Discussion
Match the Column I with Column II and select the correct answer using the codes given below :
| Column I | Column II |
|---|---|
| (P) \(Km(t) A \sin(\omega_{c}t)\) | (W) PM |
| (Q) \(A \sin[\omega_{c}t + Km(t)]\) | (X) FM |
| (R) \([1 + Km(t)] A \sin(\omega_{c}t)\) | (Y) AM |
| (S) \(A \sin\left[\omega_{c}t + K\int_{-\infty}^{t} m(t)\, d\tau\right]\) | (Z) DSBSC |
Answer: A
The correct matching is P-Z, Q-W, R-Y, S-X.
- \(Km(t)A\sin(\omega_{c}t)\) is the message times the carrier with no carrier term left: DSBSC (Z).
- \(A\sin[\omega_{c}t + Km(t)]\) has the phase proportional to the message: PM (W).
- \([1 + Km(t)]A\sin(\omega_{c}t)\) adds the carrier back, so the envelope follows the message: AM (Y).
- Phase proportional to the integral of the message means instantaneous frequency proportional to the message: FM (X).
The main trap is FM against PM: the integral inside the phase is the signature of FM, because differentiating the phase gives a frequency deviation proportional to m(t).
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