信号与系统 Lecture 4 Chapter1-4
信号与线性系统(第四版)
信号与线性系统（第四版）第一章：信号与系统概述1.1 信号的分类与特性1. 按照幅度是否连续：连续信号和离散信号2. 按照时间是否连续：连续时间信号和离散时间信号3. 按照周期性：周期信号和非周期信号4. 按照能量与功率：能量信号和功率信号连续信号：在任意时间点上都有确定值的信号，如正弦波、矩形波等。
离散信号：在离散时间点上才有确定值的信号，如采样信号、数字信号等。
连续时间信号：时间轴上连续变化的信号，如语音信号、图像信号等。
离散时间信号：时间轴上离散变化的信号，如数字音频、数字图像等。
周期信号：在一定时间间隔内重复出现的信号，如正弦波、方波等。
非周期信号：不具有周期性的信号，如爆炸声、随机信号等。
能量信号：信号的能量有限，如脉冲信号。
功率信号：信号的功率有限，如正弦波、方波等。
1.2 系统的定义与分类1. 按照输入输出关系：线性系统和非线性系统2. 按照时间特性：时变系统和时不变系统3. 按照因果特性：因果系统和非因果系统4. 按照稳定性：稳定系统和不稳定系统线性系统：满足叠加原理和齐次性原理的系统。
即输入信号的线性组合，输出信号也是相应输出的线性组合。
非线性系统：不满足线性系统条件的系统，如饱和非线性、幂次非线性等。
时变系统：系统的特性随时间变化而变化，如放大器的增益随时间衰减。
时不变系统：系统的特性不随时间变化，如理想滤波器、积分器等。
因果系统：当前输出仅取决于当前及过去的输入，与未来的输入无关。
非因果系统：当前输出与未来输入有关，如预测滤波器等。
稳定系统：对于有界输入，输出也有界；或者输入趋于零时，输出也趋于零。
不稳定系统：对于有界输入，输出无界；或者输入趋于零时，输出不趋于零。
第二章：线性时不变系统2.1 线性时不变系统的基本性质2.1.1 叠加性LTI系统对多个输入信号的叠加响应，等于这些输入信号单独作用于系统时的响应之和。
这意味着系统可以处理多个信号而不会相互干扰。
2.1.2 齐次性如果输入信号放大或缩小一个常数倍，那么系统的输出也会相应地放大或缩小同样的倍数。
信号与系统目录(Signal and system directory)
信号与系统目录（Signal and system directory）Chapter 1 signals and systems1.1 INTRODUCTION1.2 signalContinuous signals and discrete signalsTwo. Periodic signals and aperiodic signalsThree, real signal and complex signalFour. Energy signal and power signalThe basic operation of 1.3 signalAddition and multiplicationTwo, inversion and TranslationThree, scale transformation (abscissa expansion)1.4 step function and impulse functionFirst, step function and impulse functionTwo. Definition of generalized function of impulse functionThree. The derivative and integral of the impulse functionFour. Properties of the impulse functionDescription of 1.5 systemFirst, the mathematical model of the systemTwo. The block diagram of the systemCharacteristics and analysis methods of 1.6 systemLinearTwo, time invarianceThree, causalityFour, stabilityOverview of five and LTI system analysis methodsExercise 1.32The second chapter is the time domain analysis of continuous systemsThe response of 2.1LTI continuous systemFirst, the classical solution of differential equationTwo, about 0- and 0+ valuesThree, zero input responseFour, zero state responseFive, full response2.2 impulse response and step responseImpulse responseTwo, step response2.3 convolution integralConvolution integralTwo. The convolution diagramThe properties of 2.4 convolution integralAlgebraic operations of convolutionTwo. Convolution of function and impulse function Three. Differential and integral of convolutionFour. Correlation functionExercise 2.34The third chapter is the time domain analysis of discretesystemsThe response of 3.1LTI discrete systemsDifference and difference equationsTwo. Classical solutions of difference equationsThree, zero input responseFour, zero state response3.2 unit sequence and unit sequence responseUnit sequence and unit step sequenceTwo, unit sequence response and step response3.3 convolution sumConvolution sumTwo. The diagram of convolution sumThree. The nature of convolution sum3.4 deconvolutionExercise 3.27The fourth chapter is Fourier transform and frequency domainanalysis of the systemThe 4.1 signal is decomposed into orthogonal functions Orthogonal function setTwo. The signal is decomposed into orthogonal functions 4.2 Fourier seriesDecomposition of periodic signalsTwo, Fourier series of odd even functionThree. Exponential form of Fu Liye seriesThe spectrum of 4.3 period signalFrequency spectrum of periodic signalTwo, the spectrum of periodic matrix pulseThree. The power of periodic signal4.4 the spectrum of aperiodic signalsFirst, Fu Liye transformTwo. Fourier transform of singular functionsProperties of 4.5 Fourier transformLinearTwo, parityThree, symmetryFour, scale transformationFive, time shift characteristicsSix, frequency shift characteristicsSeven. Convolution theoremEight, time domain differential and integral Nine, frequency domain differential and integral Ten. Correlation theorem4.6 energy spectrum and power spectrumEnergy spectrumTwo. Power spectrumFourier transform of 4.7 periodic signals Fourier transform of sine and cosine functionsTwo. Fourier transform of general periodic functionsThree 、 Fu Liye coefficient and Fu Liye transformFrequency domain analysis of 4.8 LTI systemFrequency responseTwo. Distortionless transmissionThree. The response of ideal low-pass filter4.9 sampling theoremSampling of signalsTwo. Time domain sampling theoremThree. Sampling theorem in frequency domainFourier analysis of 4.10 sequencesDiscrete Fourier series DFS of periodic sequencesTwo. Discrete time Fourier transform of non periodic sequences DTFT4.11 discrete Fu Liye and its propertiesDiscrete Fourier transform (DFT)Two. The properties of discrete Fourier transformExercise 4.60The fifth chapter is the S domain analysis of continuous systems 5.1 Laplasse transformFirst, from Fu Liye transform to Laplasse transformTwo. Convergence domainThree, (Dan Bian) Laplasse transformThe properties of 5.2 Laplasse transformLinearTwo, scale transformationThree, time shift characteristicsFour, complex translation characteristicsFive, time domain differential characteristicsSix, time domain integral characteristicsSeven. Convolution theoremEight, s domain differential and integralNine, initial value theorem and terminal value theorem5.3 Laplasse inverse transformationFirst, look-up table methodTwo, partial fraction expansion method5.4 complex frequency domain analysisFirst, the transformation solution of differential equation Two. System functionThree. The s block diagram of the systemFour 、 s domain model of circuitFive, Laplasse transform and Fu Liye transform5.5 bilateral Laplasse transformExercise 5.50The sixth chapter is the Z domain analysis of discrete systems 6.1 Z transformFirst, transform from Laplasse transform to Z transformTwo, z transformThree. Convergence domainProperties of 6.2 Z transformLinearTwo. Displacement characteristicsThree, Z domain scale transformFour. Convolution theoremFive, Z domain differentiationSix, Z domain integralSeven, K domain inversionEight, part sumNine, initial value theorem and terminal value theorem 6.3 inverse Z transformFirst, power series expansion methodTwo, partial fraction expansion method6.4 Z domain analysisThe Z domain solution of difference equationTwo. System functionThree. The Z block diagram of the systemFour 、 the relation between s domain and Z domainFive. Seeking the frequency response of discrete system by means of DTFTExercise 6.50The seventh chapter system function7.1 system functions and system characteristicsFirst, zeros and poles of the system functionTwo. System function and time domain responseThree. System function and frequency domain responseCausality and stability of 7.2 systemsFirst, the causality of the systemTwo, the stability of the system7.3 information flow graphSignal flow graphTwo, Mason formulaStructure of 7.4 systemFirst, direct implementationTwo. Implementation of cascade and parallel connectionExercise 7.39The eighth chapter is the analysis of the state variables of the system8.1 state variables and state equationsConcepts of state and state variablesTwo. State equation and output equationEstablishment of state equation for 8.2 continuous systemFirst, the equation is directly established by the circuit diagramTwo. The equation of state is established by the input-output equationEstablishment and Simulation of state equations for 8.3discrete systemsFirst, the equation of state is established by the input-output equationTwo. The system simulation is made by the state equationSolution of state equation of 8.4 continuous systemFirst, the Laplasse transform method is used to solve the equation of stateTwo, the system function matrix H (z) and the stability of the systemThree. Solving state equation by time domain methodSolution of state equation for 8.5 discrete systemsFirst, the time domain method is used to solve the state equations of discrete systemsTwo. Solving the state equation of discrete system by Z transformThree, the system function matrix H (z) and the stability of the systemControllability and observability of 8.6 systemsFirst, the linear transformation of state vectorTwo, the controllability and observability of the systemExercise 8.32Appendix a convolution integral tableAppendix two convolution and tableAppendix three Fourier coefficients table of commonly used periodic signalsAppendix four Fourier transform tables of commonly used signalsAppendix five Laplasse inverse exchange tableAppendix six sequence of the Z transform table。
《信号与系统(第四版)》习题详解图文
故f(t)与{c0, c1, …, cN}一一对应。
7
3.3 设
第3章 连续信号与系统的频域分析
试问函数组｛ξ1(t),ξ2(t),ξ3(t),ξ4(t)｝在(0，4)区间上是否 为正交函数组，是否为归一化正交函数组，是否为完备正交函 数组，并用它们的线性组合精确地表示题图 3.2 所示函数f(t)。
题图 3.10
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第3章 连续信号与系统的频域分析 52
第3章 连续信号与系统的频域分析 53
第3章 连续信号与系统的频域分析 54
第3章 连续信号与系统的频域分析 55
第3章 连续信号与系统的频域分析 56
第3章 连续信号与系统的频域分析 57
第3章 连续信号与系统的频域分析
题解图 3.19-1
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第3章 连续信号与系统的频域分析
题图 3.2
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第3章 连续信号与系统的频域分析
解 据ξi(t)的定义式可知ξ1(t)、ξ2(t)、ξ3(t)、ξ4(t)的波形如题 解图3.3-1所示。
题解图 3.3-1
10
不难得到：
第3章 连续信号与系统的频域分析
可知在(0,4)区间ξi(t)为归一化正交函数集,从而有
激励信号为f(t)。试证明系统的响应y(t)=－f(t)。
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证 因为
第3章 连续信号与系统的频域分析
所以
即
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系统函数
第3章 连续信号与系统的频域分析
故
因此
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第3章 连续信号与系统的频域分析
3.23 设f(t)的傅里叶变换为F(jω)，且 试在K≥ωm条件下化简下式：
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第3章 连续信号与系统的频域分析 73
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信号与系统第四章1
0<t<1 1< t < 2
1
2
4.5
思考题4.4 思考题4.4
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4.5 周期信号的频谱与功率谱
一.频谱 频谱
辐频 Ak ~ kω 0 关系
相频 θ k ~ k ω 0 关系
x ( t ) = c 0 + 2 ∑ Ak cos( k ω 0 t + θ k )
k =1
∞
---三角函数形式 三角函数形式
2 2 Ak = Bk + Dk
tgθ k = Dk / Bk
− Dk = − I m {ck }, k > 0
11
复指数——> 正余弦的转换： 正余弦的转换： 复指数
B k = Re {ck }
4.4 波形对称性与傅里叶系数
1.偶对称：x(t)=x(-t) 偶对称： 偶对称
− 2 Dk = 0
4 2 Bk = T0
8
将这两者相加， 式中基波角频率 ω 0 = 2π / T0 。将这两者相加，即 为所求x(t)的傅里叶级数。所以 的傅里叶级数。 为所求 的傅里叶级数
x( t ) = Ev{ x( t )} + Od { x( t )}
4 8 = sinω0 t − 2 cosω0 t + sin3ω0 t − 2 cos3ω0 t π π 3π 9π
第 四 章
连续时间傅立叶变换 连续时间信号的谱分析和 --频分析 时--频分析
1
4.1引言 引言 4.2复指数函数的正交性 复指数函数的正交性 4.3周期信号的表示：连续时间傅里叶级数 周期信号的表示： 周期信号的表示 4.4波形对称性与傅立叶系数 波形对称性与傅立叶系数 4.5周期信号的频谱与功率谱 周期信号的频谱与功率谱 4.6傅里叶级数的收敛性 吉伯斯现象 傅里叶级数的收敛性 4.7非周期信号的表示：连续时间傅里叶变换 非周期信号的表示： 非周期信号的表示 4.8傅里叶级数与傅里叶变换的关系 傅里叶级数与傅里叶变换的关系 4.9连续时间傅里叶变换的性质与应用 连续时间傅里叶变换的性质与应用 4.10卷积定理及其应用 卷积定理及其应用 4.11相关 相关 4.12能量谱密度与功率谱密度 能量谱密度与功率谱密度 4.13信号的时 频分析和小波分析简介 信号的时---频分析和小波分析简介 信号的时
信号与系统张晔版第四章ppt
L[u(t)] est dt est 1
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延时信号
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L[eat ] eat est dt e(as)t 1
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as
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( a)
哈尔滨工业大学图象与信息技术研究所
L f (t t0 )u(t t0 ) F (s)est0
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哈尔滨工业大学图象与信息技术研究所
4.2.6 时域微分特性
推而广之：
L
d n f (t)
dt n
sn F (s)
n 1 r 0
snr 1
f
(r) (0)
式中
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d
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L
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L
d
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s2F(s)
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在应用中，可实现复杂信号的分解。
4.2.2 时域平移特性
信号与系统》课件 第一章 四节
x″(t)+a1x′(t)+a0x(t)=f(t) y(t)=b1x′(t)+b0x(t)
b1 ＋
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图 1.5-6 式(1.5-19)的系统框图
如果已知系统的框图表示， 如果已知系统的框图表示，同样可以采用辅助函数方法写 出系统的输入输出方程。 以图1.5-6所示的框图为例 ， 设右边 所示的框图为例， 出系统的输入输出方程 。 以图 所示的框图为例 积分器的输出为辅助函数x(t)，在两个积分器的输入端得到 积分器的输出为辅助函数 ，在两个积分器的输入端得到x′(t) 和x″(t)， 再在两个加法器的输出端写出两个等效方程，即 ， 再在两个加法器的输出端写出两个等效方程，
＋
y ′′(t )
∫
－a 1 －a 0
y′(t )
∫
y(t)
图 1.5-5 式(1.5-17)的系统框图 的系统框图
例 1.5-7 某连续系统的输入输出方程为
y″(t)+a1y′(t)+a0y(t)=b1f′(t)+b0f(t)
试画出该系统的框图表示。 试画出该系统的框图表示。
(1.5-19)
则系统具有叠加性。 式中， 则系统具有叠加性 。 式中 ， {f1(·)， f2(·)}表示两个激励 ， 表示两个激励 f1(·)、 f2(·)共同作用于系统。 共同作用于系统。 、 共同作用于系统
如果系统同时具有齐次性和叠加性， 就称系统具有线性特性。 如果系统同时具有齐次性和叠加性， 就称系统具有线性特性。 或表述为
4、因果性： 、因果性： 系统在t0时刻的响应只与 时刻的响应只与t=t0和t<t0时刻的输 系统在 时刻的响应只与 和 时刻的输 入有关。即激励是响应产生的原因， 入有关。即激励是响应产生的原因，响应 是激励作用的结果。 是激励作用的结果。 如：r1(t)=e1(t-1) 因果 r2(t)=e2(t+1) 非因果 是一个非因果时变线性系统。 例：r(t)=e(-t)是一个非因果时变线性系统。 是一个非因果时变线性系统 e(t-t0) e(-t-t0)≠r(t-t0)=e(-t+t0)
信号与系统4教学ppt
上两式称为双边拉普拉斯变换对，可以表示为
f (t) F (s)
拉氏变换扩大了信号的变换范围。
变换域的内在联系
时域函数 f (t)傅氏变换 频域函数 F ()
时域函数 f (t)拉氏变换 复频域函数 F (s)
4.1.2 单边拉普拉斯变换
考虑到：1. 实际信号都是有始信号，即 t 0时，f (t) 0
作业
连续信号与系统的复频域分析概述
傅里叶变换（频域）分析法
– 在信号分析和处理方面十分有效：分析谐波成分、系统的频 率响应、波形失真、取样、滤波等
– 要求信号满足狄里赫勒条件 – 只能求零状态响应 – 反变换有时不太容易
拉普拉斯变换（复频域）分析法
– 在连续、线性、时不变系统的分析方面十分有效 – 可以看作广义的傅里叶变换 – 变换式简单 – 扩大了变换的范围 – 为分析系统响应提供了规范的方法
但反变换的积分限并不改变。
以后只讨论单边拉氏变换：
（1）f (t) 和 f (t) (t) 的拉氏正变换 F(s) 是一样的。
（2）反之，当已知 F(s) ，求原函数时，也无法得 到 t < 0 时的 f (t) 表达式。
例如，常数 1 和 (t) 的（单边）拉普拉斯变换是一
样的。
单边拉氏变换的优点：
0
可见： L[tn (t)] n L[tn1 (t)]
s
依次类推：
L[tn (t)]
n s
n
1 s
n
s
22 s
1 s
1 s
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特别是 n=1 时，有
L[t (t)]
1 s2
拉普拉斯变换与傅里叶变换的关系
1. 0 0 ：只有拉氏变换而无傅氏变换
信号与系统奥本海姆习题答案
Chapter 1 Answers1.6 (a).NoBecause when t<0, )(1t x =0.(b).NoBecause only if n=0, ][2n x has valuable.(c).Yes Because ∑∞-∞=--+--+=+k k m n k m n m n x ]}414[]44[{]4[δδ ∑∞-∞=------=k m k n m k n )]}(41[)](4[{δδ ∑∞-∞=----=k k n k n ]}41[]4[{δδ N=4.1.9 (a). T=π/5Because 0w =10, T=2π/10=π/5.(b). Not periodic.Because jt t e e t x --=)(2, while t e -is not periodic, )(2t x is not periodic.(c). N=2Because 0w =7π, N=(2π/0w )*m, and m=7.(d). N=10Because n j j e e n x )5/3(10/343)(ππ=, that is 0w =3π/5, N=(2π/0w )*m, and m=3.(e). Not periodic. Because 0w =3/5, N=(2π/0w )*m=10πm/3 , it ’s not a rational number.1.14 A1=3, t1=0, A2=-3, t2=1 or -1dtt dx )( isSolution: x(t) isBecause ∑∞-∞=-=k k t t g )2()(δ, dt t dx )(=3g(t)-3g(t-1) or dtt dx )(=3g(t)-3g(t+1) 1.15. (a). y[n]=2x[n-2]+5x[n-3]+2x[n-4]Solution:]3[21]2[][222-+-=n x n x n y ]3[21]2[11-+-=n y n y ]}4[4]3[2{21]}3[4]2[2{1111-+-+-+-=n x n x n x n x ]4[2]3[5]2[2111-+-+-=n x n x n xThen, ]4[2]3[5]2[2][-+-+-=n x n x n x n y(b).No. For it ’s linearity.the relationship between ][1n y and ][2n x is the same in-out relationship with (a). you can have a try.1.16. (a). No.For example, when n=0, y[0]=x[0]x[-2]. So the system is memory. (b). y[n]=0.When the input is ][n A δ,then, ]2[][][2-=n n A n y δδ, so y[n]=0. (c). No.For example, when x[n]=0, y[n]=0; when x[n]=][n A δ, y[n]=0. So the system is not invertible.1.17. (a). No.For example, )0()(x y =-π. So it ’s not causal.(b). Yes.Because : ))(sin()(11t x t y = , ))(sin()(22t x t y =))(sin())(sin()()(2121t bx t ax t by t ay +=+1.21. Solution:We have known:(a).(b).(c).(d).1.22. Solution:We have known:(a).(b).(e).(g)1.23. Solution:For )]()([21)}({t x t x t x E v -+= )]()([21)}({t x t x t x O d --= then,(a).(b).(c).1.24.For: ])[][(21]}[{n x n x n x E v -+= ])[][(21]}[{n x n x n x O d --=then,(a).(b).1.25. (a). Periodic. T=π/2.Solution: T=2π/4=π/2.(b). Periodic. T=2.Solution: T=2π/π=2.(d). Periodic. T=0.5. Solution: )}()4{cos()(t u t E t x v π=)}())(4cos()()4{cos(21t u t t u t --+=ππ )}()(){4cos(21t u t u t -+=π )4cos(21t π= So, T=2π/4π=0.51.26. (a). Periodic. N=7Solution: N=m *7/62ππ=7, m=3.(b). Aperriodic.Solution: N=ππm m 16*8/12=, it ’s not rational number.(e). Periodic. N=16 Solution as follow:)62cos(2)8sin()4cos(2][ππππ+-+=n n n n x in this equation,)4cos(2n π, it ’s period is N=2π*m/(π/4)=8, m=1.)8sin(n π, it ’s period is N=2π*m/(π/8)=16, m=1.)62cos(2ππ+-n , it ’s period is N=2π*m/(π/2)=4, m=1. So, the fundamental period of ][n x is N=(8,16,4)=16.1.31. SolutionBecause )()1()(),2()()(113112t x t x t x t x t x t x ++=--=. According to LTI property ,)()1()(),2()()(113112t y t y t y t y t y t y ++=--=Extra problems:Sketch ⎰∞-=t dt t x t y )()(. 1. SupposeSolution:2. SupposeSketch:(1). )]1(2)1()3()[(--+++t t t t g δδδ(2). ∑∞-∞=-k k t t g )2()(δ(2).Chapter 22.1 Solution:Because x[n]=(1 2 0 –1)0, h[n]=(2 0 2)1-, then(a).So, ]4[2]2[2]1[2][4]1[2][1---+-+++=n n n n n n y δδδδδ (b). according to the property of convolutioin:]2[][12+=n y n y(c). ]2[][13+=n y n y][*][][n h n x n y =][][k n h k x k -=∑∞-∞= ∑∞-∞=-+--=k k k n u k u ]2[]2[)21(2 ][211)21()21(][)21(12)2(0222n u n u n n k k --==+-++=-∑ ][])21(1[21n u n +-= the figure of the y[n] is:2.5 Solution:We have known: ⎩⎨⎧≤≤=elsewhere n n x ....090....1][，，, ⎩⎨⎧≤≤=elsewhere N n n h ....00....1][，，,(9≤N ) Then, ]10[][][--=n u n u n x , ]1[][][---=N n u n u n h∑∞-∞=-==k k n u k h n h n x n y ][][][*][][ ∑∞-∞=-------=k k n u k n u N k u k u ])10[][])(1[][(So, y[4] ∑∞-∞=-------=k k u k u N k u k u ])6[]4[])(1[][( ⎪⎪⎩⎪⎪⎨⎧≥≤=∑∑==4,...14, (140)0N N k Nk =5, then 4≥N And y[14] ∑∞-∞=------=k k u k u N k u k u ])4[]14[])(1[][(⎪⎪⎩⎪⎪⎨⎧≥≤=∑∑==14,...114, (1145)5N N k Nk =0, then 5<N ∴4=N2.7 Solution:[][][2]k y n x k g n k ∞=-∞=-∑（a ）[][1]x n n δ=-,[][][2][1][2][2]k k y n x k g n k k g n k g n δ∞∞=-∞=-∞=-=--=-∑∑(b) [][2]x n n δ=-,[][][2][2][2][4]k k y n x k g n k k g n k g n δ∞∞=-∞=-∞=-=--=-∑∑ (c) S is not LTI system..(d) [][]x n u n =,0[][][2][][2][2]k k k y n x k g n k u k g n k g n k ∞∞∞=-∞=-∞==-=-=-∑∑∑2.8 Solution: )]1(2)2([*)()(*)()(+++==t t t x t h t x t y δδ )1(2)2(+++=t x t xThen,That is, ⎪⎪⎪⎩⎪⎪⎪⎨⎧≤<-≤<-+-=-<<-+=others t t t t t t t t y ,........010,....2201,.....41..,.........412,.....3)(2.10 Solution:(a). We know:Then,)()()(αδδ--='t t t h)]()([*)()(*)()(αδδ--='='t t t x t h t x t y )()(α--=t x t xthat is,So, ⎪⎪⎩⎪⎪⎨⎧+≤≤-+≤≤≤≤=others t t t t t t y ,.....011,.....11,....0,.....)(ααααα(b). From the figure of )(t y ', only if 1=α, )(t y ' would contain merely therediscontinuities.2.11 Solution:(a). )(*)]5()3([)(*)()(3t u et u t u t h t x t y t----==⎰⎰∞∞---∞∞--------=ττττττττd t u e u d t u eu t t )()5()()3()(3)(3⎰⎰-------=tt t t d e t u d et u 5)(33)(3)5()3(ττττ⎪⎪⎪⎪⎩⎪⎪⎪⎪⎨⎧≥+-=-<≤-=<=---------⎰⎰⎰5,.......353,.....313.........,.........0315395)(33)(3393)(3t e e d e d e t e d e t tt t t t t t t t ττττττ(b). )(*)]5()3([)(*)/)(()(3t u e t t t h dt t dx t g t ----==δδ)5()3()5(3)3(3---=----t u e t u e t t(c). It ’s obvious that dt t dy t g /)()(=.2.12 Solution∑∑∞-∞=-∞-∞=--=-=k tk tk t t u ek t t u e t y )]3(*)([)3(*)()(δδ∑∞-∞=---=k k t k t u e)3()3(Considering for 30<≤t ,we can obtain33311])3([)(---∞=-∞-∞=--==-=∑∑ee e ek t u e e t y tk k tk kt. (Because k must be negetive ,1)3(=-k t u for 30<≤t ).2.19 Solution:(a). We have known:][]1[21][n x n w n w +-=(1) ][]1[][n w n y n y βα+-=(2)from (1), 21)(1-=E EE Hfrom (2), αβ-=E EE H )(2then, 212212)21(1)21)(()()()(--++-=--==E E E E E E H E H E H ααβαβ∴][]2[2]1[)21(][n x n y n y n y βαα=-+-+-but, ][]1[43]2[81][n x n y n y n y +-+--=∴⎪⎩⎪⎨⎧=⎪⎭⎫ ⎝⎛=+=143)21(:....812βααor ∴⎪⎩⎪⎨⎧==141βα(b). from (a), we know )21)(41()()()(221--==E E E E H E H E H21241-+--=E EE E ∴][)41()21(2][n u n h n n ⎥⎦⎤⎢⎣⎡-=2.20 (a). 1⎰⎰∞∞-∞∞-===1)0cos()cos()()cos()(0dt t t dt t t u δ(b). 0dt t t )3()2sin(5+⎰δπ has value only on 3-=t , but ]5,0[3∉-∴dt t t )3()2sin(5+⎰δπ=0(c). 0⎰⎰---=-641551)2cos()()2cos()1(dt t t u d u πτπττ⎰-'-=64)2cos()(dt t t πδ0|)2(s co ='=t t π 0|)2sin(20=-==t t ππ∑∞-∞=-==k t h kT t t h t x t y )(*)()(*)()(δ∑∞-∞=-=k kT t h )(∴2.27Solution()y A y t dt ∞-∞=⎰,()xA x t dt ∞-∞=⎰,()hA h t dt ∞-∞=⎰.()()*()()()y t x t h t x x t d τττ∞-∞==-⎰()()()()()()()()()(){()}y x hA y t dt x x t d dtx x t dtd x x t dtd x x d d x d x d A A ττττττττττξξτττξξ∞∞∞-∞-∞-∞∞∞∞∞-∞-∞-∞-∞∞∞∞∞-∞-∞-∞-∞==-=-=-===⎰⎰⎰⎰⎰⎰⎰⎰⎰⎰⎰(a) ()()(2)tt y t e x d τττ---∞=-⎰,Let ()()x t t δ=,then ()()y t h t =. So , 2()(2)(2)()(2)()(2)t t t t t h t ed e d e u t τξδττδξξ---------∞-∞=-==-⎰⎰(b) (2)()()*()[(1)(2)]*(2)t y t x t h t u t u t e u t --==+---(2)(2)(1)(2)(2)(2)t t u eu t d u e u t d ττττττττ∞∞-------∞-∞=+------⎰⎰22(2)(2)12(1)(4)t t t t u t e d u t e d ττττ---------=---⎰⎰(2)2(2)212(1)[]|(4)[]|t t t t u t e e u t ee ττ-------=--- (1)(4)[1](1)[1](4)t t e u t e u t ----=-----2.46 SolutionBecause)]1([2)1(]2[)(33-+-=--t u dtde t u e dt d t x dt d t t )1(2)(3)1(2)(333-+-=-+-=--t e t x t e t x t δδ.From LTI property ,we know)1(2)(3)(3-+-→-t h e t y t x dtdwhere )(t h is the impulse response of the system. So ,following equation can be derived.)()1(223t u e t h e t --=-Finally, )1(21)()1(23+=+-t u e e t h t 2.47 SoliutionAccording to the property of the linear time-invariant system: (a). )(2)(*)(2)(*)()(000t y t h t x t h t x t y ===(b). )(*)]2()([)(*)()(00t h t x t x t h t x t y --==)(*)2()(*)(0000t h t x t h t x --=012y(t)t4)2()(00--=t y t y(c). )1()1(*)(*)2()1(*)2()(*)()(00000-=+-=+-==t y t t h t x t h t x t h t x t y δ(d). The condition is not enough.(e). )(*)()(*)()(00t h t x t h t x t y --==τττd t h x )()(00+--=⎰∞∞-)()()(000t y dm m t h m x -=--=⎰∞∞-(f). )()]([)](*)([)(*)()(*)()(000000t y t y t h t x t h t x t h t x t y "=''='--'=-'-'==Extra problems:1. Solute h(t), h[n](1). )()(6)(5)(22t x t y t y dt dt y dtd =++ (2). ]1[][2]1[2]2[+=++++n x n y n y n y Solution:(1). Because 3121)3)(2(1651)(2+-++=++=++=P P P P P P P Hso )()()()3121()(32t u e e t P P t h t t ---=+-++=δ (2). Because )1)(1(1)1(22)(22i E i E EE E E E E E H -+++=++=++=iE Eii E E i -+-+++=1212 so []][)1()1(2][1212][n u i i i k i E E i i E E i n h n n +----=⎪⎪⎪⎪⎭⎫⎝⎛-+-+++=δChapter 33.1 Solution:Fundamental period 8T =.02/8/4ωππ==00000000033113333()224434cos()8sin()44j kt j t j t j t j tk k j t j t j t j tx t a e a e a e a e a e e e je je t t ωωωωωωωωωππ∞----=-∞--==+++=++-=-∑3.2 Solution:for, 10=a , 4/2πj ea --= , 4/2πj ea = , 3/42πj ea --=, 3/42πj ea =n N jk k N k e a n x )/2(][π∑>=<=n j n j n j n j e a e a e a e a a )5/8(4)5/8(4)5/4(2)5/4(20ππππ----++++=n j j n j j n j j n j j e e e e e e e e )5/8(3/)5/8(3/)5/4(4/)5/4(4/221ππππππππ----++++= )358cos(4)454cos(21ππππ++++=n n)6558sin(4)4354sin(21ππππ++++=n n3.3 Solution: for the period of )32cos(t πis 3=T , the period of )35sin(t πis 6=Tso the period of )(t x is 6 , i.e. 3/6/20ππ==w)35sin(4)32cos(2)(t t t x ππ++= )5sin(4)2cos(21200t w t w ++=)(2)(21200005522t w j t w j t w j t w j e e j e e ----++=then, 20=a , 2122==-a a , j a 25=-, j a 25-=3.5 Solution:(1). Because )1()1()(112-+-=t x t x t x , then )(2t x has the same period as )(1t x ,that is 21T T T ==, 12w w =(2). 212111()((1)(1))jkw t jkw tk T T b x t e dt x t x t e dt T--==-+-⎰⎰111111(1)(1)jkw tjkw t T Tx t e dt x t e dt T T --=-+-⎰⎰ 111)(jkw k k jkw k jkw k e a a e a e a -----+=+=3.8 Solution:kt jw k k e a t x 0)(∑∞-∞==while:)(t x is real and odd, then 00=a , k k a a --=2=T , then ππ==2/20wand0=k a for 1>kso kt jw k k e a t x 0)(∑∞-∞==t jw t jw e a e a a 00110++=--)sin(2)(11t a e e a t j t j πππ=-=-for12)(2121212120220==++=-⎰a a a a dt t x∴2/21±=a ∴)sin(2)(t t x π±=3.13 Solution:Fundamental period 8T =.02/8/4ωππ==kt jw k k e a t x 0)(∑∞-∞==∴t jkw k k e jkw H a t y 0)()(0∑∞-∞==0004, 0sin(4)()0, 0k k H jk k k ωωω=⎧==⎨≠⎩ ∴000()()4jkw t k k y t a H jkw e a ∞=-∞==∑Because 48004111()1(1)088T a x t dt dt dt T ==+-=⎰⎰⎰So ()0y t =.kt jw k k e a t x 0)(∑∞-∞==∴t jkw k k e jkw H a t y 0)()(0∑∞-∞== ∴dt e jkw H t y Ta t jkw Tk 0)()(10-⎰=for⎪⎩⎪⎨⎧>≤=100, (0100),.......1)(w w jw H ∴if 0=k a , it needs 1000>kwthat is 12100,........1006/2>>k kππand k is integer, so 8>K3.22 Solution:021)(1110===⎰⎰-tdt dt t x Ta Tdt te dt te dt e t x T a t jk t jk t jkw T k ππ-----⎰⎰⎰===1122112121)(10t jk tde jk ππ--⎰-=1121⎥⎥⎦⎤⎢⎢⎣⎡---=----111121ππππjk e te jk t jk tjk ⎥⎦⎤⎢⎣⎡---+-=--ππππππjk e e e e jk jk jk jk jk )()(21⎥⎦⎤⎢⎣⎡-+-=ππππjk k k jk )sin(2)cos(221[]πππππk jk k j k jk k)1()cos()cos(221-==-=0............≠k404402()()1184416tj tj t t j tt j t H j h t edt ee dte edt e e dtj j ωωωωωωωω∞∞----∞-∞∞----∞===+=+=-++⎰⎰⎰⎰A periodic continous-signal has Fourier Series:. 0()j kt k k x t a e ω∞=-∞=∑T is the fundamental period of ()x t .02/T ωπ=The output of LTI system with inputed ()x t is 00()()jk t k k y t a H jk e ωω∞=-∞=∑Its coefficients of Fourier Series: 0()k k b a H jk ω= (a)()()n x t t n δ∞=-∞=-∑.T=1, 02ωπ=11k a T==. 01/221/21()()1jkw t jk tk T a x t e dt t e dt Tπδ---===⎰⎰ （Note ：If ()()n x t t nT δ∞=-∞=-∑,1k a T=） So 2282(2)16(2)4()k k b a H jk k k πππ===++ (b)()(1)()n n x t t n δ∞=-∞=--∑ .T=2, 0ωπ=,11k a T== 01/23/21/21/2111()()(1)(1)221[1(1)]2jkw t jk tjk t k T k a x t e dt t e dt t e dtT ππδδ----==+--=--⎰⎰⎰So 24[1(1)]()16()k k k b a H jk k ππ--==+, (c) T=1,02ωπ=01/421/4sin()12()jk t jk tk T k a x t e dt e dt Tk ωπππ---===⎰⎰28sin()2()[16(2)]k k k b a H jk k k ππππ==+ 3.35 Solution: T= /7π,02/14T ωπ==.kt jw k k e a t x 0)(∑∞-∞==∴t jkw k k e jkw H a t y 0)()(0∑∞-∞==∴0()k k b a H jkw =for⎩⎨⎧≥=otherwise w jw H ,.......0250,.......1)(,01,. (17)()0,.......k H jkw otherwise ⎧≥⎪=⎨⎪⎩that is 0250250, (14)k k ω<<, and k is integer, so 18....17k or k <≤. Let ()()y t x t =,k k b a =, it needs 0=k a ,for 18....17k or k <≤.3.37 Solution:11()[]()212()21312411511cos 224nj j nj n n n n j nn j nn n j j j H e h n ee ee e e e ωωωωωωωωω∞∞--=-∞=-∞-∞--=-∞=-===+=+=---∑∑∑∑A periodic sequence has Fourier Series:2()[]jk n Nk k N x n a eπ=<>=∑.N is the fundamental period of []x n .The output of LTI system with inputed []x n is 22()[]()jk jk n NNk k N y n a H eeππ=<>=∑.Its coefficients of Fourier Series: 2()jk Nk k b a H eπ=(a)[][4]k x n n k δ∞=-∞=-∑.N=4, 14k a =.So 2314()524cos()44j k Nk k b a H e k ππ==-3165cos()42k b k π=-3.40 Solution: According to the property of fourier series: (a). )2cos(2)cos(20000000t Tka t kw a e a ea a k k t jkw k t jkw k k π==+='- (b). Because 2)()()}({t x t x t x E v -+=}{2k v k k k a E a a a =+='-(c). Because 2)(*)()}({t x t x t x R e +=2*kk k a a a -+='(d). k k k a Tjka jkw a 220)2()(π=='(e). first, the period of )13(-t x is 3T T ='then 3)(1)13(131213120dme m x T dt e t x T a m T jk T t T jk T k +'--'-'-'⎰⎰'=-'='ππTjkk m T jk T T jk T jk m T jk T ea dm e m x T e dm e e m x T πππππ221122211)(1)(1---------=⎥⎦⎤⎢⎣⎡==⎰⎰3.43 (a) Proof:（i ）Because ()x t is odd harmonic ,(2/)()jk T t k k x t a e π∞=-∞=∑,where 0k a = for everynon-zero even k.(2/)()2(2/)(2/)()2T jk T t k k jk jk T tk k jk T tk k T x t a ea e e a e ππππ∞+=-∞∞=-∞∞=-∞+===-∑∑∑It is noticed that k is odd integers or k=0.That means()()2Tx t x t =-+（ii ）Because of ()()2Tx t x t =-+,we get the coefficients of Fourier Series222/200/222(/2)/2/20022/2/200111()()()11()(/2)11()()(1)jk t jk t jk t T T T T T T k T jk t jk t T T T T Tjk t jk t T T k TT a x t e dt x t e dt x t e dtT T T x t e dt x t T e dt T T x t e dt x t e dt T T πππππππ-----+--==+=++=--⎰⎰⎰⎰⎰⎰⎰ 2/21[1(1)]()jk t T kT x t e dt T π-=--⎰It is obvious that 0k a = for every non-zero even k. So ()x t is odd harmonic ,(b)Extra problems:∑∞-∞=-=k kT t t x )()(δ, π=T(1). Consider )(t y , when )(jw H ist(2). Consider )(t y , when )(jw H isSolution:∑∞-∞=-=k kT t t x )()(δ↔π11=T , 220==Tw π(1).kt j k k tjkw k k e k j H a ejkw H a t y 20)2(1)()(0∑∑∞-∞=∞-∞===ππ2=(for k can only has value 0)(2).kt j k k tjkw k k e k j H a e jkw H a t y 20)2(1)()(0∑∑∞-∞=∞-∞===πππte e t j t j 2cos 2)(122=+=- (for k can only has value –1 and 1)。
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Inverse System
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Signals and Systems
LP C
x1[n]= [n]+2[n-1]+3[n-2], y1[n]= 2[n]+ [n-1] TI P x2[n]= 2[n-2]+3[n-3], y2[n]= -[n]+2[n-1] x3[n]= [n-4], y3[n]= [n-3]+2[n-4] + [n-5] If F is Time-Invariance, can F be causal? Can F be linear? x4[n]=x3[n+4]=[n] x1[n]=x4[n], n≤0 but y1[-1] ≠ y4[-1] x1[n]-x2[n+1] = [n] y1[n]-y2[n+1]=[n]+[n+1] x4[n]=x1[n]-x2[n+1], but y4[n] ≠ y1[n]- y2[n+1] y4[n]=y3[n+4]=[n+1]+2[n]+[n-1] So, F can not be causal.
y n
x
2
n ——memoryless（无记忆）
2
——identity system ,memoryless
k
t
n
x k
y n x n 1
Systems with memory
⑤ integrate y t
x d
y t tx t
Signals and Systems
y t x
2
t
y n Re x n
Example 1.20
y n 2 x n 3
15
Chapter 1
x t
x n
Signals and Systems 线性 系统
3
Chapter 1
Signals and Systems
Continuous-Time System
a
k 0
N
d y t
k k
dt
k
b
k 0
M
d x t
k k
dt
k
N-order Linear Constant-coefficient Differential Equation
Signals and Systems
Lecture 4
Systems
1
Chapter 1
Signals and Systems
§ 1.5 Continuous-time and discrete-time systems
Be constituted by some units System Contact with some rule
Chapter 1 § 1.6.4 Stability （稳定性）
Signals and Systems
BIBO Stability (Bounded Input Bounded Output Stability) Stable System
x t M
y t B
①
②
y t tx t —— not stable
Example 1.8
RC d vc (t ) dt v c (t ) v s (t )
v s t
i t
C
v c t
Example 1.9
d v (t ) dt
f t
v t
m
m
v (t ) f (t )
v t
Linear constant-coefficient differential equation
Chapter 1
Signals and Systems
x t
§ 1.6.6 Linearity （线性）
1. Initial State （初始状态） 2. Linearity ① Additivity ② Scaling 3. Linear System
m
v t
0
14
Chapter 1 Example 1.17 Example 1.18 Example 1.19
6
Chapter 1 § 1.6.2 Invertibility and Inverse Systems 可逆系统与可逆性
Signals and Systems
可逆系统: 不同的输入导致不同的输出（一一对应）。
x t
x n
y t
w t x t
System
y n
Signals and Systems
output
System 1
System 2
Parallel interconnection (并联)
input
System 1 System 2
output
Feedback interconnection (反馈)
input System 1
output
System 2
5
Chapter 1 § 1.6 Basic System Properties § 1.6.1 Systems with and without Memory 有记忆、无记忆系统
Signals and Systems
无记忆系统: 在任意时刻(t)的输出仅仅与同时刻(t)的输入有关。 ① y n 2 x n ② y t x t ③ summer ④ delay
C
________
C LP LP
So, F can not be linear.
10
Chapter 1 因果系统 ① ②
Signals and Systems
不可预测系统
物理上可实现
y t x t 1
y t
t
x d
Causal systems
的输入。 (与该时刻以后的输入无关）
x1
x2
F
F
t R,
y1
y2
If x1(t)=x2(t), t≤T,
F is causal if y1(t)=y2(t), t≤T.
8
Chapter 1
x F y
Signals and Systems
x[n]= [n], y[n]= [n+1]-[n]+2[n-1] If F is linear, is F causal? If F is Time-Invariance, is F causal? x1[n]= [n] x2[n]=0 y1[n]= [n+1]- [n]+ 2 [n-1] y2[n]= 0 x1[n]=x2[n], n≤-1 So, F is not causal. but y1[-1] ≠ y2[-1] If a system F is linear, and its output “proceedes” the input, LP C then F cannnot be causal. x1[n]= [n] x2[n]= [n-1] y1[n]= [n+1]- [n]+ 2 [n-1] y2[n]= [n]- [n-1]+ 2 [n-2] x1[n]=x2[n], n≤-1 So, F is not causal. but y1[-1] ≠ y2[-1] If a system F is time-invariance, and its output “proceedes” 9 TI P C the input, then F cannnot be causal.
The system's function
System analysis Research system （系统分析）
System synthesizes （系统综合）
2
Chapter 1 § 1.5.1 Simple Examples of Systems
Signals and Systems
R
w n x n
y n 0
y t x
t ——noninvertible systems 不可逆系统
7
Chapter 1 § 1.6.3 Causality （因果性）
Signals and Systems
因果系统: 在任意时刻(t)的输出只取决于同时刻(t)或以前(<t)
y f t
y f n
y t
y n
y x t
y x n
线性系统的三个特性 ① 微分特性 ② 积分特性 ③ 频率保持性：
f t y t
df t dt
dy t dt
t
f d
t
y d
信号通过线性系统不会产生新的频率分量
y t e
x t
—— stable
12
Chapter 1 § 1.6.5 Time Invariance （时不变性）
Signals and Systems
时不变系统: 系统参数不随时间改变(恒参系统),系统的输出波