5.1 Amplitude Modulation modulation efficiency: the percent of the total power of the modulated signal that conveys information m(t) E ×1009 1+(m2(t) The highest efficiency that can be attained for a 100% AM signal would be 50%. For the case when square- wave modulation is used). The normalized peak envelope power(pep)of the aM signal: PEP 1+mxm()]}2
11 5.1 Amplitude Modulation • modulation efficiency:the percent of the total power of the modulated signal that conveys information. 100% 1 ( ) ( ) 2 2 + = m t m t E • The highest efficiency that can be attained for a 100% AM signal would be 50%.(For the case when squarewave modulation is used). The normalized peak envelope power (PEP)of the AM signal: 2 2 {1 max[ ( )]} 2 m t A P c PEP = +
5.1 Amplitude Modulation The voltage spectrum of the aM signal is given by S()==[O(f-f∈)+M(f-f)+(+f)+M(f+f∈ The AM spectrum is just a translated version of the modulation spectral component. The bandwidth is twice that of the modulation Example 5-1 Suppose that a 5000w AM transmitter is connected to a 50o load lf the transmitter is 100%o modulated by a 1000Hz test tone. Then compute I)the total actually average power, (2)actually PeP,3) The modulation efficiency Solution: the constant A is given by(1/2)A 2150=5000 thus peak voltage across the load will be a =707V during the times when there is no modulation
12 5.1 Amplitude Modulation • The voltage spectrum of the AM signal is given by [ ( ) ( ) ( ) ( )] 2 ( ) c c c c c f f M f f f f M f f A S f = − + − + + + + • The AM spectrum is just a translated version of the modulation spectral component.The bandwidth is twice that of the modulation • Example 5-1 Suppose that a 5000w AM transmitter is connected to a 50Ω load; If the transmitter is 100% modulated by a 1000Hz test tone. Then compute (1)the total actually average power, (2)actually PEP, (3) The modulation efficiency . • Solution: the constant Ac is given by (1/2)Ac 2 /50=5000, thus peak voltage across the load will be Ac=707V during the times when there is no modulation
5.1 Amplitude Modulation (I)the actual average power (so)=(1242+1/242m)/R (1/2)4+(m(o)/50 (2)then across the 50 Q2 load, the Pep is PEP c1+max m(t/R 2 (1/2)*4*A2/R=(1/2)*4*7072/50=2000y (The modulation efficiency would be: E ×100%=33% 1+(m2(t 13
13 5.1 Amplitude Modulation (2)then across the 50 Ω load, the PEP is: (1/ 2) [1 ]/ 50 (1/ 2 1/ 2 )/ ( ) 2 ( ) 2 2 ( ) 2 2 2 t c t c c t A m s A A m R = + = + A R W m t R A P c c PEP (1/ 2)*4* / (1/ 2)*4*707 / 50 20000 {1 max[ ( )]} / 2 2 2 2 2 = = = = + (3)The modulation efficiency would be: 100% 33% 1 ( ) ( ) 2 2 = + = m t m t E (1)the actual average power
5.1 Amplitude Modulation U(r) VI cos wer Oscillator Intermediate-powe Power amplifie (carrier freq. e) lifier (IPA) (class C amp) (class C amp) AM "dc"sur voltage for PA a() Pulse width High-power m(r) 2() electronic LOw-PINs (PWM) 1 switch High voltage from dc power supply (a) Block Diagram n()- audio input PWM U3(r) AM outpu WUUAAAAAAAAAAAAAAAA. (b Waveform Figure 5-2 Generation of high-power AM by the use of PwM
14 5.1 Amplitude Modulation
5.1 Amplitude Modulation M(圳 B a) Magnitude spectrum of modulation Weight=Ac/2 IS(f Weight=Ac/2 /2 Upper sideband sideband fc-B -f-f+B fC-Bf￡+B b)Magnitude spectrum of AM signal Fig 4-2 Spectrum of AM signal 15
15 5.1 Amplitude Modulation -B 0 B f |M(f)| -fc-B -fc -fc+ B f Ac/ 2 fc-B fc fc+ B Weight=Ac/ 2 |S(f)| Weight=Ac/ 2 a) Magnitude spectrum of modulation b) Magnitude spectrum of AM signal Upper sideband low sideband Fig. 4-2 Spectrum of AM signal