written 6.1 years ago by | • modified 6.1 years ago |
$Em=10 \sin 2π \times 10^3 t \\ Em= Em \sin 2πfmt \\ ∴Em=10 volt fm =1 \times 10^3 Hz = 1 kHz \\ \text{Carrier}, ec = 20 2π \times 10^4 t \\ ∴ec=Ec \sin(2πfct ) ∴Ec=20 volts, fc = 1 \times 10^4 Hz = 10 kHz$
- $\text{ Modulation index} m = \frac{Em}{Ec}=\frac{10}{20}=0.5 \\ \% \text{modulation} = .5 ×100=50 \%$
$ f_{USB}= fc+fm =(10+1) =11 kHz \\ f_{LsB}= fc-fm = (10-1) =9 kHz$
$\text{ AMP of each side band} =\frac{mEc}{2}=\frac{0.5 \times 20}{2}$
$B.W =2fm = 2 \times 1 kHz=2kHz$
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
$n=\frac{\frac{m^2}{2}}{1+\frac{m^2}{2}}$