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数字信号处理(英文版)课后习题答案2

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(Partial) Solutions to Assignment 2pp.73-761.16In each of the following systems, let or be the input and or be the output. Determine whether each systems is (1) linear, (2) time invariant, (3) causal, (4) BIBO stable(g).(i).ans: omitted----------------------------------------------------1.17 A linear time invariant system has impulse response Determine the output sequence for each of the followign input signals:(b)(f)(b) ans:h n is given byThe z-transform of []where ROC1:x n is given byz-transform of []where ROC2:h n is given byTherefore, the z-transform of the output []y nPerform inverse z to get [](f) ans: using the same method as in (b) (details omitted )----------------------------------------------------1.18. A linear time invariant system is defined by the difference equationb. Determine the output of the system when the intpu isc. Determine the output of the system when the input isans: omitted----------------------------------------------------1.19 The following expressions define linear time invariant systems. For each one determine the impulse respnose(a)(e)(a) ans: the impulse response is(e) ans: the impulse response is----------------------------------------------------1.20 Each of the following expressions defines a linear time invariant system. For each one determine whether it is BIBO stable or not(g)(k)BIBO: Bounded input and bounded output(g) ans: omitted(k) ans: omitted----------------------------------------------------1.21. Using the geometric series, for each of the following sequence determine the z-transform and its ROC(d)(g)(i)(d) ans:where ROC:(g) ans:The first part is equal towhere ROC1 isThe second part is equal towhere ROC2 isTherefore combining both parts:where ROC={ROC1 and ROC2}:(i) ans:where ROC: whole complex domain----------------------------------------------------1.22. You know what the and are. Using theproperties only (do not reuse the definition of the z-transform.) determine the z-transform of the following signals(c)(g)where ROC1:where ROC2:(c) ans: using z-transform property:We have:where ROC:(g) ans:details omitted. The final answer isTherefore combining both parts:where ROC={ROC1 and ROC2}:----------------------------------------------------1.23 Using partial fraction expansion, determine the inverse z-transform of the following functions:(c) ,(e) ,(c) ans:(e) ans:procedures are the same as above. details omitted.----------------------------------------------------1.24. For each of the followign linear difference equations, determine the impulse response, and indicate whether the system is BIBO stable or not(a)(c)(a) ans:Take z-transform on both sideswhere ROC:Because is finiteTherefore, the system is BIBO stable(c) ans: omitted (the same as (a))----------------------------------------------------1.25. Although most of the time we assume causality, a linear difference equation can be interpreted in a number of ways. Consider the linear difference equation(a) Determine the transfer function and the impulse response. Is the system causal ? BIBO stable ?(a) omitted.----------------------------------------------------1.26. 1.26 Consider the linear difference equation(a) Determine the transfer function . Do you have enough information to determine theregion of convergenceans:Don't have enough information to determine ROC.----------------------------------------------------1.27. Given the system described by the linear difference equationDetermine the output for each of the following input signals(a)(e)(a) ans:Take z-transform on both sides:----------------------------------------------------1.28. Repeat Problem 1.27 when the system is given in terms of the impulse responseBefore you do anything, is the system stable ? Does the frequency responseexist ?ans: omitted.----------------------------------------------------1.29. Repeat Problem 1.27 when the system isgiven by the linear difference equationBefore you do anything, is the system stable ? Does the frequency response exist ?Ans: omitted.----------------------------------------------------。

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数字信号处理 英文版 课后习题

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2章(高西全丁美玉第三版)

数字信号处理课后答案+第2章(高西全丁美玉第三版)

~ X (k ) DFS[ ~ (n)] x
j k j k e 4 (e 4

n 0
3
~ (n)e x

j
2 kn 4
π 4

n 0
1
j kn e 2

1j k e 2源自j k e 4 )
2 cos(
π j k k) e 4
~ X ( k )以4为周期
证明输入x(n)=A cos(ω0n+j)的稳态响应为
y (n) A | H (e j0 ) | cos0 n j (0 )
解： 假设输入信号x(n)=ejω0n，系统单位脉冲响应为h(n)， 则系统输出为

y ( n) h( n) x ( n) e j 0 n
(9)
x(n / 2) n 偶数 x9 (n) n 奇数 0
解：（1）
FT[ x(n n0 )]
n


x(n n0 )e jn
令n′=n－n0, 即n=n′+n0， 则
FT[ x(n n0 )]
（2）
FT[ x (n)]

n
x(n)e jn
令n′=－n， 则
FT[ x(n)]
n

x(n)e jn X (e j )
（4）
FT［x(n)*y(n)］=X(ejω)Y(ejω)

下面证明上式成立：
x ( n) y ( n)
m
x ( m) y ( n m)
FT[ x(n) y (n)]
n


x(n)e j2n X (e j2 )
数字信号处理II习题解答-2004(老师版)精选全文

数字信号处理II习题解答-2004(老师版)精选全文

可编辑修改精选全文完整版1. A sinusoidal signal )2/sin()(πn n x =is applied to a second-order linear predictor as inFig.1 . Calculate the theoretical ACF(Auto-Correlation Function) of the signal and the prediction coefficients. Verify that the zeros of the FIR prediction filter are on the unit circle at the right frequency.Using the LMS algorithm with 1.0=δ, show the evolution of the coefficients from time 0=n to 10=n . How is that evolution modified if the sign algorithm is used instead.(n x (e Fig.1. Second-order prediction filter解：a)计算预测系数理论值(滤波器系数的维纳最优解)，由2cos21]2)(sin 2sin [21]2)(sin 2[sin )]()([)(10πππππk k i i k n n E k n x n x E k r i xx =-=-=-=∑=⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡=∴2/1002/1)0()1()1()0(r r r r R x ，⎥⎦⎤⎢⎣⎡-=⎥⎦⎤⎢⎣⎡==2/10)2()1()]()([r r n x n y E r yx11201opt x yx a H R r a -⎡⎤⎡⎤===⎢⎥⎢⎥-⎣⎦⎣⎦输出最小均方误差理论值可由下式计算:2min00[()]0.5011/2TTopt yx J E y n H r ⎡⎤⎡⎤=-=-=⎢⎥⎢⎥--⎣⎦⎣⎦其中5.0)0()]([)]([22===r n x E n y E b) FIR 滤波器零点：j z z z a z H i i i ±=⇒+=-=-=-∑22111)(,即零点在正弦信号x(n)频率的21,0πω±=对应的z 平面位置.c) 用LMS 算法，n ＝0～10时系数的近似值 LMS ：⎩⎨⎧+-+=++++=+)1()()1()1()1()1()()1(n X n H n y n e n e n X n H n H Tδ 在线性预测误差滤波的LMS 算法中：)1()1()]1(),([)()1()](),([)()(21+⇒+-=⇒+=⇒n x n y n x n x n X n X n a n a n A n H T T⎩⎨⎧-+=+++=+)()()1()1()1()()()1(n X n A n x n e n e n X n A n A Tδ ()⎪⎩⎪⎨⎧=<==TA n n x n n x ]0,0[)0(0,0)(),2sin()(π所以, LMS 算法下的预测误差滤波器[]⎪⎪⎩⎪⎪⎨⎧⎥⎦⎤⎢⎣⎡-++⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡++⎥⎦⎤⎢⎣⎡--+=+)1()()1()()()1()1()1()()()()1()1(212121n x n x n e n a n a n a n a n x n x n a n a n x n e δ 344.00344.00729.027.00027.0081.019.00019.009.01.0001.0010)2(0)2(0)2()2(20)1(0)1(1)1()1(10)0(0)0(0212121----------============= a a x e n a a x e n a a n d) 符号算法[]⎪⎪⎩⎪⎪⎨⎧⎥⎦⎤⎢⎣⎡-++⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡++⎥⎦⎤⎢⎣⎡--+=+)1()()]1([)()()1()1()1()()()()1()1(212121n x n x sign n e sign n a n a n a n a n x n x n a n a n x n e δ )()]([0,1)(n x n x sign n x =∴±=4.004.007.03.0003.008.02.0002.009.01.0001.0010)2(0)2(0)2()2(20)1(0)1(1)1()1(10)0(0)0(0212121----------============= a a x e n a a x e n a a n2． A second-order adaptive FIR filter has the input as )2/sin()(πn n x = and)2(5.0)1()()(-+-+=n x n x n x n yas reference signal. Calculate the coefficients, starting from zero initial values, from time n=0 to n = 10 . Calculate the theoretical residual error and the time constant and compare with the experimental results.()1.0=δ 解：取1.0=δa) 计算n ＝0～10的系数⎪⎩⎪⎨⎧+++=++-+=+-+-+=)3()1()1()()1()2()1()()1()1()1()2(5.0)1()()(n e n X n H n H n X n H n y n e n x n x n x n y Tδ 2cos 21)(πk k r xx =⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡=2/1002/1)0()1()1()0(r r r r R x ，⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡++++=2/14/1)1(5.0)0()1()2(5.0)1()0(r r r r r r r yx①⎥⎦⎤⎢⎣⎡==-12/11yx x opt r R H 2min [()]Topt yx J E y n H r =-⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡--+--+-+-+-+=2/14/112/1)]2()()2()1()1()(2)2(25.0)1()([222Tn x n x n x n x n x n x n x n x n x E08/5)2()1(3)0(25.2=-++=r r r②Theoretical residual error: 0)21()(2min =+≈∞⇒∞σδN J J③Theoretical time constant ：2211111120 (,)(0)0.10.5N e k x k x k x R r N τλλσλδλδσδ=≈======⨯∑是特征值 注意: 由于均方收敛的时间常数比均值收敛的时间常数小, 所以实际应用中采用较保守的理论估计值,即采用均值收敛的时间常数作为算法收敛的时间常数的理论估计值. ④利用（2）（3）作迭代：)0(0)(<=n n x ，0)0()0(21==h h[]⎪⎪⎩⎪⎪⎨⎧⎥⎦⎤⎢⎣⎡+++⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡++⎥⎦⎤⎢⎣⎡+-+=+)()1()1()()()1()1()()1()()()1()1(212121n x n x n e n h n h n h n h n x n x n h n h n y n e δ )1(5.0)()1()1(-+++=+n x n x n x n y5.01112624.0]0,1[]4095.0,2376.0[1010]4095.0,2376.0[]06561.0,0[6561.0]1,0[]3439.0,2376.0[5.019]3429.0,2376.0[]0,02916.0[2916.0]0,1[]3439.0,2084.0[108]3439.0,2084.0[]0729.0,0[729.0]1,0[]271.0,2084.0[5.017]271.0,2084.0[]0,0324.0[324.0]0,1[]271.0,176.0[106]271.0,176.0[]081.0,0[81.0]1,0[]19.0,176.0[5.015]19.0,176.0[]0,036.0[36.0]0,1[]19.0,14.0[104]19.0,14.0[]09.0,0[9.0]1,0[]1.0,14.0[5.013]1.0,14.0[]0,04.0[4.0]0,1[]1.0,1.0[102]1.0,1.0[]1.0,0[1]1,0[]0,1.0[111.]0,01[]0,1.0[1]0,1[]0,0[000)1()1()1()1()1()()()(------------------+++++TTTTT T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T n H n X n e n e n X n H n y n x nδb) 根据实验结果计算残差和时间常数 残差 = 0698.0)11(2=e时间常数:通过观察见,在n 为偶数时, 误差变化波动大, 因此应选择在n 为奇数时的误差值确定时间常数较合理.2916.0)9()8()9()9(=-=X H y e T 26424.0)11()10()11()11(-=-=X H y e T)0)(,0)()]([(2994.20)()11()]()9([2222)911(222=∞=∞=∞=⇒∞-=∞--⨯-e J e E e e ee e 所以在计算时可假设由于ττ所以实验结果和理论值是符合的.3． Adaptive line enhancer. Consider an adaptive third-order FIR predictor. The input signal is )()sin()(0n b n n x +=ω where )(n b is a white noise with power 2b σ.Calculate the optimal coefficients 31,,≤≤i a opt i .Give the noise power in the sequence∑=-=31,)()(i opt i i n x a n sas well as the signal power. Calculate the SNR enhancement. 解：a) 计算31,,≤≤i a opt i⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡+++=20002000221cos 212cos 21cos 2121cos 212cos 21cos 2121b b b x R σωωωσωωωσ，)1()(+=n x n y 由于窄带信号为白噪声，时延参数D 选择1。

数字信号处理 英文版 课后习题

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
: 2
The shape of the digital filter may be found by substituting 2f S tan
H ( )
1504.4 2 2f S tan 1504.4 2
2

1504.4 2 16000 tan 1504.4 2
(b)
Use the bilinear transformation to get the digital transfer function
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)