x(t) = -e-2t u(t) + 2e-3t u(t)F -1 B( jw) = (k e-l1t + ke-l2t +L+ ke-lNt )u(t)12N A( jw) B( jw) =k1+k2+L+kNA( jw)jw + l1jw +

34、 l2jw + lNCh3.13.2 Inverse Discrete-Time Fourier Transform（由X(ejW)求x(n)）e- jWM+L+ b e- jW + bbX (e jW ) = M10 e- jWN+L+ a e- jW + aaN10B(e jW )- jW- j ( M - N )W= c0 + c1e+L+ cM - N e+A(e jW ) 当MN时存在 真分式-1 B(e jW ) x(n) = c0d (n) + c1d (n -1) +L+ cM -N d (n - (M - N ) + FWjA(e)ifthenB(e jW )ki= (1 -

35、 ri ejWWhere,) 1jWA(e)- jW=rei- 5 e- jW+ 5X (e jW ) = 6Ex3.45, 求xn- 1 e- j 2W + 1 e- jW+ 166 -1n 1 nxn = 4u n + un2 3 -1 B(e jW ) nnnF A(e jW ) = (k1r1+ k2r2+L+ kN rN)unB(e jW ) =k+k+k12L+NA(e jW )1 - r e- jW1 - r e- jW1 - re- jW12NCh3.14 Multiplication Property（时域相乘）FT:DTFT:（周期卷积）x(n)z(n) DTFT 1 X

36、(e jW ) Z (e jW )2px(t) z(t) FT 1X ( jw) * Z ( jw) 2pThe effect of windowing.(a) Truncating a signal in time by using a window function w(t).(b) Convolution of the signal and window FTs resulting from truncation in time.x(t) z(t) FT 1X ( jw) * Z ( jw) 2p 1 2ppX (e jq )Z (e j (W-q ) )dq=-pThe effect

37、of truncating the impulse response of a discrete-time system.(a) Frequency response of ideal system. (b) FW(q) for W near zero. (c) FW(q) for W slightly greater than p/2.(d) Frequency response of system with truncated impulse response.x(n)z(n) DTFT 1 X (e jW ) Z (e jW )2pApplication：Amplitude modula

38、tion（幅度调制）y(t)g(t)cos(w0t)1, 1 1t，FindY ( jw) .x(t) = cos(w0t),g(t) = Ex：0,ty(t) = g(t) cos(w0 t)12pG( jw) *p d (w + w) + d (w - w)Y ( jw) =00= 1 G j(w + w)+ 1 G j(w - w)0022信号x(t)的频谱搬移到高频段，幅度减半Application：Amplitude modulation（幅度调制）Frequency domainTime domain6000140000.5200000-40-30-20-10010203040-3-2-1