mti 7604 psk / qpsk modulation and...

Contents

Aims of the Exercise

Learning about the functioning principle of PSK and DPSK modulation Reconstruction of the transmitted bit pattern by demodulation of the PSK signal Learning about the functioning principle of QPSK and DQPSK modulation Reconstruction of the transmitted bit pattern by demodulation of the QPSK signal

Overview of Exercises

Exercise 1: PSK Modulation/Demodulation

Allocation of the oscillogramme at the measuring points DATA and BITCLOCK to a set bit pattern

Build the phase-shifts by PSK and DPSK to different bit patterns Allocation of the output voltage PSK to different bit patterns Effect of different data transmission speeds Demodulation of the PSK signal

Exercise 2: QPSK Modulation/Demodulation

Allocation of the oscillogramme at the measuring points DATA and DIBITCLOCK to a set bit pattern

Illustration of the dependency of the signals DIBIT X and DIBIT Y on the respective dibits of the set bit pattern

Build the phase-shifts by QPSK and DQPSK to different bit patterns Allocation of the output voltage QPSK to different bit patterns and dibits Effect of different data transmission speeds Demodulation of the QPSK signal

MTI 7604 PSK / QPSK Modulation and

Demodulation

MTI 7604

Page 1 of 1Name: jvourv, Date: 10/13/2012

Introduction

PSK (Phase-Shift-Keying), or more precisely 2PSK (PSK with two defined states) or BPSK (Binary Phase-Shift-Keying), and QPSK (Quadrature phase-shift-keying) or 4PSK (PSK with 4 defined states) belong to the group of so-called "digital modulation methods".

Their classification in this family is as follows:

1. 1. Digital communications signal, sinewave carrier signal: Amplitude Shift Keying = ASK Frequency Shift Keying = FSK Phase Shift Keying = PSK

2. 2. Analog communications signal, pulse-shaped carrier signal: Pulse Amplitude Modulation = PAM Pulse Phase Modulation = PPM Pulse Width Modulation = PWM

3. 3. Digitised original analog communications signal, pulse-shaped carrier signal:

Pulse Code Modulation = PCM Delta Modulation = DM

In the modulation types ASK, FSK and PSK, digital information is transferred on analog channels with the aid of a sinewave carrier. The signal converter for analog transmission paths consists of a modulator at the send side and a demodulator at the receive end. This form of transmission device is known as a Modem. In the early days of data transmission, ASK (telegraphy) and FSK (acoustic coupler as an interface) were used but due to their susceptibility to interference, they were replaced by the PSK and QPSK methods (for modems up to 2400 bps, V.22) as well as higher quality methods such as 16QAM. Today, QPSK is widely used in satellite transmissions and in radio relay systems. The relationships between the terms "Baud", "bits per second" (bps) and "characters per second" (cps) should become clear with the following example: A (Q)PSK modulator allows one byte consisting of 8 bits, to be set, e.g.:

MTI 7604 PSK / QPSK Modulation and Demodulation

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Fig. 1: Duration of a byte (character)

If this signal is transmitted with a bivalent (two-value) method of modulation (e.g. 2PSK), then the modulation rate vS (or Baud rate) is the same as the data transfer rate, vD

At vS = 600 Baud here, vD = 600 BPS, i.e. the time required for transmitting one bit is

From this it follows that the duration of one byte, is

so that

(here, without Start, Stop or check bit) can be transmitted. If now QPSK (4PSK), a quadrivalent (four-valued) method of modulation is used, then the data transfer rate is doubled

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where n is the number of significant conditions, here n = 4

This is possible because the coding does not refer to a single bit but rather to a bit-pair, a so-called "Dibit" :

Fig. 2: Formation of a Dibits

Providing that the bandwidth for the transmission channel is wide enough,

it follows that,

There are thus 1200/8 cps = 150 cps transmitted. Since apparently, the transmission capacity of a channel with a given bandwidth, is doubled when 4PSK is used instead of 2PSK, it was logical to always use higher valued methods of modulation in the development of modems for analog telephone channels with a limited bandwidth (nowadays, up to vSmax =33600 bps or more, is possible).

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The block schematic diagram of a simple 2PSK modulator is shown below:

Fig. 3: 2PSK modulator

The 2PSK signal is produced by multiplying (e.g. in a ring modulator) the carrier signal (Carrier) with the serial bit pattern (DATA). The binary value 1 here, is represented by a carrier phase shift of 0 and the binary value 0, by a shift of 180. "Phase-hits" are the result:

Fig. 4: 2PSK-modulated signal

In its simplest form, demodulation is achieved as illustrated in the following block diagram:

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Fig. 5: 2PSK demodulator

Here, the received 2PSK signal is multiplied by the carrier signal of the modulator, so that after filtering and restoring the edges, the original bit pattern (DATA) is recovered. If the carrier signal cannot be transmitted with the data, then it must be re-generated (or restored), in the demodulator with the correct phase. This possibility is not dealt with in the exercise here.

The block schematic diagram of a 4PSK modulator is shown below:

Fig. 6: 4PSK modulator

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In the coder, the signal U DIBITCLOCK, is conditioned, that in the last quarter-period of

the current Dibit, the pattern of which is sampled and generated according to the table of the signals U DIBITX and U DIBITY:

DPSK and DQPSK-Signals

From the discussion above it might appear that QPSK offers advantages over the simple modulations ASK, FSK and PSK. However, the demodulation of these signals requires various degrees of difficulty and hence expense. The method of demodulation is an important factor in determining the selection of a modulation schem.

With QPSK, the processing is more complicated, and two separate demodulators are required. The demodulator complexity increases rapidly for M-ary PSK; for this reason it is rarely used.

There are two methods to prevent such an occurrence. In one, a pilot carrier signal is sent in addition to the modulated carrier. This pilot carrier is used to synchronise the local oscillator phase. The alternative is to employ another form of modulation, differential phase-shift-keying (DSPK).

Differential PSK is actually a simple form of coding. The modulating signal is not the binary code itself, but a code that records changes in the binary code. This way, the demodulator only needs to determine changes in the incoming signal phase. Because the drifts associated with local oscillators occur slowly, this is not difficult to arrange. (The simple multiplier used above is still inadequate, but the alternatives are no more complicated.)

The PSK signal is converted to a DPSK signal with two rules:

a 1 in the PSK signal is denoted by no change in the DPSK a 0 in the PSK signal is denoted by a change in the DPSK signal

The sequence is initialised with a leading 1. An example of the pattern is thus:

Analog to the DPSK a DQPSK signal is generating from QPSK signal

DATA 00 01 10 11

DIBIT X 0 0 1 1

DIBIT Y 0 1 0 1

Phase relation of U PSK 0 90 270 180

PSK _ 0 1 0 0 1 1 0 1

DPSK 1 0 0 0 1 0 1 0 0

DATA 00 01 10 11

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DIBIT X 0 0 1 1

DIBIT Y 0 1 0 1

Phase relation of U

PSK0 90 270 180

Phase-shift of U

DPSK0 +90 -90 180

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Exercise Assembly

Ensure that all voltages are removed from the UNI-TRAIN. Assemble the exercise by inserting the PSK / QPSK modulator in the left and the PSK / QPSK demodulator in the right hand slots of the Experimenter SO4203-2B. Connect the two PCB's with 2mm connection cables, at the sockets "Car" and "(Q)PSK", as shown in the illustration above. The electrical energy for the PCB's is supplied via the internal bus of the UNI-TRAIN basic unit. After switching on the operating voltage, the exercise assembly for PSK / QPSK modulation and demodulation is ready for use.

Sub-assemblies and Components Required

Operator Elements and Sockets


Qty. Description Order No.

1 UniTr@in-I Interface with virtual instruments SO4203-2A

2 Experimenter SO4103-2B

1 PSK /QPSK modulator SO4201-9J

1 PSK /QPSK demodulator SO4201-9K

1 Measuring line set 2mm UniTr@in I SO5146-1L

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ADATA signal

output

BSYNC, test

signal

CBasic clock

potentiometer

DPatchboard,

Baud rate

ECLOCK, test

signal

F X, test signal

G Y, test signal

HDIBIT-clock, test

signal

JBIT-clock, test

signal

Fig. 2: Front panel of the PSK / QPSK modulator

K Signal output a

L Signal output b

MCarrier, signal

output

N(Q)PSK, signal

output

P GND

QAmplitude

potentiometer

RDIP-switch (8-

way)

S4PSK/2PSK

switch

T SET button

UTest signal: 0,

90, 180, 270

V

Phase 0 and

90

potentiometer

ASignal input, a

and b

B Signal input, Car

CSignal input, (Q)

PSK

D GND

E Test signal, 0

F Test signal, 90

H Test signal, TP1

J Test signal, TP2

K Test signal, TP3

L Test signal, TP4

MBaud rate select

switch

N4PSK/2PSK

switch

QTest signal,

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GPotentiometer,

phase 90

Fig. 3: Front panel of the PSK / QPSK demodulator

CLOCK

PDATA signal

output

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PSK - DPSK Modulation / Demodulation

Assigning the oscillograms at the test points DATA and DIBIT-CLOCK, to a set bit pattern

Note: Before commencing the exercise, check the alignment of the PSK modulator, SO4201-9J, in accordance with the operating instructions!

On the PSK / QPSK modulator (SO4201-9J), connect the "Sync" output to the analog test input (channel B) of the UniTr@in-Interface. On channel A measure the signal at the "Data" output. Trigger on B. For more stable triggering, set the trigger point to 50% of the amplitude of the squarewave signal.

Set the following bit pattern (Byte) on the DIP switches:

Set the switch for the modulation type to "2PSK", the jumper for the transfer rate to "1200" and by pressing the "SET" button, read in the bit pattern.

Caution: The "Set" button must be pressed again, after each new change of the settings has been completed!

On an oscilloscope, display the voltages at the test points DATA and BITCLOCK. What is the sequence of the bits during transfer? How many bits are transferred for each bit-clock ?

Result:


LSB

Bit 1

Bit 2

Bit 3

Bit 4

Bit 5

Bit 6

Bit 7

MSB

Bit 8

0 1 0 0 1 1 0 1

X = 1 ms/DIV X/T (B)

Chan. A= 2 V/DIV DC

Chan. B= 2 V/DIV DC

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Fig. 1: Data signal

Fig. 2: Bit-clock

? ? ? ?

Exercise 2 Assigning the output voltage to various bit patterns


Chan. A= 2 V/DIV DC

Chan. B= 2 V/DIV DC

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On channel A, measure the signal at the "(Q)PSK" output and determine the assignment of the binary value of the bit pattern (0 and 1) to the phase of the 2PSK signal. For this, compare the 2PSK-modulated signal to the carrier signal at the "Car." output.

Result:

Fig. 3: PSK modulated signal

? ? ? ?

In the following picture from oscilloscope was transmitted a DPSK-Signal. Write the original code beginning by the MSB in the table below.


Chan. A= 0,5 V/DIV DC

Chan. B= 2 V/DIV DC



Chan. B= 2 V/DIV DC

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Result:

Exercise 3 Effect of various rates of transfer

Determine the baud rate vS, data transfer rate vD and the number of bytes transferred per second (cps) for the "600" and "1200" settings of jumper D. Compare the appearance of the 2PSK signal at either setting of the switch.

Note: For determining the transfer rate, use the time marker on the oscilloscope. In the lower part of the operating bar for the oscilloscope, the button will be seen for cursor functions. Set this for channel A. Also, two amplitude markers are available for measuring voltages and two time markers for measuring time or frequency. The marker can be moved with the mouse to the position required. The values detected are shown at the top right .

Result ("600" position):

MSB

Bit 8

Bit 7

Bit 6

Bit 5

Bit 4

Bit 3

Bit 2

LSB

Bit 1

??? ??? ??? ??? ??? ??? ??? ???



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Fig. 4: 2PSK signal at 600 Bd

? ? ? ?

Result ("1200" position ):


Chan. B= 2 V/DIV DC



Chan. B= 2 V/DIV DC

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? ? ? ?

Exercise 4 Demodulation of the 2PSK signal

Without changing the setting of the modulation, connect the "(Q)PSK" and "Car." outputs of the QPSK / PSK modulator to the corresponding inputs on the QPSK / PSK demodulator (SO4201-9K). For synchronising, the carrier signal (CAR) must be transferred with the output signal.

Set the demodulator according to the conditions for receiving modulated data: - Transfer rate switch: Position "1200", - Type of modulation switch: Position "2PSK"

On the PSK / QPSK modulator (SO4201-9J), connect the "Sync" output to the analog test input, channel B, of the UniTr@in-Interface. On channel A, measure the signal at the "Data" output of the QPSK / PSK demodulator (SO4201-9K). Trigger on B. For more stable triggering, set the trigger point to 50% of the amplitude of the squarewave signal.

Result:

Fig. 6: 2PSK demodulation


Chan. A= 2 V/DIV DC

Chan. B= 2 V/DIV DC

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? ? ? ?

Set other different bit patterns with the DIP switches and change the data transfer rate (modulator and demodulator settings must be the same).

Result:

? ? ? ?

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PSK - DPSK Modulation / Demodulation

Assigning the oscillograms at the test points DATA and DIBIT-CLOCK, to a set bit pattern






On an oscilloscope, display the voltages at the test points DATA and BITCLOCK. What is the sequence of the bits during transfer? How many bits are transferred for each bit-clock ?

Result:


LSB

Bit 1

Bit 2

Bit 3

Bit 4

Bit 5

Bit 6

Bit 7

MSB

Bit 8

0 1 0 0 1 1 0 1


Chan. A= 2 V/DIV DC

Chan. B= 2 V/DIV DC

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Fig. 1: Data signal

Fig. 2: Bit-clock

? ? ? ?

Exercise 2 Assigning the output voltage to various bit patterns


Chan. A= 2 V/DIV DC

Chan. B= 2 V/DIV DC

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On channel A, measure the signal at the "(Q)PSK" output and determine the assignment of the binary value of the bit pattern (0 and 1) to the phase of the 2PSK signal. For this, compare the 2PSK-modulated signal to the carrier signal at the "Car." output.

Result:

Fig. 3: PSK modulated signal

? ? ? ?

In the following picture from oscilloscope was transmitted a DPSK-Signal. Write the original code beginning by the MSB in the table below.



Chan. B= 2 V/DIV DC



Chan. B= 2 V/DIV DC

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Result:


Determine the baud rate vS, data transfer rate vD and the number of bytes transferred per second (cps) for the "600" and "1200" settings of jumper D. Compare the appearance of the 2PSK signal at either setting of the switch.

Note: For determining the transfer rate, use the time marker on the oscilloscope. In the lower part of the operating bar for the oscilloscope, the button will be seen for cursor functions. Set this for channel A. Also, two amplitude markers are available for measuring voltages and two time markers for measuring time or frequency. The marker can be moved with the mouse to the position required. The values detected are shown at the top right .

Result ("600" position):

MSB

Bit 8

Bit 7

Bit 6

Bit 5

Bit 4

Bit 3

Bit 2

LSB

Bit 1

??? ??? ??? ??? ??? ??? ??? ???



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? ? ? ?



Chan. B= 2 V/DIV DC



Chan. B= 2 V/DIV DC

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? ? ? ?

Exercise 4 Demodulation of the 2PSK signal




Result:



Chan. A= 2 V/DIV DC

Chan. B= 2 V/DIV DC

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? ? ? ?


Result:

? ? ? ?

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QPSK and DQPSK Modulation/Demodulation

Exercise 1 Assigning the oscillograms at the test points DATA and DIBIT-CLOCK, to a set bit pattern






On an oscilloscope, display the voltages at the test points DATA, BITCLOCK and DIBITCLOCK. What is the sequence of the bits during transfer? How many bits are transferred for each bit-clock and Dibit-clock ?

Result:


LSB

Bit 1

Bit 2

Bit 3

Bit 4

Bit 5

Bit 6

Bit 7

MSB

Bit 8

0 1 1 1 1 0 0 0


Chan. A= 2 V/DIV DC

Chan. B= 2 V/DIV DC

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Fig. 1: Data signal

Fig. 2: Bit-clock


Kannal A= 2 V/DIV DC

Kannal B= 2 V/DIV DC


Chan. A= 2 V/DIV DC

Chan. B= 2 V/DIV DC

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Fig. 3: Dibit-clock

? ? ? ?

Exercise 2 Displaying the dependency of the DIBIT X and DIBIT Y signals on the associated Dibits of the selected bit pattern

On an oscilloscope, display the signals at the test points "DIBIT X" and "DIBIT Y" as a function of the bit pattern selected.

Result:


Chan. A= 2 V/DIV DC

Chan. B= 2 V/DIV DC

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Fig. 4: Dibit formation; Dibit X

Fig. 5: Dibit formation; Dibit Y

Exercise 3 Assigning the output voltage QPSK to various bit patterns or Dibits


Chan. A= 2 V/DIV DC

Chan. B= 2 V/DIV DC

DIBIT 00 01 10 11

DIBIT X ??? ??? ??? ???

DIBIT Y ??? ??? ??? ???

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Determine the phase relationship of the QPSK signal and the set DATA signal (the Dibits in the signal). For this, compare the phase of the QPSK signal with the signals at the test points T (0, 90, 180, 270). Bear in mind that the QPSK signal compared to the reference signals, exhibits a small but recognisable delay caused by the transit time.

Result:

Fig. 6: Phase relationship of the Dibits in a QPSK

In the Fig. 6 from oscilloscope was transmitted a QPSK-Signal. Write the original code beginning by the MSB in the table below. Then write the code, if it was transmitted a DQPSK-Signal.

Result:




Chan. B= 2 V/DIV DC

DIBIT 00 01 10 11

Phase relationship QPSK ??? ??? ??? ???

Phase shift in DQPSK signal ??? ??? ??? ???

MSB

Bit 8

Bit 7

Bit 6

Bit 5

Bit 4

Bit 3

Bit 2

LSB

Bit 1

QPSK ??? ??? ??? ??? ??? ??? ??? ???

DQPSK ??? ??? ??? ??? ??? ??? ??? ???

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Determine the baud rate vS, data transfer rate vD and the number of bytes transferred

per second (cps) for the "600" and "1200" transfer rate settings. Compare the appearance of the 4PSK signal at either setting.

Note: For determining the transfer rate, use the time marker on the oscilloscope! In the lower part of the operating bar for the oscilloscope, the button will be seen for the cursor function. Set this for channel A. Also, two amplitude markers are available for measuring voltages and two time markers for measuring time or frequency. The marker can be moved with the mouse to the position required. The values detected are shown at the top right.

Result ("600"position)


? ? ? ?


X = 0,5 ms/DIV X/T (B)


Chan. B= 2 V/DIV DC



Chan. B= 2 V/DIV DC

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? ? ? ?

Exercise 5 Demodulation of the QPSK signal




Result:

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? ? ? ?



Chan. A= 2 V/DIV DC

Chan. B= 2 V/DIV DC

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mti 7604 psk / qpsk modulation and...

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