信号与系统 第三章课件
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1通信科学与工程系四用微分和差分方程描述的因果LTI 系统1. 线性常系数微分方程()()()t bx t ay dt t dy =+给出了系统的隐含特性,要得到明确表达式,需求解方程,并且还需一个或多个附加条件。
对于因果线性时不变系统,附加条件的形式特殊简单。
2通信科学与工程系一般的N 阶线性常系数微分方程:()()∑∑===M k kk kNk k k k dt t x d b dt t y d a 00()()∑==M k kkk dtt x d b a t y 001当N=0时,输出是输入及其导数的明确函数:当N>0时,输出是输入的隐含形式,需要求解。
四用微分和差分方程描述的因果LTI 系统3通信科学与工程系求解该微分方程,通常是求出通解和一个特解,则。
()p y t ()h y t ()()()p h y t y t y t =+四用微分和差分方程描述的因果LTI 系统()p y t ()x t 特解是与输入同类型的函数.()h y t 0()0k Nk k k d y t a dt==∑通解是齐次方程的解,即的解。
0Nkk k a λ==∑欲求得齐次解,可根据齐次方程建立一个特征方程:求出其特征根。
4通信科学与工程系若t ≤ t 0时x (t )=0,则t ≤ t 0 时y (t )=0,初始松弛条件1(),k Nth k k y t C e λ==∑其中是待定的常数。
k C 当特征根均为单阶根时,可得出齐次解的形式为:四用微分和差分方程描述的因果LTI 系统()()()010100====--N N dt t y d dt t dy t y 可采用如下初始条件:5通信科学与工程系()()()t x t y dtt dy =+2()()t u Ke t x t 3=例2.14:考虑输入为时,系统的解。
()()()t y t y t y h p +=5KY =()3,05t p Ky t e t =>方程的解由特解和齐次解组成:()tp Ye t y 3=求解特解:令t > 0时,根据方程可得33332t t t Ye Ye Ke +=受迫响应自然响应四用微分和差分方程描述的因果LTI 系统6通信科学与工程系()()02=+t y dtt dy 求解齐次解:根据方程,得特征方程为()23,05t t Ky t Ce e t -=+>0/5C K =+5KC =-()[]()t u e e K t y t t 235--=20λ+=2λ=-()2th y t Ce -=齐次解四用微分和差分方程描述的因果LTI 系统根据初始条件确定C :考虑因果LTI 系统,如果t<0 时x (t )=0,则t<0 时y (t )=0. 将t = 0, y (0) = 0代入有7通信科学与工程系2. 线性常系数差分方程一般的线性常系数差分方程可表示为:与微分方程一样,它的解法也可以通过求出一个特解和齐次解来进行,其过程与解微分方程类似。
Chapter 3 Fourier Series Representation of Periodic Signals第3章周期信号的傅里叶级数表示Main content :1.The Frequency Analysis of PeriodicSiganl(周期信号的频域分析)2.The Frequency Analysis of LTI(LTI系统的频域分析)3.Properties of Fourier Series(傅立叶级数的性质)3.0 Introduction(引言)⏹The basis for time domain(chapter2)1)Signal can be represented as linear combination of shift impulses。
2)System is LTI。
⏹Periodic Singal can be represented as linear combination of complex exponentials.3.1 Historical Perspective (历史的回顾)1、The concept of using trigonometric sums to describe periodic phenomena goes back to Babylonians2、Euler examined the motion of Vibrating string is a linear combination of a few normal mode in 1748.3.1 Historical Perspective (cont)3、Largange criticized the use of trigonometric series to examine vibrating string in 1759.4、Fourier claimed that any periodic signal could be represented by harmonically related sinusoids in1807.some story about Fourier•Born in France in 1768•Fourier claimed that any periodicsignal could be represented byharmonically related sinusoids in1807•Due to Lagrange …s objection his 1768—1830paper never appeared•His paper appeared in “TheAnalytical Theory of Heat” in 1822•Dirichlet provide precise conditionsin 1829傅里叶的两个最重要的贡献——•“周期信号都可以表示为成谐波关系的正弦信号的加权和”——傅里叶的第一个主要论点•“非周期信号都可以用正弦信号的加权积分来表示”——傅里叶的第二个主要论点3.2 The Response of LTI Systems to Complex Exponentials(LTI 系统对复指数信号的响应)stenz()h n ()h t ste()y t nz()y ncontinuios timediscrete timeUsing Time domain method ,()()()()()s t sts sty t eh d eh e d H s eττττττ∞∞---∞-∞===⎰⎰()()()()()n k nknk k y n zh k zh k zH z z∞∞--=-∞=-∞===∑∑EigenvalueGain is called “Eigenvalue”Eigenfunction in-> Same function out with gain Eigenfunctiondiscrete time()h n ()h t ste()stH s enz()nH z Zcontinuious timeEigenfunctionEigenvalue()()stH s h t e dt∞--∞=⎰()()nk H z h n z∞-=-∞=∑The usefulness of decomposition in term of eigenfuction is important for the analysis of LTI systems . ts kk k k es H a t y ∑=)()(ts kk k ea t x ∑=)(If :nkkk Za n x ∑=)(n kkk k ZZ H a n y ∑=)()(complex exponential signal 、are eigenfuctionof LTI systems、are eignevalue.ste nz ()H s ()H z Conclusion:How broad a class of signals could be represented as a linear combination of complex exponentials?qustionExample 1 ( 3.1):• a LTI systems y(t)=x(t-3) , now the input x(t)=cos(4t)+cos(7t), detemin y(t)?ss ed es H 3)3()(--+∞∞-=-=⎰ττδτy(t)= 1/2e -j12e j4t + 1/2e j12e -j4t + 1/2e -j21e j7t + 1/2e j21e -j7t=cos[4(t-3)}+cos[7(t-3)]x(t)= 1/2e j4t +1/2e -j4t +1/2e j7t +1/2e -j7tThe set of harmonically related complex exponentials0(){}jk tk t eωΦ=0,1,2,k =±±Each of these signals has a fundamental frequency that ismultiple of ω0,each is periodic with period 02T πω=3.3 Fourier Series Representation of Continuous-Time Periodic Signals(连续时间周期信号的傅里叶级数表示)3.3.1. Linear Combinations of Harmonically RelatedComplex exponentialsThus , is also periodic,the form is referred to as the Fourier series representation这表明用傅里叶级数可以表示连续时间周期信号,即: 连续时间周期信号可以分解成无数多个复指数谐波分量。
0(),0,1,2jk tk k x t a e k ω∞=-∞==±±∑Example 2:0()cos x t t ω=001122j t j te e ωω-=+112a ±=Example 3 :00()cos 2cos3x t t tωω=+0000331[]2j t j t j t j t e e e e ωωωω--=+++112a ±=31a ±=Some alternative form for the Fourier series 0000*()jk t jk t jk t jk t k k k k k k k k x t a e a e a e a eωωωω∞∞∞∞-***-=-∞=-∞=-∞=-∞⎡⎤====⎢⎥⎣⎦∑∑∑∑ork ka a*-∴=*kka a -=)()(t x t x *=Suppose x(t) is real ,then is expressed in polar form as kj k k a A e θ=k a 0001()()01()kk k j jk tj k t j k t kkk k k k x t A eea A eA eθωωθωθ∞-∞++=-∞=-∞===++∑∑∑Some alternative form for the Fourier series (CONT)0001[]kkjk tj jk tj k k k a A eeA ee ωθωθ-∞--==++∑*kkj j kkk k a a A eA eθθ----=∴=Q thus :k kA A -=k kθθ-=-Conclusion: is even ,and k a k θis odd0001()[]kkjk tj jk tj k k k x t a A eeA eeωθωθ-∞--=∴=++∑0012cos()k k k a A k t ωθ∞==++∑——trigonometric functions formis expressed in rectangular form as k k ka B jC =+k a 00101()()()jk tjk tkk k k k k x t a BjC eB jC eωω-∞=-∞==++++∑∑0001()()jk tjk tk k k k k a B jC eB jC eωω∞---=⎡⎤=++++⎣⎦∑*kka a -=Q k k k kB jC B jC --∴-=+thus k kB B -=k kC C -=-Conclusion: the real part of is even ,the imaginary part of is odd k a ka0001()()()jk tjk tk k k k k x t a B jC e B jC eωω∞-=⎡⎤=+++-⎣⎦∑[]00012cos sin k k k a B k t C k t ωω∞==+-∑——trigonometric functions form(another form)3.3.2. Determination of the Fourier SeriesRepresentation of a continuous-time Periodic Signal Assuming periodic signal x(t) can be represented with the Fourier series0(),jk tk k x t a eω∞=-∞=∑002T πω=00()()jn tj k n tkk x t ea eωω∞--=-∞=∑000()0()T T jn tj k n tkk x t e dt a edtωω∞--=-∞=∑⎰⎰000()000cos()sin()T T T j k n tedt k n tdt j k n tdtωωω-=-+-⎰⎰⎰{00,,T =k n ≠k n=0000()T jn tn x t edt a T ω-∴=⎰consequently00001()T jn tn a x t e dt T ω-=⎰Notice : the integration can be over anyinterval of length T01()jk tk a x t edtω-=⎰01()T a x t dtT =⎰a 0is simply the average value of x(t) over one period1T 0T -t()x t ⋅⋅⋅⋅⋅⋅⋅⋅The spectrum of periodic square waveExample4 (3.5) :11||1,()|/20,t T x t T t T <⎧=⎨<<⎩The spectrum of periodic square wave (Cont)10011101000002sin 11T jk tjk t T k T T k T a edt eT jk T k T ωωωωω----==-=⎰101111010010002sin 222Sa()sinc()T k T T T T k T k T k T T T T ωωω===sin Sa()x x x=sin sinc()xx xππ=Whereππ-()Sa x 1πx121-sin ()c x 1x1根据可绘出的频谱图。