数字信号处理(英文版)课后习题答案4
数字信号处理 英文版 课后习题
The difference equation can be found in two ways. By replacing h[n] with y[n] and [n] with x[n], the impulse response itself furnishes the difference equation y[n] = 2x[n] – 1.5x[n–1] + x[n–2] + 0.5x[n–3] Alternatively, the same difference equation can be found by using H(z) = Y(z)/X(z), cross-multiplying, and taking an inverse z transform. 6.9 n h[n] The samples of the impulse response are given in the table. 0 0.0 1 0.5625 2 0.3164 3 0.1780 4 0.0
The samples of the impulse responses are: 0 1.0 0.0 1 0.2 –3.0 2 0.04 –6.0 3 0.0 –9.0
The transfer functions of the systems are: H1(z) = 1 + 0.2z–1 + 0.04z–2 H2(z) = –3z–1 – 6z–2 – 9z–3 In cascade, the filters give the transfer function H(z) = H1(z)H2(z) = 3z 1 6.6z 2 10.32z 3 2.04z 4 0.36z 5 6.12 (b) 6.13 6.14
数字信号处理课后答案第3和4章fb
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第３章
离散傅里叶变换(DFT)及其快速算法 (FFT)
解法二
因为
由DFT共轭对称性可得同样结果。
离散傅里叶变换(DFT)及其快速算法 (FFT)
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第３章
离散傅里叶变换(DFT)及其快速算法 (FFT)
教材第3章习题与上机题解答
1． 计算以下序列的N点DFT， 在变换区间0≤n≤N－1内， 序列定义为 (1) x(n)=1 (2) x(n)=δ(n) (3) x(n)=δ(n－n0) 0<n0<N (4) x(n)=Rm(n) 0<m<N
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第３章
离散傅里叶变换(DFT)及其快速算法 (FFT)
3． 已知长度为N=10的两个有限长序列：
数字信号处理 英文版 课后习题
H(s)
p1 s p1
23946 s 23946
After the bilinear transformation, the digital transfer function is obtained:
H(z) 23946 0.6(1 z 1 ) z 1 1 0.1989z 1 16000 23946 z 1
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1504.4 0.0859(1 z 1 ) H( z) z 1 1 0.8281z 1 16000 1504.4 z 1
The difference equation is y[n] = 0.8281y[n–1] + 0.0859x[n] + 0.0859x[n–1]. 0.0859(1 e j ) (d) The frequency response is H() . The magnitude response 1 0.8281e j may be found by taking the magnitudes of this expression for several values of . It is plotted below against frequency in Hz. (c)
数字信号处理_习题与解答
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系统是线性系统 9
( 4)T [ x( n )] x( n n0） ( a )若 | x( n ) | M ，则： | T [ x( n )] || x( n n0 ) | M 系统是稳定系统 (b ) y( n ) x( n n0 )， (i )n0 0, n n0 n, y( n )与n时刻之后的输入无关 系统是因果系统 (ii )n0 0, n n0 n, y( n )与n时刻之后的输入有关 系统不是因果系统 (c ) T [ax1( n ) bx2 ( n )] ax1( n n0 ) bx2 ( n n0 ) aT[ x1( n )] bT[ x2 ( n )] 系统是稳定系统
15
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数字信号处理课后答案
⎣⎢y[3]⎦⎥ ⎣⎢1 − j −1 j ⎦⎥ ⎣⎢2 + j⎦⎥ ⎣⎢ 2 ⎦⎥
5.56 Similarly to the solution of Problem 5.55, we first calculate the DFT X[k] of x[n]: €
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数字信号处理英文版课后答案（3）
Chapter 3 Solutions3.1 (a) (i) x[0] = 3(ii) x[3] = 5(iii) x[–1] = 2(b) (i)(ii) The sketch for x[n+1] does not show the value of the sample x[5], since this information is not provided in the question.3.2 The impulse function is zero everywhere except n = 0.(a) δ[–4] = 0(b) δ[0] = 1(c) δ[2] = 03.3 (a) This function is a mirror image of the impulse function across the verticalaxis, which means no change occurs.(b) This function shifts the impulse two steps to the right and increases its amplitude to 2.(c) This function is the sum of two impulse functions.3.4 The step function is zero for n < 0 and one everywhere else.(a) u[–3] = 0(b) u[0] = 1(c) u[2] = 13.5 (a) x[n] = 4u[n–1](b) x[n] = –2u[n](c) x[n] = 2u[–n](d) x[n] = u[n–2](e) x[n] = u[2–n]3.6(a)(b)3.7 (a) This signal is a sum of shifted step functions, each with amplitude one. x[n] = 0.1u[n –1] + 0.1u[n –2] + 0.1u[n –3] + …(b) This signal is a sum of impulse functions with increasing amplitude. x[n] = 0.1δ[n –1] + 0.2δ[n –2] + 0.3δ[n –3] + …3.8 x[n] = (u[n] – u[n –2]) + (u[n –5] – u[n –7]) + (u[n –10] – u[n –12])3.9 x[n] = 2δ[n] – 3δ[n –1] + δ[n –2] – δ[n –3] + 3δ[n –4]3.10 (a)x[n] = δ[n –3] + δ[n –4] + δ[n –5] + δ[n –6] – δ[n –7] – δ[n –8] – δ[n –9] – δ[n –10] – δ[n –11] ∑∑==-δ--δ=117k 63k ]k n []k n [(b) x[n] = u[n –3] – 2u[n –7] + u[n –12] 3.11 x[n] = u[n] + 2u[n –4]3.123.13 The signal has values 1, 0.5, 0.25, 0.125, etc. These values can be generated from the function 0.5n , where each value is the amplitude of an impulse function. The signal may be expressed as ∑∞=-δ=0k k ]k n [)5.0(]n [x = δ[n] + 0.5δ[n –1] + 0.25δ[n –2] + 0.125δ[n –3] + …3.14 (a)By Euler’s identity, ⎪⎭⎫ ⎝⎛π-⎪⎭⎫ ⎝⎛π==π-4n sin j 4n cos e]n [x 4n j(b)From (a), 4π=Ω. Therefore, 18422=ππ=Ωπ. The digital period is 8 samples.3.15 (a)⎪⎭⎫⎝⎛π-⎪⎭⎫ ⎝⎛π==π-3n sin j 3n cos e ]n [x 3n j(b) The magnitude of any complex number is the square root of the sum of the squares of the real and imaginary parts. The magnitudes in the last column in (a) show that 3n j eπ- = 1. This equation is true for all n.3.16 The frequency of the analog signal is f = ω/2π = 200/2π Hz. The samplingfrequency f S = 1/T S = 1/(25x10–3) = 40 Hz. The digital frequency is Ω = 2πf/f S = 200/40 = 5 rads. The sampled signal is x[n] = 5sin(n Ω) = 5sin(5n).3.17 Check 2π/Ω for each function. The function is periodic if this ratio is rational. (a) 2π/Ω = 2π/(4/5) = 10π/4 = 5π/2This ratio is not rational, so the sinusoid is not periodic. (b) 2π/Ω = 2π/(6π/7) = 14/6 = 7/3This ratio is rational, and in lowest terms. The number in the numerator, 7, is the number of samples before the sequence repeats. (c) 2π/Ω = 2π/(2π/3) = 3This result may be seen as 3/1. Thus, the sinusoid is periodic with period 3.3.18 (a) From the equation for x(t), ω = 2πf = 1000π, so f = 500 Hz. Since seven samples are collected every three cycles, N = 7 and M = 3, so37M N 2==ΩπThis means SS f 5002f f 276π=π=π=Ω. Solving S f 500276π=π gives f S = 1167 Hz. (b) Since f S > 2f, the sampling rate is adequate to avoid aliasing.3.19 (a) The following samples are graphed below.(b) The ratio 2π/Ω is 2π/(4π/5)=10/4 = 5/2. The numerator, 5, indicates the sinusoidal sequence repeats every five samples. Because the denominator of the ratio is 2, these five samples are collected over two analog periods.(c) The analog signal is superimposed over the digital signal with a dashed line in the figure below.3.20 The analog frequency of x(t) is f = ω/(2π) = 2500π/(2π) = 1250 Hz. The digital frequency of x[n] is π/3 radians. These frequencies must be related through the equationSS f 12502f f 23π=π=π=ΩThe solution to this equation is f S = 7500 Hz. One other solution is possible, since⎪⎭⎫⎝⎛π=⎪⎭⎫ ⎝⎛π-π=⎪⎭⎫ ⎝⎛π3n 5cos 3n n 2cos 3n cos . This view givesSf 1250235π=π=Ωor f S = 1500 Hz. At this frequency, aliasing occurs. The signal appears at a frequency of 250 Hz. This explains why this second sampling rate works:315002502f f 2S π=π=π=Ω3.21 (a) From 052n 92=π+π, n = 59-. The shift moves the function left by 9/5samples. (b) The samples in the two signals do not match, because the shift is not an integer.(c) 951.052)0(92sin ]0[x 1=⎪⎭⎫ ⎝⎛π+π=927.052)1(92sin ]1[x 1=⎪⎭⎫ ⎝⎛π+π=(d) For a phase shift of two samples to the right,0)2(92n 92=θ+π=θ+π, so 94π-=θ. Thus, ⎪⎭⎫ ⎝⎛π-π=94n 92sin ]n [x 1. One period of this signal contains the same sample values as one period of x 2[n]. 3.22 (a)(b)(c)(d) 3.23(a)Since the digital sinusoid is periodic,M 10M N 2==Ωπ. Since Sf f 2π=Ω, M10f f S =. Therefore, 10Mf f S=. Possible frequencies f of the analog signal are defined by M = 1, 3, 7, 9, 11, …, that is, all integers that do not share any factors with 10. Other integers M result in a digital period less than 10. For 4 kHz sampling, the possible frequencies f are 400, 1200, 2800, 3600, 4400, … Hz. (b) The only two frequencies from (a) that lie within the Nyquist range are 400 Hz and 1200 Hz. All other frequencies f, when sampled at 4 kHz, produce aliases at 400 or 1200 Hz.3.24 A 16-bit image uses 16 bits to represent the gray scale level for each pixel in the image. A total of 65,536 gray scale values can be represented with 16 bits.3.25 Each square on the checkerboard is recorded by a 16 x 16 block of pixels. All pixels in the white squares have gray scale value 255. All pixels in the black squares have gray scale value 0.。
离散数字信号处理 奥本海姆 第三版 第四章 习题答案 英文版
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数字信号处理》课后作业参考答案
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(英文版)课后习题答案4
(Partial) Solutions to Assignment 4pp.81-82Discrete Fourier Series (DFS)Discrete Fourier Transform (DFT), k=0,1,...N-1, n=0,1,...N-1Discrete Time Fourier Transform (DTFT)is periodic with period=2πFourier Series (FS)Fourier Transform (FT)---------------------------------------------------- 2.1 Consider a sinusoidal signalQ2.1 Consider a sinusoidal signalthat is sampled at a frequency s F =2 kHza). Determine an expressoin for the sampled sequence , and determine itsdiscrete time Fourier transformb) Determinec) Re-compute ()X from ()X F and verify that you obtain the same expression as in (a)a). ans:=where andUsing the formular:b) ans:wherec). ans:Let be the sample function. The Fourier transform of isUsing the relationship orwhereConsider only the region where ( orthereforewhereEND-----------------------------2.3 For each shown, determine whereis the sampled sequence. The sampling frequence is given for each case.(b) Hz(d) Hztheory: the relationship between DTFT and FT iswhereorb. ans:d. ans:omitted (using the same method as above)----------------------------------------------------2.4 In the system shown, let the sequence be and the sampling frequency be kHz. Also let the lowpass filter be ideal, with bandwidth (a). Determine an expression for Also sketch the frequency spectrum (magnitude only) within the frequency range(b) Determine the output signal(a) ansFrom class notes, we have where is an ZOH interpolation function and We can writeFirstly, to findwhereIt can be found asSecondly, find This can be solved either by FT or DTFT.We can writewhere andUsing the formula:we haveUsing the formula,:we have from DTFT of y[n]Note the above expression is two pulses at and -the scaling factor is:whereTherefore,where(b) ans:After the ideal LPF, the Fourier transform ofTake inverse Fourier transform of , the output signal is:Note both the and θ terms are introduced by ZOH functionwhere is introduced because is non-ideal and θ represents the delay of----------------------------------------------------Q 2.5. We want to digitize and store a signal on a CD, and then reconstruct it at a later time. Let the signaland let the sampling frequency Hz.(a) Determine the continuous time signal after the reconstruction.(a) ans: Assuming (ZOH+ ideal LPF) is used. This problem can be solved by using the results directly from Q2.4. In Q2.5 there are 3 sinusoidal signals instead of only one in Q2.4. Details of the solutions are omitted.----------------------------------------------------Q 2.6 In the system shown, determine the output signal for each of the following input signal Assume the sampling frequency kHz and the low pass filter (LPF) to beideal, with bandwidth(b)(d)Ans (b) (d): same as in problem Q2.5.----------------------------------------------------2.7 Suppose in DAC you want to use a linear interpolation between samples, as shown in the accompanying figure. This reconstructor can be called a first order hold, because the equation of a line is a polynomial of degree 1(a). Show that with a triangular pulse as shownin the figure(b). Determine an expression for in terms ofand(c). In the accompanying figure, let kHz, and the filterbe ideal with bandwidth Determine the outputAns: omitted.----------------------------------------------------2.9 In the following system, let the signal be affected by some random error as shown. The error is white, zero mean, with variance Determine the variance of the error after the filter for each of the filter(b)(b) ans:The variance of the output of the filter is given byTherefore--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------。
数字信号处理丁玉美版第四章课后答案
1.解:当N =1024=210时，直接计算DFT 的复数乘法运算次数为 N 2=1024×1024=1 048 576次复数加法运算次数为 N （N －1）=1024×1023=1 047 552次直接计算所用计算时间TD 为 T D =4×10－6×10242+1 047 552×10－6=5.241 856 s用FFT 计算1024点DFT 所需计算时间T F 为快速卷积时， 需要计算一次N 点FFT （考虑到H (k )= DFT ［h (n )］已计算好存入内存）、 N 次频域复数乘法和一次N 点IFFT 。
所以， 计算1024点快速卷积的计算时间T c 约为所以， 每秒钟处理的采样点数（即采样速率）由采样定理知， 可实时处理的信号最高频率为 应当说明， 实际实现时， f max 还要小一些。
这是由于实际中要求采样频率高于奈奎斯特速率， 而且在采用重叠相加法时， 重叠部分要计算两次。
重叠部分长度与h (n )长度有关， 而且还有存取数据和指令周期等消耗的时间。
2.解:直接计算1024点DFT 所需计算时间TD 为T D =10×10－9×10242+10×10－9×1 047 552=20.961 28 ms用FFT 计算1024点DFT 所需计算时间T F 为 快速卷积计算时间T c 约为可实时处理的信号最高频率f max 为由此可见， 用DSP 专用单片机可大大提高信号处理速度。
所以， DSP 在数字信号处理领域得到广泛应用。
机器周期小于1 ns 的DSP 产品已上市，其处理速度更高。
3. 解: 因为x (n )和y (n )均为实序列， 所以， X (k )和Y (n )为共轭对称序列， j Y (k )为共轭反对称序列。
可令X (k )和j Y (k )分别作为复序列F (k )的共轭对称分量和共轭反对称分量， 即F (k )=X (k )+j Y (k )=F ep (k )+F op (k )计算一次N 点IFFT 得到 f (n )=IFFT ［F (k )］=Re ［f (n )］+j Im ［f (n )］由DFT 的共轭对称性可知Re ［f (n )］=IDFT ［F ep (k )］=IDFT ［X (k )］=x (n )j Im ［f (n )］=IDFT ［F op (k )］=IDFT ［j Y (k )］=j y (n )66F 66510lb lb 10210245101010241010230.72 ms N T N N N ----=⨯⨯+⨯=⨯⨯⨯+⨯⨯=c F 2102471680 μs 41024 μs 65536 μs TT =+=+⨯=次复数乘计算时间s 6102415 625 /6553610F -<=⨯次秒s max 156257.8125 kHz 22F f <==99F 881010l b 1010l b 2102410101010241020.1536 ms N T N N N ----=⨯⨯+⨯⨯=⨯⨯+⨯⨯=c F 3921024 20.153610101010240.317 44 ms T T --=+=⨯⨯+⨯⨯=次复数乘计算时间max s c 1110241 = 3.1158 MHz=1.6129 MHz 222f F T =故 1()[()()]2x n f n f n *=+1()[()()]2j y n f n f n *=-。
数字信号处理课后习题答案(全)1-7章
最后结果为 0
n<0或n>7
y(n)= n+1 0≤n≤3 8－n 4≤n≤7
y(n)的波形如题8解图（一）所示。 (2) y(n) =2R4(n)*［δ(n)－δ(n－2)］=2R4(n)－2R4(n－2)
=2［δ(n)+δ(n－1)－δ(n+4)－δ(n+5) y(n)的波形如题8解图（二）所示
因此系统是非线性系统。
第 1 章 时域离散信号和时域离散系统
（6） y(n)=x(n2)
令输入为
输出为
x(n－n0)
y′(n)=x((n－n0)2) y(n－n0)=x((n－n0)2)=y′(n) 故系统是非时变系统。 由于
T［ax1(n)+bx2(n)］=ax1(n2)+bx2(n2) =aT［x1(n)］+bT［x2(n)］
因此系统是非时变系统。
第 1 章 时域离散信号和时域离散系统
（5） y(n)=x2(n)
令输入为
输出为
x(n－n0)
y′(n)=x2(n－n0) y(n－n0)=x2(n－n0)=y′(n) 故系统是非时变系统。 由于
T［ax1(n)+bx2(n)］=［ax1(n)+bx2(n)］2 ≠aT［x1(n)］+bT［x2(n) =ax21(n)+bx22(n)
第 1 章 时域离散信号和时域离散系统
题4解图（一）
第 1 章 时域离散信号和时域离散系统
题4解图（二）
第 1 章 时域离散信号和时域离散系统
题4解图（三）
第 1 章 时域离散信号和时域离散系统
(4) 很容易证明： x(n)=x1(n)=xe(n)+xo(n)
数字信号处理英文版课后答案（4）
Chapter 4 Solutions4.1 (a) The pass band gain for this filter is unity. The gain drops to 0.707 of this value at 2400 Hz and 5200 Hz. Thus, the frequencies passed by the filter lie in the range 2400 to 5200 Hz.(b) The filter is a band pass filter.(c) The bandwidth is the range of frequencies for which the gain exceeds 0.707 of the maximum value, or 5200 –2400 = 2800 Hz.4.2 A low pass filter passes frequencies between DC and its cut-off frequency. The bandwidth is identical to the cut-off frequency. Thus, the cut-off frequency is 2 kHz.4.3 The maximum pass band gain of the filter is 20 dB. The bandwidth is defined as the range of frequencies for which the gain is no more than 3 dB below the pass band gain, or 17 dB. This gain occurs at the cut-off frequency of 700 Hz. For a high pass filter, the bandwidth is the range of frequencies between the cut-off frequency, 700 Hz, and the Nyquist frequency (equal to half the sampling rate), 2 kHz. The bandwidth is 1300 Hz. 4.4 The low pass filter has a cut-off frequency of 150 Hz and bandwidth 150 Hz. The band pass filter has cut-off frequencies at 250 Hz and 350 Hz for a bandwidth of 100 Hz. The high pass filter has a cut-off frequency of 400 Hz and a bandwidth of 100 Hz, which extends from its cut-off frequency to the Nyquist limit at half the sampling rate.4.5 (a) The low pass filter output is on the left. The high pass filter output is on the right.(b) An approximation to the original vowel signal can be found by adding the high and low pass waveforms together.x[n]n4.6 (a) linear (b) non-linear (c) non-linear (d) linear4.7 Since the new input is shifted to the right by two positions from the original input, the new output is shifted to the right by two positions from the original output.4.8(a) y[n] = –0.25y[n–1] + 0.75x[n] – 0.25x[n–1](b) y[n] = y[n–1] – x[n] – 0.5x[n–1]4.9 (a) The system is non-recursive.b0 = b1 = b2 = 1/3(b) The system is recursive.a0 = 1, a1 = –0.2, b0 = 1(c) The system is recursive.a0 = 1, a1 = 0.5, b0 = 1, b1 = –0.44.10 (a)(c)(d)4.114.124.13 The overall input x[n] for any sampling instant is the sum of the inputs x1[n] and x2[n]. This overall input is applied to the difference equation in the normal way to obtain outputs.4.144.154.16 y[n] = 0.5y[n–2] + 1.2x[n] – 0.6x[n–1] + 0.3x[n–2]4.17 y[n] = 2.1x[n–1] – 1.5x[n–2]4.18 w[n] = x[n] + 0.3w[n–1] – 0.1w[n–2]y[n] = 0.8w[n] – 0.4w[n–2]4.19 The difference equation for the first second-order section isy1[n] = –0.1x[n] + 0.2x[n–1] + 0.1x[n–2]The difference equation for the second second-order section isy[n] = y1[n] + 0.3y1[n–2]Substituting the first equation into the second givesy[n] = (–0.1x[n] + 0.2x[n–1] + 0.1x[n–2])+ 0.3(–0.1x[n–2] + 0.2x[n–3] + 0.1x[n–4]) = –0.1x[n] + 0.2x[n–1] + 0.07x[n–2] + 0.06x[n–3] + 0.03x[n–4] 4.204.21 The direct form 2 equations are:w[n] = x[n] + 1.2w[n–1] – 0.5w[n–2]y[n] = w[n] – 0.2 w[n–1]4.22 (a) y[n] = –0.14 y[n–1] – 0.38 y[n–2] + x[n] y[n]y[n](b) w[n] = x[n] – 0.14w[n –1] – 0.38w[n –2] y[n] = w[n]Note that the difference equation diagram for this part is the same as that for part (a).4.23 The first ten samples of the impulse response are:4.24 From the figure, the filter has a finite impulse response. It may be described as a sum of impulse function as:h[n] = 0.5δ[n] + 0.4δ[n –1] + 0.3δ[n –2] + 0.2δ[n –2] The difference equation has the parallel form:y[n] = 0.5x[n] + 0.4x[n –1] + 0.3x[n –2] + 0.2x[n –3]4.25 The impulse response is finite, with samples as listed in the table.The impulse response samples for a FIR filter serve directly as b k coefficients, so y[n] = x[n] + 0.3x[n –1] + 0.09x[n –2] + 0.027x[n –3]This result may also be seen by writing the impulse response in terms of impulse functions: h[n] = δ[n] + 0.3δ[n –1] + 0.09δ[n –2] + 0.027δ[n –3]y[n]4.264.27 The impulse response may be found from the difference equation ash[n] = – 0.5h[n –1] + δ[n] – 0.8δ[n –1]The step response may be found froms[n] = – 0.5s[n –1] + u[n] – 0.8u[n –1]or by finding a cumulative sum of the impulse response samples.4.28 The difference equation for a five-term moving average filter is()]4n [x ]3n [x ]2n [x ]1n [x ]n [x 51]n [y -+-+-+-+=The impulse response,()]4n []3n []2n []1n []n [51]n [h -δ+-δ+-δ+-δ+δ=is plotted below.4.29 The impulse response belongs to a non-recursive filter because, after a finite number of samples, the output settles to zero permanently.4.30 (a) The response to an impulse function is, by definition, the impulse response. Therefore, the answer to (a) is provided in the question.(b) The signal x[n] consists of two impulse functions with different amplitudes and locations. The response to this input will be the same combination of impulse responses, that is,y[n] = 0.8h[n] + 0.5h[n–1]The output samples are listed in the following table:4.31 The step response can be obtained froms[n] = u[n] – 0.5u[n–1] – 0.7u[n–2]The first ten samples are:4.32 (a) The impulse response can be obtained fromh[n] = –0.75h[n–1] + δ[n] – 0.5δ[n–1](b) The step response may be obtained froms[n] = –0.75s[n–1] + u[n] – 0.5u[n–1]or by finding a cumulative sum of the impulse response samples.。
数字信号处理 Mitra 第三版 答案
4.
Page 290, Problem 5.12:
Replace
“
x1[n]
=
⎜⎝⎛
x[n]
+
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+ n]⎟⎠⎞WNn ” with
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+
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”.
5.
Page 291, Problem 5.15: Replace “ h[n] = ⎩⎨⎧x[0n,],
3. Page 418, Problem 7.53: Replace “ − j0.3 ” with “ − j ”.
4. Page 421, Problem 7.79: Replace “ 0.5 − 0.4z −1 + 0.8z −2 + 0.8z −3 − 0.4z −4 + 0.5z −5 ” with “ − 0.1 + 0.5z −1 + 0.05z −2 + 0.05z −3 + 0.5z −4 − 0.1z −5 ”.
4.
Page 161, Problem 3.6(d):
Replace
“
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⎜⎛ ⎝
Ω a
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”
with
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5. Page 162, Problem 3.16: Replace “1.. ” with “1. ”.
1
6.
Page 163, Problem 3.21(d):
12. Page 167, Problem 3.61(d): Replace “ Hb (e jω ) ” with “ Hd (e jω ) ”.
数字信号处理课后答案
A → x( t ) 2
Signals and Signal Processing 8
Typical Signal Processing Operations
• Multiplexing (复用) and Demultiplexing
– For an efficient utilization or a wideband transmission channel
云南大学课程：数字信号处理
Typical Signal Processing Operations
• Filtering — convolutional integral
– To improve the quality of the signal – To pass / block certain frequency components in a signal through the filter without any distortion
“DSP”
云南大学课程：数字信号处理
Signals and Signal Processing 18
Course Outline
• • • • • • • • • • Signals and Signal Processing (2hr) Discrete-Time Signals and Systems (4hr) Discrete-Time Fourier Transform (6hr) Digital Processing of Continuous-Time Signals (4hr) Finite-Length Discrete Transform (8hr) LTI Discrete-Time Systems in the Transform Domain (6hr) Digital Filter Structures (4hr) IIR Digital Filter Design (6hr) FIR Digital Filter Design (4hr) DFT Algorithm Implementation (4hr)
数字信号处理课后答案第3和4章
用DFT/FFT对信号进行谱分析的误差表现在三个方面， 即混叠现象、 栅栏效应和截断效应。 截断效应包括泄漏和 谱间干扰。
第３章 离散傅里叶变换(DFT)及其快速算法
第３章 离散傅里叶变换(DFT)及其快速算法
(FFT)
xN(n)=IDFT［X(k)］为x(n)的周期延拓序列（以N为延拓周期） 的主值序列。 以后这一结论可以直接引用。
［例3.4.2］ 已知 x(n)=R8(n), X(ejω)=FT［x(n)］
对X(ejω)采样得到X(k)，
X(k)X(ej)|2πk, k0,1, ,5 6
第３章 离散傅里叶变换(DFT)及其快速算法
(FFT)
当然， 截取信号的长度要足够长。 但如果截取的长度 不够长， 而依靠在所截取的序列尾部加零点， 增加变换区 间长度， 也不会提高分辨率。 例如， 分析周期序列的频谱， 只观察了一个周期的1/4长度， 用这些数据进行DFT， 再通 过尾部增加零点， 加大DFT的变换区间N， 也不能分辨出是 周期序列， 更不能得到周期序列的精确频率。
令m=N－1－n， 则上式可写成
0
N1
X(k) x(m )W N k(n1) x(m )W N km
m N1
m 0
W N k(N 1 )X ( (k)N )R N (k)
第３章 离散傅里叶变换(DFT)及其快速算法
(FFT)
当 k N 时（N为偶数）， 2
因为
X N 2 W N N 2(N 1 )X N 2 NW N N 2(N 1 )X N 2
数字信号处理课后习题Ch4
= ( n−2 α )π sin[( n − α ) wc ] sin[( n − α ) w0] 因为 h(n)=hd (n)W R (n)= hd(n) R N (n) 所以（1）当 N 为奇数时 h（n）= { (0 ≤ n ≤ N − 1) ; 0 , 其他 （2）当 N 为偶数时 h（n）的表达式与 N 为奇数时相同 （3）若采用汉明窗设计 h(n)= h d (n)W(n) = ( n−2 α )π sin[( n − α ) wc ] sin[( n − α ) w0] [0.54-0.46cos (
N ⎧ 2 1 ⎪ H (ω ) = b ( n ) c o s [ω ( n − )] ∑ ⎪ 2 n =1 ∴ ⎨ ⎪ N b (n ) = 2 h ( − 1 + n) ⎪ 2 ⎩
∴ （1）当 N 为奇数时
2
sinωc (n − α ) ⎧ RN (n), 0 ≤ n ≤ N − 1 ⎪2cosω0 (n-α ) π (n − α ) h( n) = ⎨ ⎪0, 其他 ⎩
（2）当 N 为偶数时 h(n)的表达式与 N 为奇数时的相同； （3）若用汉明窗设计
h( n) = hd (n)ω (n) = 2 cos ω0 ( n − α ) sin ωc ( n − α ) 2π n [0.54 − 0.46 cos( )]RN (n) π (n − α ) N −1
(1 ).h d ( n ) = =
1 2π
∫
2π 0
H d (e
jω
)e
jn ω
dω
1 π + ω c − j ( ω − π ) α jn ω e e dω 2 π ∫π − ω c π +ωc 1 = e jπ α ∫ e j ( n −α )ω d ω π −ω c 2π 1 1 = e jπ α e j ( n − α ) ω 2π j ( n − α ) 1 e jπ α [e = 2π j ( n − α )
数字信号处理试题(英文版)
Test Of Digital Signal Processing 1Give the answer of the following problems as possible in English.1. Supposing when and .Please give its ZT , DTFTand N-point DFT.Explain the relations among these three transforms .2.A real 8-point sequence .Its 8-point DFT is X[k].(1) Supposing.Sketch .(2) Determine the value of .3. A sequence has the 128-point DFT . Give the expression of .4. A causal LTI discrete-time system has the transfer function.(1) Determine the difference equation of the system. (2) Give the pole-zero plot and the ROC of the system. (3) Is this system stable?(4) Develop a realization of the system in any canonic form. 5. Design a FIR linear phase causal LPF which has cutoff frequency, transition bandwidthand minimum stopband attenuation .Develop the impulse response in closed form by Windowed Fourier Series method.6.If you use following MA TLAB program to design a lowpass filter ,which has the specifications:passband edge 800Hz ,stopband edge 1kHz, passband ripple 0.5dB, minimunstopband attenuation 40dB ,and sampling rate is 5kHz.What data should you input to the computer? % Program For Elliptic IIR Lowpass Filter Design %Wp = input('Normalized passband edge = '); Ws = input('Normalized stopband edge = '); Rp = input('Passband ripple in dB = ');Rs = input('Minimum stopband attenuation in dB = '); [N,Wn] = ellipord(Wp,Ws,Rp,Rs) [b,a] = ellip(N,Wn); [h,omega] = freqz(b,a,256);plot (omega/pi,20*log10(abs(h)));grid; xlabel('\omega/\pi'); ylabel('Gain, dB'); title('IIR Elliptic Lowpass Filter'); 7. In the system of Figure (a), has a bandlimited spectrumas sketched in Figure (b) and is being,0][=n x 0<n 1->N n )(z X )(ωj eX ][k X ⎭⎬⎫⎩⎨⎧≤≤≤≤=76,050,1][n n n x ][][4k X W k Y k=]}[{][k Y IDFT n y =]4[Y ⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧≤≤=otherwise n n n x ,01270),645cos(][π][k X ][k X )5.0)(5.0(1)(22+---=z z z z z H πω25.0=c πω1.0≤∆dB 40)(t x a )(Ωj X asampled at the Nyquist rate. The digital ideal LPF has the frequency response as shown in Figure (c) .Sketchthe spectrum ,and of ,and respectively.Test Of Digital Signal Processing2Give the answer of the following problems in English as possible as you can.1. Determine the fundamental period of sequence .2. Consider the z - transform.(1) (1) Give the pole-zero plot of .(2) (2) Determine all possible ROCs of . Discuss the type of inverse z -transform (left-sided, right-sided, ortwo-sided sequences) associated with each of these ROCs . (3) (3) Give the expression of right-sided sequence.3. A real 8-point sequence.Its 8-point DFT is X[k].(1) Determine the values of,.(2) (2) Supposing.Sketch.)(ωj e H )(ωj e X )(ωj eY )(Ωj Y a ][n x ][n y )(t y aF i g u re (a ) T h e Di g i t a l P r oc e s s i n g o f A C o n t i n uo u s Si g n a lx )(t y aF i g u re (b )-F i g u re (c )mm21()cos()3sin()32f n n n ππ=+))(()(2112814314311----+-++=zzzzz X )(z X )(z X 1,05[]0,67n n x n n +≤≤⎧⎫=⎨⎬≤≤⎩⎭[0]X ∑=7k k X ][][][4k X W k Y k=]}[{][k Y IDFT n y =][][n x n y N=][n h(3) (3) Determine the value of.4. Let ={1 2 1 1},and ={1 1 1},.(1)(1) Sketch the linear convolution.(2)(2) denote the N-point circularconvolution .Determine the values of,.Explain which value is equal to the value of.Digital Signal Processing3Note: Give the answer of the following problems as possible in English.1.（20 points ）A length-8 sequence is given by. Please give its expressions of ZT, DTFTand 8-point DFT in closed-form.2.（20 points ）Consider a length-8 real sequence, defined for,with 8-point DFT,.(1) Calculate the following values of without computing the DFT itself:(a), (b), (c).(2) Sketch the sequence whose 8-point DFT is given by.3. (18points) A causal LTI discrete-time system has the difference equation.(4) (1) Determine the transfer function of the system. (5) (2) Give the pole-zero plot of the system. (6) (3) Determine the impulse response of the system.4. (15 points) The frequency response of a linear phase FIR filter is given by ]4[Y ][n h 30≤≤n ][n x 20≤≤n ][*][][n h n x n y =][55y ][56y ][5y 1,07[]0,n x n otherw ise≤≤⎧=⎨⎩)(z X )(ωj e X ][k X 07n ≤≤{}[]31243012x n =---[]X k 07k ≤≤[]X k [0]X 7[]k X k =∑72[]k X k =∑[]y n 34[][]kY k W X k =1[][2][]4[2]4y n y n x n x n --=--.(1) (1) Determine and sketch the impulse response of the filter.(2) (2) Which type is this linear phase FIR filter?5. (15 points) Using bilinear method , design a digital 2nd-order Butterworth lowpass filter which has the cut-offfrequency.(1) (1) Give the transfer function of the designed filter.(2) (2) Develop a realization structure of the designed filter in cannonic form.Note: The transfer function of analog 2nd-order Butterworth lowpass filter with 3-dB cut-off frequencyis.6. (12 points) Let be an order-63 FIR filter, andbe a length-64 sequence. The input-output relationshipof the filter is shown as following figure.Please write a MATLAB program to computeby FFT , plot the magnitude and phase spectrum of.5. 5. A sequencehas DTFTshown as following figure. Sketch DTFTof. Ifand,whereand.Give the expression between and .2()[14cos(2)]j j H eeωωω-=+[]h n 2c πω=cΩ21()())1a ccH s s s =++ΩΩ[]h n []x n []y n []y n ][n h )(ωj eH )(ωj eY =][n y (1)()n h n -)(][kj eH k H ω=)(][kj eY k Y ω=kNk π2ω=10-≤≤N k ][k Y ][k H ωTest Of Digital Signal Processing4（Give the answer of the following problems in English as possible as you can ）1． 1． A sequenceis known as below:, where(1) Sketch the real part of.(2) If A=1, is it periodic? If it is, give its period.2． 2． For each of the following discrete-time systems, where y[n] and x[n] are, respectively, the output and theinput sequences, determine whether or not the system is (1)linear, (2)causal ,(3) stable,(4)shift-invariant.Where .3． 3． Calculate the corresponding transformation of following functions:(1) Known, its 1024-point DFT is wanted.(2) Known, its DTFT is wanted.(3) Known, its IDTFT is wanted.4． 4． X[k], 0≤ k ≤7, is the 8-point DFT of sequence {x[n]}={ -1, 2, -3, 2, 0, -4, 6, 2}, 0≤ k ≤7. Pleasecalculate the following values without computing DFT:(a); (b) ; (c); (d) .5． 5． Given two sequences: {g[n]}={1 2 3 4}(), {h[n]}={5 0 3 }().(1) Determine the linear convolution of the two sequence *.[]xn [](1)nx n A j =++1()4nA =[]x n 2()()sin (1)()a y n n x n =+2()[][1]b y n a x n b=-+, is a nonzero constanta b 15()cos(),0102364x n n n π=≤≤21,015(){20,n x n o th erw ise≤≤=322313()(4cos sin )cos()32jj X eeωωωωω=++[4]X 7(/4)[]j k k eX k π-=∑70[]k X k =∑72|[]|k X k =∑03n ≤≤02n ≤≤[][]L y n g n =[]h n(2) Determine the circular convolution of the two sequenceNfor N=4 andN=6, While the lengths of sequences less than N, do the zero-padding.(3) From the results of (1) and (2), make the conclusion of how to use circular convolution to help calculating linear convolution.6. Consider the z - transform as below:.(7) (1) Give the pole-zero plot of .(8) (2) Determine all possible ROCs of. Discuss the type of inverse z -transform (left-sided, right-sided,or two-sided sequences) associated with each of these ROCs . (9) (3) Can you find the? If it is possible, what is it? Test Of Digital Signal Processing（Give the answer of the following problems in English as possible as you can ）6． 1． A sequenceis known as below:, where(1) Sketch the real part of.(2) If A=1, is it periodic? If it is, give its period.7． 2． For each of the following discrete-time systems, where y[n] and x[n] are, respectively, the output and theinput sequences, determine whether or not the system is (1)linear, (2)causal ,(3) stable,(4)shift-invariant.Where.8． 3． Calculate the corresponding transformation of following functions:(1) Known, its 1024-point DFT is wanted.(2) Known , its DTFT is wanted.[][]C e y n g n =[]e h n *21121()3(1)(156)4z X z z zz----+=+-+)(z X )(z X ()j X eω[]xn [](1)nx n A j =++1()4nA =[]x n 2()()sin (1)()a y n n x n =+2()[][1]b y n a x n b=-+, is a nonzero constanta b 15()cos(),0102364x n n n π=≤≤21,015(){20,n x n o th erw ise≤≤=(3) Known, its IDTFT is wanted.9． 4． X[k], 0≤ k ≤7, is the 8-point DFT of sequence {x[n]}={ -1, 2, -3, 2, 0, -4, 6, 2}, 0≤ k ≤7. Pleasecalculate the following values without computing DFT:(a); (b); (c); (d) .10．5． Given two sequences: {g[n]}={1 2 3 4}(), {h[n]}={5 0 3 }().(1) Determine the linear convolution of the two sequence*.(2) Determine the circular convolution of the two sequenceNfor N=4 andN=6, While the lengths of sequences less than N, do the zero-padding.(3) From the results of (1) and (2), make the conclusion of how to use circular convolution to help calculating linear convolution.11．6． Consider the z - transform as below:.(10) (1) Give the pole-zero plot of .(11) (2) Determine all possible ROCs of. Discuss the type of inverse z -transform (left-sided, right-sided,or two-sided sequences) associated with each of these ROCs . (12) (3) Can you find the? If it is possible, what is it?322313()(4cos sin )cos()32jj X eeωωωωω=++[4]X 7(/4)[]j k k eX k π-=∑70[]k X k =∑72|[]|k X k =∑03n ≤≤02n ≤≤[][]L y n g n =[]h n [][]C e y n g n =[]e h n *21121()3(1)(156)4z X z z zz----+=+-+)(z X )(z X ()j X eω。
数字信号处理米特拉第四版实验四答案
Q4.5
The plots of the first 100 samples of the impulse responses of the two filters of Questions 4.2
3
2
1
0
-1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 /
From this plot we make the following observations: Usually, it is desirable for a filer to have an approximately linear phase in the passband, which is the same thing as an approximately constant group delay in the passband. This filter is notch filter; it is a bandstop filter with a narrow stopband centered at a normalized frequency just below 0.3. From the graph above, we see that the group delay is approximately constant over much of the passband.
M=3
Magnitude Spectrum |H(ej)| 1
0.5
0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 / Phase Spectrum arg[H(ej)]
4 2 0 -2 -4
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 /
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(Partial) Solutions to Assignment 4
pp.81-82
Discrete Fourier Series (DFS)
Discrete Fourier Transform (DFT)
, k=0,1,...N-1
, n=0,1,...N-1
Discrete Time Fourier Transform (DTFT)
is periodic with period=2πFourier Series (FS)
Fourier Transform (FT)
---------------------------------------------------- 2.1 Consider a sinusoidal signal
Q2.1 Consider a sinusoidal signal
that is sampled at a frequency s F =2 kHz
a). Determine an expressoin for the sampled sequence , and determine its
discrete time Fourier transform
b) Determine
c) Re-compute ()X from ()X F and verify that you obtain the same expression as in (a)
a). ans:
=
where and
Using the formular:
b) ans:
where
c). ans:
Let be the sample function. The Fourier transform of is
Using the relationship or
where
Consider only the region where ( or
therefore
where
END
-----------------------------
2.3 For each shown, determine where
is the sampled sequence. The sampling frequence is given for each case.
(b) Hz
(d) Hz
theory: the relationship between DTFT and FT is
where
or
b. ans:
d. ans:
omitted (using the same method as above)
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2.4 In the system shown, let the sequence be and the sampling frequency be kHz. Also let the lowpass filter be ideal, with bandwidth (a). Determine an expression for Also sketch the frequency spectrum (magnitude only) within the frequency range
(b) Determine the output signal
(a) ans
From class notes, we have where is an ZOH interpolation function and We can write
Firstly, to find
where
It can be found as
Secondly, find This can be solved either by FT or DTFT.
We can write
where and
Using the formula:
we have
Using the formula,:
we have from DTFT of y[n]
Note the above expression is two pulses at and -the scaling factor is:
where
Therefore,
where
(b) ans:
After the ideal LPF, the Fourier transform of
Take inverse Fourier transform of , the output signal is:
Note both the and θ terms are introduced by ZOH function
where is introduced because is non-ideal and θ represents the delay of
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Q 2.5. We want to digitize and store a signal on a CD, and then reconstruct it at a later time. Let the signal
and let the sampling frequency Hz.
(a) Determine the continuous time signal after the reconstruction.
(a) ans: Assuming (ZOH+ ideal LPF) is used. This problem can be solved by using the results directly from Q2.4. In Q2.5 there are 3 sinusoidal signals instead of only one in Q2.4. Details of the solutions are omitted.
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Q 2.6 In the system shown, determine the output signal for each of the following input signal Assume the sampling frequency kHz and the low pass filter (LPF) to be
ideal, with bandwidth
(b)
(d)
Ans (b) (d): same as in problem Q2.5.
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2.7 Suppose in DAC you want to use a linear interpolation between samples, as shown in the accompanying figure. This reconstructor can be called a first order hold, because the equation of a line is a polynomial of degree 1
(a). Show that with a triangular pulse as shown
in the figure
(b). Determine an expression for in terms of
and
(c). In the accompanying figure, let kHz, and the filter
be ideal with bandwidth Determine the output
Ans: omitted.
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2.9 In the following system, let the signal be affected by some random error as shown. The error is white, zero mean, with variance Determine the variance of the error after the filter for each of the filter
(b)
(b) ans:
The variance of the output of the filter is given by
Therefore
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