written 8.4 years ago by | • modified 8.4 years ago |
Mumbai University > Electronics Engineering > Sem4 > Fundamentals of Communication Engineering
Marks: 10M
Year: May2014
written 8.4 years ago by | • modified 8.4 years ago |
Mumbai University > Electronics Engineering > Sem4 > Fundamentals of Communication Engineering
Marks: 10M
Year: May2014
written 8.4 years ago by |
The output of the balanced modulator contains two sidebands and sum of the harmonic components.
As indicated in the Fig7, the input voltage at diode $D_1$ is $(v_c+v_m)$ and input voltage at diode $D_2$ is $(v_c-v_m)$.
Fig1. Generation of DSBSC signal using balanced modulator
The primary current of the output transformer is $i_1=i_{d1}-i_{d2}$.
Where,
$i_{d1}=a+b(v_c+v_m )+ c(v_c+v_m)^2 \ \ and \\ i_{d2}=a+b(v_c-v_m )+ c(v_c-v_m)^2$
Thus, we get,
$i_1=i_{d1}-i_{d2} = 2bv_m + 4cv_m v_c$
The modulating and carrier voltage are represented as,
$v_m=V_m sinω_m t \ \ \ and \\ v_c=V_c sinω_c t$
Substituting for $v_m$ and $v_c$ and simplifying, we get,
$i_1=2bV_m \text{sin}ω_m t+4 c \frac{mV_c}2 cos(ω_c-ω_m )t-4 c \frac{mV_c}2 cos(ω_c+ω_m )t $
The output voltage $v_0$ is proportional to primary current $i_1$ and assume constant of proportionality as α, which can be expressed as,
$v_0= αi_1=2αbV_m sinω_m t+4αc \frac{mV_c}2 cos(ω_c-ω_m )t-4αc \frac{mV_c}2 cos(ω_c+ω_m )t$
Let $P=2αbV_m$ and $Q=2αc \frac{mV_c}2.$
Thus we have,
$v_0= Psinω_m t+2Q cos(ω_c-ω_m )t-2Q cos(ω_c+ω_m )t$
The above equation shows that carrier has been cancelled out , leaving only two sidebands and the modulating frequencies.
The modulating frequencies from the output is eliminated by the tuning of the output transformer, which results in the below equation of the generated DSBSC wave.
$v_0= 2Q cos(ω_c-ω_m )t-2Q cos(ω_c+ω_m )t$