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Modulating signal $10 \sin 2 \pi \times 10^3 t$ is used to modulate carrier signal $20 \sin 2 \pi \times 10^4 t$. Find the modulation index $\%$ modulation frequency. and amp. B.W and spectrum
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$Em=10 \sin 2π \times 10^3 t \\ Em= Em \sin 2πfmt \\ ∴Em=10 volt fm =1 \times 10^3 Hz = 1 kHz \\ \text{Carrier}, ec = 20 2π \times 10^4 t \\ ∴ec=Ec \sin⁡(2πfct ) ∴Ec=20 volts, fc = 1 \times 10^4 Hz = 10 kHz$

1. $\text{ Modulation index} m = \frac{Em}{Ec}=\frac{10}{20}=0.5 \\ \% \text{modulation} = .5 ×100=50 \%$
2. $f_{USB}= fc+fm =(10+1) =11 kHz \\ f_{LsB}= fc-fm = (10-1) =9 kHz$

3. $\text{ AMP of each side band} =\frac{mEc}{2}=\frac{0.5 \times 20}{2}$

4. $B.W =2fm = 2 \times 1 kHz=2kHz$

5. Spectrum

Total Power :

$\text{Pt = carrier power + power in USB + P in LSB}$

$Pt. = \frac{E^2 \cos}{R}+\frac{E^2 usb}{R}+\frac{E^2 lsb}{R}$

$P_c→ \text{carrier power}$

$P_c= \frac{E^2 \cos}{R}$

$\text{But, Ecos-Rms value of carrier} \frac{EC}{\sqrt{2}}$

$\therefore Pc = \frac{\bigg(\big(\frac{EC}{\sqrt{2}}\big)^2\bigg)}{R}$

$\text{Ec = Peak carrier amp}$

$→\text{Pusb=Plsb}=\frac{E^2 lsb}{R}$

$\text{Peak amp} – \frac{mEc}{2}$

$\text{Pusb = Plsb} = \frac{\big[\frac{mEc}{2\sqrt{2}}\big]^2}{R}$

$=\frac{m^2 Ec^2}{8R}$

$=\frac{m^2}{4} \times \frac{Ec^2}{2R}→Pc$

$Pt = Pc + Pusb + Plsb$

$= Pc + \frac{m^2}{4} Pc+ \frac{m^2}{4} Pc$

$Pt = \big[1+\frac{m^2}{2}\big]Pc$

Or $\frac{Pt}{Pc}=1+\frac{m^2}{2}$

$→\frac{Pt}{Pc}=1+\frac{m^2}{2}$

$∴m=\bigg[2\big(\frac{Pt}{Pc}-1 \big)\bigg]^{\frac{1}{2}}$

$→n=\frac{Plsb+Pusb}{Pt}=\frac{\frac{m^2}{4} Pc + \frac{m^2}{4} Pc}{ \big[1+\frac{m^2}{2}\big]Pc}$

Efficiency

$n=\frac{\frac{m^2}{2}}{1+\frac{m^2}{2}}$

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