Agenda
AM Modulator circuit. AM Demodulator circuit.
AM Modulator Type
Non-Linear Modulator. Linear Modulator.
Non Linear modulator
Use non-linear devices such as diode and transistor. Fig below (next slide) show that:
Non-Liner Modulator
N.L.D
fm
RL
s AM (t )
fc
Linear Modulator
Linear Anode such as (class A, class B, class C) amplifiers. Fig below (next slide) show that:
Linear Modulator
RF I/P
Antenna
Class B AF amp
Class C RF amp
Product Modulators
s AM (t )
sm (t )
S c (t ) Ac cos wc t
Product Modulators
Consist of Analog product multiplier and adder. As shown below:
sm (t )
s AM (t )
sc (t )
Modulators using the Squared Law
give highest frequency than Product Modulators. As shown below:
sm (t )
S c (t ) Ac cos wc t
Non-Liner Element
Filter
Vo (t )
S AM (t )
Modulators using the Squared Law
Verifying the following relation:
Vo (t ) a1Vin(t ) a2Vin (t )
2
This relation is verified by using Diodes and Transistors.
Balance Modulators
Simple method to generate a DSB-SC signals
Use two conventional AM modulators arranged in the configuration .
Balance Modulators
Use two square-Law AM Modulators. Select modulators must be approximately identical characteristics so that the carrier component cancels out at the summing junction.
Balance Modulator Diagram
Balance Modulator circuit using two Diodes
D1
i2 (t )
V1 (t )
R1 C1 L1
sm (t )
V2 (t )
R2
Vo (t )
C1
L2
i1 (t )
sc (t ) cos wc t
D2
Analysis
The characteristic curve for each Diodes verify the following relation:
i1 (t ) a1V1 (t ) a V (t )
2 2 1
--------------------- (1)
the amplitude of V1 (t ) is given by:
Analysis
V1 (t ) cos wct S m (t )
--------------------- (2)
While the amplitude of V2 (t ) is:
V2 (t ) cos wc t S m (t )
--------------------- (3)
so
i1 (t ) a1[cos wct S m (t )] a2 [cos wct S m (t )]
2
---------- (4)
Analysis
And
i2 (t ) a1[cos wct S m (t )] a2 [cos wct S m (t )]
2
------- (5)
So we can find the amplitude of
Vo (t ) [i1 (t ) i2 (t )] R
---------- (6)
Vo (t )
Is:
Vo (t ) R [2a1Sm (t ) 4a 2sm(t ) cos c t ]
---------- (7)
Analysis
Since the resonance circuit operates at the carrier frequency ( f c ) so the output Vo (t ) is:
Vo (t ) 4a2 R S m (t ) cos wct
---------- (8)
I.e. the circuit works like DSB-SC Modulator. I.e. it works like Product Modulator.
AM Demodulators (Detector )
The restoration of the massage signal from the modulated signal is called Demodulation and detection. There are two Method of demodulation: - Asynchronous (Envelope Detector)-Synchronous.
Demodulators (Detectors) for AM signals
s AM (t )
AM DeModulation
sm (t )
Asynchronous Envelope Detector)
Consists of a diode and an RC circuit, which is basically a simple low pass filter. Circuit diagram for an envelope detector is shown in Figure below:
Asynchronous (Envelope Detector)
R2 12 V Vin R1 741 R4 12 V
s AM (t )
R3
741 C1
C2
D1
R5
C5
C3
C4
R6
-12 V -12 V
Asynchronous(Envelope Detector) Analysis
During the positive half-cycle of the input signal, the diode is conducting and the capacitor charges up to the peak value of the input signal. When the input falls below the voltage on the capacitor, the diode becomes reverse-biased and the input becomes disconnected from the output.
During this period, the capacitor discharges slowly through the load resistor R. On the next cycle of the carrier, the diode conducts again when the input signal exceeds the voltage across the capacitor.
The capacitor charges up again to the peak value of the input signal and the process is repeated again. The time constant RC must be selected so as to follow the variations in the envelope of the carriermodulated signal.
1 1 RC wc wm
In such a case, the capacitor discharges slowly through the resistor and, thus, the output of the envelope detector closely follows the message signal.
Synchronous Detector
We can also demodulate the signal using Balance Modulator (BM), which is called Synchronous Detector as shown below:
Synchronous Detector
s AM (t )
y1 (t )
Filter
y(t )
Local Oscillator ALo cos wc t
Synchronous Detector Analysis
Depending on multiply the modulated signal with sine wave signal that generated form (Local Oscillator), which is denoted by .
VLo (t ) ALo cos wct
Suppose the following:
S AM (t ) E (t ) cos wct
---------- (1)
E (t ) Ac [1 Sm (t )]
VLo (t ) ALo cos wct ---------- (2)
Where ALo is the carrier amplitude.
For simplicity suppose the
ALo
=2
2
---------- (3)
y1 (t ) 2 E (t ) cos wct
1 2 E (t )[ (1 cos2wct )] 2
E (t )[(1 cos 2wct )]
E (t ) E (t ) cos 2wct
---------- (4)
We can eliminate the frequency 2 wc using low pass filter (LPF),
so we get the following o/p:
y (t ) E (t )
But
---------- (5)
E (t ) Ac [1 Sm (t )]
---------- (6)
y(t ) Ac Ac Sm (t )
Synchronous Detector Analysis
We can also suppress the DC component by using
capacitor, the final o/p ell be:
Ac S m (t )
