The amplitude of upper and lower side bands of $A.M.$ wave where a carrier signal with frequency $11.21 MHz$ , peak voltage $15 V$ is amplitude modulated by a $7.7 kHz$ sine wave of $5 V$ amplitude are $\frac{a}{10} V$ and $\frac{b}{10} V$ respectively. Then the value of $\frac{a}{b}$ is _____.

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Solution

The upper sideband frequency is the sum of carrier wave and modulated wave frequency while the lower side band frequency is the difference in carrier and modulated wave frequency.

The required ratio is,

$\frac{a}{10} = \frac{b}{10} = \frac{μ A_{C}}{2}$

$\Rightarrow \frac{a}{b} = 1$

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The amplitude of upper and lower side bands of A.M. wave where a carrier signal with frequency 11.21 MHz, peak voltage 15 V is amplitude modulated by a 7.7 kHz sine wave of 5 V amplitude are a10 V and b10 V respectively. Then the value of ab is _____.

The upper sideband frequency is the sum of carrier wave and modulated wave frequency while the lower side band frequency is the difference in carrier and modulated wave frequency. The required ratio is, a10=b10=μAC2 ⇒ ab=1

The amplitude of upper and lower side bands of $A.M.$ wave where a carrier signal with frequency $11.21 MHz$ , peak voltage $15 V$ is amplitude modulated by a $7.7 kHz$ sine wave of $5 V$ amplitude are $\frac{a}{10} V$ and $\frac{b}{10} V$ respectively. Then the value of $\frac{a}{b}$ is _____.

The upper sideband frequency is the sum of carrier wave and modulated wave frequency while the lower side band frequency is the difference in carrier and modulated wave frequency.

The required ratio is,

$\frac{a}{10} = \frac{b}{10} = \frac{μ A_{C}}{2}$

$\Rightarrow \frac{a}{b} = 1$