First slide

Modulation

Question

An AM wave is given by $v = 500 [1 + 0.5 \sin 12560 t] \sin [5.26 x 10^{5} t]$ . The modulating frequency is

Easy

A

2.0 kHz
B

1.0 kHz
C

12.5 kHz
D

50 kHz

Solution

$\begin{array}{l} Given with standard form \\ C_{m} (t) = A_{c} (1 + μ \sin ω_{m} t) \sin ω_{c} t \\ ∴ ω_{m} = 2 π f_{m} = 12560 \\ \Rightarrow f_{m} = \frac{12560}{2 π} = 2 kHz \end{array}$

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