    Question

# An amplitude-modulated wave is represented by, ${C}_{m}=10\left(1+0.2\mathrm{sin}12560t\right)\mathrm{sin}\left(1.11×{10}^{4}t\right)V$, the modulating frequency in $kHz$ will be

Open in App
Solution

## Step 1. Given data:An amplitude-modulated wave $={C}_{m}=10\left(1+0.2\mathrm{sin}12560t\right)\mathrm{sin}\left(1.11×{10}^{4}t\right)V..........\left(1\right)$Step 2. Formula used:In general,An amplitude-modulated wave $={C}_{m}={A}_{c}\left(1+\frac{{A}_{m}}{{A}_{c}}\mathrm{sin}{\omega }_{m}t\right)\mathrm{sin}{\omega }_{c}tV..........\left(2\right)$where, ${A}_{m}$is the Amplitude of modulating wave in $m$ ${A}_{c}$is the Amplitude of carrier wave in $m$ $\mathrm{sin}{\omega }_{m}t$and $\mathrm{sin}{\omega }_{c}t$ is the phase of modulating wave and carrier wave respectively. Step 3. Find the modulating frequency:By comparing equations $\left(1\right)$and $\left(2\right)$, we get ${\omega }_{m}=12560rad/sec$ ${f}_{m}=\frac{{\omega }_{m}}{2\pi }Hz\left\{\because {\omega }_{m}=2{\mathrm{\pi f}}_{\mathrm{m}}\right\}$where, ${f}_{m}$ is the frequency of modulation ${f}_{m}=\frac{12560}{2×3.14}Hz\left\{Assume\mathrm{\pi }=3.14\right\}$ ${f}_{m}=2000Hz=2kHz$ Hence, modulating frequency is $2kHz$.  Suggest Corrections  6      Similar questions  Explore more