⟹ Frequency deviation = 3 . (AM Bandwidth) = 6 fm
and ?= 6 is given in (d) only.
?c + n ?m = (1008 x 103)2p
⟹ 2p x 106 + n. 4p x 103 = (1008 x 103).2p
⟹ n = 4
So, ? = 6, n = 4 hence answer is 5j4(6).
List I | List II | ||
---|---|---|---|
A. | 200 MHz | 1. | SHF |
B. | 20 GHz | 2. | UHF |
C. | 500 kHz | 3. | VHF |
D. | 500 MHz | 4. | MF |
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