重大信号与系统作业
北交20春《信号与系统》在线作业二_答案

(单选题)1: 当输入信号的复频率等于系统函数的零点时,系统的强迫响应分量为()。
A: 无穷大
B: 不为零的常数
C: 0
D: 随输入信号而定
正确答案: C
(单选题)2: 满足傅氏级数收敛条件时,周期信号f(t)的平均功率()。
A: 大于各谐波分量平均功率之和
B: 不等于各谐波分量平均功率之和
C: 小于各谐波分量平均功率之和
D: 等于各谐波分量平均功率之和
正确答案: D
(单选题)3: 卷积δ(t)*f(t)*δ(t)的结果为()。
A: δ(t)
B: δ(2t)
C: f(t)
D: f(2t)
正确答案: C
(单选题)4: 信号的时宽与信号的频宽之间呈()。
A: 正比关系
B: 反比关系
C: 平方关系
D: 没有关系
正确答案: B
(单选题)5: 设一个矩形脉冲的面积为S,则矩形脉冲的傅氏变换在原点处的函数值等于()。
A: S/2
B: S/3
C: S/4
D: S
正确答案: D
(单选题)6: 线性系统具有()。
A: 分解特性
B: 零状态线性
C: 零输入线性
D: 以上全对
正确答案: D
(单选题)7: 如果一连续时间二阶系统的系统函数H(s)的共轭极点在虚轴上,则它的h(t)应是()。
信号与系统 陈后金 第二版 课后习题答案(完整版)

(1) f (t) = 3sin 2t + 6 sinπ t
(2) f (t) = (a sin t) 2
(8)
f
(k)
=
cos⎜⎛ ⎝
πk 4
⎟⎞ ⎠
+
sin⎜⎛ ⎝
πk 8
⎟⎞ ⎠
−
2
cos⎜⎛ ⎝
πk 2
⎟⎞ ⎠
解:(1)因为 sin 2t 的周期为π ,而 sin πt 的周期为 2 。
显然,使方程
−∞
0
2-10 已知信号 f (t) 的波形如题 2-10 图所示,绘出下列信号的波形。
f (t)
2
1
−1 0
t 2
题 2-10 图
(3) f (5 − 3t) (7) f ′(t) 解:(3)将 f (t) 表示成如下的数学表达式
(5) f (t)u(1 − t)
由此得
⎧2
f
(t)
=
⎪ ⎨ ⎪ ⎩
f (t)u(1− t) 2
1
0.5
t
−1 0
1
(7)方法 1:几何法。由于 f (t) 的波形在 t = −1处有一个幅度为 2 的正跳变,所以 f ′(t) 在 此处会形成一个强度为 2 的冲激信号。同理,在 t = 0 处 f ′(t) 会形成一个强度为 1 的冲激信 号(方向向下,因为是负跳变),而在 0 < t < 2 的区间内有 f ′(t) = −0.5 (由 f (t) 的表达式可
第 1 页 共 27 页
《信号与系统》(陈后金等编)作业参考解答
(2)显然,该系统为非线性系统。 由于
T{f (t − t0 )}= Kf (t − t0 ) + f 2 (t − t0 ) = y(t − t0 )
信号与系统 于敏慧(第二版)第二周作业答案

y0(t)
1
t
0
2
4
(6) x(t) = dx0 (t) , h(t) = dh0 (t) 。
dt
dt
x(t) * h(t) = dx0 (t) * dh0 (t) = d 2 y0 (t)
dt dt
dt 2
x(t) ∗ h(t) = 0.5δ(t) − 0.5δ(t − 2)
2.10 求 y[n] = x1[n]* x2[n]* x3[n] 。 其 中 x1[n] = (0.5)n u[n] , x2[n] = u[n + 3] 和
(2)利用(1)的结果,求系统的逆系统的单位样值(脉冲)响应。
(3)利用(2)的结果,结合卷积性质,求一信号 x[n],使之满足
x[n]* h[n] = 2n (u[n] − u[n − 4])
解:(1) h[n] − Ah[n −1] = δ [n],其中 h[n] = (1 )n u[n] , 2
(通项: an = a1q n−1 )
n
∑ 此题: a1 = 1, q = 2 ; x[n]* h[n] = 2nu[n]*u[n] = ( 2k )u[n] = (2n+1 −1)u[n] k =0
2.6 计算图 2-45(b)与(c)所示信号 x(n)与 h(n)的卷积和,注意:N=4。 解:(b)利用脉冲信号δ(n)的卷积性质以及卷积的延时性质计算:
k =−∞
+ 3] =
u[n + 3] 0.5k
k =0
;
= 2(1 − 0.5n+4 )u[n + 3]
(2) x1[n]* x2[n]* x3[n] = 2(1 − 0.5n+4 )u[n + 3]* (δ [n] − δ [n −1]) ; = 2(1 − 0.5n+4 )u[n + 3] − 2(1 − 0.5n+3 )u[n + 2]
信号与系统仿真作业

nGDOU-B—11—112广东海洋大学学生实验报告书(学生用表)课程名称课程号学院(系)信息学院专业班级学生姓名学号实验地点04002 实验日期实验一连时间信号的MATLAB表示和连续时间LTI系统的时域分析一、实验目的1.掌握MA TLAB产生常用连续时间信号的编程方法,并熟悉常用连续时间信号的波形和特性;2.运用MATLAB符号求解连续系统的零输入响应和零状态响应;3.运用MATLAB数值求解连续系统的零状态响应;4.运用MATLAB求解连续系统的冲激响应和阶跃响应;5.运用MATLAB卷积积分法求解系统的零状态响应。
二、实验原理1. 连续信号MATLAB实现原理从严格意义上讲,MA TLAB数值计算的方法并不能处理连续时间信号.然而,可用连续信号在等时间间隔点的取样值来近似表示连续信号,即当取样时间间隔足够小时,这些离散样值能够被MATLAB处理,并且能较好地近似表示连续信号.MATLAB提供了大量生成基本信号的函数.比如常用的指数信号、正余弦信号等都是MATLAB的内部函数。
为了表示连续时间信号,需定义某一时间或自变量的范围和取样时间间隔,然后调用该函数计算这些点的函数值,最后画出其波形图.三、实验内容1.实例分析与验证根据以上典型信号的MA TLAB函数,分析与验证下列典型信号MA TLAB程序,并实现各信号波形图的显示,连续信号的图形显示使用连续二维图函数plot().(1)正弦信号:用MA TLAB命令产生正弦信号2sin(2/4)ππ+,并会出时间0≤t≤3的波形图。
程序如下:K=2;w=2*pi ;phi=pi/4;t=0:0.01:3;ft=K*sin (w*t+phi );plot(t,ft ),grid on ;axis ([0,3,-2。
2,2.2])title (’正弦信号’)(2) 抽样信号:用MA TLAB 中的sinc(t)函数命令产生抽样信号Sa(t),并会出时间为66t ππ-≤≤的波形图。
信号与系统第一章习题及作业(1,2)

(2)(余弦序列是否为周期信号,取决于2л/Ω0是正整 (余弦序列是否为周期信号,取决于 Ω 有理数还是无理数。) 数、有理数还是无理数。) 因此, 因此, 2л/Ω0=2л·7/8л=7/4=N/m Ω =2л·7/8л 所以基波周期为N=7; 所以基波周期为N=7; N=7
因为2л/Ω =16л 为无理数, (4) 因为 Ω0=16л,为无理数,则此信号不是周期 信号. 信号. (5) 因为周期信号在[-∞,+∞]的区间上,而本题的重 因为周期信号在[ ∞,+∞]的区间上, 的区间上 复区间是[0, +∞],则此信号为非周期信号 则此信号为非周期信号, 复区间是[0, +∞],则此信号为非周期信号,
f(n) 1 0 3 6 … n
9、判断是否为线性系统?为什么? 、判断是否为线性系统?为什么?
( 3) ( 5) (7 )
y( t ) = ln y( t 0 ) + 3t 2 f ( t ) y( t ) = y( t 0 ) + f 2 ( t ) y( t ) = sin t ⋅ f ( t )
8、一个连续时间系统的输入-输出关系为 、一个连续时间系统的输入 输出关系为
1 t+T y ( t ) = T [ f ( t ) ] = ∫ T2 f (τ )d τ T t− 2 试确定系统是否为线性的?非时变的?因果的? 试确定系统是否为线性的?非时变的?因果的?
解:积分系统是线性的,因此系统是线性系统。 积分系统是线性的,因此系统是线性系统。
sin ω 0 tε ( t )
sin ω 0 ( t − t 0 )ε ( t )tt0 Nhomakorabeat
sin ω 0 tε ( t − t 0 )
北理工-信号与系统-第三版-第三章-作业参考答案

k
| u[k ] | ,有界
是非稳定系统
(e) 显然n<0时,h[n]=0,所以是因果系统;
k
| h[k ] | | u[k ] / n | ,无界
k
是非稳定系统
(f) 显然n<0时,h[n]=0,所以是因果系统;
| h[k ] |
(d)
y[n] x[n] h[n]
k
[k n ] [n k n ]
1 2
[n n1 n2 ]
3.11在LTI离散时间系统中 已知x[n]=u[n]时的零状态响应(单位阶跃响应)为s[n],求单位抽样响应h[n]; 已知h[n],求s[n].
y[n] - 4y[n-1] =2x[n]+3x[n-1];
令x[n]=δ[n],则有 h[n] – 4h[n-1] =2 δ[n]+3 δ[n-1];当n<0时,h[n]=0,得h[0]=2,h[1]=11,
特征方程为 λ-4=0, 得λ=4,
h[n]=c(4)nu[n],由h[1]=4c=11,c=11/4得 h[n]=(11/4)(4)nu[n-1]=11 (4)n-1u[n-1],考虑h[0]=2=2 δ[n],得 h[n]=2 δ[n]+11 (4)n-1u[n-1]。(n>0的解) (b).据图有同(a)一样的结果…。 (c).据图 y[n]=3y[n-1]- 2y[n-2]+ x[n]+2x[n-1]+x[n-2] ,即差分方程为 y[n] -3y[n-1]+2y[n-2] = x[n]+2x[n-1]+x[n-2], 先求
信号与系统课后答案第八章作业答案后半部分

频率响应为
H
(e jΩ
)
=
H
(z)
|z = e jΩ
=
4 ⎡⎣ejΩ −1⎤⎦
3
⎡⎢⎣e
jΩ
−
1 3
⎤ ⎥⎦
经计算得极点为 p = 1 ,零点为 z = 1。 3
H(e jΩ)
(Ω)
幅频响应图(横坐标进行了归一化处理)
(c)Yx (z) =
y(−1) + 2 y(−2) + 2 y(−1)z−1 1− z−1 − 2z−2
=
8⋅ z +1⋅ 3 z−2 3
z, z +1
z
>2
其逆
z
变换即零输入响应为
yx
(n)
=
8 3
⋅
2n
u(n)
+
1 3
⋅
(−1)n
u(n)
(d)根据上面计算的零输入和零状态响应可知系统的完全响应为
f (n) = (−1)n u(n) , y(−1) = 0 , y(−2) = 1;
解:(1)将原式两边取单边 Z 变换得,
Y (z) −[z−1Y (z) + y(−1)] − 2[z−2Y (z) + y(−2) + y(−1)z−1] = F (z) + z−1F (z)
整理得:
Y (z)
=
题图 8-23
根据系统框图可得 h(n) = h1(n) ∗[h2 (n) + h3 (n)] ,故 h(n) = δ (n) ∗[h2 (n) + h3(n)] = u(n) + u(n − 2)
信号与系统课后答案(第二版)+曾禹村+第二章作业参考答案

i1(t) = i2 (t) + i3 (t) , i2 (t) R2 − L 有 8i2 `(t) + 3i2 (t) = 2e`(t) ˆ ˆ 由 h`(t) + 3h(t) = 2δ (t)
0
h
(−1) t 3
T
t
t 3E − τ E (t) = ∫ δ (τ )dτ − ∫ e 8 u(τ )dτ −∞ 4 −∞ 32
x(t)
1
2 t
yx(t)
1 2 3 4 t
0
1
0
Qh(0) = 0, t ≤ 0, 有 0 ≤ t <1 , h(t) + h(t −1) + h(t − 2) = h(t) = t 时 1≤ t < 2时 h(t) + h(t −1) + h(t − 2) = h(t) + h(t −1) =1 , h(t) =1− h(t −1) =1− (t −1) = 2 −t 2 ≤ t < 3 , h(t) + h(t −1) + h(t − 2) =1 时 h(t) =1− h(t −1) − h(t − 2) =1− (2 − (t −1)) − (t − 2) = 0 3 ≤ t < 4时 h(t) = 4 − t − h(t −1) − h(t − 2) =4 −t − 0 − (2 − (t − 2)) = 0 , t, 0 ≤ t < 1 ∴h(t) = 2 − t, 1 ≤ t ≤ 2 0, t < 0,2 < t
解: (e) 特征方程为 λ2+4λ+4=0 得 λ1=-2, λ2=-2。 则 h(t)= (c1eλ1 t+ c2eλ2t)u(t)=( c1e- 3 t+ c2e-2 t)u(t) h`(t)= (c1+ c2)δ(t)+(-3c1e- 3 t-2c2e- 2t)u(t) h``(t)= (c1+ c2)δ`(t)+(-3c1-2c2) δ(t)+ (9c1e- 3 t+4c2e- 2t)u(t) 将x(t)= δ(t), y(t)=h(t)代入原方程得:
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Signals and SystemChap11.6 Determine whether or not each of the following signals is periodic: (a): (/4)1()2()j t x t e u t π+= (b): 2[][][]x n u n u n =+- (c): 3[]{[4][14]}k x n n k n k δδ∞=-∞=----∑1.9 Determine whether or not each of the following signals is periodic, If a signal is periodic , specify its fundamental period:(a): 101()j t x t je = (b): (1)2()j t x t e -+= (c):73[]j n x n e π=(d): 3(1/2)/54[]3j n x n e π+= (e): 3/5(1/2)5[]3j n x n e += 1.14 considera periodic signal 1,01()2,12t x t t ≤≤⎧=⎨-<<⎩with period T=2. Thederivative of this signal is related to the “impulse train”()(2)k g t t k δ∞=-∞=-∑,with period T=2. It can be shown that 1122()()()dx t A g t t A g t t dt=-+-. Determine the values of 1A , 1t , 2A , 2t .1.15.Consider a system S with input x[n] and output y[n].This system is obtained through a series interconnection of a system S 1 followed by a system S2. The input-output relationships for S 1 and S 2 are S 1: ],1[4][2][111-+=n x n x n y S 2: ]3[21]2[][222-+-=n x n x n yWhere ][1n x and ][2n x denote input signals.(a) Determine the input-output relationship for system S.(b)Does the input-output relationship of system S change if the order in which S 1 and S 2 are connected in series is reversed(i e ,if S 2 follows S 1)? 1.16.Consider a discrete-time system with input x[n] and output y[n].The input-output relationship for this system is]2[][][-=n x n x n y(a) Is the system memoryless?(b) Determine the output of the system when the input is ][n A δ, where A is any real or complex number. (c) Is the system invertible?1.17.Consider a continuous-time system with input x(t) and output y(t) related by))(sin()(t x t y =(a) Is this system causal? (b) Is this system linear?1.21.A continous-time signal ()x t is shown in Figure P1.21. Sketch and label carefully each of the following signals:(a): (1)x t - (b): (2)x t - (c): (21)x t + (d): (4/2)x t - (e): [()()]()x t x t u t +- (f):()[(3/2)(3/2)]x t t t δδ+--1.22. A discrete-time signal ()x t is shown in Figure P1.22. Sketch andlabel carefully each of the following signals:(a): [4]x n - (b): [3]x n - (c): [3]x n (d): [31]x n + (e): [][3]x n u n - (f): [2][2]x n n δ-- (g):11[](1)[]22n x n x n +- (h): 2[(1)]x n - 1.25.Determine whether or not each of the following continuous-time signals is periodic. If the signal is periodic, determine its fundamental period.(a): ()3cos(4)3x t t π=+ (b): (1)()j t x t e π-= (c):2()[cos(2)]3x t t π=-(d): (){cos(4)()}x t t u t ενπ= (e): (){sin(4)()}x t t u t ενπ= (f): (2)()t n n x t e∞--=-∞=∑1.26. Determine whether or not each of the following discrete-time signals is periodic. If the signal is periodic, determine its fundamental period. (a):6[]sin(1)7x n n π=+ (b): []cos()8nx n π=- (c): 2[]cos()8x n n π= (d):[]cos()cos()24x n n n ππ=(e):[]2cos()sin()2cos()4826x n n n n ππππ=+-+Chap 22.1 Let]3[]1[2][][---+=n n n n x δδδ and ]1[2]1[2][-++=n n n h δδCompute and plot each of the following convolutions: (a)][*][][1n h n x n y = (b)][*]2[][2n h n x n y += (c)]2[*][][3+=n h n x n y2.3 Consider an input x[n] and a unit impulse response h[n] given by],2[)21(][2-=-n u n x n].2[][+=n u n hDetermine and plot the output ].[*][][n h n x n y = 2.7 A linear system S has the relationship[][][2]k y n x k g n k ∞=-∞=-∑Between its input x[n] and its output y[n], where g[n]=u[n]-u[n-4]. (a) Determine y[n] where ]1[][-=n n x δ (b) Determine y[n] where ]2[][-=n n x δ (c) Is S LTI?(d) Determine y[n] when x[n]=u[n] 2.10 Suppose that⎩⎨⎧≤≤=elsewhere t t x ,010,1)( And )/()(αt x t h =,where 10≤<α.(a) Determine and sketch )(*)()(t h t x t y =(b) If dt t dy /)( contains only three discontinuities, what is the value ofα?2.11 Let)5()3()(---=t u t u t x and )()(3t u e t h t -=(a) Compute )(*)()(t h t x t y =. (b) Compute )(*)/)(()(t h dt t dx t g =. (c) How is g(t) related to y(t)? 2.20 Evaluate the following integrals: (adt t t u )cos()(0⎰∞∞-(b)⎰+50)3()2sin(dt t t δπ (c)⎰--551)2cos()1(τπττd u2.27 We define the area under a continuous-time signal )(t v as⎰∞∞-=dt t v A v )(Show that if )(*)()(t h t x t y =, thenh x y A A A =2.40 (a) an LTI system with input and output related through the equationτττd x e t y tt )2()()(-=⎰∞---What is the impulse response h(t) for this system?(b) Determine the response of the system when the input x(t) is as shown in Figure P2.40.Chap 33.1 A continuous-time periodic signal x(t) is real value and has a fundamental period T=8. The nonzero Fourier series coefficients for x(t) arej a a a a 4,2*3311====--.Express x(t) in the form)cos()(0k k k k t A t x φω+=∑∞=3.2 A discrete-time periodic signal x[n] is real valued and has a fundamental period N=5.The nonzero Fourier series coefficients for x[n] are10=a ,4/2πj e a --=,4/2πj e a =,3/*442πj ea a ==- Express x[n] in the form)sin(][10k k k k n A A n x φω++=∑∞=3.3 For the continuous-time periodic signal)35sin(4)32cos(2)(t t t x ππ++= Determine the fundamental frequency 0ω and the Fourier series coefficients k a such thattjk k kea t x 0)(ω∑∞-∞==.3.5 Let 1()x t be a continuous-time periodic signal with fundamental frequency 1ω and Fourier coefficients k a . Given that211()(1)(1)x t x t x t =-+-How is the fundamental frequency 2ω of 2()x t related to ?Also, find a relationship between the Fourier series coefficients k b of 2()x t and the coefficients k a You may use the properties listed in Table 3.1. 3.8 Suppose we are given the following information about a signal x(t): 1. x(t) is real and odd.2. x(t) is periodic with period T=2 and has Fourier coefficients k a .3. 0=k a for 0||>k .4 1|)(|21202=⎰dt t x .Specify two different signals that satisfy these conditions.3.13 Consider a continuous-time LTI system whose frequency response is⎰∞∞--==ωωωω)4sin()()(dt e t h j H t jIf the input to this system is a periodic signal⎩⎨⎧<≤-<≤=84,140,1)(t t t x With period T=8,determine the corresponding system output y(t). 3.15 Consider a continuous-time ideal lowpass filter S whose frequency response is1, (100)()0, (100)H j ωωω⎧≤⎪=⎨>⎪⎩When the input to this filter is a signal x(t) with fundamental period6/π=T and Fourier series coefficients k a , it is found that)()()(t x t y t x S=→.For what values of k is it guaranteed that 0=k a ?3.35.Consider a continuous-time LTI system S whose frequency response is 1,||250()0,H j otherwise ωω≥⎧=⎨⎩When the input to this system is a signal x(t) with fundamental period/7T π= and Fourier series coefficients k a ,it is found that the output y(t)is identical to x(t).For what values of k is it guaranteed that 0k a =?Chap 44.1 Use the Fourier transform analysis equation(4.9)to calculate the Fourier transforms of;(a))1()1(2---t u e t (b)|1|2--t eSketch and label the magnitude of each Fourier transform.4.2 Use the Fourier transform analysis equation(4.9) to calculate the Fourier transforms of: (a))1()1(-++t t δδ (b))}2()2({-+--t u t u dtdSketch and label the magnitude of each Fourier transform.4.5 Use the Fourier transform synthesis equation(4.8) to determine the inverse Fourier transform of ()()|()|j X j X j X j e ωωω= ,where|()|2{(3)(3)}X j u u ωωω=+-- 3()2X j ωωπ=-+Use your answer to determine the values of t for which x(t)=0. 4.6 Given that x(t) has the Fourier transform ()X j ω, express the Fourier transforms of the signals listed below in the terms of ()X j ω.You may find useful the Fourier transform properties listed in Table4.1. (a))1()1()(1t x t x t x --+-= (b))63()(2-=t x t x(c) )1()(223-=t x dtd t x4.11 Given the relationships)()()(t h t x t y *=And)3()3()(t h t x t g *=And given that x(t) has Fourier transform )(ωj X and h(t) has Fouriertransform )(ωj H ,use Fourier transform properties to show that g(t) has the form)()(Bt Ay t g =Determine the values of A and B.4.13 Let x(t) be a signal whose Fourier transform is()()()(5)X j ωδωδωπδω=+-+-And let()()(2)h t u t u t =--(a) Is x(t) periodic? (b) Is ()*()x t h t periodic?(c) Can the convolution of two aperiodic signals be periodic?4.14 Consider a signal x(t) with Fourier transform )(ωj X .Suppose we are given the following facts: 1. x(t) is real and nonnegative.2. ),()}()1{21t u Ae j X j F t --=+ωωwhere A is independent of t.3.⎰∞∞-=πωω2|)(|d j X .Determine a closed-form expression for x(t).Chap 66.1 Consider a continuous-time LTI system with frequency response()()|()|H j H j H j e ωωω= and real impulse response h(t). Supposethat we apply an input 00()cos()x t t ωφ=+ to this system .Theresulting output can be shown to be of the form0()()y t Ax t t =-Where A is a nonnegative real number representing an amplitude-scaling factor and 0t is a time delay.(a)Express A in terms of |()|H j ω.(b)Express 0t in terms of 0()H j ω6.3 Consider the following frequency response for a causal and stable LTI system:1()1j H j j ωωω-=+ (a) Show that |()|H j A ω=,and determine the values of A.(b)Determine which of the following statements is true about ()τω,the group delay of the system.(Note ()(())/d H j d τωωω=- ,where ()H j ω is expressed in a form that does not contain any discontinuities.)1.()0 0for τωω=>2.()0 0for τωω>>3 ()0 0for τωω<>6.5 Consider a continuous-time ideal bandpass filter whose frequency response is⎩⎨⎧≤≤=elsewherej H c c ,03||,1)(ωωωω (a) If h(t) is the impulse response of this filter, determine a function g(t)such that)(sin )(t g tt t h c πω= (b) As c ω is increased, dose the impulse response of the filter get moreconcentrated or less concentrated about the origin?Chap 77.1 A real-valued signal x(t) is know to be uniquely determined by its samples when the sampling frequency is 10,000s ωπ=.For what values of ω is ()X j ω guaranteed to be zero?7.2 A continuous-time signal x(t) is obtained at the output of an ideal lowpass filter with cutoff frequency 1,000c ωπ=.If impulse-train sampling is performed on x(t), which of the following sampling periods would guarantee that x(t) can be recovered from its sampled version using an appropriate lowpass filter?(a) 30.510T -=⨯(b) 3210T -=⨯(c) 410T -=7.3 The frequency which, under the sampling theorem, must be exceeded by the sampling frequency is called the Nyquist rate. Determine the Nyquist rate corresponding to each of the following signals:(a)()1cos(2,000)sin(4,000)x t t t ππ=++ (b)sin(4,000)()t x t tππ=(c) 2sin(4,000)()()t x t t ππ= 7.4 Let x(t) be a signal with Nyquist rate 0ω. Determine the Nyquist ratefor each of the following signals:(a)()(1)x t x t +- (b)()dx t dt(c)2()x t(d)0()cos x t t ω 7.9 Consider the signal2sin 50()()t x t tππ= Which we wish to sample with a sampling frequency of 150s ωπ= to obtain a signal g(t) with Fourier transform ()G j ω.Determine the maximum value of 0ω for which it is guaranteed that0()75() ||G j X j for ωωωω=≤Where ()X j ω is the Fourier transform of x(t).Chap 88.1 Let x(t) be a signal for which M ()0 when ||> X j ωωω=.Another signal y(t) is specified as having the Fourier transform ()2(())c Y j X j ωωω=-.Determine a signal m(t) such that8.3 Let x(t) be a real-valued signal for which()0 ||2,000X j when ωωπ=>.Amplitude modulation is performed toproduce the signal()()sin(2,000)g t x t t π=A proposed demodulation technique is illustrated in Figure P8.3 where g(t) is the input, y(t) is the output, and the ideal lowpass filter has cutoff frequency 2,000πand passband gain of 2. Determine y(t).FigureP8.38.22 In Figure P8.22(a), a system is shown with input signal x(t) and output signal y(t).The input signal has the Fourier transform ()X j ω shown in Figure P8.22(b). Determine and sketch ()Y j ω, the spectrum of y(t).8.28 In Section 8.4 we discussed the implementation of single-sideband modulation using 090 phase-shift networks, and in Figure8.21 and 8.22 we specifically illustrated the system and associated()()()x t y t m t =spectra required to retain the lower sidebands.Figure P8.28(a) shows the corresponding system required to retain the upper sidebands.(a) With the same ()X j ω illustrated in Figure8.22, sketch12(),()Y j Y j ωω,and ()Y j ω for the system in FigureP8.28(a), and demonstrate that only the upper sidebands are retained.(b) F or ()X j ω imaginary, as illustrated in FigureP8.28(b), sketch12(),()Y j Y j ωω and ()Y j ω for the system in FigureP8.28(a), and demonstrate that , for this case also, only the upper sidebands are retained.Chap 99.2 Consider the signal 5()(1)t x t e u t -=- and denote its Laplace transform by X(s).(a)Using eq.(9.3),evaluate X(s) and specify its region of convergence. (b)Determine the values of the finite numbers A and 0t such that theLaplace transform G(s) of 50()()t g t Ae u t t -=-- has the same algebraic form as X(s).what is the region of convergence corresponding to G(s)?9.5 For each of the following algebraic expressions for the Laplace transform of a signal, determine the number of zeros located in the finite s-plane and the number of zeros located at infinity: (a)1113s s +++ (b) 211s s +- (c) 3211s s s -++9.7 How many signals have a Laplace transform that may be expressed as2(1)(2)(3)(1)s s s s s -++++ in its region of convergence? Solution:There are 4 poles in the expression, but only 3 of them have different real part.∴ The s-plane will be divided into 4 strips which parallel to the jw-axis and have no cut-across.∴ There are 4 signals having the same Laplace transform expression.9.8 Let x(t) be a signal that has a rational Laplace transform with exactly two poles located at s=-1 and s=-3. If 2()() ()t g t e x t and G j ω=[ the Fourier transform of g(t)] converges, determine whether x(t) is left sided, right sided, or two sided. 9.9 Given that1(),{}Re{}sat e u t Re s a s a -↔>-+ Determine the inverse Laplace transform of22(2)(),Re{}3712s X s s s s +=>-++ 9.10 Using geometric evaluation of the magnitude of the Fourier transform from the corresponding pole-zero plot ,determine, for each of the following Laplace transforms, whether the magnitude of the corresponding Fourier transform is approximately lowpass, highpass, or bandpass. (a): 11(),{}1(2)(3)H s e s s s =ℜ>-++ (b): 221(),{}12s H s e s s s =ℜ>-++ (c): 232(),{}121s H s e s s s =ℜ>-++ 9.13 Let ()()()g t x t x t α=+- ,Where ()()t x t e u t β-=. And the Laplace transform of g(t) is 2(),1{}11s G s e s s =-<ℜ<-. Determine the values of the constants αand β.。