EXPERIMENT 6 : ENVELOPES AND ENVELOPE RECOVERY (Demod. of DSB)
Objective:
Examining and recovering envelopes of AM modulated signals.
Equipment:
¾
¾
NI ELVIS (National Instruments Educational Laboratory Virtual Instrumentation Suite)
LabVIEW 7.0
General Information:
The envelope of a signal is that boundary within which the signal is contained, when viewed
in the time domain. It is an imaginary curve. This boundary has an upper and lower part and
these parts are mirror images of each other. It is customary to consider only one of them as
‘the envelope’ (typically the upper boundary). It is possible to generate a signal having the
same shape as this imaginary curve. The circuit which does this is commonly called an
envelope detector.
Every signal has an envelope. For a narrow band signal y (t ) = a (t ). cos( wc t + ϕ (t )) , the
envelope signal is e(t) = a(t)or e(t) = -a(t).
For DSB signals, e(t ) = A(1 + m. cos wm t ) . In the case 0 ≤ m ≤ 1, the absolute sign has no
effect, and so there is a direct relationship between the message and the envelope as can be
seen in Figure 1. However, when m > 1, the absolute value of the message and the message
itself have the same modulated signal; hence it is not possible to recover the message signal
uniquely (a(t) or a(t)?) as can be seen in Figure 2.
Figure 1. DSB, with m = 1
e(t)
Figure 2. DSB with m= 1.5
For the DSBSC signal, where a(t) = cos(ωmt), there is no DC component and the envelope is
shown in Figure 3.
e(t)
Figure 3. DSBSC
The diode detector is the prime example of an envelope detector. The diode detector is an
approximation to the ideal envelope detector which is a circuit that takes the absolute value of
its input and then passes the result through a low-pass filter. The output from this low-pass
filter is the required envelope signal. The ideal envelope recovery arrangement is shown in
Figure 4.
Figure 4 The Ideal Envelope Recovery Arrangement
The absolute value operation, being non-linear, must generate some new frequency
components. Among them are those of the desired envelope. It is the purpose of the low-pass
filter to separate the desired components generated by the absolute value operation.
A circuit which takes an absolute value is a full-wave rectifier. The operation of rectification
is non-linear. Because the aim of the rectifier is to take the absolute value of the signal being
rectified, it should multiply the signal with +1 when it is positive and with –1 when it is
negative.
An analysis of the ideal envelope detector is not a trivial mathematical exercise, except for
DSB with m < 1. In this case, the zero crossings of a DSB signal are uniform and are spaced
at half the period of the carrier. Then, the action of an ideal rectifier on such a signal is
equivalent to multiplying it by a square wave s(t) as shown in Figure 5. It is important to
ensure that the phases of the DSB and s(t) are matched correctly.
Figure 5 The Function s(t) and its Operation upon an AM Signal
As s(t) has the same period with the carrier s (t ) = 4 cos(ω t ) − 1 cos(3ω t ) + 1 cos(5 ω t ) K
c
c
c

the Fourier series expansion of s(t) is:
3
π 
5
and the AM signal: x DSB (t ) = Ac .(1 + mx(t )) cos(ω c t ) has the Fourier transform
X DSB (ω ) =
Ac
[δ (ω − ω c ) + δ (ω + ω c )] + m. Ac [X (ω + ω c ) + X (ω − ω c )]
2
2
The input to the low-pass filter will be the rectifier output y(t) = s(t)• xDSB(t) and the rectifier
output will consist of:
• a term at DC and another centered at ±2ωc due to the 1st harmonic in s(t),
• a term at ±2ωc and ±4ωc due to the 3rd harmonic in s(t),
• a term at ±4ωc and ±6ωc due to the 5th harmonic in s(t) and so on.
The rectifier output spectrum is shown in Figure 6.
ωc
desired
2ωc
4ωc
6ωc
undesired
Figure 6. The Rectifier Output Spectrum
For the case of a DSB signal, modulated with music, the DC term is of little interest to the
listener, but it is a direct measure of the strength of the carrier term, and so is used as an
automatic gain control signal in receivers.
Figure 7. DSB AM Signal
The detector we will examine is by far the simplest. Let us observe the transmitted carrier AM
waveform of Fig 7. If Ac+mx(t) never goes negative, the upper outline, or envelope, of the
AM wave is exactly equal to Ac+mx(t). If we can build a circuit that follows this outline, we
will have built a demodulator. The circuit of Fig. 8 is a simple envelope detector and serves as
a demodulator. The output follows an exponential curve, corresponding to the discharge of
the capacitor, between the peaks of the AM wave as shown in Figure 9.
Figure 8. Envelope Detector
Figure 9. Output waveform of AM Detector
If the time constant of the RC circuit is appropriately chosen, the output approximately
follows the outline of the input curve, and the circuit acts as a demodulator. The output
contains a ripple, but this does not cause a problem since the ripple has the carrier’s frequency
(residual radio frequency) and we are interested only in frequencies below fm (the highest
frequency present in the message signal). The RC time constant must be short enough that the
envelope can track the changes in peak values of the AM waveform. The peaks are spaced at
intervals equal to the reciprocal of the carrier frequency, while the heights of these peaks
follow the information, x(t). We can consider a worst case where x(t) is a pure sinusoid at a
frequency of fm. This would provide the fastest possible change in peak values. At this
frequency, the peaks vary from a maximum to minimum in 1/2fm second.. It takes an
exponential function 5 time constants to get within 0.7 percent of its final value. Therefore, if
we set the RC time constant to 10 percent of 1/fm, the envelope detector can follow even the
highest frequency. For example, with an fm of 5 kHz, the time constant would be set to 1/50
msec, or 20 µsec. This rule of thumb represents a first cut at envelope detector design
(a)
Figure 10. Envelope Detector Circuit
W<RC<<fc
xc(t) is DSB AM signal
e(t) is the envelope of xc(t)
R1C1 is a low-pass filter
The capacitor C2 removes the DC component of e(t).
(b)
Procedure:
1. Build the DSB Demodulator shown in Figure 10.Let x(t)=cos(2πfmt), where fm is 400
Hz. Carrier frequency is 5 kHz. Assume that the phase of carrier signal is zero and the
amplitude of the carrier signal is 1.
For m=0.2, m=0.5, m=1, m=1.2 and m=1.5 display the modulated signal in time
domain. Then demodulate the modulated signal. Display the demodulated signal in
both time domain and frequency domain and compare them with those of the message.
Is there any differences, if yes explain why?