A 1000 KHz carrier is simultaneously modulated with $f_{m 1} = 300 H z$ , $f_{m 2} = 800 H z$ and $f_{m 3} = 1 k H z$ audio sine waves. What will be the frequencies present in the output?
(a) $\begin{array}{l} f_{L S B 1} = 999.7 k H z \\ f_{U S B 1} = 1000.3 k H z \end{array}$
(b) $\begin{array}{l} f_{L S B 2} = 999.2 k H z \\ f_{U S B 2} = 1000.8 k H z \end{array}$
(c) $\begin{array}{l} f_{L S B 3} = 999 k H z \\ f_{U S B 3} = 1001 k H z \end{array}$
(d) $\begin{array}{l} f_{L S B 3} = 990 k H z \\ f_{U S B 3} = 1010 k H z \end{array}$

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a only

b, c and d

a, b and c

a, c only

answer is C.

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Detailed Solution

$F r e q u e n c i e s p r e s e n t i n t h e o u t p u t a r e l o w e r f r e q u e n c y m o d u l a t i o n f_{c} - f_{m} a n d u p p e r f r e q u e n c y m o d u l a t i o n f_{c} + f_{m} . (a) f_{c} = 1000 k H z f_{m 1} = 300 H z = 0.3 k H z f_{L S B 1} = f_{c} - f_{m 1} = 1000 k H z - 0.3 k H z = 999.7 k H z f_{U S B 1} = f_{c} + f_{m 1} = 1000 k H z + 0.3 k H z = 1000.3 k H z (b) f_{c} = 1000 k H z f_{m 2} = 800 H z = 0.8 k H z f_{L S B 2} = f_{c} - f_{m 2} = 1000 k H z - 0.8 k H z = 999.2 k H z f_{U S B 2} = f_{c} + f_{m 1} = 1000 k H z + 0.8 k H z = 1000.8 k H z (c) f_{c} = 1000 k H z f_{m 3} = 1 k H z f_{L S B 3} = f_{c} - f_{m 3} = 1000 k H z - 1 k H z = 999 k H z f_{U S B 3} = f_{c} + f_{m 3} = 1000 k H z + 1 k H z = 1001 k H z$