数字信号处理第八章答案史林赵树杰编著
(完整版)数字信号处理课后答案_史林版_科学出版社
第一章 作业题 答案############################################################################### 1.2一个采样周期为T 的采样器，开关导通时间为()0T ττ<<，若采样器的输入信号为()a x t ，求采样器的输出信号()()()a a x t x t p t ∧=的频谱结构。
式中()()01,()0,n p t r t n t r t ττ∞=-∞=-≤≤⎧=⎨⎩∑其他解：实际的采样脉冲信号为：()()n p t r t n τ∞=-∞=-∑其傅里叶级数表达式为：()000()jk tn p t Sa k T eTωωτω∞=-∞=∑采样后的信号可以表示为：()()()ˆa a xt x t p t δ= 因此，对采样后的信号频谱有如下推导：()()()()()()()()()()()()()0000000000000ˆˆsin 1j t a a jk t j t a n jk t j t a k j k ta k ak a k X j x t e dtx t Sa k T e e dtTSa k T x t e e dtTSa k T x t edtTSa k T X j jk Tk T X j jk T kωωωωωωωωτωωτωωτωωτωωωωωω∞--∞∞∞--∞=-∞∞∞--∞=-∞∞∞---∞=-∞∞=-∞∞=-∞Ω=====-=-⎰∑⎰∑⎰∑⎰∑∑%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 1.5有一个理想采样系统，对连续时间信号()a x t 进行等间隔T 采样，采样频率8s πΩ=rad/s ，采样后所得采样信号()a x t ∧经理想低通滤波器()G j Ω进行恢复，已知()41/4,,4G j ππ⎧Ω≤⎪Ω=⎨Ω>⎪⎩今有两个输入信号12()cos(2)()cos(5)a a x t t x t t ππ==和，对应的输出信号分别为12()()a a y t y t 和，如题1.5图所示，问12()()a a y t y t 、有没有失真，为什么？题1.5图 理想采样系统与恢复理想低通滤波器解：因为是理想采样系统，因此采样后的信号频谱可以表示为：()()1ˆa a s k X j X j jk T ∞=-∞Ω=Ω-Ω∑8s πΩ=，12πΩ=，25πΩ=，折叠频率为2s Ω，而滤波器对4πΩ≤的信号通过，因此有如下图：结论：1）1()a y t 不失真、2()a y t 失真。
数字信号处理及MATLAB实现第八章习题答案
1. 数字信号处理器有哪些结构特点？答：1. DSP内部采用的是程序空间和数据空间分开的哈佛结构；2. 多总线结构；3. 流水线结构：在DSP中，执行一条指令，需要通过取指、译码、取操作数和执行四个阶段。
在程序运行过程中，这四个阶段不是依次进行的，而是重叠的进行的；4. 多处理单元：DSP内部通常包括有算术逻辑运算单元(ALU)、辅助寄存器运算单元(ARAU)、累加器(ACC)以及硬件乘法器(MUL)等多个处理单元。
它们可以在一个指令周期内同时进行运算；5. 硬件配置强：新一代DSP的接口功能愈来愈强，如：TMS320C5000系列芯片片内具有串行口、主机接口(HPl)、DMA控制器、软件控制的等待状态产生器、锁相环时钟产生器以及实现在片仿真符合IEEEll49．1．标准的测试访问口，更易于完成系统设计。
许多DSP芯片都可以工作在省电方式，使系统功耗降低。
2. TMS320C54x系列DSP片内有多少条总线，具体说明有哪些？答：TMS320C54x系列DSP是TI公司于1996年推出的新一代定点数字信号处理器。
其采用先进的修正哈佛结构，片内有8条总线，分别为1条程序存储器总线，3条数据存储器总线和4条地址总线。
3. TMS320VC5402的在片外围电路有哪些？答：1. 软件可编程等待状态发生器；2. 可编程分区转换逻辑电路；3.带有内部振荡器或用外部时钟源的片内锁相环(PLL)时钟发生器；4. 时分多路(TDM)串行口；5. 缓冲串行口(BSP)；6. 2个16位定时器；7. 8位并行主机接口(HPl)；8. 外部总线关断控制，以断开外部的数据总线、地址总线和控制信号。
4. TMS320VC5402有哪些片内资源？答：TMS320VC5402的片内资源按功能包括运算单元、寄存器、片内RAM 和ROM、片外存储器接口、DMA控制器、主机接口、串口、定时器、时钟产生器和中断控制器。
5. CCS有几种工作模式？具体说明。
数字信号处理英文版课后答案（8）
Chapter 8 Solutions 8.1 The Fourier transform gives the spectrum of this non-periodic signal:Ω-Ω-++=Ω2j j e 25.0e 5.01)(X8.2 The samples for the signal are:The spectrum for the signal is given byΩ-Ω-Ω-++-=Ω4j 3j j e e e 5.0)(X8.3 Ω-Ω-Ω-Ω-∞-∞=Ω-++++==Ω∑8j 6j 4j 2j n jn e e e e 1e ]n [x )(X8.4 The first eight sample values for the signal are shown in the table.The signal contains an infinite number of non-zero samples, but the first 5, shown above, should be sufficient to approximate the DTFT reasonably well.Ω-Ω-Ω-Ω-∞-∞=Ω-+-+-≈=Ω∑4j 3j 2j j n jn e 0039.0e 0156.0e 0625.0e 25.01e ]n [x )(X8.5 The DTFT for x 1[n] isΩ-Ω-∞-∞=Ω-++==Ω∑2j j n jn 11e 3e 21e ]n [x )(XThe DTFT for x 2[n] isΩ-Ω-Ω-∞-∞=Ω-++==Ω∑4j 3j 2j n jn 22e 3e 2e 1e ]n [x )(XThe DTFT for x 3[n] isΩ-Ω-∞-∞=Ω-++==Ω∑2j j n jn 33e 1e 23e ]n [x )(XAll three signals have identical magnitude spectra, shown below.|X(The phase spectra of the three signals differ. They are shown in the figure below. From the DTFT expressions above, it is easy to see that )(X e )(X 12j 2Ω=ΩΩ- and)(X e )(X 12j 3Ω-=ΩΩ-. The first relationship means that the phase for x 2[n] will always be 2Ω less than the phase for x 1[n]. The second relationship means that the phase for x 3[n] is always 2Ω less than the negative of the phase for x 1[n], since X 1(-Ω) produces phases that are the negatives of the phases for X 1(Ω), following the odd phase spectrum rule. Both of these two relationships can be confirmed in the table or the plot, remembering that θ ± 2π = θ.8.6 The samples of the signal are shown in the table:The DTFT isΩ-Ω-Ω-∞-∞=Ω-+-==Ω∑4j 2j j n jn e 866.0e 866.0e 866.0e ]n [x )(X8.7 The period of the signal is N = 10. The sample values are listed in the table:The N = 10 DFS coefficients are given by:510k 2j 410k 2j 310k 2j 210k 2j 9n n10k 2j k e e e e e ]n [x c π-π-π-π-=π-+++==∑= 1|–2πk/5 + 1|–3πk/5 + 1|–4πk/5 + 1|–πkBecause of the symmetry of the spectrum, it is enough to calculate the coefficients for k = 0 to k = N/2 = 5, and to produce the other parts of the spectrum from this data.The magnitudes in the second half of the spectrum are a mirror image of those in the first. The phases in the second half are the negatives of the phases in the first half.|X(8.8 (a) The signal x[n] has the digital period N = 4. Its spectrum can be found using the discrete Fourier series:34k 2j 4k2j 3n n4k 2j k ee2e]n [x c π-π-=π--+==∑= 2 + 1|–πk/2 – 1|–3πk/2(b) The magnitude spectrum appears to repeat every second sample, while the phase spectrum repeats every four samples. The period of both spectra must be the same, so the overall period must be 4. As for all periodic signals, the period of the spectrum matches the period of the signal. 8.9 The signal has a period of N = 4, so the DFS coefficients are given by:24k2j 3n n4k 2j k e1e ]n [x c π-=π--==∑= 1 – 1|–πkSeveral cycles of the spectra are shown in the figures below.(8.10 The spectra for non-periodic signals are produced using the DTFT. The spectra, X(Ω), are smooth, continuous functions of frequency with a period of 2π. They are plotted against digital frequency Ω. If desired, the spectra can be plotted as X(f) versus frequency in Hz, using f = Ωf S /(2π).The spectra for periodic signals having a period N are produced using the DFS. The spectra, c k , are line functions of frequency with a period of N. They are plottedagainst the index k. If desired, the spectra can be plotted against frequency in Hz, using f = kf S /N.For both non-periodic and periodic signals, magnitude spectra are even and phase spectra are odd.8.11 (a) The signal is non-periodic. Its spectrum is given by the DTFT:Ω-Ω-Ω-∞-∞=Ω-++==Ω∑3j 2j j n jn e 3e 2e e ]n [x )(XThe magnitude and phase spectra appear as dashed lines in the figure in part (b). (b) The signal is a periodic version of the signal in (a), with period N = 4. Its spectrum is given by the DFS:34k2j 24k 2j 4k2j 3n n4k 2j k e3e 2ee]n [x c π-π-π-=π-++==∑= 1|–πk/2 + 2|–πk + 3|–3πk/2 The magnitude and phase spectrum are plotted below, dashed lines for the DTFT and solid for the DFS. Note that the DFS samples the DTFT.8.12 The harmonic frequencies are given by f = kf S /N. For f S = 12 kHz and N = 72, the first five harmonics are: 166.7, 333.3, 500.0, 666.7, and 833.3 Hz.8.13 For the first cosine, N/M = 2π/Ω = 2π/(2π/3) = 3, so the period is 3. For the second cosine, N/M = 2π/Ω = 2π/(π/3) = 6, so the period is 6. The lowest common multiple of these two periods is 6, so this is the overall period of the waveform.The signal samples are given by The Fourier coefficients are calculated as:56k2j 46k2j 36k2j 26k 2j 6k2j 5n n6k 2j k e5.0e 5.1e e 5.1e5.03e]n [x c π-π-π-π-π-=π---+--==∑= 3 – 0.5|–k π/3 –1.5|–2k π/3 + 1|–k π –1.5|–4k π/3 – 0.5|–5k π/3 The magnitude spectrum is periodic with period 6. Six spectrum samples cover the range from 0 to the sampling frequency, 4 kHz. Therefore, each point of the spectrum covers 4000/6 = 2000/3 Hz. As a result, k = 3 corresponds to the Nyquist frequency of 2 kHz. Using kf S /N, the two spikes in the spectrum below the Nyquist frequency, at k =1 and k = 2, map to frequencies of 2000/3 = 666.7 and 2(2000)/3 = 1333.3 Hz. Using Ω = 2πf/f S , these analog frequencies correspond to the digital frequencies π/3 and 2π/3 rads. Thespike at the higher frequency is twice the height of the other because its amplitude in the signal is double that of the other component. The other two spikes in the spectrum, at k = 6 – 1 and k = 6 – 2 map to imaged versions of the baseband frequencies.8.14 (a) Since the magnitude spectrum is periodic with period 14, the digital signal is periodic with the same period. (b) Fourteen points of the magnitude spectrum cover the sampling frequency, 16 kHz. Each point covers an interval 16/14 = 1.143 kHz wide. For a 16 kHz sampling frequency, the Nyquist frequency is 8 kHz. The first seven points of the magnitude spectrum cover this range. Three spikes occur within the Nyquist range, at k = 1, 2 and 3, or, using kf S /N, 1143, 2286 and 3429 Hz.(The magnitude spectrum belongs to the signal x[n] = sin(n π/7) + 2sin(n2π/7) +3sin(n3π/7). The digital frequencies π/7, 2π/7 and 3π/7 rads are, for a 16 kHz sampling rate, obtained from the analog frequencies 1143, 2286 and 3429 Hz.)8.15 (a) Since the magnitude spectrum has a period of 24, the digital signal also has a period of 24 samples. (b) Twenty-four samples cover 12 kHz, which means each point of the magnitude spectrum covers 0.5 kHz. The spikes at k = 2, 4 and 9 map to frequencies of kf S /N = 1, 2 and 4.5 kHz. The other three spikes are occur above the Nyquist frequency, at k = 24 –2 = 22, k = 24 – 4 = 20 and k = 24 – 9 = 15. The frequencies that correspond to these values of k are imaged copies of the baseband frequencies. (c) Using Ω = 2πf/f S , the digital frequencies are π/6, π/3 and 3π/4 rads.(The signal whose magnitude spectrum is shown is x[n] = cos(n π/6) + cos(n π/3) + cos(n3π/4).)8.16 The Fourier expansion can be matched to ∑-=π=1N 0k n N k2j k e c N 1]n [x . Since N = 16,n 1622j n1612j n 1612j n 1622j e e 2j 1e 2j e ]n [x πππ-π-+++-=⎪⎪⎭⎫ ⎝⎛+++-=πππ-π-n 1622j n 1612j n 1612j n 1622j e 8e 4j 8e 4j e 881The only non-zero coefficients c k are: c –2 = 8, c –1 = –j4, c 0 = 8, c 1 = j4, c 2 = 8. The other 11 coefficients in each period must be zero. The magnitudes of the non-zero coefficients are 8, 4, 8, 4, 8 and the phases are 0, –π/2, 0, π/2 and 0. The magnitude and phase spectra constructed using this information are shown below. Remember that the sequence of magnitudes and phases repeats every 16 points.8.17 (a)(i) Since 2π/Ω = 14π/(6π) = 7/3, the digital period is 7.(ii) The signal contains the frequency f = Ωf S /(2π) = 30000/7 Hz. For a digital period of 7, each point of the magnitude spectrum covers f S /N = 10000/7 Hz. Since each frequency is represented by kf S /N, a spike occurs in this magnitude spectrum at k = 3. Due to imaging, a second, symmetrically-placed spike occurs at N – 3 = 7 – 3 = 4.(b)(i) Since 2π/Ω = 10π/(3π) = 10/3, the digital period is 10.(ii) The signal contains the frequency f = Ωf S /(2π) = 3000 Hz. For a digital period of 10, each point of the magnitude spectrum covers f S /N = 1000 Hz. Therefore, a spike occurs in this magnitude spectrum at k = 3. Due to imaging, a second, symmetrically placed spike occurs at N – 3 = 10 – 3 = 7.(c)(i) For the first component 2π/Ω = 6, and for the second component 2π/Ω = 16. The digital period is the lowest common multiple of these two periods, or 48.(ii) The signal contains the frequencies f = Ωf S/(2π) = 10000/6 = 5000/3 Hz and f = Ωf S/(2π) = 10000/16 = 625 Hz. For a digital period of 48, each point of the magnitude spectrum covers f S/N = 10000/48 = 625/3 Hz. Therefore, spikes occur in this magnitude spectrum at k = 3 and k = 8. Symmetrically placed spikes occur at N – 3 = 48 – 3 = 45 and N – 8 = 48 – 8 = 40 as a result of imaging.(d)(i) As in part (c), the digital period is 48.(ii) The spikes occur in the same locations as in (c), but the higher frequency spike is twice as tall as the lower frequency spike.8.18 As evidenced by the zeros in the magnitude spectrum, some frequencies are excluded from this signal. The most significant contribution lies at a digital frequency of about 0.1 radian. The exact value is 0.113 radians. With f S = 20 kHz, f = Ωf S/(2π) = (0.113)(20000)/(2π), or about 360 Hz. The next biggest peak occurs at about 0.3 radians. The exact value is 0.336 radians, which corresponds to a frequency of 1070 Hz. Most of the important signal content lies below the fourth zero in the spectrum, at 0.395 radians or 1257 Hz.8.19 The number of points in the DFS spectrum gives the digital period of the underlying signal. The digital period in this case is N = 23. The periodic signal whose magnitude spectrum is shown has a large DC component and contributions at all harmonic frequencies, kf S/N = k(20000)/23 = 869.6k Hz. The first few harmonics are 869.6 Hz, 1739.1 Hz, 2608.7 Hz, …. The amplitudes of the harmonics decrease rapidly with frequency. The fundamental frequency of the signal is 869.6 Hz, so the period of the signal is NT S = 1.15 msec.8.20 (a) Ω-)X5.0(e5.0+Ωj-=(b) For a sampling frequency of f S= 16 kHz, the Nyquist frequency is 8 kHz. A cut-off of 2 kHz corresponds to a digital frequency of Ω = 2πf/f S = π/4 radians. The low pass filter extracts the lowest frequency elements in the signal.(c) Cut-off frequencies of 3 and 6 kHz correspond to digital frequencies of 0.375πand 0.75π radians. The band pass filter extracts the mid-range frequencies.(d) A cut-off of 7 kHz corresponds to a digital frequency of 0.875π radians. The high pass filter extracts only the highest frequency elements in the signal.8.21 (a) Each of the three terms is periodic. The digital period for each is 14, 3 and 16. The lowest common multiple for these integers is N = 336, the digital period for x[n]. The analog frequencies of the three terms are given by f = Ωf S/(2π). They are 1143, 5333, and 1000 Hz. The DFS frequencies are given by f = kf S/N = 47.6k, so the magnitude spectrum for the signal will contain peaks at k = 24, 112 and 21. These three peaks are shown below. Note the images of these peaks in the second half of the spectrum, at k = 363 – 21 = 342, k = 363 – 24 = 339, and k = 363 – 112 = 251.(b) The low pass filter will extract the two lowest-frequency peaks, at 1000 and 1143 Hz. The DFS magnitude spectrum will contain a peak at k = 21 and one at k = 24, plus imaged peaks at k = 363 – 21 = 342 and k = 363 – 24 = 339.(c) The band pass filter will extract the high frequency peak, at 5333 Hz. The DFS magnitude spectrum for the filtered output will contain a peak at k =112, plus an imaged peak at k = 363 – 112 = 251.(d) The high pass filter output will contain no peaks.8.22 (a) The spectrum has 64 points, so N = 64 is the digital period of the square wave. The fundamental frequency is f S/N = 4000/64 = 62.5 Hz.(b) The period in seconds is the reciprocal of the fundamental frequency, or NT S = 16 msec.(c) The DC component gives the average value of the signal. For this signal, the average is zero.(d) The harmonics present in the signal are odd multiples of the fundamental frequency. The only ones that lie below 500 Hz are 62.5k = 62.5, 187.5, 312.5, and 437.5 Hz. These frequencies correspond to the indices k = 1, 3, 5, 7.。
数字信号处理习题及解答
只有在如上周期延拓序列中无混叠的点上， 才满足f(n)=fl(n),所以 f(n)=fl(n)=x(n)*y(n) 7≤n≤19
令
数字信号处理习题及解答
第二章Z变换及离散时间系统分析
3 解答
n≥0时， 因为c内无极点，x(n)=0; n≤－1时， c内有极点0 ， 但z=0是一个n阶极点， 改为求
圆外极点留数， 圆外极点有z1=0.5, z2=2， 那么
数字信号处理习题及解答
第二章Z变换及离散时间系统分析 3 解答 (2) 收敛域0.5<|z|<2:
数字信号处理习题及解答
第三章信号的傅里叶变换 1 解答
(1) (2) (3)
数字信号处理习题及解答
第三章信号的傅里叶变换 2 试求如下序列的傅里叶变换：
(1) x1(n)=δ(n－3)
（2）
数字信号处理习题及解答
第三章信号的傅里叶变换 2 解答
(1) (2)
数字信号处理习题及解答
第三章信号的傅里叶变换
第一章离散时间信号与离散时间系统
4 解答
数字信号处理习题及解答
第二章Z变换及离散时间系统分析 1
数字信号处理习题及解答
第二章Z变换及离散时间系统分析 1 解答
数字信号处理习题及解答
第二章Z变换及离散时间系统分析 1 解答
数字信号处理习题及解答
第二章Z变换及离散时间系统分析 2
数字信号处理习题及解答
第二章Z变换及离散时间系统分析 2 解答
数字信号处理习题及解答
第二章Z变换及离散时间系统分析 2 解答
数字信号处理习题及解答
第二章Z变换及离散时间系统分析 3 已知
求出对应X(z)的各种可能的序列表达式。
数字信号处理课后答案-史林版-科学出版社
(2)试确定采样信号频谱不混叠的最低采样频率，并画出此时 的频谱图形。
(3)画出由(3)中的序列 恢复 的框图（可用复理想低通滤波器）。
题1.7图 的频谱图形
解：采样间隔为 ，因此采样频率为 。
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其傅里叶级数表达式为：
采样后的信号可以表示为：
因此，对采样后的信号频谱有如下推导：
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1.5有一个理想采样系统，对连续时间信号 进行等间隔T采样，采样频率 rad/s，采样后所得采样信号 经理想低通滤波器 进行恢复，已知
第二章
作业题答案
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2.1将序列 表示为 及 延迟的和。
解：首先将 表示为单位脉冲序列的形式：
对于单位脉冲函数 ，用单位阶跃序列 表示，可得：
将上式带入到 的单位脉冲序列表达式中，可得：
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今有两个输入信号 ，对应的输出信号分别为 ，如题1.5图所示，问 有没有失真，为什么？
题1.5图理想采样系统与恢复理想低通滤波器
解：因为是理想采样系统，因此采样后的信号频谱可以表示为：
， ， ，折叠频率为 ，而滤波器对 的信号通过，因此有如下图：
结论：1） 不失真、 失真。2）输出信号中存在两种频率： 、
数字信号处理—原理、实现及应用(第4版)第8章 时域离散系统的实现 学习要点及习题答案
·185·第8章 时域离散系统的实现本章学习要点第8章研究数字信号处理系统的实现方法。
数字信号处理系统设计完成后得到的是该系统的系统函数或者差分方程，要实现还需要设计一种具体的算法，这些算法会影响系统的成本以及运算误差等。
本章介绍常用的几种系统结构，即系统算法，同时简明扼要地介绍数字信号处理中的量化效应，最后介绍了MA TLAB 语言中的滤波器设计和分析工具。
本章学习要点如下：(1) 由系统流图写出系统的系统函数或者差分方程。
(2) 按照FIR 系统的系统函数或者差分方程画出其直接型、级联型和频率采样结构，FIR 线性相位结构，以及用快速卷积法实现FIR 系统。
(3) 按照IIR 系统的系统函数或者差分方程画出其直接型、级联型、并联型。
(4) 一般了解格型网络结构，包括全零点格型网络结构系统函数、由FIR 直接型转换成全零点格型网络结构、全极点格型网络结构及其系统函数。
(5) 一般了解如何用软件实现各种网络结构，并排出运算次序。
(6) 数字信号处理中的量化效应，包括A/D 变换器中的量化效应、系数量化效应、运算中的量化效应及其影响。
(7) 了解用MA TLAB 语言设计、分析滤波器。
8.5 习题与上机题解答8.1 已知系统用下面差分方程描述311()(1)(2)()(1)483y n y n y n x n x n =---++- 试分别画出系统的直接型、级联型和并联型结构。
差分方程中()x n 和()y n 分别表示系统的输入和输出信号。
解：311()(1)(2)()(1)483y n y n y n x n x n --+-=+- 将上式进行Z 变换，得到121311()()()()()483Y z Y z z Y z z X z X z z ----+=+ 112113()31148z H z z z ---+=-+ (1) 按照系统函数()H z ，画出直接型结构如图S8.1.1所示。
数字信号处理(第三版)课后答案及学习指导(高西全-丁玉美)第八章
第8章 上机实验
x2n=ones(1， 128)； %产生信号x2n=un hn=impz(B， A， 58)； %求系统单位脉冲响应h(n) subplot(2， 2， 1)； y=′h(n)′； tstem(hn， y)；
%谐振器对正弦信号的响应y32n figure(3) subplot(2， 1， 1)； y=′y31(n)′； tstem(y31n， y) title(′(h) 谐振器对u(n)的响应y31(n)′) subplot(2， 1， 2)； y=′y32(n)′； tstem(y32n， y)； title(′(i) 谐振器对正弦信号的响应y32(n)′)
%调用函数tstem title(′(d) 系统单位脉冲响应h1(n)′) subplot(2， 2， 2)； y=′y21(n)′； tstem(y21n， y)；
第8章 上机实验
title(′(e) h1(n)与R8(n)的卷积y21(n)′)
subplot(2， 2， 3)； y=′h2(n)′； tstem(h2n， y)；
注意在以下实验中均假设系统的初始状态为零
第8章 上机实验
3． （1） 编制程序， 包括产生输入信号、 单位脉冲响应 序列的子程序， 用filter函数或conv函数求解系统输出响应 的主程序。 程序中要有绘制信号波形的功能。 （2） 给定一个低通滤波器的差分方程为
y(n)=0.05x(n)+0.05x(n－1)+0.9y(n－1) 输入信号
第8章 上机实验
8.1.3
实验结果与波形如图8.1.1所示。
第8章 上机实验
数字信号处理课后答案课件
傅里叶变换的性质
线性性质
若离散信号x(n)和y(n)的 傅里叶变换分别为 X(e^jωn)和Y(e^jωn)， 则对于任意实数a和b，有 aX(e^jωn) + bY(e^jωn) 的傅里叶变换等于 aX(e^jωn)和bY(e^jωn) 的傅里叶变换之和。
从而实现信号的分离、抑制或提 取。
滤波器分类
根据不同的特性，滤波器可分为 低通、高通、带通和带阻滤波器，
每种滤波器都有各自的应用场景 和特点。
滤波器原理
滤波器的原理是基于频率响应， 即不同频率的信号经过滤波器后， 其幅度和相位会发生不同的变化。
IIR滤波器设计
IIR滤波器概述
IIR滤波器设计方法
IIR滤波器稳定性
在设计IIR滤波器时，需要考虑其稳定 性。如果系统函数的极点位于单位圆 外，则系统不稳定，可能会导致无穷 大的输出。因此，在设计过程中需要 进行稳定性分析。
FIR滤波器设计
FIR滤波器概述
FIR（Finite Impulse Response）滤 波器是一种具有有限冲击响应的数字 滤波器，其系统函数可以表示为有限 项之和。
插值法
对于非周期性的连续时间信号，可以通过插值法得到离散时间信号。常用的插值方法包括 线性插值、多项式插值、样条插值等。
傅里叶变换法
对于任何连续时间信号，可以通过傅里叶变换将其转换为频域表示形式，然后对频域表示 形式进行采样，得到离散时间信号。再通过逆傅里叶变换将其转换回时域表示形式。
05 第五章 信号的分 析与合成
抽样定理的充分性
对于任何连续时间信号，如果其最高频率分量小于等于fmax，则可 以通过其抽样信号无失真地重建出原信号。
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第八章练习题答案
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 8.1 设线性时不变系统的输入序列为()x n ，输出序列为()y n ，其差分方程为 311
()(1)(2)()(1)483
y n y n y n x n x n =
---++- 求其系统函数()H z ，并分别画出该系统的直接型、级联型和并联型算法结构。
解：
移项 311
()(1)(2)()(1)483y n y n y n x n x n -
-+-=+- Z 变换 121
311()()()()()483y z y z z y z z x z x z z ----+=+
1
12
113()31148
z H z z z
---+=
--+ （1） 直接型 1
12
113()31148z H z z z
---+=--+ （2） 级联型1
11113()11(1)(1)
24
z H z z z ---+=-- （3） 并联型，将()H z 进行部分分式展开
1
()3
1111()()()()2424z H z A B
z z z z z +
==+
---- 111103()11223()()24z A z z z z +
=-==--
11173()11443()()24z B z z z z +
=-==---
11
1071073333()1111()()112424
z z H z z z z z ----
=+=+
----
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 8.6 题8.6图画出了10中不同的系统算法结构流图，试分别求出它们的系统函数()H z 。
()x n ()
y n (a)
1
1()1H z az -=
- 直接1型 1
110.5()10.3H z z --+=-直接2型
1-1
-()
x n ()
y n (c)
12()H z a bz cz --=++ 直接1型 11
11
()11H z az bz
--=
+-- 并联型，内直接1型 ()
x n ()
y n (e)
1
1220.24()10.250.2z H z z z ---+=-- 转置型 11
11()10.510.75H z z z --=-+ 级联型 直接1型
()
x n ()
y n (g)
()
x n ()
y n (b)
()
x n ()
y
n (d)
()
x n ()
y n
(f)
()
x n ()
y n (h)
1
12
10.25()10.250.4z
H z z z ---+=
-+直接2型 1
12
3sin 4()312cos 4
z H z z z
---=-+直接分析图 ()
x n ()y n 1a 1
b 2
a 2
b 3
a 1
z
-1
z
-1
z -(i)
12012121
1231
()11b b z b z H z a z a z a z
-----++=--- 级联型 直接2型，右边分子项 常数项不加负号 直接1型 右边分母项 常数项加负号
121
01234121
123()11b b z b z b b z H z a z a z a z ------+++=+
--- 并联型 直接2型 题8.6图 10种不同系统的算法结构
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 8.14 已知FIR 数字滤波器的系统函数为 12341
()(10.9 2.10.9)10
H z z z z z ----=
++++ 试画出该滤波器的直接型算法结构和线性相位算法结构。
()
x n ()
y n 1a 1
b 2
a 2
b 3
a 1z -1z -1
z
-3
b 4
b (j)。