In an amplitude modulated signal,the maximum | vedclass

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(A) Let $A_c$ be the amplitude of the carrier wave and $A_m$ be the amplitude of the modulating wave.Given that the maximum amplitude $A_{max} = A_c +

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$5$

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(A) Let $A_c$ be the amplitude of the carrier wave and $A_m$ be the amplitude of the modulating wave.Given that the maximum amplitude $A_{max} = A_c + A_m = 15 V$.Given that the minimum amplitude $A_{min} = A_c - A_m = 5 V$.Adding the two equations: $(A_c + A_m) + (A_c - A_m) = 15 V + 5 V$.$2 A_c = 20 V \Rightarrow A_c = 10 V$.Subtracting the two equations: $(A_c + A_m) - (A_c - A_m) = 15 V - 5 V$.$2 A_m = 10 V \Rightarrow A_m = 5 V$.Thus,the amplitude of the modulating wave is $5 V$.

In an amplitude modulated signal,the maximum amplitude is $15 V$ and the minimum amplitude is $5 V$. The amplitude of the modulating wave will be (in $V$)

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$A$ carrier wave $v_{c}(t) = 160 \sin(2 \pi \times 10^{6} t) \text{ V}$ is made to vary between $V_{\max} = 200 \text{ V}$ and $V_{\min} = 120 \text{ V}$ by a message signal $v_{m}(t) = A_{m} \sin(2 \pi \times 10^{3} t) \text{ V}$. The peak voltage $A_{m}$ of the modulating signal is ..... .

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