Let’s have a look at the block diagram of the DSB-SC system shown below:
Here, by observing the above figure, we can say that a product modulator generates a DSB-SC signal.
The signal is obtained by the multiplication of baseband signal x(t) with carrier signal cos ωct
By frequency shifting property of Fourier transform:
From the above equation, it is clear that only 2 components are present in the spectrum. These two are the two sidebands that are placed at +ωc and -ωc.
Let’s have a look at the pictorial representation of a waveform for DSB-SC system:
In order to get an exact an idea about the suppression of carrier in DSB-SC system.
Consider the baseband or modulating signal,
x(t) = Ax cos (2πfxt)
and the carrier signal,
c(t) = Ac cos (2πfct)
The mathematical representation of the signal at the output of the product modulator is given as:
s(t) = x(t).c(t)
Further,
The maximum frequency is fc + fx
The minimum frequency is fc – fx
As we know,
Bandwidth is given as
BW = fmax – fmin
BW = fc + fx – (fc – fx)
BW = 2fx
Thus, at the output, the DSB-SC wave contains a signal whose frequency is twice the frequency of the baseband signal.
The carrier without any information content is suppressed by a balanced modulator. Its principle of operation is such that, when two signals of the different frequency are passed through a non-linear resistance then an amplitude modulated signal with the suppressed carrier is achieved at the output.
It can be a diode, JFET or BJT that possess non-linear resistance characteristic.
A non-linear device has the capability to produce 2 sidebands with a carrier. But, a balanced mode connection of 2 non-linear devices produces a DSB-SC signal.
Let’s have a look at the circuit of the balanced modulator using diodes:
As we can see that the baseband input signal is applied at the input of 2 diodes that are 180⁰ phase reversed with each other through a centre tapped transformer.
Hence the input at D1,
v1 = cos ωct + x(t)
and input at D2,
v2 = cos ωc t – x(t)
At the output side tuned bandpass filter is obtained by parallel connection of RLC circuit.
So, the current through D1 is given as
Similarly,
the output voltage is given by
vo = i1 R – i2 R
On substituting the above-given value of i1 and i2 in the output equation, we will have,
vo = R [2 a x(t)] (+ 4b x (t) cos ωct)]
Therefore, the output is,
vo = 2aR x(t) + 4bRx (t) cos ωct
2aR x(t) = modulating signal
4bR x(t) cos ωct = DSB-SC signal
Thus, from the above expression, it is clear that output voltage is a combination of modulating signal along with the DSB-SC signal. After the elimination of the modulating signal, the DSB-SC signal is then passed to the LC bandpass and is received at the output.
Thus we will have,
4bR x(t) cos ωct = K x(t) cos ωct
at the output.
DSB-SC technique allows us to have a transmission that reduces overall power consumption rate, thereby ensuring a stronger signal at the output.
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