Question

An audio signal $u_m=20\sin 2\mathrm{\pi }\left(1500\mathrm{t}\right)$ modulates a carrier $u_c=80\sin 2\mathrm{\pi }\left(100000\mathrm{t}\right)$. The value of percent modulation is ________.

Answer 1

Step 1. Given data:

The modulating signal, $u_m=20\sin 2\mathrm{\pi }\left(1500\mathrm{t}\right).....\left(1\right)$

The carrier signal, $u_c=80\sin 2\mathrm{\pi }\left(100000\mathrm{t}\right).....\left(2\right)$

Step 2. The formula used:

In general, The modulating signal, $u_m=A_m\cos \left({\omega }_{m}t\right)......\left(3\right)$

The carrier signal, $u_c=A_c\cos \left({\omega }_{c}t\right)......\left(4\right)$

where $A_m$and $A_c$ are Amplitudes of signal in $metres$

By comparing $\left(1\right)$ with $\left(3\right)$, and $\left(2\right)$ with $\left(4\right)$, we get

$A_m=20m$

$A_c=80m$

The percentage modulation is given by, $\frac{A_m}{A_c}\times 100$

Step 3. Finding percent modulation:

$\%modulation=\frac{20}{80}\times 100=0.25\times 100=25$

Hence, the percent modulation is $25$