Laboratory guide for communications 2 di

Purpose: To become familiar with the Modulation module and wave analyzer 1: Modulation background: In communication, modulation is the process of modifying the properties amplitude, f

Communications II: Digital Communications

Francisco G Glover SJ Josef Rene L Villanueva, ECE

Ateneo de Davao University

Davao City June, 2011

This text is a laboratory manual for Communications II: Digital Communications

It is the second installment in the series of Communications courses in the Electronics Engineering Program, following the first course, Communications 1: AM and FM

It includes nine experiments which focus on the concepts of Digital Modulation and Transmission Systems The first experiment is a refresher for the basic but important concept of signal synthesis, which is also covered in Communications I: AM and FM The succeeding experiments introduce the different digital modulation schemes and the electronic circuits needed for its operation

Each experiment is accompanied by a hardware module specifically designed to help the students understand more clearly the theory behind digital communications systems These modules are also made to be easy to use on the part of both students and instructors as only minimal external connections will have to be made, and test points are labeled accordingly with the experiment procedures This way, the students can focus more on understanding the concepts and seeing them in action before stress

on the circuitry is given

The schematic diagram, list of components, and components placement guide is placed inside each of the modules as guide for easy troubleshooting and also as reference for those interested in the design of the modules

Ateneo de Davao University

June, 2011

Table of Contents

Experiment #1 Modulation Materials: Modulator module, Dual Function Generator, oscilloscope, Wave

Analyzer, analog and digital voltmeters

Purpose: To become familiar with the Modulation module and wave analyzer

1:

Modulation background:

In communication, modulation is the process of modifying the properties

(amplitude, frequency or phase) of a sinusoidal (sine or cosine) electromagnetic wave to

convey information; de-modulation is the reverse process, recovering the original

information from the modulated wave The original unmodified electromagnetic wave is

called a carrier, with a given fixed amplitude, A, frequency, f, and phase,

carrier = A cos(2 ft + )

The carrier is an example of a periodic function with period T (T = 1/f) For any

periodic function, f(t) = f(t + nT), where n = 0, ±1, ±2, , that is, the wave form repeats

itself over and over again every T seconds

Fourier’s Theorem states that any periodic function may be represented by

sinusoidal waves of varying amplitude and with periods of T, T/2, T/3, , or with frequencies of 0, f, 2f, 3f,

(1)

where = 2 f

So if the carrier is modified in any way it is no longer a pure sinusoid, and must

contain additional frequency components or harmonics, multiples of the original carrier

t n B t

n A A

t f

n n n

ncos sin)

(

1 1

0

COMMUNICATIONS II

C2-1 000000000000010 0

This Modulation module has two basic functions:

Provide adder and multiplier circuits to illustrate basic

modulation / de-modulation concepts

2: Accept and condition AC and DC voltages from the Dual Function Generator for use by other modules

To use, the Modulation module‟s 9-pin DB-9 connector must be plugged into the Dual Function Generator (included in the Communications 1: AM and FM package)

frequency A wave analyzer may be used to detect the presence of these components

In this module we explore the use of such a wave analyzer

A circuit element is said to be linear if the instantaneous output is proportional to the input; otherwise it is said to be non-linear If the input to a linear element is

sinusoidal (single frequency) the output is also sinusoidal with the same frequency and phase but possibly different amplitude If a sinusoidal signal is passed through a non-

linear element, the instantaneous output is no longer proportional the input; this means

that the wave shape has change In is still has the same period but additional frequency components (harmonics) have been introduced

Two sinusoidal signals may be combined by addition or by multiplication

(2a) (2b)

An oscilloscope can display a time-varying signal as amplitude vs time graph; a

spectrum analyzer can present the same signal as amplitude vs frequency graph

(Amplitude vs frequency graph of a sine wave consists of a single vertical line segment; for any periodic

wave the display is a series of equally-spaced vertical lines of various heights) A wave analyzer can detect the presence of individual frequency components but does not produce a graphical display Such a wave analyzer will be used extensively in following experiments, so details on its theory and operation follow

The wave analyzer

The face plate of the Wave Analyzer is shown

in Fig 1 (The analyzer is included in the Communications 1:

analyzed is connected to the BNC terminal labeled

IN The other BNC terminal marked OUT is

connected to the oscilloscope An analog voltmeter

may be connected to the pair of red and black jacks

also labeled OUT The block diagram shows the

basic components of the instrument; a sinusoidal

oscillator, wave multiplier, amplifier, low-pass filter

and a peak detector The lower-panel knobs control the amplitude and frequency of the internal oscillator The ABCD selector switch connects the oscilloscope to various key

test points in the circuit; the input signal at A, the local oscillator at B, the amplified

product of the input and local oscillator signals at C and the output of the low-pass filter

at D Amplifier gain may be increased by the x10 switch

The periodic input signal of fundamental angular frequency, 0, may contain any number of harmonics, 1, 2, , n, of various amplitudes To detect the presence

of any particular harmonic, n, first set the local oscillator frequency, osc, to the desired

cos A + cos B = 2 cos ½(A+B) cos ½(cos A–B) cos C cos D = ½ { cos (C+D) + cos (C–D) }

Fig 1 Wave Analyzer faceplate

n so the amplified output of the multiplier is the product of the input harmonics and

local oscillator signals For any particular harmonic j, Eq 2b may be applied:

cos jt cos osct = ½ { cos ( j + osc)t + cos ( j – osc )t } (3)

For the special case, j = osc= n, the second cosine term on the right becomes cos (0)t = 1, a constant (if j osc this term gives a slowly varying signal between +1 and –1) The

signals of all the product terms then enter the low pass filter but only cosine terms with

( j – osc ) 0 are passed Of course if ( j – osc ) exactly equals zero, the cosine term

is a constant 1, and wave filters are not designed to handle constants As ( j – osc )

moves away from zero, the output signal varies more rapidly while its amplitude

decreases This signal moves on to the peak detector In use, slowly vary osc through values a bit above and below n and the peak detector will give the maximum input voltage, which may be read from an attached analog voltmeter.

The wave multiplier unit is quite sensitive, and can detect harmonics whose amplitude is too small to appear on the analog meter Here is where the oscilloscope can be helpful to obtain an approximate measure of small amplitudes Set the wave analyzer selector switch to D to monitor the signal before it enters the peak detector If clipping is seen, reduce the local oscillator gain Use a convenient oscilloscope vertical gain Set the TIME/DIV to 10 ms As you vary osc about j you can easily see the changes in amplitude and frequency Estimate the number of scale divisions for signal peak-to-peak value, and multiply this by the VOLTS/DIV setting Alternately set TIME/DIV

to 10 s The display then shows horizontal lines moving up and down on the screen, from which the number of screen divisions may be estimated (use either the AC or DC input)

Note that the analog voltmeter gives a peak value, the oscilloscope method gives a

peak-to-peak value

If we multiply Eq 2b by any positive constant, AB, the result is:

Acos C Bcos D = ½ AB { cos (C+D) + cos (C–D) } (4) This implies that increasing the amplitudes of the input signal and local oscillator increases the signals at points C, D and the output of the peak detector For maximum

detection sensitivity, increase the local oscillator amplitude and use the X10 gain

setting Always check the wave shape at test point C to assure that no clipping occurs When comparing the relative intensity of various harmonics, it is important to maintain constant the local oscillator amplitude

Activity #1: Detecting frequency components

1: Connect the Modulation module to the Dual Function

Generator using the 9-pin connector Connect the FG-1

terminal on the Modulation module to the Wave Analyzer

input Connect an oscilloscope to the Wave Analyzer BNC

OUT connector and an analog voltmeter to the red and black

output jacks Turn ON power for all units

Freq Amplitude

10.00 khz 20.00 khz 30.00 kHz Table 1 10 kHz sine

2: Set the FG-1 signal to 10.00 kHz, sine, 3.0 VP-P as measured on the oscilloscope (Wave Analyzer selector switch at position A) With the Wave Analyzer determine the amplitudes (to within 2 significant figures) of the frequency components listed in Table 1 Use the largest oscillator amplitude that does not cause clipping, as viewed at test point C If the input is not a perfect sine wave, some harmonics may be present

3: Change FG-1 from sine to square and repeat step 2

above Use the frequencies of Table 2 Should a perfect

square wave contain odd harmonics? _

Adding or Multiplying Sinusoidal Signals

Equations 2a and 2b are not really separate statements, but rather a single statement in two forms To show this, let C = ½(A+B) and D = ½(A–B), and make this substitution in (2b); the result is (2a) Try it and see for yourself!

The Modulation module contains an adder circuit and a multiplier circuit, each

with two inputs and a single output The multiplier has a balance control that is needed

for current stability In this activity we use these elements to give us more insight into the addition and multiplication of sinusoidal signals

Activity #2 Addition

1: Equation 2a is repeated here:

cos A + cos B = 2 cos ½(A+B) cos ½(cos A–B)

1a: Set FG-1 at 39.00 kHz so A = 2 39000 t,

FG-2 at 41.000 kHz so B = 2 41000 t, each

sinusoidal, at 4.0VP_P and apply these to the

adder input

1b: Connect the adder output to the oscilloscope

through the Wave Analyzer

1c: Set the oscilloscope TIME/DIV at 0.1 ms and adjust the variable sweep,

trigger level and the channel gains for a stable display At different sweep speeds

various stationary displays may appear After a careful adjustment the screen display should appear somewhat like Fig 2 Obtain a similar display on your oscilloscope From Eq 2a we have:

cos 2 39,000t + cos 2 41,000 t = {2cos 2 1,000t} cos 2 40,000 t (5)

Freq Amplitude

10.00 kHz 20.00 kHz 30.00 kHz 40.00 kHz 50.00 kHz

190.00 kHz

Table 2 10 kHz square

Fig 2 Addition: 39 + 41

The right side of the equation suggests a 40 kHz wave with amplitude varying at 1 kHz

Is this a reasonable description of your display? _

2: In Eq.2a the coefficients of the two cosines are the same How might we interpret

an expression with different coefficients: A cos + B cos if A > B? How about (A–B) cos + B (cos + cos )? Does this suggest a constant-amplitude wave added to the

pattern of Fig 2? _

3: Change slightly the amplitude of either FG-1 or FG-2 and describe and explain the

resulting pattern:

4: The sum of two non-negative numbers is zero only if both numbers are zero

Describe the pattern if either FG-1 or FG-2 amplitude is set to zero: _ _

5: The left side of Eq 5 refers to the frequencies 39 kHz and

41 kHz, while the right side refers to 1 kHz and 40 kHz Does

the adder add or remove frequencies? Use the Wave

Analyzer to measure the amplitudes (to two significant

figures) of the frequencies in Table 3 for the signal leaving

the adder

Does addition introduce any new frequencies? _ Does addition remove any frequencies already present?

Activity #3 Multiplication

1: Eq 2b is repeated here:

cos C cos D = ½ { cos (C+D) + cos (C–D) }

1a: Set FG-1 at 40.00 kHz so C = 2 40000 t, FG-2 at 1.00 kHz so D= 2 1000 t,

each sinusoidal, with 4.0VP_P and apply these to the multiplier input

1b: Connect the multiplier output to the oscilloscope through the Wave Analyzer

1c: Set the oscilloscope TIME/DIV at 0.1 ms and adjust the variable sweep,

trigger level and the channel gains for a stable display Is it possible to obtain a

display similar to Fig 2?

2: Describe what happens if you vary slightly the balance control:

_

3: From Eq 2b we have:

cos 2 40.00 kHz t cos 2 1.00 kHz t = ½ {cos 2 41.00 kHz t + cos 2 39.00 kHz t } (6)

Fr

eq

Am plitude

1.

00 kHz 39 00 kHz 40 00 kHz 41 00 kHz Table 3 Addition

Compare the right side of (6) with the left side of (5) Does this explain why our

multiplication display is similar to Fig 2 for the addition display? _ Just by looking at the oscilloscope screen explain how you might know if it resulted from

the multiplication of 40kHx and 1 kHz signals or the addition of 39 and 41 kHz signals:

_

_

4: Change slightly the amplitude of either FG-1 or FG-2 and describe and explain the

resulting pattern:

_

5: The product of two non-negative is zero if either of the numbers are zero Describe

the pattern if either FG-1 or FG-2 amplitude is set to zero:

6: The left side of Eq 6 refers to the frequencies 40 kHz and

1 kHz, while the right side refers to 39 kHz and 41 kHz Does

the multiplier add or remove frequencies? Use the Wave

Analyzer to measure the amplitudes (to two significant

figures) of the frequencies in Table 4 for the signal leaving

the multiplier (Before measurement, set the Wave Analyzer oscillator

to 40.00 and vary the balance to obtain minimum amplitude.)

Does multiplication introduce any new frequencies?

_

Does multiplication remove any frequencies already present?

We have made much use of Eqs 2a and 2b Mathematically, they are

statements from Trigonometry The equal sign state that both sides of the equation

have the same value, but expressed in different form It is helpful to give a physical

interpretation for these important equations When we apply Eqs 2a and 2b to

time-varying signals, cos A becomes cos 2 ft or cos t Every such sinusoidal signal has a

definite frequency, f, and period T ( T = 1/f) No matter what the amplitude or phase, the instantaneous value of the signal (voltage, current, displacement, etc.) is zero 2f times a

second, and the time between these zero crossings is ½ T seconds

Physically, cos 1 t + cos 2 t states that the signal contains two separate

frequencies but tells us nothing about the time between the instantaneous zero values

of the signal For an instantaneous zero value (the graph line crosses the horizontal axis)

either both signals are zero at the same instant, or are of equal and opposite algebraic

value On the other hand, cos 3 t cos 4 t tells us exactly when the instantaneous

zero values occur, but gives us no information about the frequency content of the signal Therefore a particular oscilloscope pattern of zero values may be the result either

Freq Amplitude

1.00 kHz 38.00 kHz 39.00 kHz 40.00 kHz 41.00 kHz 42.00 kHz

Table 4

Multiplication

adding two particular frequencies or by multiplying two other particular frequencies So,

given the two frequencies, Eq 2a tells us about the zero crossings Given the zero

crossings, Eq.2b tells us about the frequencies present

Adding or Multiplying a Sinusoidal and Constant Signal

Activity #4 Adding a constant

1: In symbols, adding a constant to a sinusoidal wave is expressed as K + cos t

Does this change the amplitude of the cos t term? Does this change the frequency of the cos t term?

Twice each cycle cos t = 0, the zero crossing Is it possible to select K so that the sum never equals zero? Explain: _

Connect FG-1 (any convenient sine frequency) to one terminal of the adder and the

adjustable DC source to the other terminal View the adder output on the oscilloscope

as the DC is varied Describe and explain the results:

2: The oscilloscope vertical position control moves the screen pattern up or down

without changing the shape Does this suggest an adder circuit? _

Activity # 5 Multiplying by a constant

1: In symbols, multiplying a constant and a sinusoidal wave is expressed as K cos t

Does this change the amplitude of the cos t term? Does this change the frequency of the cos t term?

Twice each cycle cos t = 0 Is it possible to select K so that the product never equals zero? Explain:

Connect FG-1 (any convenient sine frequency) to the multiplier (use terminal marked with a capacitor symbol) and the adjustable DC source to the other terminal View the multiplier output on the oscilloscope as the DC voltage is varied Describe and explain the results:

2: The oscilloscope vertical gain control enlarges or decreases the screen pattern

without moving it up or down Does this suggest an adder circuit? _

Linear or non-Linear?

In terms of the definitions presented above the adder is a linear device, since

the same sinusoidal waves we entered came out as sinusoidal waves of the same frequency (provided there is no clipping) The multiplier is a non-linear device since the

frequencies of the outgoing signals are different from the incoming However it is

possible to operate the linear adder in a non-linear mode (when clipping occurs)

1: Connect FG-1, sine, 500 Hz, to one input of the adder and a DC source (initially set

to zero) to the other Connect the output to the oscilloscope through the Wave Analyzer

Increase FG-1 to provide a maximum amplitude undistorted screen pattern (Notice the

on-off switch to listen to the sound output of the adder.)

Slowly increase the DC voltage until you see clipping in the wave display At this point

can you notice a difference in the sound quality? Describe what you hear:

_

_

2: For the values in Table 5 measure the amplitude of the

harmonics that are introduced by the clipping

In general whenever there is clipping, the process is

non-linear, the periodic output is not similar to the periodic

input This means the frequency content of the signal has

changed The process of changing the frequency content of a periodic signal is basic in electronic communication, transmission and reception, modulation and de-modulation

To optimize available power, unwanted harmonics are to be

avoided The following experiments will examine various

modulation and de-modulation methods As an introduction

we use a simple diode, even though many unwanted

frequencies will be generated The connections are shown in

Fig 3

3: Modulation As carrier set FG-1 to 40.00 kHz, sine, and maximum amplitude; as

modulation set FG-2 to 2.00 kHz, sine, maximum amplitude Connect output to the

oscilloscope through the Wave Analyzer, as in previous activities Use the balance control for minimum 40.00 kHz signal Examine the output frequencies of Table 6 Notice that the desired sum and difference frequencies, 38 and 42 kHz, are present but the original 2 and 40 kHz frequencies are not removed, as would be done by the

multiplier unit Also additional harmonics are present In any application these unwanted

frequencies would waste power The diode is a rather imperfect multiplier

Freq Amplitude

1,000 Hz 1,500 Hz 2,000 Hz 5,000 Hz 10,000 Hz 42,000 Hz

Table 5 500 Hz distortion

Fig 3 Diode Mod-Demod

4: De-modulation Basically modulation and

de-modulation are the same process, transforming one set of

frequencies into a different set If one frequency is low

(modulation frequency) and the other quite high (carrier frequency)

it is called modulation If both frequencies are high (carrier,

side-bands) but relatively close together it is called

de-modulation Here we take FG-1 and FG2 as 40.00 and 44.00

and check on frequencies of Table 6

Again we see the original frequencies still remain, the

sum and difference frequencies are present but weaker than expected, and additional

weaker harmonics are also present The following modules examine much better means for modulation and de-modulation

The Modulation Module

As noted above, the Modulation module‟s 9-pin DB-9 connector must be plugged into the dual Function Generator (included in the Communications 1: AM and FM package) Individual test points are available for the FG-1 and FG-2 signals, and for ±12.0 VDC,

5.0 VDC, 0-10 VDC and ground, and a locally generates binary modulation signal at

TTL level All these voltages are also contained is a single cable for easy connection to

other modules The module also contains an adder, multiplier and diode for performing

the activities described above

In the discussions above the carrier is sinusoidal while the modulation may be

either analog or digital The Communications 1: AM and FM package considered only

analog modulating signals In this course we consider only digital modulating signals In

digital amplitude modulation it is desirable to switch the amplitude between binary “0” and “1” levels only at the moments when the signal is instantaneously zero; otherwise voltage transients can occur due to inductance For this reason the module shown in

block diagram in Fig 5 includes a wave shaper to convert the sinusoidal carrier from

FG-1 at TP1 into a square wave at TP2 , and a frequency divider, to provide a binary modulation signal (at TP4) of ½, ¼, 1/8 or 1/16 the carrier frequency, synchronized with the carrier

Recall that both the continuously repeating binary 11110000 and 10101010 signals have the form of a square wave with 50% duty cycle The signals differ only in

the bit period, the duration of the signal held high or low to represent a single bit As a

help to comparing the original modulating signal and the recovered de-modulated

signal, the module includes a pattern generator that encodes together groups of 10 bit

periods in the form of two high start bits followed by 8 user-selectable data bitsTP3 as

shown in Fig 4 The user selects as output either square or encoded A signal is

provided atTP5 for connection to the oscilloscope EXT synchronization input to provide

a more stable display A block diagram of the circuit is presented in Fig 5

Freq Amplitude

2.00 kHz 4.00 kHz 36.00 kHz 38.00 kHz 40.00 kHz 42.00 kHz 44.00 kHz

Table 6 2 and 40 kHz

Fig 4 Typical modulation pattern

Fig 5 Modulation generator

Modulation Module Parts List

Part Value Description

IC2 7805 positive voltage regulator (+5V)

IC3 7905 negative voltage regulator (-5V)

IC4 74LS93N Decade, divide by twelve and binary counter

IC5 74LS00N Quad 2-input NAND gate

IC6 74LS165N 8-bit parallel load SHIFT REGISTER

IC7 74LS90N Decade, divide by twelve and binary counter

IC8 74LS32N Quad 2-input OR gate

IC9 LM317T Positive VOLTAGE REGULATOR

IC10 LM1496N IC balanced modulator

IC11, IC12 LM318N operational amplifier

Experiment #2 Amplitude Shift Keying

analog and digital voltmeters

the appropriate circuits to accomplish this

Background:

Historically, Amplitude Shift Keying (ASK) is perhaps the simplest and oldest form of modulation, for it basically turns on and off the carrier The Morse code “dots and dashes” were the earliest form of ASK: a dash duration is three times a dot duration and each numeral and letter is represented by its own pattern of dots and dashes

Present-day ASK is based on bits; a binary “1” is represented by the carrier on,

a binary “0” by the carrier off During the on period the actual carrier amplitude value is not significant but should remain constant Normally the bit rate is some sub-multiple of the carrier frequency Although during the on periods the carrier frequency does not

change, this carrier interruption produces additional frequencies in the modulated

signal For ASK modulation we need fixed-frequency carrier and a practical on-off

switch

A practical on-off switch

For high-speed switching mechanical relays are not practical A simple

solid-state switch may be built around the properties of a silicon diode When reverse-biased,

(the Positive terminal at a lower potential than the banded or Negative terminal) a real diode presents

a resistance on the order of several megohms A change to forward-bias gives an

exponential decrease in resistance with a corresponding exponential increase in

current In the first activity we investigate the current-voltage relationship of a biased silicon diode to be used as a switch and from this data we determine its resistance

forward-Activity 1: A diode as a switch

1: Set up Fig.1 from the modulation part of the

ASK module using the external DC voltage

source and external meters Since the test

voltages shown in Table 1 are small compared to

the adjustable voltage source, it is convenient to

use any one of the series dropping resistors (1kΩ

or 10kΩ)

Fig 1 Diode resistance measurement

COMMUNICATIONS II

C2-1 000000000000010 0

2: Set the voltage across the diode as indicated in

Table 1, record the diode current, and calculate the

corresponding diode resistance

To calculate the diode resistance, use voltage divider

principle

3: What is the ratio of the largest to smallest

calculated diode resistance over the given control

voltage range:

The circuit of Fig 2 uses diodes to switch on

and off the current between two identical

transformers There are two inputs, the signal in

and the diode control If the diode control voltage is

sufficient to make the diodes conduct, the signal out

should match the signal in If the control voltage is

too small ( << 0.6 volts ) neither diode conducts, and signal out is zero So just by varying the diode control voltage, the signal out is turned on or off Behold a diode

switch!

Ideal transformers should introduce no distortion between the primary and

secondary currents, although the relative amplitudes may be quite different Just how

ideal are our transformers? Since similar transformers are used in this and subsequent

modules, we first look more closely at the transformer‟s properties

Activity 2: Transformer properties

The coils of the transformers used on the module are wound on toroidal ferrite cores The untapped coil has 30 turns; the center-tapped coil has 20 turns

1: With an ohmmeter, measure the resistance of the untapped coil, XL: _

2: To measure input reactance with output

open, use the circuit shown in Fig 3 Use the

same gain setting for each channel Adjust the

frequency, f, of FG-1 so that the trace for each

channel shows the same amplitude Since the

same current flows in each, at this frequency, XL

= R (R = 10.0 k on the module)

3: Use this method to find the inductance of the primary winding: Recall that inductance, L, is a property of the coil; reactance, XL, also depends of frequency, f Adjust f and the amplitude of the signal source, FG-1, so that Channel 2 signal amplitude (coil) is 1 division while the channel 1 signal amplitude (resistor) is 4 divisions At this frequency XL = ¼ R What is this frequency? _

Volts Current,

mA

Resistance,

rD0.400

0.450 0.500 0.550 0.600 0.650 0.700 0.750

Table 1

Fig 2 Diode switching circuit

Fig 3 Circuit to measure impedance

4: Use a similar procedure to find the frequency for XL = 4R:

There is a certain amount of capacitance, C, between adjacent windings of any inductor, so that real inductors actually behave as a parallel LC circuit, providing an extremely high impedance at the resonant frequency For the circuit shown in Fig 3, at this resonant frequency the Channel 1 trace (resistor) is small compared to Channel 2 (coil)

5: Find the parallel resonant frequency: _

6: Calculate L, the coil reactance at resonance: _

7: The coil‟s distributed capacitance cannot be measured directly However use the resonant frequency formula, f = (1/2 ) (LC)-1/2 , to find an approximate value for C; Estimated distributed capacitance, C:

Activity 3: Full Circuit Behavior

Next reconnect the circuit as in Fig 4 Use the DC source (the 0-10V DC of the module) as

diode control at TP3 For signal in, use FG-1 at TP2 (sine, 40 kHz, variable amplitude)

Monitor signal in (TP2) and signal out (TP4) with the oscilloscope (for both channels set gain

as 0.5 Volts/Division)

1: At zero diode control voltage is there a detectable signal out? 2: Gradually increase the diode control DC voltage

At what voltage is a signal out first detected? _

3: At what control voltage is signal out = signal in ? 4: If the diode control voltage is further increased, does signal out increase?

5: Use the oscilloscope vertical position control to overlay the two channel signals

Do they have the same amplitude? _ the same phase?

6: Increase the signal in frequency to 200 kHz Are the above results the same?

7: With sufficient diode control voltage, is the circuit of Fig 4 transparent to the signal voltage? _ Is this circuit a practical on-off switch?

8: Look again at the circuit of Fig 4 If the diode control terminals are short-circuited, should there be any signal out? Explain your answer:

_

9: If the two diodes are not perfectly matched, or the center-tapped coils are not

symmetric, should there is a signal out for sufficiently large values of signal in? Explain:

Activity 4: Amplitude Shift Keying modulation of a carrier

1: Our next step is to modulate a carrier by

ASK Fig 4 is a suitable circuit As carrier use

FG-1, (40.00 kHz, sine, 4.00 VP-P) connected to

TP2

2: Connect modulation (clock, ÷16) to TP3

Connect oscilloscope Channel 1 (0.5 VOLT / DIV)

to TP5 to display amplified modulated carrier

and Channel 2 (5 VOLTS/DIV) to TP3 to display

modulating signal Set oscilloscope TIME/DIV at

0.1 mS, and sync on Channel 2

The display should appear as shown in

Fig 5 The upper trace is the modulated carrier;

the lower trace is the modulating signal

3: Change the modulating signal to (40.00kHz, ÷4,

20 sec/DIV) The signal should appear as in Fig 6

For modulation ÷16 we see packets of 8 cycles,

not 16; for ÷4 we see packets of 2 cycles, not 4

_

_

_

4: Change the modulation signal to (÷16, 8-bit,

encode “0011000”) and describe the display: (when using

8-bit remember to use the SYNC signal from the modulation unit)

_ _ 5: Do the modulating signal and the modulated carrier have the same form?

Activity 5: Frequency content with ASK

1: To explore the frequency content of the modulated carrier use the circuit of the previous activity (Fig.4) Set the FG-1 generator to 40.00 kHz and the modulation

pattern to (square, ÷16) This produces a square-wave modulation with a frequency 1/16 that of the carrier, or 2.50 kHz The modulated wave should contain the original

Fig 4 ASK modulator circuit

Fig 5 40 kHz, ÷16, 0.1 mS/DIV

Fig 6 40 kHz, ÷4, 20 S/DIV

carrier as well as the sum and difference frequencies: 40.00 kHz and 40.00 ± 2.50 kHz (Take the Wave Meter input from TP5)Use the wave analyzer to verify these expectations and describe your results: _ _ 2: What is the approximate amplitude of the carrier frequency as compared with the two side-bands?

3: Recall from Fourier analysis that a

square wave contains the fundamental

frequency and only odd harmonics

This suggests the presence of

additional components at 40.00 ± 7.50

kHz, 40.00 ± 12.50 kHz, and so on, but

minimum signal at 40.00 ± 5.00 kHz,

40.00 ± 10.00 kHz and so on Measure

the possible components for the

frequencies listed in Table 2

4: For a final trial maintain the carrier

frequency at 40.00 kHz, but change the

modulation divisor to ÷8 This produces a

square-wave modulation with a frequency

1/8 that of the carrier, or 5.00 kHz

Measure the possible components for the

frequencies listed in Table 3

Activity 6: De-modulation: Peak detector

Modulation without an accompanying de-modulation is

really useless For ASK the major components of the signal to be

de-modulated are the carrier and the upper and lower sidebands,

separated from the carrier by the modulation frequency Pass

these through a non-linear device, as a diode, and a difference

frequency appears, just equal to the original modulation

frequency The sum frequencies may be removed by a low-pass

filter This is the basis of a peak detector, as pictured in Fig 7

For this circuit the RC time constant is adjustable from 2.5 sec

to 52 sec Increasing the RC time constant decreases the size

of the spikes However if RC is too large the difference between

a binary 0 and 1 will be blurred It is possible to pass the output

through a Schmitt-trigger shaping circuit, but the resulting pulse

width depends partly on RC Peak detection is more effective

where the “divide by” factor is quite large, e.g., for a modulation bit-rate of 1 MHz and a carrier frequency of 250 MHz

Freq,

kHz

Amplitude, Volts

Freq, kHz

Amplitude, Volts

Freq, kHz

Amplitude, Volts

Table 3 40.00 kHz carrier, modulation ÷8

Fig 7 Envelope detector

Fig 8 Effect of RC

1: Connect TP5, the output of the ASK Modulator, (40 kHz, ÷16) to TP6, the input of the

Envelope Detector, vary the R value and describe the results: _

Activity 7: De-modulation: Retriggerable One-Shot

1: An alternate approach is to use a retriggerable one-shot

such as a 74LS125 chip The rising edge of an input pulse

gives an output that remains high for a time interval set by

an external resistor, R, and capacitor, C Assume a 40.00

MHz carrier (period = 25 sec) which triggers such a

retriggerable one-shot chip For a RC time constant < 25

sec, the output is a series of pulses of width < 25 sec

and repetition rate of 40 MHz However for RC > 25 sec,

the output is high, and remains high as long as the carrier is turned on (plus a brief additional time, RC)

Fig 10 shows oscilloscope displays of the

modulated carrier and the de-modulated signal for

different RC settings of the one-shot circuit The

modulating signal is a square-wave with 50% duty

cycle Pattern (a) displays the signal TP8 when

connected to TP5 The carrier is 40 kHz with ÷16

modulation The effect of the input diode is clearly

seen, indicating the original amplitude is 1.5

volts, sufficient to trigger the TTL input of a

74LS125 chip Pattern (b) shows the

de-modulated signal at TP9 for minimum RC time

constant Each signal cycle triggers the one-shot,

but the triggered output returns to ground before

the next cycle arrives Pattern (c) is for a RC

time-constant only slightly less than one period of the

carrier, 25 sec In pattern (d) the time-constant is

just adequate Notice in pattern (e) the output is

high for more than 50% of the cycle, indicating that

the time-constant is significantly longer than the

period of the carrier For different carrier

frequencies, the one-shot time-constant must be

re-adjusted Also note that the amplitude of the signal must be sufficient to trigger the one-shot chip The output signal level follows TTL standards so there is no need for additional pulse shaping

1: Connect TP5, the output of the ASK Modulator, (40 kHz, ÷16) to TP8, the input of the

Retriggerable One-Shot Adjust the carrier amplitude to give at TP5 the maximum signal amplitude without distortion Vary the R value of the one-shot circuit and compare the

Fig 9 One-shot circuit

Fig 10 One-shot views

observed oscilloscope patterns (Use the SWEEP VARIABLE and SYNC level controls, and sync

on the chip output to obtain a stationary pattern) Describe your observations:

_

2: Change the modulation pattern from ÷16 to ÷4, vary the R value and describe your observations: _ _

3: Change the modulation pattern from square-wave to various 8-bit patterns How well

does the de-modulated output compare with the original modulating signal? (Notice the effect of an excessive RC time constant):

Amplitude Shift Keying Module Parts List

primary – 30 turns

Experiment 3 Frequency Shift Keying

Materials : FSK module, Modulator module, oscilloscope, Wave Analyzer, analog and digital

frequencies These are the mark frequency, f m, and the

space frequency, f s

Fig 1 shows the ideal frequency domain

representation of an FSK signal A binary “1” is

represented by the mark frequency, while a binary “0” by

the space frequency Fig.1 FSK signal in Frequency Domain

2 ASK modulators = 1 FSK modulator

In our previous experiment of

Amplitude Shift Keying (ASK), we recall that

the transmit signal is a single frequency

carrier turned on and off by the binary data

Conceptually, by using two separate ASK

modulators, one for the mark frequency and

the other for the space frequency, we can

come up with an FSK modulator

Fig.2 Conceptual Model of an FSK Modulator

However, a more practical approach to this conceptual model is using a 2-Channel Analog Multiplexer (MUX) We apply both the mark and space frequencies to the inputs of the Analog MUX, and take the binary data as the channel selector

Activity 1: FSK Modulator: 2-Channel Analog MUX

1 Set FG1 to sine, 40.00 kHz, 2.0 Vp-p and FG2

to sine, 20.00 kHz, 2.0 Vp-p Set the

Modulation to direct, ÷16 (2.50 kHz)

2 In the 2-Channel Analog MUX block on the

module (as shown on the right), connect A 0

3 View the modulator OUT on Channel 1 of the

oscilloscope ( 1 V/div, 50µs/div) and A 0 on Channel 2 ( 2

V/div ) Set the trigger source to EXT and the trigger

LEVEL to (+) slope (Remember to use the SYNC on the

Modulation unit.) Sketch both displays on the graph

provided

What do you observe in this first set up? How does the

modulator behave?

_

4 This time connect A o to ground and A 1 to FG1 Repeat

5 Now connect A o to FG2 and A 1 to FG1 Repeat step 3

above and record your observations

such a model: the mark frequency and space frequency generators must be synchronized with the input binary data Otherwise, there would be phase discontinuities in the signal whenever

the multiplexer switches channels In Fig 7, notice that the space frequency, f s, is out of phase

with the mark frequency, f m, thus there is a rough transition in frequency during the switching Abrupt phase discontinuities may introduce unwanted frequencies in the signal As a result, the demodulator may have a trouble with tracking the frequency shifts and errors may occur

Adding synchronizing circuits in the modulator would make it rather complicated However, instead of using two separate signal generators to generate our carrier frequencies,

we can use a single oscillator to generate both In this module we are using an XR2206 monolithic function generator chip It is a versatile chip that generates different types of waveforms: sine waves, triangular waves, square waves, AM, FM and FSK signals More detail

on this chip is discussed in Experiment 7: Angle Modulation in Communications I

In generating FSK signals, it uses current switches to select which frequency, mark or space, to produce A portion of the block diagram of the XR2206 is shown below

Fig 5 XR2206 Block Diagram

The heart of the function generator is a Voltage Controlled Oscillator (VCO) When used as a plain function generator, only one timing resistor is used But when used as an FSK modulator, each timing resistor along with the timing capacitor determines the output frequency The output frequency is as function of the timing resistor and timing capacitor as given below

Equation 1 Output frequency of XR2206

Activity 2: Generating f m and f s using XR2206

1 Set TR1 to minimum and TR2 to maximum Measure each resistance, solve for the mark

and space frequencies (Remember to turn off the power when measuring resistances.)

C = 1 nF TR1 = TR2 =

1/(TR1 x C) = 1/(TR2 x C) =

2 Connect the input pin 9 to the 0-10 VDC voltage source from the modulation module and first set it to 0 V With a digital multimeter, measure the frequency of the signal at the output

3 Now slowly increase the voltage until the frequency changes At what voltage did the shift occur?

Measure the output frequency f = Which frequency is it?

Is there a range of voltage where the frequency is quite indeterminate? What is that range?

Now that we have a VCO that generates our mark and space frequencies, is it enough to produce a Continuous-Phase FSK? We will soon find out

Activity 3: CP-FSK

Fig 6 (a) Space frequency; (b) Mark frequency; (c) Modulation signal (d) Phase-Continuous FSK signal with inconsistent transitions (e) Phase-Continuous FSK signal with consistent transitions

1 Connect Channel 1 of the oscilloscope to the XR2206 modulator output pin 2 and Channel 2 to the Modulation signal, and choose DUAL mode Set Horizontal Sweep to

0.1 ms/div, NORMAL triggering and trigger SOURCE to Channel 2, (+) Slope

2 With TR1 minimum and TR2 maximum as in the previous activity, connect the pin 9 to

the Modulation (direct, 40.00 kHz, ÷16) Were you able to achieve a stable oscilloscope

display?

3 Next set both TR1 and TR2 values at almost midpoint Were you able to achieve a

stable oscilloscope display? _

4 Finally, set TR1 to maximum and TR2 to minimum Again, viewing the traces on the

oscilloscope, is there a stable display?

Looking at Fig 6d, we have a Phase-Continuous FSK signal This is somehow what we are getting in our oscilloscope display It is erratic because the scope sweeps the display continuously with time But what we are actually getting is a CP-FSK signal with inconsistent transitions This means that as the Modulation signal changes, the XR2206 modulator switches between the two frequencies at different points each time

To have a consistent transition, the mark and space frequencies should be an exact multiple of Modulation signal frequency for a square wave data, or one-half the bit rate for a train of 1‟s and 0‟s

5 View the Modulation signal on Channel 2, with

Horizontal Sweep set to 20 µs/div, NORMAL triggering

and SOURCE to Channel 2 For easier adjustment,

change the Modulation module setting to direct, ÷4

6 Slowly adjust TR1 and TR2 until you get an exact

number of cycles of the mark and space frequencies in

one bit frame Look at Fig 6e for your reference,

although the transition points may differ (it may be at

the peak points rather than the zero-crossing points)

7 If you are now able to get a stable display, set the Modulation module setting back to direct, ÷16 Sketch the display

8 Determine the two FSK frequencies using a digital multimeter For the mark frequency,

disconnect the XR2206 pin 9 from the Modulation signal and leave it open For the space frequency, connect pin 9 to ground.

Frequency Spectrum of an FSK Signal

Recall that for Amplitude Shift Keying, the frequency components of the modulated signal are that of the original carrier and a number of sidebands, which are basically the products of the carrier and the components of the square wave data

Fig 7 Frequency Spectrum of an ASK signal with 40.00 kHz carrier and 2.50 kHz Modulation

In our earlier model, we generated an FSK signal by combining two ASK signals We can therefore assume that the frequency spectrum of an FSK signal is the superimposition of the two ASK spectra of the mark and space frequencies

Activity 4: FSK Frequency Spectrum

Using the wave analyzer, record the approximate amplitudes of the expected frequency components of the FSK signal as indicated in the table Instead of an analog meter, use an

oscilloscope for the amplitude measurements Set FG1 to sine, 40.00 kHz and Modulation to

direct, ÷16 from the modulation module Use the Modulation signal as input to the XR2206

modulator to generate a continuous-phase FSK Sketch the frequency spectrum of the FSK signal

Does the graph show that the FSK spectrum is indeed a superimposition of the ASK spectra of the mark and space frequencies?

Noncoherent Demodulation

In noncoherent demodulation, no local frequencies are generated in the demodulator circuit It makes use of a peak detector, just like in ASK demodulation A simplified block diagram is shown below

Fig 8 Noncoherent FSK demodulator

Activity 5: Noncoherent FSK Demodulation

To be able to understand how a Noncoherent FSK demodulator works, let us first examine each block: (1) the Bandpass Filter (BPF), (2) Peak Detector, and (3) the Comparator

5.1 Bandpass Filter

Bandpass filters are filters that pass a certain range of frequencies and attenuate those

outside this range Generally a bandpass filter is characterized by its center frequency, f c, bandwidth, Q factor, gain, and filter response, etc In this activity we will only investigate the center frequency and bandwidth

1 Connect FG1 to TP1, and monitor it on

the oscilloscope Channel 1 Set FG1

range to 200 kHz, sine with minimum

frequency and amplitude of 4.0 Vp-p

Monitor TP2 on Channel 2

2 Slowly adjust FG1 to determine the

frequency at which the TP2 amplitude

is at maximum This is the filter‟s center

frequency, f C

Output signal amplitude

3 To determine the bandwidth, refer to Fig.9 Take the output signal amplitude at the filter‟s f C

as the 0-dB point The -3dB points are

the two side frequencies whose

amplitude is 0.707 that of the amplitude of the f C Determine these side frequencies, and calculate the filter bandwidth

5 Now use the FSK signal in Activity 4 (FG1 to sine, 40.00 kHz; FG2 to sine, 20.00 kHz; Modulation to direct, ÷16) as input to the two bandpass filters What do you notice with each filter output? Explain

_ _ _

5.2 Envelope Detector

The envelope detector in this module is the same as the

one we used in the ASK module The diode cuts off the

negative portion of the input signal, and the R-C combination

holds the peak of the envelope depending on the time constant

It is important to note that in adjusting the R-C time constant, we

should consider the period of one bit of information

Using the same FSK signal as in the

previous activity as input, adjust the R-C time

constant of each envelope detector to bring the

output waves as close as possible to those of Fig

11 Set the oscilloscope Horizontal Sweep to 0.2

ms/div and the vertical gain to a convenient

setting It would be helpful to monitor the original

data input on Channel 2 of the oscilloscope and

choose this channel also for the trigger source

Count how many peaks are enclosed in

one bit frame for both the mark and space signals

Mark Space

Recall that in a 2.50-kHz data, we can get 8 cycles of the mark frequency, and 4 cycles

of the space frequency Does this correspond to the number of peaks enclosed in one bit frame

in the envelope detector output?

5.3 Comparator

The final block in the Noncoherent Demodulator is the level comparator The mark signal

is applied to the non-inverting input of the comparator, while the space signal to its inverting input In Activity 5.2 we observed that whenever the mark signal is high, the space signal is low, and vice versa What we need is something to determine If the

mark signal has larger amplitude, the comparator swings to its

positive rail (positive supply voltage), thus interpreting a “1” bit

And if the space signal has larger amplitude, the comparator

swings to its negative rail (negative power supply), interpreting a

“0” bit

1: On Channel 1 of the oscilloscope, monitor the Modulation

signal, and the Demodulated Output on Channel 2 Set the

trigger source to Channel 1 You may need to readjust the R-C

Fig.10 Envelope Detector

Fig.11 Envelope Detector Output

time constants of the peak detectors to come up with a good output Take note of the duty cycle

of the demodulated output, it should be the same as that of the original data Sketch the oscilloscope display

2: What have you noticed with the demodulated output? Is it exactly identical to the original Modulation signal? Explain

_ 3: We have already tried demodulating a square wave input using the Noncoherent Demodulator Let us now try using a number of different bit patterns on the Modulation module

Set FG1 to sine, 40.00 kHz, and Modulation to encode, ÷16 This gives us a 10-bit string

consisting of two start bits followed the 8-bit data Does the demodulator recover the data effectively?

_

PLL Demodulation

As you may recall, we have gone through with the details of a Phase-Lock Loop, or PLL,

in Experiment 13 of Electronics 3 We have also explored its use in demodulating FM signals in Experiment 7 of Communications 1 AM and FM In this activity, we will investigate how a PLL can demodulate FSK signals, as it is the more common method of FSK demodulation used nowadays

Activity 6: PLL FSK Demodulation

In this module, we are using

an LM565 Phase-Lock Loop Chip Its

free-running frequency output at pin

4 is determined by the R-C

combination connected at pins 7 and

8 With a digital multimeter, determine

the VCO‟s free-running frequency, f n ,

at pin 4 _

Recall that when a frequency

lock is achieved, the PLL can follow

the changes in the input frequency as

long as it remains in its tracking

range Let us see whether the PLL

can track our FSK signal

frequencies

1 Use the XR2206 modulator to generate the FSK frequencies Connect the modulator

OUT to the PLL demodulator input pin 2 Monitor the VCO output frequency at pin 4

using a digital multimeter

Fig.12 FSK Demodulator using an LM565 PLL Chip

2 First connect the XR2206 input pin 9 to ground for a “0” bit input The modulator should

generate the space frequency What is the VCO output frequency?

Measure the VCO input voltage at the LM565 pin 7

3 For a “1” bit input, leave the input the XR2206 pin 9 open What is the VCO output frequency? Measure again the VCO input voltage at pin 7

Can the PLL track the FSK mark and space frequencies?

Notice that when the PLL input changes between the mark and space frequencies, the phase comparator gives an error voltage at pin 7, which is applied internally to the VCO input This is how the loop tracks the changes in the frequency However when we have a constantly varying input, as in a train of data, this voltage changes as well

4 Set FG1 to sine, 40.00 kHz and Modulation to direct,

÷16 Connect the Modulation signal to the XR2206 input

pin 9

5 Connect the modulator OUT to the PLL demodulator

input pin 2

6 Set the oscilloscope horizontal sweep to 0.2 ms/div, and

both channels to 2 volts/div Adjust the vertical position of

the two channels such that the ground is at the bottom of

oscilloscope display Monitor the two inputs of the comparator (make sure both channels

are in DC mode) Use the Modulation sync to achieve a stable display Sketch the

display What do you notice with the two traces? _

7 Next, monitor the original data input on Channel 1 and the comparator output Channel 2 Does this indicate that the PLL demodulator recovered the original information? Explain _

Compare the PLL output pin 7 and the Comparator output Is there any need for the

external comparator?

_

Again, let us try using a number of different bit patterns like we did in the Activity 5 Does the demodulator effectively recover the information?

_

Frequency Shift Keying Module Parts List

Part Value Description C1, C2 0.1μF ceramic capacitor

C3, C6, C7, C9, C10, C27 0.001μF ceramic capacitor

C4 1μF/25V electrolytic capacitor

C5 22μF/16V electrolytic capacitor

C8, C11 5pF ceramic capacitor C12, C13 5nF mylar capacitor

IC5, IC7 LM311N single comparator

IC6 LM565HCN phase-lock loop

IC8 79L05 Negative VOLTAGE REGULATOR (-5V) R1, R2, R16 2kΩ ¼ W resistor

Experiment 4 Binary Phase Shift Keying

Materials: BPSK module, Modulation module, dual function generator, oscilloscope, wave analyzer

Purpose: Investigate the modulation and demodulation techniques of BPSK and the appropriate circuits to accomplish this

Background:

The value of a sinusoidal function (sine or cosine) ranges from –1 to +1, is a function

of its dimensionless argument, , and is periodic with a period of 2 ; sin( + 2N ) = sin

The argument may be a function of only time, t, or of both distance and time as in

wave motion The function itself may be multiplied by any constant, A, or amplitude, to

modify its range A frequently met form is A sin( t + ), in which the angular frequency, , measured in radians per second, represents 2 f The frequency, f, measured in units

of hertz, is the number of times the function repeats itself in one second (provided is constant) The symbol is called the phase It gives the value of the function at time t=0

Frequency or cycles per second may be compared with speed or distance per

second To move at a speed of 100 meters per second does not mean that the object

has actually covered 100 meters or moved for a full second Therefore new concepts

are needed: average speed, the total distance traveled divided by the total time taken, and instantaneous speed, the distance/time ratio as the time interval becomes

vanishingly small A similar distinction is made for average and instantaneous frequency Of course if the speed or frequency is constant, the average and instantaneous values are equal

For an oscillator running at 50 MHz (fifty million cycles per second) the value of is of little interest as long as it is constant However if is changing with time, the

instantaneous oscillator frequency depends on both f and the change in Suppose initially f=100 and =0 Now increase f so f=101 so each second the system goes

through one additional cycle However we would get the same increase in oscillations if

f were constant but increased at a constant rate of 2 radians per second For

constant oscillator frequency both f and d /dt must remain constant (or changes in either must mutually cancel) Changing only f to convey information is termed frequency modulation;

changing only with change in time (d /dt) is phase modulation The end result may be the same in either case; the general process is called angular modulation

In phase modulation as noted, the frequency, f, is held constant and information

is conveyed by changes in the time derivative of phase, d /dt In Phase Shift Keying

(PSK), a form of phase modulation, phase changes are discrete, not continuous; can

assume only a fixed number of constant values (this implies that d /dt is zero between phase

changes and infinite just at the moment of change! ) In Binary Phase Shift Keying (BPSK) only two

phase values are used; may be either 0 or (0o or 180o)

COMMUNICATIONS II C2-4

000000000000010 0

BPSK Modulation

The properties of the trigonometric functions state that cos( + ) = – cos so

the two possible binary states in BPSK are cos( ct) and –cos ( ct) where c is the angular frequency of the carrier wave The modulating signal is a train of mixed “0” and

“1” pulses with a pre-determined pulse width or bit frame A “1” indicates a 180o

carrier phase shift, a “0” indicates an un-shifted carrier What hardware can do such modulation?

Recall that the primary and secondary currents in a transformer have the same frequency The phase difference between them is 0

or 180o depending on the polarity of the secondary

connections This suggests passing the carrier

signal through a transformer with a simple reversing

switch at the output, as suggested by Fig 1 Of

course the switch must be controlled not by hand but

by the modulating pulse train

In the module on Amplitude Shift Keying (ASK) we use a pair of diodes as an

on-off switch Here we use a two pairs of diodes to form reversing switch, as

shown in Fig 2

The AC signal and the DC

control flow in parallel through the

diodes With an adequate forward

control voltage the AC signal freely

flows in either direction through the

diode, while with zero or reverse control

voltage there is no conduction (assuming

the amplitude of the AC signal is insufficient to

change the diode conducting state.)

The diode control voltage at A and B is applied to the center tap of each

transformer, so the same voltage is applied to all diodes If this voltage is quite small,

none of the diodes conduct so the signal out is effectively isolated from the signal in If

A is sufficiently positive ( 0.6 volts) with respect to B, only diodes a and d conduct

Current leaving point P flows through the right transformer from Q to R However if B is sufficiently positive ( 0.6 volts) with respect to A, only diodes b and c conduct Current leaving point P flows in the right transformer from R to Q So just by changing the

polarity of the control voltage, the signal out is in-phase or 180o out-of-phase with the

signal in The amplitude of signal in must be sufficiently small so as not to turn on any

diode After modulation signal out may be amplified to the desired level

Fig 1 Basic phase reversal

Fig.2 A balanced modulator using diodes

Activity #1 A BPSK Modulator

Fig 3 shows the block

diagram of the BPSK transmitter

used in this experiment It is built

around a diode balanced

modulator Equal positive and

negative voltages must be

applied to the diodes pairs

shown in Fig 2 This balancing is done by the Offset Adjust block The carrier signal entering the Diode Balanced Modulator block is reduced to a suitable level within the block, and after modulation is amplified by the Gain block

1: For carrier input

Use the FG-1 signal and set to 40.00 kHz, 4.0 VP-P, sine

Connect FG-1 to TP2

2: Modulate this carrier with a “1”,”0” alternating pattern, viewed on Channel 2

Connect CLOCK, set at ÷ 16, to TP1 and also to Channel 2

Set Channel 2 to the following: DC, 2 VOLTS/DIV and adjust POSITION to place

signal on lower portion of the screen Set TIME/DIV = 0.1 ms, SYNC on CH2, adjust

SWEEP VAR and Trigger LEVEL to display five cycles on the screen

3: View the modulated signal on Channel 1

Connect TP3 to Channel 1 and set to: DC, 2 VOLTS/DIV Adjust POSITION to

place signal on upper portion of the screen Adjust module GAIN to give a 4.0 VP-P signal

4: Slowly adjust the Offset Adjust from full CCW to CW and

observe the changes in the TP3 pattern Fig 4 shown here

reproduces Fig.5 of the ASK module Compare the present

oscilloscope pattern with this figure: _

5: Describe any difference for Offset Adjust settings of full

CCW, mid-range and full CW: _

_

6: What is the relation between the diode currents and the Offset Adjust knob? _

_

7: Set Offset Adjust so the pattern appears as in Fig 5 This is

the balance position Does the upper trace frequency appear

constant as Offset Adjust is varied?

8: Change the TIME/DIV to 20 s and adjust SWEEP VAR so

slightly more than one-half cycle of the lower trace fits the

screen Observe and changes in the upper trace (modulated

output) as Offset Adjust (modulation signal) changes Then

Fig.3 A binary phase shift key transmitter

Fig 4 Deja Vue on ASK

Fig 5 BPSK balance

vary the “÷“ control to 8, 4 and 2 and describe changes in the upper trace:

Activity #2: Frequency content with BPSK

1: Attach the Wave Analyzer to the modulated signal at TP3 Use the same settings for

the Carrier and Modulation settings as in the previous activity Set the Wave Analyzer to

the carrier frequency, 40.00 kHz Vary the Offset Adjust knob over its full range and

describe and changes in the strength of the 40.00 kHz frequency component: _ _

2: Before each frequency measurement, check the

signal for clipping prior to the Low Pass Filter (knob

position C) If clipping is present reduce the local

oscillator setting (viewed at knob position B) Determine

the approximate voltage for the frequencies and Offset

Adjust values listed in Table 1

3: Compare these values with Table 2 of the ASK

module, and list differences and similarities:

_ _ _

4: Explain why the BPSK results for CCW and CW settings are so similar to those of

ASK: _

5: Compare Figs 4, 5 and Table 1 data With ASK there is zero output 50% of the time,

yet the 40.00 kHz is the dominant frequency component With BPSK the screen display

shows a 40.00 kHz pattern yet the Wave Analyzer indicated zero carrier just at balance

Are these results reasonable and to be expected? Explain your answer: _ _ _

What do the equations say?

Why are the sidebands similar for BPSK and ASK, and why is the carrier present

in ASK and absent in BPSK? Modulation and the appearance of new frequencies

results from multiplication, not addition In conventional amplitude modulation we seem

to be multiplying two factors, cos ct and cos mt and expect from Eq 2b of the

Modulation module the result:

cos mt cos ct = ½ { cos ( c+ m)t + cos ( c– m)t } (1)

freq CCW Bal CW 45.00

42.50 40.00 37.50 35.00 32.50

Table 1

The result gives sum and difference terms but the carrier term, cos ct , does not

appear Referring again to Fig 2, the carrier is applied to the terminals marked Signal

in, the modulation is applied to the terminals marked diode control and the modulated

carrier appears at the signal out terminals Apart from a constant amplitude factor the

Signal in term is cos ct. But which of the possible forms shown in Fig 6 should we use

as modulation?

The signal shown at left of Fig 6, (A + M cos m t), keeps one diode permanently

conducting and the other permanently off, so signal out is only the un-modulated

carrier, cos ct

The signal at center of Fig 6, (B + M cos m t), allows one diode to conduct for

more than a half period and the other for less If this is used as the modulation term in Eq.1, a carrier frequency component will be found in the resulting modulated signal

This is the near balance condition:

(B + M cos mt) cos ct = B (1 + M / B cos mt) cos ct

= B cos ct + M cos mt cos ct

= B cos ct + M / 2 { cos ( c+ m)t + cos ( c– m)t } (2)

If M = B, the result is that of 100% amplitude modulation

The signal at right of Fig 6, (M cos m t), has no DC offset, so if this is the form

of the modulation term in Eq 2, the B cos ct will be zero and no carrier component is

found in signal-out This is the balance condition The function of the Offset Adjust is to

assure that there is no DC component (the A and B factors in Fig 6) in the modulation signal Although our discussion here treats only of sinusoidal signals, you should recall that the square-wave modulation signals we actually used may be represented by

Fourier‟s theorem as the sum of a fundamental and harmonics, all of which are

Fig 6 Which form should be used for modulation?