Electronics and Communication Engineering - Communication Systems - Discussion
Discussion Forum : Communication Systems - Section 1 (Q.No. 11)
11.
A 400 W carrier is amplitude modulated with m = 0.75. The total power in AM is
Answer: Option
Explanation:
.
Discussion:
6 comments Page 1 of 1.
Irfan said:
4 years ago
Thanks @Athul.
BHAVANI.SIGADANA said:
7 years ago
I can understand how to find out the total power. Plese tell me.
Harika said:
9 years ago
Thank you so much @Athul.
Athul said:
10 years ago
Total power in AM = Carrier power(Pc) + Upper side band power(Pusb) + Lower sideband power(Plsb).
i.e., Pt = Pc+Pusb+Plsb.
= Ac^2/2+Ac^2m^2/8+Ac^2m^2/8.
= Ac^2/2(1+m^2/4+m^2/4).
= Ac^2/2(1+2(m^2/4).
Pt = Pc(1+(m^2)/2).
So Pt = 400(1+(0.75)^2/2).
= 400(1+0.5625/2).
= 400(2.5625/2).
= 400(1.28125).
= 512.5w.
i.e., Pt = Pc+Pusb+Plsb.
= Ac^2/2+Ac^2m^2/8+Ac^2m^2/8.
= Ac^2/2(1+m^2/4+m^2/4).
= Ac^2/2(1+2(m^2/4).
Pt = Pc(1+(m^2)/2).
So Pt = 400(1+(0.75)^2/2).
= 400(1+0.5625/2).
= 400(2.5625/2).
= 400(1.28125).
= 512.5w.
(4)
Rainy said:
1 decade ago
Pt = Pc(1+m^2/2).
(1)
Muhammad zain said:
1 decade ago
What is this formula?
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