written 6.1 years ago by | • modified 5.0 years ago |
Similar Question
Draw the dig of phase cancellation SSB generation and explain the carrier and unwanted sideband are supposed. What change is necessary to suppress the other side band?
written 6.1 years ago by | • modified 5.0 years ago |
Similar Question
Draw the dig of phase cancellation SSB generation and explain the carrier and unwanted sideband are supposed. What change is necessary to suppress the other side band?
written 6.1 years ago by |
Generation of SSB using phase shift method:
Operation:
The balanced modulator M1 has two inputs, the modulating signal without any phase shift and the RF carrier with a 90 degree phase shift.
The other balanced modulator M2 receives the modulating signal with a 90degree phase shift and carrier without any phase shift.
At the output of both the balanced modulators there is DSB-SC signal consisting of both sidebands and carrier is completely removed.
The Upper Sidebands (USBs) at the output of both the balanced modulator lead the carrier by 90degrees.But LSB at output of M1 leads the carrier by 90 degrees and LSB at output of M2 lags behind the carrier by 90degrees.Thus the LSBs are out of phase.
So when the outputs of M1 and M2 are applied to the adder the LSBs are cancelled out and output of adder consists of only upper side band.
The linear amplifiers will follow the adder. They are class B or AB type amplifiers used to amplify the USB without introducing any distortion.
Mathematical proof
$i/ps$ to $M1$ $- \cos wmt \\ \cos (wct + 90^∘)$
$i/ps$ to $M2$ $- \cos (wmt + 90^∘ ) \\ \cos wct$
$o/p of M1 = \cos (wct). \cos wmt$
USB wid 90 degree advance LSB – 90 degree delay
$o/p of M2 = \cos wct. \cos(wmt + 90^∘)$
o/p of adder = o/p of M1 + o/p of M2
$= \cos (wct + wmt + 90^∘ )$