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22 - Q uadrature am plitude m odulation

Experiment 22 – Quadrature amplitude modulation

Preliminary discussion
Quadrature amplitude modulation (QAM) using analog message signals is a method of double-
sideband suppressed carrier (DSBSC) modulation that permits multiple users to transmit their
signal over the same portion of the radio frequency spectrum. This is also known as phase
division multiplex (PDM).

Figure 1 below shows the block diagram of the mathematical implementation of QAM. Message-
1 in the upper arm of the arrangement in Figure 1 below is multiplied with a carrier to create a
DSBSC signal (in this case, called DSBSCI). Simultaneous to this, Message-2 in the lower arm is
multiplied with the same carrier, but having a 90° phase shift, to create another DSBSC signal
(called DSBSCQ).

D S BS C I
M e ssage 1

C arrier QAM

O
output

M e ssage 2
DSBSCQ

Figure 1

The two DSBSC signals are then simply added together for transmission and, as they have the
same carrier frequency, their spectral compositions overlap in the radio-frequency spectrum.
While this suggests that the two sets of signals would be irretrievably mixed, the orthogonal
relationship between the carriers allows the sidebands to be separated by the receiver using
phase discrimination (introduced in Experiment 8 and revisited here).

22-2 © Emona Instruments Experiment 22 – Quadrature amplitude modulation

Figure 2 below shows the block diagram of the mathematical implementation of QAM
demodulation.

QAM R ecovered
M essage 1 or
input
M essage 2

Local carrier

O
Figure 2

Notice that the arrangement multiplies the QAM signal with a local carrier, the frequency of
which must be identical to modulator’s carrier frequency at the transmitter. Importantly, the
local carrier is connected to the multiplier via a phase shifter.

To understand how this arrangement can be used to recover either one of the messages in the
QAM signal and reject the other, consider the math involved.

The QAM signal consists of two modulated signals which are:

DSBSCI = m1 × sin(c) and DSBSCQ = m2 × cos(c)

Where: m1 is Message-1, m2 is Message-2 and c is the carrier.

Now suppose the demodulator’s local carrier is adjusted so that it is in phase with the carrier
used to modulate Message-1 - That is, it is set to sin(c). As these signals are multiplied
together, this can be expressed mathematically as m 1.sin(c) x sin(c).

Using trigonometric identities, this results in:

m1.sin(c) x sin(c) = m1/2 [cos(c-c) – cos(c+c)]

Which is the same as:

m1.sin(c) x sin(c) = m1/2 [cos(0) – cos(2c)]

Experiment 22 – Quadrature amplitude modulation © Emona Instruments 22-3

And, as cos(0) = 1, this can be rewritten as:

m1.sin(c) x sin(c) = [m1/2 x 1] – [m 1/2 x cos(2c)]

Importantly, these expressions show that there will be a non-zero component at the same
frequency as Message-1 (m1) and a non-zero component at twice the carrier frequency. The
component at twice the carrier frequency is removed using the low-pass filter on the
multiplier’s output (see Figure 2). In other words, Message-1 in the QAM signal has been
recovered.

Simultaneous to this, the local carrier sin(c) is multiplied by the DSBSC Q signal on the output
of the modulator’s other branch. Mathematically, this can be expressed as m 2.cos(c) x sin(c).

Using trigonometric identities, this results in:

m2.cos(c) x sin(c) = m2/2 [sin(c+c) + sin(c-c)]

Which is the same as:

m2.cos(c) x sin(c) = m2/2 [sin(2c) + sin(0)]

And, as sin(0) = 0, then this can be rewritten as:

m2.cos(c) x sin(c) = m2/2 [sin(2c)] + 0

These expressions directly above show that there will only be a non-zero component at twice
the carrier which is removed using the low-pass filter on the multiplier’s output. In other
words, Message-2 has been rejected and so the output due to this branch is zero.

Finally, suppose that the local carrier is adjusted so that it is in phase with the carrier used to
modulate Message-2 – That is, it is set to be cos(c). It’s the same frequency but a different
phase.

Mathematically, this would result in:

m1.sin(c) x cos(c) = m1/2 [sin(c+c) + sin(c-c)]

m2.cos(c) x cos(c) = m2/2 [cos(c-c) – cos(c+c)]

Using the same reduction method as above, these expressions show that there will be a non-
zero component at the same frequency as Message-2 (m 2) and a non-zero component at twice
the carrier frequency. In other words, Message-2 in the QAM signal has been recovered and
Message-1 has been rejected. Again, the component at twice the carrier frequency is removed
using the low-pass filter on the multiplier’s output.

In effect, by adjusting the phase of the carrier used to demodulate the QAM, we are able to
“tune” to one message or the other.

22-4 © Emona Instruments Experiment 22 – Quadrature amplitude modulation

This interesting relationship between a sin and cos signal is the basis for the concept of
“orthogonality” – That is, that signals can be independent of each other which allows us to
transmit multiple messages on the same frequency.

The experiment
In this experiment you’ll use the Emona Telecoms-Trainer 101 to generate a QAM signal by
generating two DSBSC signals having orthogonal carriers then adding them together. Then,
you’ll explore how phase discrimination implemented using product detection can be used to
demodulate one DSBSC signal or the other.

It should take you about 60 minutes to complete this experiment.

Equipment

 Emona Telecoms-Trainer 101 (plus power-pack)

 Dual channel 20MHz oscilloscope
 three Emona Telecoms-Trainer 101 oscilloscope leads
 assorted Emona Telecoms-Trainer 101 patch leads
 one set of headphones (stereo)

Experiment 22 – Quadrature amplitude modulation © Emona Instruments 22-5

Procedure

Part A – Generating a QAM signal

1. Gather a set of the equipment listed on the previous page.

2. Set up the scope per the instructions in Experiment 1.

3. Locate the VCO module and turn both its Gain and Frequency Adjust controls fully
counter-clockwise.

4. Set the VCO module’s Range control to the LO position.

5. Connect the set-up shown in Figure 3 below.

Note: Insert the black plug of the oscilloscope lead into a ground ( GND) socket.

VCO

D IG ITA L

G A IN

FREQ

HI

LO

V CO S IN E
IN P U T

Figure 3

6. Activate the scope’s measurement tools and make the necessary adjustments to measure
the frequency of the signal on Channel 1.

7. Use the VCO module’s Frequency Adjust control to set its output to about 1kHz.

22-6 © Emona Instruments Experiment 22 – Quadrature amplitude modulation

8. Set the scope’s Timebase control to 200µs/div (if it’s not already).

9. Modify the set-up as shown in Figure 4 below.

Remember: Dotted lines show leads already in place.

Note: Consider using patch leads in the one colour for Part A to distinguish the QAM
modulator from the demodulator that you’ll patch later in the experiment.

VCO M ASTER M U LT IP L IE R
S IG N A L S

D IG ITA L
DC
X
AC

G A IN DC
Y
100kHz
S IN E AC

100kHz kX Y
FREQ
COS
M U LT IP L IE R
100kHz
D IG ITA L
HI
8kHz
LO D IG ITA L
2kHz X DC
D IG ITA L
VCO S IN E 2kHz
IN P U T S IN E
Y DC kXY

Figure 4

Experiment 22 – Quadrature amplitude modulation © Emona Instruments 22-7

The set-up in Figure 4 on the previous page can be represented by the block diagram in Figure
5 below. The VCO module’s 1kHz sinusoidal output is used to model Message-1 and the Master
Signals module’s 2kHz SINE output is used to model Message-2. The two messages are
connected to one input (y) of a Multiplier module. The other input to the Multiplier modules is a
100kHz sinewave (the carrier). However, the two carriers are out of phase with each other by
90° which is a requirement of QAM.

M essage 1
T o Ch.1

V CO M ultiplier

1kH z Y
DSBSCI
T o Ch.2
X

100 kH z
S IN E

M aster
S ignals

100 kH z
M aster COS
S ignals X
2kH z
S IN E Y
DSBSCQ

M ultiplier

Figure 5

10. Compare the signals.

Question 1
What feature of the Multiplier module’s output indicates that the signal is DSBSC?

22-8 © Emona Instruments Experiment 22 – Quadrature amplitude modulation

 Question 2
Given the Multiplier module’s inputs, how many significant spectral components are there
in the DSBSCI signal and what are their frequencies?

If the scope that you’re using has an FFT math option then perform Steps 11 to 14. Otherwise,
consider Figure 6 (which shows the two components in the DSBSC I signal) then go directly to
Step 15.

11. Activate the FFT option.

12. Make the necessary adjustments to view the spectrum of the DSBSC I signal (on Channel
2) using a frequency span from DC to about 200kHz and using a linear scale for the
vertical axis. You should obtain a display similar to Figure 6 below.

Figure 6

Experiment 22 – Quadrature amplitude modulation © Emona Instruments 22-9

13. Us the cursors to verify your answer to Question 2.

14. Deactivate the scope’s FFT option (for now).

15. Move the scope’s connections to the set-up as shown in Figure 7 below.

VCO M ASTER M U LT IP L IE R
S IG N A L S

D IG ITA L
DC
X
AC

G A IN DC
Y
10 0kH z AC
S IN E
10 0kH z kXY
FREQ COS
M U LT IP L IE R
10 0kH z
D IG ITA L
HI
8kHz
LO D IG ITA L
2kHz X DC
D IG ITA L
VCO S IN E 2kHz
IN P U T S IN E
Y DC kXY

Figure 7

22-10 © Emona Instruments Experiment 22 – Quadrature amplitude modulation

This change can be represented by the block diagram in Figure 8 below. The scope has been
adjusted to view Message-2 together with the DSBSC Q signal.

1kH z Y
DSBSCI

10 0kH z
S IN E

10 0kH z
COS
X
2 kH z
S IN E Y
DSBSCQ
T o C h.2

M essage 2
T o C h.1

Figure 8

16. Compare the signals.

Question 3
Given the Multiplier module’s inputs, how many significant spectral components are there
in the DSBSCQ signal and what are their frequencies?

Experiment 22 – Quadrature amplitude modulation © Emona Instruments 22-11

If the scope that you’re using has an FFT math option then perform Steps 17 and 18.
Otherwise, go directly to Step 19.

17. Reactivate the scope’s FFT option and use the cursors to verify your answer to Question
3.

18. Deactivate the scope’s FFT option.

19. Locate the Adder module and set both of its gain controls fully counter-clockwise.

20. Modify the set-up as shown in Figure 9 below.

VCO MASTER M U LT IP L IE R ADDER

S IG N A L S

D IG ITA L
DC
X
AC

G A IN DC
Y
10 0 k H z
S IN E AC

10 0 k H z kX Y G
FREQ
COS
M U LT IP L IE R
10 0 k H z A
D IG ITA L
HI
8kHz
LO D IG ITA L
2kHz X DC
D IG ITA L
VCO S IN E 2kHz g
IN P U T S IN E
Y DC kXY B G A+gB

Figure 9

22-12 © Emona Instruments Experiment 22 – Quadrature amplitude modulation

The set-up in Figure 9 can be represented by the block diagram in Figure 10 below. The Adder
module is used to add the DSBSCI and DSBSCQ signals. This turns the set-up into a complete
QAM modulator.

1kH z Y DSBSCI

10 0kH z
S IN E A A d d er

QAM
signal
T o Ch .1
10 0kH z B
COS
X
2 kH z
S IN E Y D S BS C Q

M ultiplier

Figure 10

21. Adjust the Adder module’s G control to set the DSBSCI signal to 4Vpp.

22. Disconnect the patch lead to the Adder module’s A input.

Note: This temporarily removes the DSBSCI signal from the signal on the Adder
module’s output.

23. Adjust the Adder’s g control to set the DSBSC Q signal to 4Vpp.

24. Reconnect the patch lead to the Adder’s A input.

25. Observe the signal on the Adder module’s output.

Experiment 22 – Quadrature amplitude modulation © Emona Instruments 22-13

 Question 4
According to the theory, what type of signal is now present on the Adder’s output?

Question 5
Given the spectral composition of the two DSBSC signals on the Adder module’s inputs,
what are the significant spectral components in the signal on its output?

If the scope that you’re using has an FFT math option then perform Steps 26 to 28.
Otherwise, go directly to Step 29.

26. Reactivate the scope’s FFT option and make the necessary adjustments to view the
spectrum of the QAM signal (on Channel 1). Again, use a frequency span from DC to
about 200kHz and continue to use a linear scale for the vertical axis.

27. Use the cursors to verify your answer to Question 5.

28. Deactivate the scope’s FFT option.

22-14 © Emona Instruments Experiment 22 – Quadrature amplitude modulation

Part B – Using phase discrimination to recover demodulate the QAM signal
It’s possible to use phase discrimination to recover one of the message signals while rejecting
the other. The next part of the experiment lets you do this.

29. Locate the Phase Shifter module and set its Phase Change control to the 180° position.

30. Set the Phase Shifter module’s Phase Adjust control to about the middle of its travel.

31. Modify the set-up as shown in Figure 11 below.

Note: As you’re now implementing the demodulator, use patch leads of a different colour
to distinguish it from the modulator.

VCO M ASTER M U LT IP L IE R ADDER

S IG N A L S

D IG ITA L
DC
X
AC

G A IN DC
Y
1 0 0k H z
S IN E AC

1 0 0k H z kXY G
FREQ
COS
M U LT IP L IE R
1 0 0k H z A
D IG ITA L
HI
8kH z
LO D IG ITA L
2kH z X DC
D IG ITA L
VCO S IN E 2kH z g
IN P U T S IN E
Y DC kX Y B GA+gB

PHASE M U LT IP L IE R U T IL IT IE S
S H IF T E R
C O M PA R AT O R
REF

LO
X DC

Y DC kX Y IN OUT
PHASE
S E R IA L T O R E C T IF IE R
PA R A L L E L
O
0
S /P

O D IO D E & R C L P F
180

S E R IA L X1
R C LPF
IN OUT

C LK X2

Figure 11

Experiment 22 – Quadrature amplitude modulation © Emona Instruments 22-15

The additions to the set-up can be represented by the block diagram in Figure 12 below. If you
compare it to Figure 2 in the preliminary discussion, you’ll notice that it implements a QAM
demodulator to recover one of the messages while rejecting the other (depending on the phase
of the local carrier).

M ultiplier
m odule R C LPF

QAM X M essage 1 or
M essage 2
input
T o Ch.2
Y

"S tolen" local

carrier

100 kH z

O
Phase M aster
S hifter S ignals

Figure 12

32. Compare the waveforms.

Note: You should still be viewing the original Message-1 on Channel 1. But the signal on
the RC LPF module’s output connected to Chanel 2 should look nothing like it. This is
because both Message-1 and Message-2 are being recovered at once. To check this for
yourself using your hearing…

22-16 © Emona Instruments Experiment 22 – Quadrature amplitude modulation

33. Locate the Buffer module and turn its Gain control fully counter-clockwise.

34. Modify the set-up to include the Buffer module as shown in Figure 13 below.

VCO MASTER M U LT IP L IE R ADDER N O IS E

S IG N A L S G E N E R AT O R

D IG ITA L
DC 0dB
X
AC -6 d B

G A IN DC -2 0d B
Y
100kH z
S IN E AC
BUFFER
100kH z kX Y G
FREQ
COS
M U LT IP L IE R
100kH z A
D IG ITA L
HI
8kH z G A IN
LO D IG ITA L
2kH z X DC
D IG ITA L
VCO S IN E 2kH z g IN OUT
IN P U T S IN E
Y DC kXY B G A+gB

PHASE M U LT IP L IE R U T IL IT IE S
S H IF T E R
C O M PA R AT O R
REF

LO
X DC

Y DC kX Y IN OUT
P HASE
S E R IA L T O R E C T IF IE R
PA R A L L E L
O
0
S /P

O D IO D E & R C L P F
180

S E R IA L X1
R C LP F
IN OUT

C LK X2

Figure 13

35. Without wearing the headphones, plug them into the Buffer module’s headphone socket.

36. Put the headphones on.

37. Adjust the Buffer module’s Gain control to set it to a comfortable sound level.

Note: You should now hear both Message-1 (which, at 1kHz, is the lower sounding of the
two tones) and Message-2 (the higher sounding of the two tones).

Experiment 22 – Quadrature amplitude modulation © Emona Instruments 22-17

38. While listening to the RC LPF module’s output and watching it on the scope’s display,
adjust the Phase Shifter module’s Phase Adjust control until the demodulator recovers
only Message-1.

Note: Once done, you should hear only the 1kHz tone and the RC LPF module’s output
should be an undistorted copy of Message-1.

Question 6
Show mathematically why the Message-1 has been recovered and what other component
is also produced (but filtered by the low-pass filter).

Question 7
Show mathematically why the Message-2 has been rejected and what other component is
produced (but filtered by the low-pass filter).

22-18 © Emona Instruments Experiment 22 – Quadrature amplitude modulation

If the scope that you’re using has an FFT math option then perform Steps 39 to 42.
Otherwise, go directly to Step 43.

39. Reactivate the scope’s FFT option and adjust it to view the spectrum of the
demodulator’s output (on Channel 2).

40. Make the necessary adjustments to view the demodulator’s output using a frequency
span from DC to about 10kHz and using a logarithmic scale for the vertical axis.

Note: You should notice that Message-2 has not been fully rejected.

41. Use the cursors to obtain the difference in magnitude (in decibels) between the two
messages and record this figure in Table 1 on Page 22-21.

42. Deactivate the scope’s FFT option.

43. Move the scope’s Channel 1 connection to the set-up as shown in Figure 14 below.

VCO MASTER M U LT IP L IE R ADDER N O IS E

S IG N A L S G E N E R AT O R

D IG ITA L
DC 0dB
X
AC -6 d B

G A IN DC -2 0d B
Y
1 0 0 kH z
S IN E AC
BUFFER
1 0 0 kH z kX Y G
FREQ
COS
M U LT IP L IE R
1 0 0 kH z A
D IG ITA L
HI
8kHz G A IN
LO D IG ITA L
2kHz X DC
D IG ITA L
VCO S IN E 2kHz g IN OUT
IN P U T S IN E
Y DC kXY B G A+gB

PHASE M U LT IP L IE R U T IL IT IE S
S H IF T E R
C O M PA R AT O R
REF

LO
X DC

Y DC kX Y IN OUT
PHASE
S E R IA L T O R E C T IF IE R

O
PA R A L L E L
0
S /P

O D IO D E & R C L P F
180

S E R IA L X1
R C LP F
IN OUT

CLK X2

Figure 14

Experiment 22 – Quadrature amplitude modulation © Emona Instruments 22-19

44. While listening to the RC LPF module’s output and watching it on the scope’s display,
adjust the Phase Shifter module’s Phase Adjust control until the demodulator recovers
only Message-2.

Note: Once done, you should hear only the 2kHz tone and the RC LPF module’s output
should be an undistorted copy of Message-2.

Question 8
Show mathematically why the Message-2 has been recovered and what other component
is also produced (but filtered by the low-pass filter).

Question 9
Show mathematically why the Message-1 has been rejected and what other component is
produced (but filtered by the low-pass filter).

22-20 © Emona Instruments Experiment 22 – Quadrature amplitude modulation

If the scope that you’re using has an FFT math option then perform Steps 45 to 47.
Otherwise, go directly to Question 10.

45. Reactivate the scope’s FFT option and check that you are still viewing the spectrum of
the demodulator’s output.

Note: You should notice that Message-1 has not been fully rejected.

46. Use the cursors to obtain the difference in magnitude (in decibels) between the two
messages and record this figure in Table 1.

47. Deactivate the scope’s FFT option.

Table 1

How much smaller Message-2 How much smaller Message-1

is relative to Message-1 is relative to Message-2

Question 10
As you can see from the results in Table 1, the unwanted message is not fully rejected.
(If you have not been able to perform the necessary steps, the unwanted messages are
about -33dB relative to the wanted message). Speculate on why this is happening.

Question 11
What potential problem would this lack of full rejection of the unwanted message cause
for a receiver?

Experiment 22 – Quadrature amplitude modulation © Emona Instruments 22-21

It’s possible to hear the crosstalk caused by the imperfect rejection of the unwanted message
by increasing its size. As the demodulator is presently tuned to recover Message-2, then to do
so, Message-1 must be replaced. The Speech module will be used for this purpose as it has a
much larger output voltage than the VCO module when it picks-up nearby sounds.

48. Replace the 1kHz sinewave being used to model Message-1 the signal from the Speech
module by modifying the set-up as shown in Figure 15 below.

VA R IA B L E M ASTER M U LT IP L IE R ADDER N O IS E
DCV S IG N A L S G E N E R AT O R
+5V

VDC DC 0 dB
X
GND AC -6 d B

VDC DC -2 0 d B
Y
1 00 k H z AC
S IN E
BUFFER
1 00 k H z kXY G
COS
SPEECH M U LT IP L IE R
1 00 k H z A
D IG ITA L
8kHz G A IN
A EXOR D IG ITA L
2kHz X DC
D IG ITA L
2kHz g IN OUT
S IN E
B A+B Y DC kX Y B GA+gB

PHASE M U LT IP L IE R U T IL IT IE S
S H IF T E R
C O M PA R ATO R
REF

LO
X DC

Y DC kXY IN OUT
PHASE
S E R IA L T O R E C T IF IE R

O
PA R A L L E L
0
S /P

O D IO D E & R C L P F
1 80

S E R IA L X1
R C LPF
IN OUT

C LK X2

Figure 15

22-22 © Emona Instruments Experiment 22 – Quadrature amplitude modulation

 VCO MA STER M U LT IP L IE R ADDER N O IS E
S IG N A L S G E N E R AT O R

D IG ITA L
DC 0dB
X
AC -6 d B

G A IN DC -2 0d B
Y
1 00 k H z
S IN E AC
BUFFER
1 00 k H z kXY G
FREQ
COS
M U LT IP L IE R
1 00 k H z A
D IG ITA L
HI
8 kH z G A IN
LO D IG ITA L
2 kH z X DC
D IG ITA L
VCO S IN E 2 kH z g IN OUT
IN P U T S IN E
Y DC kX Y B G A+gB

PHASE M U LT IP L IE R U T IL IT IE S
S H IF T E R
C O M PA R AT O R
REF

LO
X DC

Y DC kX Y IN O UT
P HASE
S E R IA L T O R E C T IF IE R

O
PA R A L L E L
0
S /P

O D IO D E & R C L P F
180

S E R IA L X1
RC LPF
IN OUT

CLK X2

49. Gently run your fingernail over the microphone while listening to the recovered 2kHz
tone.

Note: Although the system is adjusted to recover Message-2, you should be able to hear
Message-1 being recovered too. And there should be an indication of this crosstalk
visible on the scope’s display of the recovered Message-2.

50. Modify the scope’s connection to the set-up as shown in Figure 16 below.

Figure 16

Experiment 22 – Quadrature amplitude modulation © Emona Instruments 22-23

51. Set the scope’s Timebase control to 2µs/div.

52. Compare the two carriers.

Question 12
You may be surprised to see that the modulator’s cos(c) carrier and the demodulator’s
local carrier are not in phase even though Message-2 is being recovered. Speculate on
why this is.

Question 13
As its not critical for the two DSBSC signals of a two-channel analog QAM signal to be
phased exactly 90º apart for this system to work, why not use three channels and put
them at 60º apart? Discuss whether such a system would work.

22-24 © Emona Instruments Experiment 22 – Quadrature amplitude modulation