e 9 of 66 -
3.1.3.2 AM:
m
2
Average power: Pa = (1+ ------- )
・
Pc
2
where, Pc: carrier power, m: depth of modulation.
(A/
√
2)
2
A
2
While, Pc = -------------- = ----------
R
L
2R
L
and, Vpp = 2 (1 + m)
・
A
1 Vpp Vpp
2
Therefore, Pc = --------
・
( --------------- )
2
= ----------------------
2R
L
2(1 + m ) 8(1 + m)
2
・
R
L
m
2
Vpp
2
Therefore, Pa = (1 + -------- )
・
--------------------
2 8(1 + m)
2
・
R
L
(Vpp/2
√
2)
2
Vpp
2
PEP: Pp = ------------------- = ---------
R
L
8R
L
m
2
Pp
Therefore, Pa = (1 + -------- )
・
-------------
2 (1 + m)
2
(1 + m)
2
or, Pp = -----------------
・
Pa
1+ m
2
/2
m= 1: Pp = 2.67 Pa
m= 0.8 : Pp = 2.47 Pa
m= 0.5 : Pp = 2.1 Pa
Vpp
m
・
A
A