Question 2: The output signal from an AM modulator is: OL I noite9n 9aM (t) o ngiz 1uqiuo orT = A cos(1000nt) + cos(800t) + cos(1200tt) Find the value of "A" such that the power efficiency of this AM becomes 33.33%. modulaton index enitubon'srli onim 37.33% ).3333 A z.? -Amplitu

Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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/8
2.
/2
t.
2.
Question 2:
The output signal from an AM modulator is:
gaM (t) = A cos(1000t) + cos(800tt) + cos(1200rt) ot ngiz uqiuo orT
Find the value of "A" such that the power efficiency of this AM becomes 33.33%.
modulation index
D= 33-33=0.3333
4.
ugey
sec
-Acos (1000TE)
+i ws (20T
2.
(5)
2.
2.
2.
A
2+1
aomad
38
Transcribed Image Text:/8 2. /2 t. 2. Question 2: The output signal from an AM modulator is: gaM (t) = A cos(1000t) + cos(800tt) + cos(1200rt) ot ngiz uqiuo orT Find the value of "A" such that the power efficiency of this AM becomes 33.33%. modulation index D= 33-33=0.3333 4. ugey sec -Acos (1000TE) +i ws (20T 2. (5) 2. 2. 2. A 2+1 aomad 38
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