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What are the minimum and maximum Amplitude of an AM wave represented by the expression: $v = 5(1+0.6\cos 6280 t) \sin 211 \times 10^4 \; \text{volts}$?

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Given an AM wave - $v = 5(1+0.6\cos 6280 t) \sin 211 \times 10^4 \; \text{volts}$
This can be compared to a standard AM wave of the form: $v = V_c (1+m\cos \omega_s t) \sin \omega_c t$
$(1)$ Carrier Amplitude: $V_c = 5\;V$
$(2)$ Modulation factor: $m = 0.6$
$(3)$ Signal Frequency: $f_s = \large\frac{\omega_s}{2\pi}$$ = \large\frac{6280}{2\pi}$$ = 1 \; kHz$
$(4)$ Carrier Frequency: $f_c = \large\frac{\omega_c}{2\pi}$$ = \large\frac{211 \times 10^4}{2\pi}$$ = 336 \; kHz$
Minimum Amplitude $ = V_c - mV_c = 5 - 0.6 \times 5 = 2\; V$
Maximum Amplitude $ = V_c + mV_c = 5 + 0.6 \times 5 = 8\; V$
answered Jul 13, 2014 by balaji.thirumalai
 

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