第三章1 234 56 6.16.2 78910第4章 (1)(2)()()()sin(2)sin(2)m c s t m t c t f t Ac f t ππ==[cos 2()cos 2()]2c m c m Acf f t f f t ππ=--+ (){[()][()]}4c m c m AcS f f f f f f f δδ=+-+--{[()][()]}4c m c m Acf f f f f f δδ-+++-+(3)相干解调相干解调:将接收信号与载波信号sin(2)fct π相乘,得到()sin(2)()sin(2)sin(2)c c c c r t f t A m t f t f t πππ=()[1cos(4)]2cc A m t f t π=- 通过低通滤波器抑制载频的二倍频分量,得到解调信号为0()()2cA y t m t =2解:(1)444)4cos()cos(2 1.210)()cos(2102 1.110t t t s t πππ++=⨯⨯⨯⨯⨯ 444cos(2 1.110)[10.5cos(20.110)]t t ππ=+⨯⨯⨯⨯调制系数是a=0.5; 信号频率是f=1000Hz(2)44441()[(10)(10)]2[( 1.110)( 1.110)]2S f f f f f δδδδ=++-+++-⨯⨯441[( 1.210)( 1.210)]2f f δδ+++-⨯⨯(3)3解:(1)已调信号无法用包络检波解调,因为能包络检波的条件是()1m t ≤, 这里的max ()151A m t ==>,用包络检波将造成解调波形失真。
(2)相干解调:将接收信号与载波提取电路的信号cos()c t ω相乘,得到()cos()()cos()cos()()[1cos(2)]2c c c c c cA r t t A m t t t m t t ωωωω==+ 通过低通滤波器抑制载频的二倍频分量,得到解调信号0()()2cA y t m t = (3)c发端加导频的DSB-SC AM 信号产生框图如上图:在DSB-SC 信号上加上导频,在接收时就可以提取导频作为解调波 4解:(1)()2cos[2()]c m s t f f t π+=()()()c m c m S f f f f f f f δδ=+++--(2)调制方式为上边带调制。
(3)5解:(1) ()cos20002sin 2000m t t t ππ=+ ()sin 20002cos2000m t t t ππ∧=-(2)调制信号:m(t)载波:5()cos(2810)c c t A t π=⨯⨯⨯下边带信号为:55ˆ()()cos(2810)()sin(2810)22c cA A s t m t t mt t ππ=+⨯⨯⨯⨯⨯⨯下 550(cos(2810)cos 20002sin 2000)t t t πππ=+⨯⨯⨯ 550(sin(2810)sin 20002cos 2000)t t t πππ+-⨯⨯⨯5550[cos(2810)sin(2810)]cos 2000sin 2000t t t t ππππ=+⨯⨯⨯⨯⨯⨯ 55100[cos(2810)sin(2810)]sin 2000cos 2000t t t t ππππ+-⨯⨯⨯⨯⨯⨯ 3350[cos(279910)2sin(279910)]t t ππ=-⨯⨯⨯⨯11()50[(79900)(79900)(79900)22S f f f j f δδδ=++--+下(79900)]j f δ+-1150[()(79900)()(79900)]22j f j f δδ=-+++-()(79900)(79900)]S f f f δδ=++-下(3)S(f)6 解:(1)A:ˆ()()cos 2()sin 2c c c c s t A m t f t A mt f t ππ=-上33ˆ()cos(245510)()sin(245510)c c A m t t A mt t ππ=-⨯⨯⨯⨯⨯⨯ B:ˆ()()cos(2)cos(2)()sin(2)cos(2)B c c c c c c s t A m t f t f t A mt f t f t ππππ=- 1ˆ[()()cos(4)()sin(4)]2c c c c c A m t A m t f t A mt f t ππ=+- C:1()()2C c s t A m t =D:ˆ()()cos(2)sin(2)()sin(2)sin(2)D c c c c c c s t A m t f t f t A mt f t f t ππππ=- 1[()sin(4)()(4)()]2c c c c c A m t f t A m t cos f t A m t ππ∧∧=+-E:1ˆ()()2E c s t A m t =- F:1()()2F c s t A m t =G:()()G c s t A m t =(2)只需最末端的相加改为相减即可,如图:A:ˆ()()cos 2()sin 2c c c c t A m t f t A mt f t ππ=+下s 33ˆ()cos(245510)()sin(245510)c c A m t t A mt t ππ=+⨯⨯⨯⨯⨯⨯ B:ˆ()()cos(2)cos(2)()sin(2)cos(2)B c c c c c c s t A m t f t f t A mt f t f t ππππ=+ 1ˆ[()()cos(4)()sin(4)]2c c c c c A m t A m t f t A mt f t ππ=++ C:1()()2C c s t A m t =D:ˆ()()cos(2)sin(2)()sin(2)sin(2)D c c c c c c s t A m t f t f t A mt f t f t ππππ=+ 1[()sin(4)()()cos(4)]2c c c c c A m t f t A m t A m t f t ππ∧∧=+-E:1ˆ()()2E c s t A m t = F:1()()2F c s t A m t =-G:()()G c s t A m t = 7解:解调框图:()[()()]2c c AR f M f f M f f =++- 11()[()()]*[()()]22c c c c A Y f M f f M f f f f f f δδ=++-++-[(2)()()()()()4c c c c AM f f H f f M f H f f M f H f f =+++++-(2)()]c c M f f H f f +-- 通过低通滤波器后0()()[()()]4c c AY f M f H f f H f f =++- 由图可看出()()c c H f f H f f ++-在[-W,W]内为常数,所以不失真。
H(f)8解:(1) 已调信号平均功率:2cA Ps ==50W (2) 调制指数:4f β= (3) 最大频偏:max4100400f m ff Hz Hz β===⨯⨯V(4) 调频信号带宽:2(1)251001000f B W Hz Hz β=+==⨯⨯9解:(1) 调制指数:4f β=,信号带宽:2(1)25100010000f m B f Hz β=+=⨯⨯= (2) 调制指数:max ()f f mK m t f β=,若fm 加倍,则2f β=信号带宽:2(1)23200012000f m B f Hz β=+=⨯⨯=(3) 调制指数:4p β=,信号带宽:2(1)25100010000f m B f Hz β=+=⨯⨯= (4) 调制指数:max ()p p K m t β=,若fp 加倍,4p β= 信号带宽:2(1)25200020000f m B f Hz β=+=⨯⨯=10解:(1)因为max ()m t K >,()]()cos [1c m t Ks t K t ω=+,所以调制为有离散大载波的双边带调制。
(2)()()cos sin t c c m d s t t t ττωω-∞=-⎰,为窄带FM 调制。
(3)ˆ()()cos ()sin c c t m t t mt t ωω=+下s ,为下边带调制。
(4)FM 调制。
(5)PM 调制。
11解:已调双边带调幅波(DSB-AM)信号:()()cos(2)c s t m t f t π=解调接收信号:()()cos(2)()cos(2)()sin(2)c s c c c r t m t f t n t f t n t f t πππ=+- 已调双边带调幅波(DSB-AM)信号的功率:20011()42222M P Ps m t P W P W ====⨯则04M P P W =输入加性白高斯噪声的功:2220011()()()42222c s N Pn n t n t n t W N W =+===解调后信号:011()()()22c s t m t n t =+解调后信号功率:20011()44M P m t P P W ===解调后噪声功率:001142Pn Pn N W ==则输出信噪比:00000212P W P N N W N S ==⎛⎫⎪⎝⎭ 12解:(1)解调输入信号:ˆ()[()cos ()sin ]()cos ()sin c c c c c s c r t A m t t mt t n t t n t t ωωωω=-+- 解调输入信号的平均功率:2R M c P A P =由10lg80T R P dB P =得810TRP P =,所以4410R P W -=⨯ 解调输入噪声功率:60022210n N P W N W W -===⨯解调信号:011()()()22c s t m t n t =+解调输出信号的平均功率:2201()44M c A P m t P ==解调输出噪声功率:001144n n P P N W == 解调输出信噪比:22000000420014M M R n c c A P P A P P P N W N W N W N S=====⎛⎫⎪⎝⎭ (2) 解调输入信号:()[1()]cos ()cos ()sin c c n c c s c r t A am t t n t t n t t ωωω=++-解调输入信号的平均功率:2221122R Mn c c P A A a P =+ 由10lg80T R P dB P =得810TRP P =,所以4410R P W -=⨯ 解调输入噪声功率:6002224210n N P W N W W -=⨯⨯==⨯解调信号:011()()()22n c c s t A am t n t =+解调输出信号的平均功率:22201()44Mn c A P m t a P ==解调输出噪声功率:001142n n P P N W == 解调输出信噪比:2222000004122Mn Mn n c c A a P P A a P P N W N W N S ===⎛⎫ ⎪⎝⎭ 22222200(1)211Mn Mn Mn R Mn Mn Ac a P a P a P P a P N W a P N W+==++ 因为Mn P =0.2W ,a=0.85所以220000.12625.251Mn R RMna P P P a P N W N WN S ===+⎛⎫ ⎪⎝⎭ 13解:(1)()m t 的总带宽fm 为48KHz ,max ()4801048f f m K m t KHzf KHzβ=== FM 的带宽,2(1)211481056f m B f KHz KHz β=+=⨯⨯= (2)(3)鉴频器输出的噪声功率谱为02220n cN f P Kf A == 所以1路和12路的平均功率比值为4824442121397Kf df P P Kf df==⎰⎰15答:(1)前置射频放大器的的等效噪声系数;(1)e T F T =-Q (上P120)------ e T 为等效噪声系数 T 为热噪声源温度 F 为噪声系数,由题目已知,得:()0.2101300175.5e T K =-≈。