信号教学课件华中科技大学chapter2

1、chapter 2 linear time-chapter 2 linear time-invariant systemsinvariant systems 2.0 introduction2.0 introduction representation of signals as linear combination of delayed impulses. convolution sum(卷积和卷积和) or convolution integral(卷卷积积分积积分) representation of lti systems. impulse response and systems p

2、roperties solutions to linear constant-coefficient difference and differential equations (线性常系数差分或微分方程线性常系数差分或微分方程). 2.1 discrete-time systems: the convolution sumderivation steps:step 1: representing discrete-time signals in terms of unit samples: kknkxnxstep 2: defining unit sample response hn : r

3、esponse of the lti system to the unit sample n. n hn step 3: writing any arbitrary input xn as: 2 2 1 1 0 1 1 2 2nxnxnxnxnxnxstep 4: by taking use of linearity and time-invariance, we can get the response yn to xn which is the weighted linear combination of delayed unit sample responses as following

4、:22 1 1 0 1 122nhxnhxnhxnhxnhxnykknhkxthe convolution sum representation of lti systems convolution sum or superposition sum :kknhkxnyconvolution operation symbol: nhnxnylti system is completely characterized by its response to the unit sample -hn .example 2.1 nxn 0 1 2nhn-2 0 2(a) consider a lti sy

5、stem with unit sample response hn and input xn, as illustrated in figure (a). calculate the convolution sum of these two sequences graphically.kknhkxnykxk 0 1 2kh-k -2 0 2 (b)h-2h2110.5kxk 0 1 2khn-kn-2 n+2if n-4, graph of yn in example 2.1 from example 2.1, we can draw the following table:thus, we

6、obtain a method for the computation of convolution sum, that is suitable for two short sequences. xn = 1,1,10hn = 0.5, 1, 0.5, 1, 0.5-2xn*hn = 0.5,1.5,2,2.5,2,1.5,0.5-2 0.5 1.5 2 2.5 2 1.5 0.50.5 1 0.5 1 0.5xn h-2 h-1 h0 h1 h2 hnx0 x1 x2：x0h-2 x0h-1 x0h0 x0h1 x0h2 0 0 x1h-2 x1h-1 x1h0 x1h1 x1h2 0 0

7、x2h-2 x2h-1 x2h0 x2h1 x2h2 0 0 0 0 0 0 0 0 0y-2 y-1 y0 y1 y3y2y40.5 1 0.5 1 0.5 1 1 10.5 1 0.5 1 0.5+ 0.5 1 0.5 1 0.5example 2.2 consider an input xn and a unit sample response hn given by.2,2212nunhnunxndetermine and plot the output nhnxny22212knukuknhkxnykkk22221knknyusing the geometric sum formul

8、a to evaluate the equation, we have 1122221122112114142121nnknkny21121nunynn21yn graph of yn in example 2.2 2.2 continuous-time lti systems: the convolution integralthe representation of continuous-time signals in terms of impulses: t -02 k x(t)staircase approximation to a continuous-time signal x(t

9、)x(-2 )t -2-)2()2(txx(0) 0 t)()0(tx 2x()t)()(tx-0 x(-)t)()(tx mathematical representation for the rectangular pulseskktkxtx)()()( as , the summation approaches an integral and is the unit impulse function 0)(t)(tdtxtx)()()(compared with the sampling property of the unit impulse: dttttxtx)()()(00give

10、 the as the response of a continuous-time lti system to the input , then the response of the system to pulse is )(th)(t)(kt)(kth thus, the response to is)( ty)( txkkthkxty)()()( as , 0kkthkxty)()()(lim0 in addition, the summing becomes an integral. therefore, dthxty)()()(convolution integral or supe

11、rposition integral :unit impulse response h(t) : the response to the input . （单位冲激响应）（单位冲激响应）)(tconvolution integral symbol:)()()(thtxtya continuous-time lti system is completely characterized by its unit impulse response h(t) .example 2.3consider the convolution of the following two signals, which

12、are depicted in (a):0 t1x(t)t0 2t2th(t)t (a)2th(t)t-2t t 0 t1x()for t 0 )(x)(th.)t ( y0 interval 1. for t 0, there is no overlap between the nonzero portions of and , and consequently, from the definition of the convolution integral of two continuous-time signals,dthxty)()()(otherwisettthx, 00,)()(2

13、021)()(tdttyt2th(t) t-2t 0 t t1x()for 0 t t .thus, for 0 t t .interval 2. for 0 t t,otherwisettthx, 00,)()(2021)()(tttdttyt2th(t) t-2t t t1x()for t t 2t thus, for t t t but t-2t 0, i.e. t t 0, but t-2t t, i.e. 2t t 3t thus, for 2t t 3t.for 2t t t, or equivalently, t 3t, there is no overlap between t

14、he nonzero portions of and hence, x()2th(t) 0 t t-2t t1for t 3ttttttttttttttttttttty3, 032,23212,210,210, 0)(2222summarizing, 2.3 properties of convolution operation 2.3.1 the commutative property(交换律交换律) in discrete time : in continuous time: kkknhkxknxkhnxnhnhnxdthxdtxhtxththtx)()()()()()()()(2.3.

15、2 the distributive property (分配律分配律)(2121nhnxnhnxnhnhnx)()()()()()()(2121thtxthtxththtx in discrete time : in continuous time: y(t)y2(t)y1(t) h2(t)x(t) h1(t)x(t) h1(t)+h2(t)y(t) two equivalent systems: having same impulse responses 2.3.3 the associative property (结合律结合律) in discrete time :)()(2121nh

16、nhnxnhnhnx in continuous time: )()()()()()(2121ththtxththtxxnh1nynh2nxnhn=h1n*h2nynxnhn=h2n*h1nynxnh2nynh1nfour equivalent systems2.3.4 convolving with impulse)()()()()()(00ttxtttxtxttx00nnxnnnxnxnnx2.3.5 differentiation and integration of convolution integral)()()()()(thtxthtxty)()()()()(dhtxthdxdy

17、tttcombining the two properties, we have tttdhxdhtxthdxty)()()()()()()(2.3.6 first difference and accumulation of convolution sum nhnxnhnxny nknknkkhnxnhkxky nknkkhnxnhkxny2.4.1 lti systems with and without memoryfor a discrete-time lti system without memory: nknh for a continuous-time lti system wi

18、thout memory:)()(tkth2.4.2 invertibility of lti systemsthe impulse responses of a system and its inverse system satisfy the following condition: in discrete-time :1nnhnhin continuous-time: )()()(1tthth txththtxtxthtytythtx11since2.4 properties of lti systems2.4.3 causality for lti systems for a caus

19、al discrete-time lti system: 00nfornhfor a causal continuous-time lti system: 00)(tforth2.4.4 stability for lti systemskkhfor a stable discrete-time lti system: for a stable continuous-time lti system: dtth )(absolutely summable absolutely integrable supposeproof: kkhknxnhnxny bnx kkhbny kkh ny kkhk

20、nxthenifthentherefore, the absolutely summable is a sufficient condition to guarantee the stability of a discrete-time lti system. 0, 00,*nhnhnhnhnx kkh kkh 1nx 0y kkhkxy 0to show that the absolutely summable is also a necessary condition for the stability of a discrete-time lti system, letwhere, is

21、 conjugate .nh nh *then, xn is bounded by 1, that ishowever, kkhkh2 ifthen2.5 the unit step response(单位阶跃响应单位阶跃响应) of an lti system the unit step response, sn or s(t), is the output of an lti system when input xn=un or x(t)=u(t). the unit step response of a discrete-time lti system is the running su

22、m of its unit sample response: nkkkhknukhnhnunsthe unit sample response of a discrete-time lti system is the first difference of its unit step response: nsnsnsnh 1the unit step response of a continuous-time lti system is the running integral of its unit impulse response: dhdtuhthtutst)()()()()()(the

23、 unit impulse response of a continuous-time lti system is the first derivative of the unit step response :)()()(tsdttdsth2.6 causal lti systems described by differential and difference equations rtvtvtics)()()( dttdvctic)()( )(1)(1)(tvrctvrcdttdvscc linear constant-coefficient differential equation)

24、(tvs)(tvc: input signal; : output signal. ci (t)vsr + example 2.4linear constant-coefficient difference equation is the mathematical representation of a discrete-time lti system. linear constant-coefficient differential equation is the mathematical representation of a continuous-time lti system. we

25、must specify one or more auxiliary conditions to solve a differential (difference) equation . initial rest(初始静止初始静止): for a causal lti system, if x(t)=0 for tt0, then y(t) must also equal 0 for t0, so let)(typttteaeae 5 . 15 . 0 taking x(t) and for t 0 into the original equation yields 3 athusso the

26、 solution of the differential equation for t0 is 0,3)(5 .0 tebetyttin example 2.4,taking use of the condition initial rest, we obtain 3 btteety 33)(5 . 0)()(3)(5 . 0tueetytt consequently, or for t0example 2.5jack saves money every month. it is known that at the beginning of the nth month the amount

27、he saves into the bank is rmb xn yuan, and the rate of interest is per month. suppose jack wouldnt withdraw his bank deposits in whatever situation, try to give the difference equation relating xn and yn, which is the deposits of jack at the end of the nth month. (before the bank calculates the inte

28、rest)solution:yn is consists of the sum of the following three parts: xn saved at the beginning of the nth month yn-1 interest at the end of the (n-1)th month yn-1 deposit of the (n-1)th monthso the difference equation is nxnyny11 11nynynxnyalsodifference:for sequence xn, its first forward differenc

29、e(一阶前向差分一阶前向差分) is defined as xn = xn+1 xnsecond backward difference as 2x n = xn xn-1 = xn-2xn-1+xn-2 analogously, second forward difference can be constructed as 2xn = xn = xn+1 xn = xn+2-2xn+1+xn its first backward difference (一阶后向差分一阶后向差分) is defined as xn = xn xn-1general nth-order linear const

30、ant-coefficient difference equation:mjjniijnxbinya00nynynyhpfirst resolution:n auxiliary conditions: 01210000 nnynynynysecond resolution: (recursive method(迭代法)mjniijinyajnxbany0101example 2.6solve the difference equation and the initial condition is y0=1.02 athe eigen equation is so the eigenvalue

31、is a = 2nhcny2we can write2 12nnyny21dndnyp let22) 1(22121ndnddndnyp taking into the original equation yields 94,3121ddthus913c9431)2(ncnynynynphthe solution for the given equation isfrom the initial condition of y0=1, we have 43)2(1391nnynconsequently, example 2.7consider the difference equation3 1

32、5 . 0nxnyny .1nnxdetermine the output recursively with the condition of initial rest and 15 . 03 nynxnyrewrite the given difference equation asstarting from initial condition, we can solve for successive values of yn for n1:305 . 0 1 3 1 yxy5 . 031 5 . 0232 yxy2)5 . 0(325 . 0333 yxy3)5 . 0(335 . 043

33、4 yxy1)5 . 0(3 15 . 03 nnynxnyconsidering yn =0 for n0, the solution is 1)5 . 0(31 nunyn)()()()()(nynytytynynynytytytyzszizszihphpwhat are the relationships between yp, yh, yzi and yzs ?question:example 2.8consider a continuous-time lti system described by the following differential equationwith ini

34、tial conditions of and input 30, 20yy tuetxt43 0,3221tecectyttzi3000, 2000yyyyyyzizizizi txtxtytyty 651, 321cc 2121323020ccyccyfrom the definition of the zero-input response, we havein the case of zero input, thus we can writeequivalently, tueetyttzi323consequently, 0,3221teaeathtt tbtath tath 0,0hh

35、next, solve for h(t). ) 1 (65ttththth from the definition of the unit impulse response, we haveand for t0, it becomes 065 ththth h(t) is the solution for the homogeneous equation. thus,and because the system is a causal one, there should be 000hhthe initial conditions used to determine a1 and a2 are

36、?0?0hhbutlet (2) then (3)4, 1ba 4000000 bhhthdtth1010hh4040hh tttabta5taking equations (2) and (3) into equation (1), we havecomparing the coefficients of the corresponding terms on each sidecompute the integral in the interval of 0-, 0+ on both sides of equation (2) to obtainanalogously to equation

37、 (3) to obtain 1000000ahhthdtthconsequently,2, 121aa tueeettt43229623ttttdeedee0240436deeett230423 dhtxthtxtyzs 2121324010aahaahthen fromwe obtainso tueethtt232then0 15 . 0nhnh0 1h30h05 . 0nforcnhnexample 2.9consider a causal lti system described by3 15 . 0nxnyny determine the unit sample response h

38、n. for n0, hn satisfies the difference equationand there should be substituting hn for yn and n for xn in the original difference equation, and let n=0, we obtainits obvious thattaking use of h0, we make out the coefficient in hn:3cso05 . 03nfornhnin fact, for n=0, h0 also satisfy nnh5 . 03thus, we

39、can write nunhn5 . 03you may also try the recursive method to obtain the hn for this system! 2.7 block diagram representations of first-order systems described by differential and difference equations first-order difference equation : 1nbxnayny addition delay multiplication three basic elements in block diagram: adder, multiplier and delayer. (方框图) (加法器)(乘法器) (延时器) basic elements for the block diagra