数字信号处理第三章习题
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3-1 画出)5.01)(25.01()264.524.14)(379.02()(211211------+--+--=z zz z z z z H 级联型网络结构。
解:243-2 画出112112(23)(465)()(17)(18)z z z H z z z z --------+=--+级联型网络结构。
解:()x n ()y n 243-3 已知某三阶数字滤波器的系统函数为1211252333()111(1)(1)322z z H z z z z -----++=-++,试画出其并联型网络结构。
解:将系统函数()H z 表达为实系数一阶,二阶子系统之和,即:()H z 11122111111322z z z z ----+=+-++ 由上式可以画出并联型结构如题3-3图所示:)题3-3图3-4 已知一FIR 滤波器的系统函数为121()(10.70.5)(12)H z zz z ---=-++,画出该FIR滤波器的线性相位结构。
解: 因为121123()(10.70.5)(12)1 1.30.9H z zz z z z z ------=-++=+-+,所以由第二类线性相位结构画出该滤波器的线性相位结构,如题3-4图所示:()x n 1-1-1z -题3-4图3-5 已知一个FIR 系统的转移函数为:12345()1 1.25 2.75 2.75 1.23H z z z z z z -----=+--++求用级联形式实现的结构流图并用MATLAB 画出其零点分布及其频率响应曲线。
解: 由转移函数可知,6=N ,且)(n h 偶对称,故为线性相位系统,共有5个零点,为5阶系统,因而必存在一个一阶系统,即1±=z 为系统的零点。
而最高阶5-z 的系数为+1,所以1-=z 为其零点。
)(z H 中包含11-+z 项。
所以:11()()(1)H z H z z -=+。
1()H z 为一四阶子系统,设12341()1H z bzcz bz z ----=++++,代入等式,两边相等求得12341()10.2530.25H z zz z z ----=+-++,得出系统全部零点,如图3-5(b )所示。
第三章离散傅里叶变换及其快速算法习题答案参考3.1 图P3.1所示的序列(xn 是周期为4的周期性序列。
请确定其傅里叶级数的系数(X k。
解:(111*0((((((N N N nk nk nk N N N n n n X k x n W x n W x n W X k X k −−−−−=====−= =−=∑∑∑3.2 (1设(xn 为实周期序列,证明(x n 的傅里叶级数(X k 是共轭对称的,即*((X k X k =− 。
(2证明当(xn 为实偶函数时,(X k 也是实偶函数。
证明:(1 111**((([(]((N nk N n N N nk nkNNn n Xk x n W Xk x n W xn W X−−=−−−==−=−===∑∑∑ k(2因(xn 为实函数,故由(1知有 *((Xk X k =− 或*((X k X k −= 又因(xn 为偶函数,即((x n x n =− ,所以有(111*0((((((N N N nk nk nk N N N n n n X k x n W x n W x n W X k X k −−−−−=====−= =−=∑∑∑3.3 图P3.3所示的是一个实数周期信号(xn 。
利用DFS 的特性及3.2题的结果,不直接计算其傅里叶级数的系数(Xk ,确定以下式子是否正确。
(1,对于所有的k; ((10Xk X k =+ (2((Xk X k =− ,对于所有的k; (3; (00X=(425(jkX k eπ,对所有的k是实函数。
解:(1正确。
因为(x n 一个周期为N =10的周期序列,故(X k 也是一个周期为N=10的周期序列。
(2不正确。
因为(xn 一个实数周期序列,由例3.2中的(1知,(X k 是共轭对称的,即应有*((Xk X = k −,这里(X k 不一定是实数序列。
(3正确。
因为(xn (0n ==在一个周期内正取样值的个数与负取样值的个数相等,所以有 10(0N n Xx −=∑ (4不正确。
数字信号处理刘顺兰第三章完整版习题解答一、题目解答1. 题目利用时域抽样、频域抽样、零填充、插值法等,实现信号的变换。
1.1 时域抽样时域抽样是指将一个连续时间信号在时间轴上的等间隔位置上进行采样,可以得到一个离散时间信号。
时域抽样的原理是,将时间轴上的信号按照一定的时间间隔进行采样,每个采样点的振幅值就是该点对应的连续时间信号的振幅值。
时域抽样可以通过以下步骤进行实现:1.假设连续时间信号为x(t),采样频率为Fs(采样频率是指每秒采样的次数),采样间隔为Ts(采样间隔是指相邻两个采样点之间的时间间隔)。
2.根据采样频率和采样间隔,计算出采样点数N:N =Fs * T,其中T为采样时长。
为Ts。
4.在每段的中点位置进行采样,得到N个采样点。
5.将N个采样点按照时域顺序排列,即可得到离散时间信号。
1.2 频域抽样频域抽样是指将一个连续频谱信号在频率轴上的等间隔位置上进行采样,可以得到一个离散频谱信号。
频域抽样的原理是,将频率轴上的信号按照一定的频率间隔进行采样,每个采样频率点上的能量值就是该频率点对应的连续频谱信号的能量值。
频域抽样可以通过以下步骤进行实现:1.假设连续频谱信号为X(f),采样频率为Fs(采样频率是指每秒采样的次数),采样间隔为Δf(采样间隔是指相邻两个采样频率点之间的频率间隔)。
2.根据采样频率和采样间隔,计算出采样点数N:N =Fs / Δf,其中Δf为采样频率点之间的频率间隔。
为Δf。
4.在每段的中点位置进行采样,得到N个采样频率点。
5.将N个采样频率点按照频域顺序排列,即可得到离散频谱信号。
1.3 零填充零填充是指在信号的末尾添加一些零值样本,使得信号的长度变长。
零填充的原理是,通过增加信号的长度,可以在时域和频域上提高信号的分辨率,从而更精确地观察信号的特征。
零填充可以通过以下步骤进行实现:1.假设原始信号为x(n),长度为N。
2.计算需要填充的长度L,L > 0。
习题课第三章
3-17:x1(n)=R5(n), X1(ejw)=DTFT[x1(n)],幅频,相频特性
已知是50点的有限长序列,非零值范围为是15点的有限长序列,非零值范围,对两序列做50点圆周卷积,即试问,y(n)中哪个n 值范围对应于的结果。
)(1n x )
(2n x 49
0≤≤n 19n 5≤≤)
())(()()()()(5050249
0121n R m n x m x n x n x n y m −=Θ=∑=)()(21n x n x ∗
•要求会用圆卷积代替线性卷积,因为圆卷积可以用快速算法实现
•两序列线性卷积长度分别等于N1,N2
•起始非零点为两序列各自非零点之和;终点非零值点位两序列各自非零点终点序号之和
•这里就是5-68(64个点)为线性卷积结果的非零点
•这里做50点的圆卷积,就是对这64个点做以周期为50的延拓,所以必然有重合相加点,不等于线性卷积结构。
第三章习题1. Consider a Wiener filtering problem characterized by the following valuesfor the correlation matrix R of the tap-input vector x (n) and cross-correlation vector p between x (n) and the desired response d(n):⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡=25.05.015.05.01P R (a) Suggest a suitable value for the step-size parameter μ that wouldensure convergence of the method of steepest descent, based on the given value for matrix R .(b) Using the value proposed in part (a), determine the recursions forcomputing the elements )(1n w and )(2n w of the tap-weight vector w (n). For this computation, you may assume the initial values0)0()0(21==w w .(c) Investigate the effect of varying the step-size parameter μ on thetrajectory of the tap-weight vector w (n) as a varies from zero to infinity.2. The error performance of a real-valued filter, using a single tap weight w ,is defined by,))(0(20min w w r J J -+=where r(0) is the autocorrelation function of the tap input x (n) for zero lag,min J is the minimum mean-square error, and o w is the Wiener solution for the optimum value of the tap weight w .(a) Determine the bounds on the step-size parameterμof the steepest-descent algorithm used to recursively compute the optimum solution o w .(b) Plot the curve for cost function of the filter.3. Continuing with Problem 2, do the following:(a) Formulate the learning curve of the filter in the terms of its onlynatural mode )(n υ.(b) Determine the first derivative of the mean-square error J with respectto the natural mode of the filter.4. Consider an autoregressive (AR) process of order one, described bydifference equation),()1()(n n x n x να+--=where α is the AR parameter of the process and )(n ν is a zero-meanwhite noise of variance 2νσ.(a) Set up a liner predictor of order one to compute the parameter α.Specifically, use the method of steepest descent for the recursive computation of the Wiener solution for the parameter α.(b) Plot the error-performance curve for this problem, identifying theminimum point of the curve in terms of known parameters.(c) What is the condition on the step-size parameter μ to ensurestability? Justify your answer.5. An autoregressive (AR) process is described by the second-order differenceequation),()2()1(5.0)(n n x n x n x ν+-+--=where )(n ν is a zero-mean white noise of unit variance. The method ofsteepest descent is used for the recursive computation of the optimum weight vector of the forward linear predictor applied to the process x(n). Find the bounds on the step-size parameter μ that ensure stability of the steepest-descent algorithm.6. Repeat Problem 5 for the backward predictor applied to the second-order ARprocess x(n).7. The LMS algorithm is used to implement a dual-input, single-weight adaptive noise canceller. Set up the equations that define the operation of this algorithm.8. Consider the use of a white-noise sequence of zero mean and variance2σas the input to the LMS algorithm. Evaluate(a) the condition for convergence of the algorithm in the mean square, (b) The excess mean-square error.9. The leaky LMS algorithm.Consider the time-varying cost function,)()()(22n w n e n J α+=where w (n) is the tap-weight vector of a transversal filter, e(n) is theestimation error, and α is a constant. As usual,),()()()(n n n d n e H x w -=where d(n) is the desired response and x (n) is the tap-input vector. In theleaky LMS algorithm, the cost function J (n) is minimized with respect to the weight vector w (n). Show that the time update for the tap-weight vector )(ˆn wis defined by ).()()(ˆ)1()1(ˆn e n n n *+-=+x w wμμα 10. The tandem connection of adaptive filters arises in some applications (e.g.,acoustic echo cancellation). Consider, then, Figure P3.1,which shows a simplified tandem configuration involving a pair of LMS adaptive filters. The input vector x (n) is applied simultaneously to both filters, and the error signal )(1n e produced by filter I serves the purpose of having a desired response for filter II.(a) Formulate the update equations for the tandem configuration of thefigure.(b) Show that this tandem configuration converges in the mean square ifboth adaptive filters 1 and 2 converge in the mean squareindividually.()n e Adapter Filter 1Adapter Filter 2Figure P3.111. Consider the adaptive FIR filter shown in Figure P3.2. The system )(z Cis characterized by the system function19.011)(-+=zz C Determine the optimum coefficients of the adaptive FIR filter 110)(-+=z b b z B that minimize the MSE. The additive noise is white with variance 1.02=ωσ.Figure P3.2。