Question

Figure 3 shows a typical AM signal, showing the parts. Note that the information modulates the envelope Message Signal Times Carrier Signal Times Amplitude Modulated Signal Time of the carrier signal. Figure 3. Amplitude modulated signal Exercise From the Figure 3, deduce: (a) mo) (b) v.0 (c) (1 (d) 4 Page 3 of 6

Answer 1

We deduce from the above Fig's. following

the given symbol and their value deduced from given graph as follows

1. m(t) = Message signal

• m(t) = Am.cos (wmt)
• m(t) = Am.cos (2$\pi$fct)
• m(t) = Am.cos (2$\pi$/Tt)
• From Above Fig we deduce the value of Am & T of message signal as
• Am = 1 & T =  9/50 sec Or fm =1/T= 50/9 or wm = 100$\pi$/9 hence m(t) given as
• m(t) = 1.cos (100$\pi$t/9)
• m(t) = 1.cos (2$\pi$(50/9)t)

2. vc(t) = Carrier Signal signal

• vc(t) = Ac.cos (wct)
• From Above Fig we deduce the value of Am & T of carrier signal as  
• Ac = 2 & T =  9/5 sec Or fc =1/T= 5/9 Hz  or wm = 10$\pi$/9 hence vc(t) given as
• vc(t) = 2.cos(10$\pi$t/9)
• vc(t) = 2.cos(2$\pi$(5/9)t)

3. s(t) = Modulated signal

• s(t) = Ac .m(t).cos (wct)
• when putting value of m(t) from equ. 1 it becomes
• s(t) = Ac.Am cos (wmt) .cos (wct)
• From Above Fig. we deduce the value of Ac,Ac,wm,wc of Modulated signal as
• s(t) = 2.1.cos (100$\pi$t/9t) .cos (2$\pi$(5/9t) which became as
• s(t) = 2.cos (100$\pi$t/9t) .cos (2$\pi$(5/9t)

4. $\mu$ = Modulation Index which defined as here Maximum amplitude of

• Here Maximum Amplitude Peak Volatge of Modulated signal is defined as
• A = Ac ( 1 + $\mu$ m(t) )
• Amax = Maximum Amplitude ( Peak Volatge) of Modulated signal
• Amin = Minimum Amplitude of Modulated signal
• From Above Fig. we deduce the value of Amax= +3V & Amin= +1V & Ac= +2V
• Amax = Ac.( 1 +  $\mu$)
• Amin = Ac.( 1 -  $\mu$) also
• Amax - Amin = 2 .Ac.$\mu$ which gives us
• $\mu$ = (Amax - Amin)/ 2 .Ac.
• here putting values of Amax =+3 , Amin = +1, And Ac= +2 we get
• $\mu$= [ (+3) -(+1)]/2.2=  (3 - 1)/4 = 2/4 =0.5
• $\mu$ = Modulation Index= 0.5
• hich shows perfect Modulation