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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。
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.。
数字信号处理知到章节测试答案智慧树2023年最新西安工程大学绪论单元测试1.请判断下面说法是否正确：为了有效地传播和利用信息，常常需要将信息转换成信号，因此信号是信息的载体，通过信号传递信息。
（）参考答案:对2.请判断下面说法是否正确：模拟信号预处理的主要作用是滤除输入模拟信号中的无用频率成分和噪声，避免采样后发生频谱混叠失真。
（）参考答案:对3.下列关于信号分类方式的选项正确的是（）。
参考答案:按信号幅度的统计特性分类;按信号的维数分类;按信号自变量与参量的连续性分类4.下列不属于数字信号处理软件处理方法特点的选项是（）。
参考答案:处理速度快5.下列关于数字系统处理精度描述正确的选项是（）。
参考答案:精度由系统字长与算法决定第一章测试1.请判断下面说法是否正确：时域离散信号通过量化编码转换为数字信号，是一种无损变换。
( )参考答案:错2.下列信号是周期信号的有（）。
参考答案:;;3.信号的最小周期是（）。
参考答案:24.请判断下面说法是否正确：线性时不变时域离散系统具有线性性质和时不变特性。
（）参考答案:对5.以下序列是系统的单位脉冲响应h(n)，则是稳定系统的有（）。
参考答案:;第二章测试1.请判断下面说法是否正确：时域离散信号和系统分析可以通过傅里叶变换和Z变换两种数学工具（）。
参考答案:对2.请判断下面说法是否正确：周期序列的傅里叶变换以为周期，而且一个周期内只有N个冲激函数表示的谱线（）。
参考答案:错3.实序列的傅里叶变换具有（）。
参考答案:共轭对称性质4.已知序列，其Z变换和收敛域为（）。
参考答案:;5.序列，其傅里叶变换为（）。
参考答案:第三章测试1.在变换区间0≤n≤N-1内，序列的N点DFT在k=0的值为（）。
参考答案:N2.在变换区间0≤n≤N-1内，序列的N点DFT的值为（）参考答案:13.已知，求=（）参考答案:1/N4.已知，求=（）参考答案:5.已知，求=（）参考答案:第四章测试1.请判断下面说法是否正确：模拟信号数字处理中，模拟信号与数字信号之间的相互转换中要求不能丢失有用信息（）。
合工大《数字信号处理》习题答案第2章习 题2.4 设系统分别用下面的差分方程描述，)(n x 与)(n y 分别表示系统输入和输出，判断系统是否是线性非时变的。
（1）)()(0n n x n y -= （3）)sin()()(n n x n y ω=解： （1）)()()()()]()([21020121n by n ay n n bx n n ax n bx n ax T +=-+-=+所以是线性系统。
由于)()]([0n n x n x T -= 所以是时不变系统。
（3）)()()sin()]()([)]()([212121n by n ay n n bx n ax n bx n ax T +=+=+ω，所以是线性系统。
)()sin()()]([m n y n m n x m n x T -≠-=-ω，所以不是时不变系统。
2.5 给定下述系统的差分方程，试判定系统是否是因果稳定系统，并说明理由。
（1）)1()()(++=n x n x n y （3）)()(n x e n y =解：（1）该系统是非因果系统，因为n 时刻的输出还和n 时刻以后（)1(+n 时间）的输入有关。
如果M n x ≤|)(|，则M n x n x n y 2|)1(||)(||)(|≤++≤，因此系统是稳定系统。
（3）系统是因果系统，因为n 时刻的输出不取决于)(n x 的未来值。
如果M n x ≤|)(|，则M n x n x e e e n y ≤≤≤)|(|)(|||)(|，因此系统是稳定系统。
2.6 以下序列是系统的单位冲激响应)(n h ，试说明该系统是否是因果、稳定的。
（1）)(2)(n u n h n= （3）)2()(+=n n h δ解：（1）当0<n 时，0)(=n h ，所以系统是因果的。
由于所以系统不稳定。
（3）当0<n 时，0)(≠n h ，所以系统是非因果的。
由于所以系统稳定。
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。
数字信号处理课后答案1.2 教材第一章习题解答1. 用单位脉冲序列()n δ及其加权和表示题1图所示的序列。
解：()(4)2(2)(1)2()(1)2(2)4(3) 0.5(4)2(6)x n n n n n n n n n n δδδδδδδδδ=+++-+++-+-+-+-+-2. 给定信号：25,41()6,040,n n x n n +-≤≤-⎧⎪=≤≤⎨⎪⎩其它（1）画出()x n 序列的波形，标上各序列的值；（2）试用延迟单位脉冲序列及其加权和表示()x n 序列； （3）令1()2(2)x n x n =-，试画出1()x n 波形； （4）令2()2(2)x n x n =+，试画出2()x n 波形； （5）令3()2(2)x n x n =-，试画出3()x n 波形。
解：（1）x(n)的波形如题2解图（一）所示。
（2）()3(4)(3)(2)3(1)6() 6(1)6(2)6(3)6(4)x n n n n n n n n n n δδδδδδδδδ=-+-+++++++-+-+-+-（3）1()x n 的波形是x(n)的波形右移2位，在乘以2，画出图形如题2解图（二）所示。
（4）2()x n 的波形是x(n)的波形左移2位，在乘以2，画出图形如题2解图（三）所示。
（5）画3()x n 时，先画x(-n)的波形，然后再右移2位，3()x n 波形如题2解图（四）所示。
3. 判断下面的序列是否是周期的，若是周期的，确定其周期。
（1）3()cos()78x n A n ππ=-，A 是常数；（2）1()8()j n x n e π-=。
解：（1）3214,73w w ππ==，这是有理数，因此是周期序列，周期是T=14； （2）12,168w wππ==，这是无理数，因此是非周期序列。
5. 设系统分别用下面的差分方程描述，()x n 与()y n 分别表示系统输入和输出，判断系统是否是线性非时变的。
数字信号处理课后答案1.2 教材第一章习题解答1. 用单位脉冲序列()n 及其加权和表示题1图所示的序列。
解：()(4)2(2)(1)2()(1)2(2)4(3)0.5(4)2(6)x n nn n n n nnn n 2. 给定信号：25,41()6,040,nnx n n其它（1）画出()x n 序列的波形，标上各序列的值；（2）试用延迟单位脉冲序列及其加权和表示()x n 序列；（3）令1()2(2)x n x n ，试画出1()x n 波形；（4）令2()2(2)x n x n ，试画出2()x n 波形；（5）令3()2(2)x n x n ，试画出3()x n 波形。
解：（1）x(n)的波形如题2解图（一）所示。
（2）()3(4)(3)(2)3(1)6()6(1)6(2)6(3)6(4)x n nnnn n n n n n （3）1()x n 的波形是x(n)的波形右移2位，在乘以2，画出图形如题2解图（二）所示。
（4）2()x n 的波形是x(n)的波形左移2位，在乘以2，画出图形如题2解图（三）所示。
（5）画3()x n 时，先画x(-n)的波形，然后再右移2位，3()x n 波形如题2解图（四）所示。
3. 判断下面的序列是否是周期的，若是周期的，确定其周期。
（1）3()cos()78x n A n，A 是常数；（2）1()8()j n x n e 。
解：（1）3214,73w w ，这是有理数，因此是周期序列，周期是T=14；（2）12,168ww，这是无理数，因此是非周期序列。
5. 设系统分别用下面的差分方程描述，()x n 与()y n 分别表示系统输入和输出，判断系统是否是线性非时变的。
（1）()()2(1)3(2)y n x n x n x n；（3）0()()y n x n n ，0n 为整常数；（5）2()()y n x n ；（7）0()()n m y n x m 。
