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1、2-1 画出下列各时间函数的波形图，注意它们的区别1）x1(t) = sinWtu(t)1-123tx1(t)042）x2(t) = sinW(t t0) u(t)01t0tx2(t)-13）x3(t) = sinWtu(t t0)01t0tx3(t)4）x2(t) = sinW(t t0) u(t t0)01t0tx4(t)-12-2 已知波形图如图2-76所示，试画出经下列各种运算后的波形图01tx(t)-1123图 2-76（1）x(t-2 )01tx ( t-2 )-11234（2）x(t+2 )-31tx ( t+2 )-4-2-101（3）x(2t)01tx(2t)-1123（4）
2、x(t/2 )01tx ( t/2 )-11234-2（5）x(-t)-31tx (-t)2-2-101（6）x(-t-2)x (-t-2)-51t0-4-3-2-11（7）x( -t/2-2 )-51t0-4-3-2-11x ( -t/2-2 )-7-6-8（8）dx/dt01tdx/dt-1123-2- (t-2)2-3应用脉冲函数的抽样特性，求下列表达式的函数值(1)(t) dt = x(-t0)(2)(t) dt = x(t0)(3)u(t -) dt = u()(4)u(t 2t0) dt = u(-t0)(5)(t+2) dt = e2-2(6)(t-) dt = +(7)= 1-
3、 = 1 cost0+jsint02-4求下列各函数x1(t)与x2(t) 之卷积，x1(t)*x2(t)(1) x1(t) = u(t), x2(t) = e-at u(t) ( a0 )x1(t)*x2(t) = = = (2) x1(t) =(t+1) -(t-1) , x2(t) = cos(t + ) u(t)矚慫润厲钐瘗睞枥庑赖。x1(t)*x2(t) = cos(t+1)+u(t+1) cos(t-1)+u(t-1)(3) x1(t) = u(t) u(t-1) , x2(t) = u(t) u(t-2)x1(t)*x2(t) =当 t 0时，x1(t)*x2(t) = 0当 0
4、t 1时，x1(t)*x2(t) = = t当 1t 2时，x1(t)*x2(t) = = 1当 2t3时，x1(t)*x2(t) = =3-t当 3t时，x1(t)*x2(t) = 0x1(t)* x2(t)01t123(4)x1(t) = u(t-1) , x2(t) = sin t u(t)x1(t)*x2(t) = = 1- cos(t-1)2-5已知周期函数x(t)前1/4周期的波形如图2-77所示，根据下列各种情况的要求画出x(t)在一个周期( 0tT )的波形聞創沟燴鐺險爱氇谴净。(1) x(t)是偶函数，只含有偶次谐波分量f(t) = f(-t), f(t) = f(tT/2)
5、tT/203T/4T/4T-T/4-T/2f(t)(2) x(t)是偶函数，只含有奇次谐波分量f(t) = f(-t), f(t) = -f(tT/2)tT/203T/4T/4T-T/4-T/2f(t)(3) x(t)是偶函数，含有偶次和奇次谐波分量f(t) = f(-t)tT/203T/4T/4T-T/4-T/2f(t)(4) x(t)是奇函数，只含有奇次谐波分量f(t) = -f(-t), f(t) = -f(tT/2)tT/203T/4T/4T-T/4-T/2f(t)(5) x(t)是奇函数，只含有偶次谐波分量f(t) = -f(-t), f(t) = f(tT/2)tT/203T/4T
6、/4T-T/4-T/2f(t)(6) x(t)是奇函数，含有偶次和奇次谐波分量f(t) = -f(-t)tT/203T/4T/4T-T/4-T/2f(t)tT/203T/4T/4T-T/4-T/2f(t)2-6利用信号x(t)的对称性，定性判断图2-78所示各周期信号的傅里叶级数中所含有的频率分量(a) t2T-2T-T0x(t)T这是一个非奇、非偶、非奇偶谐波函数，且正负半波不对称，所以含有直流、正弦等所有谐波分量，因为去除直流后为奇函数。残骛楼諍锩瀨濟溆塹籟。(b)t0Tx (t)-T这是一个奇函数。也是一个奇谐波函数，所以只含有基波、奇次正弦谐波分量。(c)tT-T-T/20x(t)T/
7、2除去直流分量后是奇函数，又f(t) = f(tT/2)，是偶谐波函数，所以含有直流、偶次正弦谐波。(d)t0T/2x (t)-T/2T-T正负半波对称，偶函数，奇谐波函数，所以只含有基波、奇次余弦分量。(e)t0T/2x (t)-T/2T奇函数、正负半波对称，所以只含有正弦分量（基、谐）(f)tT-T-T/20x(t)T/2正负半波对称、奇函数、奇谐波函数，所以只含有基波和奇次正弦谐波。2-7 试画出x(t) = 3cos1t + 5sin21t的复数谱图（幅度谱和相位谱）解：a0 = 0, a1 = 3, b2 = 5, c1 = 3, c2 = 5|x1| = |(a1-jb1)| =
8、, |x2| = c2= 1 = arctan (-) = 0, -1= 02 = arctan (-) = -, -2= 3-1121n1|xn|0-2112-1121n10-21/2-/22-8求图2-8所示对称周期矩形信号的傅里叶级数t0Tx (t)-TE/2-E/2T/2-T/2解：这是一个正负半波对称的奇函数，奇谐函数，所以只含有基波和奇次正弦谐波。bn = = = = = ，n为奇数，n = 1，3，5 = 0 ，n 为偶数，n = 2，4，6 x(t) = 指数形式的傅里叶级数 0 , n = 0,2, 4 Xn= (an-jbn) = , n = 1, 3, 5 x(t) =
9、a0 + 2-9求图2-9所示周期信号的傅里叶级数t0T/2x (t)-T/2TT/43T/4E解：此函数是一个偶函数 x(t) = x(-t) 其傅里叶级数含有直流分量和余弦分量ao = + + = + + E = = an = = = = , n = 1, 2, x(t) = 2-10若已知Fx(t) = X()利用傅里叶变换的性质确定下列信号的傅里叶变换(1) x(2t5)(2) x(1t)(3) x(t) cos t解：(1) 由时移特性和尺度变换特性可得Fx( 2t - 5) =(2) 由时移特性和尺度变换特性Fx(at) = Fx(t-t0) = Fx(1t) = (3) 由欧拉公
10、式和频移特性cos t = F = X(0)0= 1Fx(t) cos t = X(1) + X(+1)2-11已知升余弦脉冲x(t) = 求其傅里叶变换解：x(t) = u( t +)u( t)求微分 = = =+ = + 由微分特性可得：( j)3X() = X() = 2-12已知一信号如图2-81所示，求其傅里叶变换t-/20x(t)/2解：(1) 由卷积定理求x(t) = * = = 由时域卷积定理X() = = (2) 由微分特性求 ， t 0 = ，0 t = ( t +) +( t)2(t)由微分特性( j)2X() = X() = 2-13已知矩形脉冲的傅里叶变换，利用时移特
11、性求图2-82所示信号的傅里叶变换，并大致画出幅度谱解： = E u( t + )u( t) = x(t) = ( t + )( t)由时移特性和线性性X() = = 2j = 2j02E-2-14已知三角脉冲x1(t)的傅里叶变换为X1() = 试利用有关性质和定理求x2(t) = x1(t) cos0t的傅里叶变换酽锕极額閉镇桧猪訣锥。解：由时移性质和频域卷积定理可解得此题由时移性质Fx1(t) = 由频移特性和频域卷积定理可知：Fx(t )cos0t=X(0)+X(+0)X2 () = Fx1 (t)cos0t= X1 (0) + X(+0)= Sa2+Sa22-15求图2-82所示X()的傅里叶逆变换x(t)-00|X()|0A-00|X()|0A-00()0/2-00()0a)b)/2-/2-/2彈贸摄尔霁毙攬砖卤庑。解：a) X() = |X()| = 由定义：x(t) = = = = = = b) =+=+=+=2-16确定下列信号的最低抽样频率与抽样间隔(1) Sa(100t)(2)Sa2(100t)(3)Sa(100t)+ Sa2(100t)解：(1)由对偶性质可知：Sa(100t)的频谱是个矩形脉冲
