数字信号处理 第四章 答案
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第二章习题解答1、求下列序列的z 变换()X z ,并标明收敛域,绘出()X z 的零极点图。
(1) 1()()2nu n (2) 1()()4nu n - (3) (0.5)(1)nu n --- (4) (1)n δ+(5) 1()[()(10)]2nu n u n -- (6) ,01na a <<解:(1) 00.5()0.50.5nn n n zZ u n z z ∞-=⎡⎤==⎣⎦-∑,收敛域为0.5z >,零极点图如题1解图(1)。
(2) ()()014()1414n nn n z Z u n z z ∞-=⎡⎤-=-=⎣⎦+∑,收敛域为14z >,零极点图如题1解图(2)。
(3) ()1(0.5)(1)0.50.5nnn n zZ u n z z --=-∞-⎡⎤---=-=⎣⎦+∑,收敛域为0.5z <,零极点图如题1解图(3)。
(4) [](1Z n z δ+=,收敛域为z <∞,零极点图如题1解图(4)。
(5) 由题可知,101010910109(0.5)[()(10)](0.5)()(0.5)(10)0.50.50.50.50.50.5(0.5)n n nZ u n u n Z u n Z u n z z z z z z z z z z z --⎡⎤⎡⎤⎡⎤--=--⎣⎦⎣⎦⎣⎦⋅=-----==--收敛域为0z >,零极点图如题1解图(5)。
(6) 由于()(1)nn n a a u n a u n -=+--那么,111()(1)()()()nn n Z a Z a u n Z a u n z z z a z a z a a z a z a ----⎡⎤⎡⎤⎡⎤=---⎣⎦⎣⎦⎣⎦=----=-- 收敛域为1a z a <<,零极点图如题1解图(6)。
(1) (2) (3)(4) (5) (6)题1解图2、求下列)(z X 的反变换。
1. 请推导出三阶巴特沃思低通滤波器的系统函数,设1/c rad s Ω=。
解:幅度平方函数是:2261()()1A H j Ω=Ω=+Ω令: 22s Ω=- ,则有:61()()1a a H s H s s-=- 各极点满足121[]261,26k j k s ek π-+==所得出的6个 k s 为:15==j es 2321321jes j +-==π12-==πj e s 2321343jes j --==π2321354j es j -==π2321316j es j +==π15==j e s 2321321je s j +-==π12-==πj e s 2321343je s j --==π2321354j es j -==π2321316j es j +==π122))()(()(233210+++=---=s s s k s s s s s s k s H a 1221)(23+++==s s s s H a 代入s=0时, ,可得,故:1=)s (H a 10=k2. 设计一个满足下列指标的模拟Butterworth 低通滤波器,要求通带的截止频率6,p f kHz =,通带最大衰减3,p A dB =,阻带截止频率12,s f kHz =,阻带的最小衰减25s A dB =,求出滤波器的系统函数。
解: 2,2s s p p f f ππΩ=Ω=0.10.1101lg 101N 2lg()s pA A s p⎛⎫- ⎪-⎝⎭≥ΩΩ=4.15取N=5,查表得H(p)为:221()(0.6181)( 1.6181)(1)H p p p p p p =+++++ 因为3,p A dB =所以c p Ω=Ω[]52222()()0.618 1.618cs p c c c c c c H s H p s s s s s =Ω=Ω=⎡⎤⎡⎤+Ω-Ω+Ω-Ω+Ω⎣⎦⎣⎦3. 设计一个模拟切比雪夫低通滤波器,要求通带的截止频率 f p =3kHz ,通带衰减要不大于0.2dB ,阻带截止频率 f s = 12kHz ,阻带衰减不小于 50dB 。
第一章 离散时间信号与系统2.任意序列x(n)与δ(n)线性卷积都等于序列本身x(n),与δ(n-n 0)卷积x(n- n 0),所以(1)结果为h(n) (3)结果h(n-2) (2(4)3 .已知 10,)1()(<<--=-a n u a n h n,通过直接计算卷积和的办法,试确定单位抽样响应为 )(n h 的线性移不变系统的阶跃响应。
4. 判断下列每个序列是否是周期性的,若是周期性的,试确定其周期:)6()( )( )n 313si n()( )()873cos()( )(ππππ-==-=n j e n x c A n x b n A n x a分析:序列为)cos()(0ψω+=n A n x 或)sin()(0ψω+=n A n x 时,不一定是周期序列,nmm m n n y n - - -∞ = - ⋅ = = ≥ ∑ 2 31 2 5 . 0 ) ( 01当 3 4n m nm m n n y n 2 2 5 . 0 ) ( 1⋅ = = - ≤ ∑ -∞ = - 当 aa a n y n a a an y n n h n x n y a n u a n h n u n x m m nnm mn -==->-==-≤=<<--==∑∑--∞=---∞=--1)(11)(1)(*)()(10,)1()()()(:1时当时当解①当=0/2ωπ整数,则周期为0/2ωπ;②;为为互素的整数)则周期、(有理数当 , 2 0Q Q P QP =ωπ ③当=0/2ωπ无理数 ,则)(n x 不是周期序列。
解:(1)0142/3πω=,周期为14 (2)062/13πω=,周期为6 (2)02/12πωπ=,不是周期的 7.(1)[][]12121212()()()()()()[()()]()()()()[()][()]T x n g n x n T ax n bx n g n ax n bx n g n ax n g n bx n aT x n bT x n =+=+=⨯+⨯=+所以是线性的T[x(n-m)]=g(n)x(n-m) y(n-m)=g(n-m)x(n-m) 两者不相等,所以是移变的y(n)=g(n)x(n) y 和x 括号内相等,所以是因果的。
第3章 离散时间信号与系统时域分析3.1画出下列序列的波形(2)1()0.5(1)n x n u n -=- n=0:8; x=(1/2).^n;n1=n+1; stem(n1,x);axis([-2,9,-0.5,3]); ylabel('x(n)'); xlabel('n');(3) ()0.5()nx n u n =-()n=0:8; x=(-1/2).^n;stem(n,x);axis([-2,9,-0.5,3]); ylabel('x(n)'); xlabel('n');3.8 已知1,020,36(),2,780,..n n x n n other n≤≤⎧⎪≤≤⎪=⎨≤≤⎪⎪⎩,14()0..n n h n other n≤≤⎧=⎨⎩,求卷积()()*()y n x n h n =并用Matlab 检查结果。
解:竖式乘法计算线性卷积: 1 1 1 0 0 0 0 2 2)01 2 3 4)14 4 4 0 0 0 0 8 83 3 3 0 0 0 0 6 62 2 2 0 0 0 0 4 41 1 1 0 0 0 02 21 3 6 9 7 4 02 6 10 14 8)1x (n )nx (n )nMatlab 程序:x1=[1 1 1 0 0 0 0 2 2]; n1=0:8; x2=[1 2 3 4]; n2=1:4; n0=n1(1)+n2(1);N=length(n1)+length(n2)-1; n=n0:n0+N-1; x=conv(x1,x2); stem(n,x);ylabel('x(n)=x1(n)*x2(n)');xlabel('n'); 结果:x = 1 3 6 9 7 4 0 2 6 10 14 83.12 (1) 37πx (n )=5sin(n) 解:2214337w πππ==,所以N=14 (2) 326n ππ-x (n )=sin()-sin(n)解:22211213322212,2122612T N w T N w N ππππππ=========,所以(6) 3228n π-x (n )=5sin()-cos(n) 解:22161116313822222()T N w T w x n ππππππ=======,为无理数,所以不是周期序列所以不是周期序列3.20 已知差分方程2()3(1)(2)2()y n y n y n x n --+-=,()4()nx n u n -=,(1)4y -=,(2)10,y -=用Mtalab 编程求系统的完全响应和零状态响应,并画出图形。
数字信号处理第四章习题第四章习题4.1 (a) By expanding the equation()()[]()==?--∞→∞→2200021T T Ft j T xx T xx dt e t x T E lim F P E lim F 00πΓ taking the expected value, and finally taking the limit as ∞→0T , show that the right-hand side converges to )(f xx Γ.(b) Prove that2102211)(1)(∑∑-=---+-==N n fn j fm j N N m xx en x N e m r ππ.4.2 For zero-mean, jointly Gaussian random variables, X 1, X 2, X 3, X 4, itis well known that)()()()()()()(3241423143214321X X E X X E X X E X X E X X E X X E X X X X E ++=. Use this result to derive the mean-square value of ()m r xx and the variance, given by()[][]()()()[]∑∞-∞=+-+-≈n xx xx xx xx m n m n n m N N m r γγγ*22varwhich is defined as[][][]22(()(var m r E m r E m r xx xx xx -=. 4.3 By use of the expression for the fourth joint moment for Gaussianrandom variables, show that(a)()()[]??--++++=2212122121421)(sin )(sin )(sin )(sin 1f f N N f f f f N N f f f P f P E x xx xx ππππσ (b)[]??--+++=2212122121421)(sin )(sin )(sin )(sin )()(cov f f NN f f f f N N f f f P f P x xx xx ππππσ(c)[]??+=242sin 2sin 1)(var f N fN f P x xx ππσ under the condition that the sequence ()n x is a zero-mean white Gaussian noise sequence with variance 2x σ.4.4 Generalize the results in Problem 4.3 to a zero-mean Gaussian noiseprocess with power density spectrum )(f xx Γ, as given by()[]()+Γ=222sin 2sin 1var f N fN f f P xx xx ππ (Hint: Assume that the colored Gaussian noise process is the output of a linear system excited by white Gaussian noise.)4.5 Show that the periodogram values at frequencies,1,1,0,/-==L k L k f k given by (4.1.35), can be computed by passing the sequence through a bank of L IIR filters, where each filter has an impulse resp onse )()(/2n u e n h N nk j k π-= and then computing the magnitude-squared value of the filter outputs at n=N. Note that each filter has a pole on the unit circle at the frequency f k .4.6 The Bartlett method is used to estimate the power spectrum of asignal x(n). We know that the power spectrum consists of a single peak with a 3 dB bandwidth of 0.01 cycle per sample, but we do not know the location of the peak.(a) Assuming that N is large, determine the value of M=N/K so thatthe spectral window is narrower than the peak.(b) Explain why it is not advantageous to increase M beyond thevalue obtained in part (a).4.7 The N-point DFT of a random sequence x(n) is ∑-=-=10/2)()(N n N nk j e n x k X π.Assume that E[x(n)]=0 and E[x(n)x(n+m)]=)(2m w δσ (in other words,x(n) is a white noise process).(a) Determine the variance of X(k).(b) Determine the autocorrelation of X(k).4.8 An AR(2) process is described by the difference equation)()2(81.0)(n n x n x ω+-=, where w(n) is a white noise process withvarian ce 2ωσ.(a) Determine the parameters of the MA(2), MA(4), and MA(8)models that provide a minimum mean-sequare error fit to thedata x(n).(b) Plot the true spectrum and those of the MA (q), q=2,4,8spectra and compare the results. Comment on how well the MA(q) models approximate the AR (2) process.4.9 An MA (2) process is described by the difference equation )2(81.0)()(-+=n n n x ωω, where w(n) is a white noise process withvariance 2ωσ.(a) Determine the parameters of the AR(2), AR(4), and AR(8)models that provide a minimum mean-square error fit to the data x(n).(b) Plot the true spectrum and those of the AR(p), p=2,4,8, andcompare the results. Comment on how well the AR(p) models appoximate the MA (2) process.4.10 The autocorrelation sequence for an AR process x(n) ismxx m ??? ??=41)(γ (a) Determine the difference equation for x(n)(b) Is your answer unique? If not, give any other possiblesolutions.4.11 Suppose that we represent an ARMA(p,q) process as a cascade ofan MA(q) followed by an AR(p) model. The input-output equation for the MA(q) model is ∑=-=qk k k n w b n v 0)()(, where w(n) is a whitenoise process. The input-output equation for the AR(p) model is∑==-+pk k n v k n x a n x 1)()()((a) By computing the autocorrelation of v(n), show thatq m d b b m mq k m w m k k w vv ≤≤==∑-=+0)(022σσγ(b) Show that 1)()(00=+=∑=a k m a m pk vx k vv γγ4.12 Suppose that the AR(2) process in Problem 4.8 is corrupted by anadditive white noise process v(n) with variance 2v σ. Thus, we havey(n)=x(n)+v(n)(a) Determine the difference equation for y(n) and thusdemonstrate that y(n) is an ARMA(2,2) process. Determine the coefficients of the ARMA process.(b) Generalize the result in part (a) to an AR(p) process∑=+--=pk k n w k n x a n x 1)()()( and )()()(n v n x n y +=.4.13 The harmonic decomposition problem considered by Pisarenko maybe expressed as the solution to the equationa a a Γa H w yy H 2σ=The solution for a may be obtained by minimizing the quadratic form a Γa yy H subject to th e constraint that a a H =1. The constraint can be incorporated into the performance index by means of a Lagrange multiplier. Thus the performance index becomes()a a a Γa H yy H 1-+=λζ.By minimizing ζ with respect to a , show that this formulation is equivalent to the Pisarenko eigenvalue problem given in (4.4.9), with the Lagrange multiplier playing the role of the eigenvalue. Thus,show that the minimum of ζ is the minimum eigenvalue 2w σ.4.14 The autocorrelation of a sequence consisting of a sinusoid withrandom phase in noise is)(2cos )(21m m f P m w xx δσπγ+=where 1f is the frequency of the sinusoidal, P its power, and 2w σthe variance of the noise. Suppose that we attempt to fit an AR(2) model to the data.(a) Determine the optimum coefficients of the AR(2) model as afunction of 2w σ and 1f .(b) Determine the reflection coefficients 1K and 2K correspondingto the AR(2) model parameters.(c) Determine the limiting values of the AR(2) parameters and (1K ,2K )as 02→w σ.4.15 This problem involves the use of cross-correlation to detect a signalin noise and estimate the time delay in the signal. A signal x(n) consists of a pulsed sinusoid corrupted by a stationary zero-mean white noise sequence. That is, 10),()()(0-≤≤+-=N n n w n n y n x ,where )(n w is the noise with variance 2w σ and the signal is -≤≤=otherwise M n n A n y ,010,cos )(0ω. The frequency 0ω is known, but the delay 0n , which is a positiveinteger, is unknown, and is to be determined by cross-correlating x(n) with y(n). Assume that 0n M N +>. Let ∑-=-=10)()()(N n xy n x m n y m rdenote the cross-correlation sequence between x(n) and y(n). In the absence of noise, this function exhibits a peak at delay 0n m =. Thus,0n is determined with no error. The presence of noise can lead toerrors in determining the unknown delay.(a) For 0n m =, determine ()[]0n r E xy . Also, determine thevariance ()[]0var n r xy , due to the presence of the noise. In bothcalculations, assume that the double-frequency term averages to zero. That is, 0/2ωπ>>M .(b) Determine the signal-to-noise ratio, defined as[]{}[])(var )(020n r n r E SNR xy xy = (c) What is the effect of the pulse duration M on the SNR?。
数字信号处理(姚天任江太辉)第三版课后习题答案第二章2.1 判断下列序列是否是周期序列。
若是,请确定它的最小周期。
(1)x(n)=Acos(685ππ+n ) (2)x(n)=)8(π-ne j(3)x(n)=Asin(343ππ+n )解 (1)对照正弦型序列的一般公式x(n)=Acos(ϕω+n ),得出=ω85π。
因此5162=ωπ是有理数,所以是周期序列。
最小周期等于N=)5(16516取k k =。
(2)对照复指数序列的一般公式x(n)=exp[ωσj +]n,得出81=ω。
因此πωπ162=是无理数,所以不是周期序列。
(3)对照正弦型序列的一般公式x(n)=Acos(ϕω+n ),又x(n)=Asin(343ππ+n )=Acos(-2π343ππ-n )=Acos(6143-n π),得出=ω43π。
因此382=ωπ是有理数,所以是周期序列。
最小周期等于N=)3(838取k k =2.2在图2.2中,x(n)和h(n)分别是线性非移变系统的输入和单位取样响应。
计算并列的x(n)和h(n)的线性卷积以得到系统的输出y(n),并画出y(n)的图形。
(a)1111(b)(c)111110 0-1-1-1-1-1-1-1-1222222 3333 3444………nnn nnnx(n)x(n)x(n)h(n)h(n)h(n)21u(n)u(n)u(n)a n ===22解 利用线性卷积公式y(n)=∑∞-∞=-k k n h k x )()(按照折叠、移位、相乘、相加、的作图方法,计算y(n)的每一个取样值。
(a) y(0)=x(O)h(0)=1y(l)=x(O)h(1)+x(1)h(O)=3y(n)=x(O)h(n)+x(1)h(n-1)+x(2)h(n-2)=4,n ≥2 (b) x(n)=2δ(n)-δ(n-1)h(n)=-δ(n)+2δ(n-1)+ δ(n-2)y(n)=-2δ(n)+5δ(n-1)= δ(n-3) (c) y(n)=∑∞-∞=--k kn k n u k u a)()(=∑∞-∞=-k kn a=aa n --+111u(n)2.3 计算线性线性卷积 (1) y(n)=u(n)*u(n) (2) y(n)=λnu(n)*u(n)解:(1) y(n)=∑∞-∞=-k k n u k u )()(=∑∞=-0)()(k k n u k u =(n+1),n ≥0即y(n)=(n+1)u(n) (2) y(n)=∑∞-∞=-k k k n u k u )()(λ=∑∞=-0)()(k kk n u k u λ=λλ--+111n ,n ≥0即y(n)=λλ--+111n u(n)2.4 图P2.4所示的是单位取样响应分别为h 1(n)和h 2(n)的两个线性非移变系统的级联,已知x(n)=u(n), h 1(n)=δ(n)-δ(n-4), h 2(n)=a n u(n),|a|<1,求系统的输出y(n).解 ω(n)=x(n)*h 1(n) =∑∞-∞=k k u )([δ(n-k)-δ(n-k-4)]=u(n)-u(n-4)y(n)=ω(n)*h 2(n) =∑∞-∞=k kk u a )([u(n-k)-u(n-k-4)]=∑∞-=3n k ka,n ≥32.5 已知一个线性非移变系统的单位取样响应为h(n)=an-u(-n),0<a<1 用直接计算线性卷积的方法,求系统的单位阶跃响应。
第一章 离散时间信号与系统2.任意序列x(n)与δ(n)线性卷积都等于序列本身x(n),与δ(n-n 0)卷积x(n- n 0),所以(1)结果为h(n) (3)结果h(n-2) (2(4)3 .已知 10,)1()(<<--=-a n u a n h n,通过直接计算卷积和的办法,试确定单位抽样响应为 )(n h 的线性移不变系统的阶跃响应。
4. 判断下列每个序列是否是周期性的,若是周期性的,试确定其周期:)6()( )( )n 313si n()( )()873cos()( )(ππππ-==-=n j e n x c A n x b n A n x a分析:序列为)cos()(0ψω+=n A n x 或)sin()(0ψω+=n A n x 时,不一定是周期序列,nmm m n n y n - - -∞ = - ⋅ = = ≥ ∑ 2 31 2 5 . 0 ) ( 01当 3 4n m nm m n n y n 2 2 5 . 0 ) ( 1⋅ = = - ≤ ∑ -∞ = - 当 aa a n y n a a an y n n h n x n y a n u a n h n u n x m m nnm mn -==->-==-≤=<<--==∑∑--∞=---∞=--1)(11)(1)(*)()(10,)1()()()(:1时当时当解①当=0/2ωπ整数,则周期为0/2ωπ;②;为为互素的整数)则周期、(有理数当 , 2 0Q Q P QP =ωπ ③当=0/2ωπ无理数 ,则)(n x 不是周期序列。
解:(1)0142/3πω=,周期为14 (2)062/13πω=,周期为6 (2)02/12πωπ=,不是周期的 7.(1)[][]12121212()()()()()()[()()]()()()()[()][()]T x n g n x n T ax n bx n g n ax n bx n g n ax n g n bx n aT x n bT x n =+=+=⨯+⨯=+所以是线性的T[x(n-m)]=g(n)x(n-m) y(n-m)=g(n-m)x(n-m) 两者不相等,所以是移变的y(n)=g(n)x(n) y 和x 括号内相等,所以是因果的。
第一章数字信号处理概述简答题:1.在A/D变换之前和D/A变换之后都要让信号通过一个低通滤波器,它们分别起什么作用?答:在A/D变化之前为了限制信号的最高频率,使其满足当采样频率一定时,采样频率应大于等于信号最高频率2倍的条件。
此滤波器亦称为“抗混叠”滤波器.在D/A变换之后为了滤除高频延拓谱,以便把抽样保持的阶梯形输出波平滑化,故又称之为“平滑”滤波器.判断说明题:2.模拟信号也可以与数字信号一样在计算机上进行数字信号处理,自己要增加一道采样的工序就可以了。
( )答:错.需要增加采样和量化两道工序。
3.一个模拟信号处理系统总可以转换成功能相同的数字系统,然后基于数字信号处理理论,对信号进行等效的数字处理.( ) 答:受采样频率、有限字长效应的约束,与模拟信号处理系统完全等效的数字系统未必一定能找到。
因此数字信号处理系统的分析方法是先对抽样信号及系统进行分析,再考虑幅度量化及实现过程中有限字长所造成的影响。
故离散时间信号和系统理论是数字信号处理的理论基础.第二章 离散时间信号与系统分析基础一、连续时间信号取样与取样定理计算题:1.过滤限带的模拟数据时,常采用数字滤波器,如图所示,图中T 表示采样周期(假设T 足够小,足以防止混叠效应),把从)()(t y t x 到的整个系统等效为一个模拟滤波器.(a ) 如果kHz rad n h 101,8)(=π截止于,求整个系统的截止频率. (b)对于kHz T 201=,重复(a )的计算.解 (a )因为当0)(8=≥ωπωj e H rad 时,在数 — 模变换中)(1)(1)(Tj X Tj X Te Y a a j ωω=Ω=所以)(n h 得截止频率8πω=c 对应于模拟信号的角频率c Ω为8π=ΩT c因此 Hz Tf c c 6251612==Ω=π 由于最后一级的低通滤波器的截止频率为Tπ,因此对T8π没有影响,故整个系统的截止频率由)(ωj e H 决定,是625Hz 。