Q.1: The following normal AM signal has a carrier frequency equal to 100 kHz. x(t) = 20 cos(2 x 105nt) + 4 cos(1.8 × 10 nt) + 4 cos(2.2 x 10 nt) where the message signal m(t) is a tone signal.
Q.1: The following normal AM signal has a carrier frequency equal to 100 kHz. x(t) = 20 cos(2 x 105nt) + 4 cos(1.8 × 10 nt) + 4 cos(2.2 x 10 nt) where the message signal m(t) is a tone signal.
Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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![Q.1: The following normal AM signal has a carrier frequency equal to 100 kHz.
x.(t) = 20 cos(2 × 105nt) + 4 cos(1.8 x 105nt) +4 cos(2.2 x 105nt)
where the message signal m(t) is a tone signal.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7b6685cc-4385-4277-8430-e1d436e5a204%2Fdbe3dd07-f39e-44f7-8c56-fe90e5f77979%2F68hbr97_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Q.1: The following normal AM signal has a carrier frequency equal to 100 kHz.
x.(t) = 20 cos(2 × 105nt) + 4 cos(1.8 x 105nt) +4 cos(2.2 x 105nt)
where the message signal m(t) is a tone signal.
![The modulation index is equal to:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7b6685cc-4385-4277-8430-e1d436e5a204%2Fdbe3dd07-f39e-44f7-8c56-fe90e5f77979%2Fwsqjpp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The modulation index is equal to:
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