Laboratory guide for communications 2 di

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 i Table of Contents 1: Modulation 2: Amplitude Shift Keying (ASK) 12 3: Frequency Shift Keying (FSK) 21 4: Binary Phase Shift Keying (BPSK) 33 5: Bit-splitting 42 6: Clock Recovery 53 7: Quadrature Phase Shift Keying (QPSK) 65 8: 16 Quadrature Amplitude Modulation 78 9: Pulse Modulation 91 ii COMMUNICATIONS II C2-1 000000000000010 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 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) 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, f (t ) A0 An cos n t n where Bn sin n t (1) n = 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 1: Modulation 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 nonlinear 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 cos A + cos B = cos ½(A+B) cos ½(cos A–B) (2a) cos C (2b) cos D = ½ { cos (C+D) + cos (C–D) } 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 (The analyzer is included in the Communications 1: AM and FM package) In use the periodic signal to be 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 Fig Wave Analyzer faceplate 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 1: Modulation 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: n 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) 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 1: Modulation Freq Amplitude 10.00 khz 20.00 khz 30.00 kHz Table 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 significant figures) of the frequency components listed in Table 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 above Use the frequencies of Table Should a perfect square wave contain odd harmonics? _ Freq 10.00 kHz 20.00 kHz 30.00 kHz 40.00 kHz 50.00 kHz Amplitude 190.00 kHz Table 10 kHz square 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 = cos ½(A+B) cos ½(cos A–B) 1a: Set FG-1 at 39.00 kHz so A = 39000 t, FG-2 at 41.000 kHz so B = 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 Fig Addition: 39 + 41 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 Obtain a similar display on your oscilloscope From Eq 2a we have: cos 39,000t + cos 41,000 t = {2cos 1,000t} cos 40,000 t (5) 1: Modulation The right side of the equation suggests a 40 kHz wave with amplitude varying at 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 refers to the frequencies 39 kHz and 41 kHz, while the right side refers to 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 for the signal leaving the adder Fr eq Am plitude 00 kHz 39 00 kHz 40 Does addition introduce any new frequencies? _ 00 kHz 41 Does addition remove any frequencies already present? 00 kHz Table Addition 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 = 40000 t, FG-2 at 1.00 kHz so D= 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 40.00 kHz t cos 1.00 kHz t = ½ {cos 41.00 kHz t + cos 39.00 kHz t } 1: Modulation (6) Compare the right side of (6) with the left side of (5) Does this explain why our multiplication display is similar to Fig 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 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 refers to the frequencies 40 kHz and 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 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? _ Freq Amplitude 1.00 kHz 38.00 kHz 39.00 kHz 40.00 kHz 41.00 kHz 42.00 kHz Table Multiplication 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 timevarying signals, cos A becomes cos 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 1t + cos 2t 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 3t cos 4t 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 1: Modulation 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 # 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 = 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? _ 1: Modulation 4: Overlay one trace on the other by adjusting the channel vertical position Describe any differences between the two traces for various N values: Observation: Sampling is basically multiplying a sine wave of frequency fm by an amplitude-shifted square wave, Eq The resulting frequency content is the original content, fm, plus the sum and difference products of fm with fs and its harmonics, as listed in Eq Now if the original analog wave contains two sine waves, fm1 and fm2, then after multiplication, these two original frequencies as well as their sums and differences with fs and its harmonics are found in the sampled wave And if the larger of fm1 and fm2 satisfies the Nyquist requirement, Eq 5, we can be sure that all sum and difference products are greater that the two original frequencies To go further, suppose the original analog signal contains a spread of frequencies, the greatest of which is fm as determined by Fourier analysis Sampling this signal results in the original frequency spread, to fm, as well as the products of this spread with the sums and differences of fs and its harmonics And as long as fs > 2fm the original frequencies remain in the output, un-contaminated by all the new product terms The diagram below in Fig.6 illustrates this arrangement Fig No aliasing since fs > fm Activity 4: PAM Demodulation Demodulating a PAM signal is rather simple Since the flat-top samples basically follow the shape of the original signal, all we need is a smoothing circuit to reduce the edges of the PAM signal, which can easily be done a low-pass filter However, Fig presents an alternate way of viewing the same demodulation process The spread of frequencies from to fm is the frequency domain picture of the original signal This spread is repeated over and over The spread on the far left is the easiest to obtain: use a low-pass filter to remove all frequency components greater than fm With a bit more effort, any of the other spreads could be recovered by using an appropriate band-pass filter and mixing the output with fs or one of its higher harmonics First, let us look at the Low Pass Filter (LPF) filter characteristics and determine the cutoff frequency, f o 1: Connect TP8, the LPF output, to Channel #2 Set trigger source to Channel #2 Fig Low Pass Filter 2: Connect FG2 directly to TP7, the LPF input Set FG2 mode to sine Starting from a minimum value, gradually increase FG2 frequency until the filter output (TP8, Channel #2) is maximum (Because of coupling capacitance the filter output is not 9: Pulse Modulation 95 Record maximum at Hz.) this frequency for maximum amplitude: 3: At this frequency adjust FG2 amplitude to 4.00 VP–P Record the VP–P filter output at the following frequencies: 1.0 kHz 2.0 kHz 3.0 kHz 4.0 kHz 5.0 kHz 6.0 kHz 7.0 kHz 4: Describe any phase changes between the LPT input and output signals: _ _ 5: Recall that at the cutoff frequency, fo, the output amplitude is -3 dB at 0.707 less than maximum Record fo at 2.8 VP–P (4.0 x 0.707): _ 6: Next, we use this LPF to de-modulate the PAM signal We are involved with wire communication In the present case our wire runs from TP6, the output of the flat-top PAM modulator at the transmitting end, to TP7, the input to the low-pass filter demodulator at the receiving end 7: Set FG1 (2.000 kHz, sine, 4.0VP–P) Connect Channel #1 to the sample-and-hold fm input, TP4 Connect Channel #2 to LPF output, TP9 Trigger on EXT For both channels set VOLTS/DIV = Vary the xN setting and compare the two channel displays: How they differ in amplitude? How they differ in phase? Which N value gives the best signal recovery? _ What is the fs value for this setting? For all N values is the Nyquist relation, fs ≥ 2fm, satisfied? If our low-pass filter were ideal (zero output for all f > fo), would the demodulated output be the same for all the N values? Explain: _ Do you think the Nyquist relation assumes an ideal LPF? _ Activity 5: PAM for a compound signal Our module contains an adder circuit (inputs at B and C, output at D) Recall that adder outputs contain all and only frequencies present at the inputs, unlike mixers which reject the input frequencies and provide new sum-and-difference frequencies at the output Here we use as input to the sample-and-hold circuit the sum of sine waves from FG1 and FG2 and view the demodulated output 1: Use the same connections as in the previous activity, with the following exceptions: Connect adder input B to FG1, set at 2.000 kHz, sine Connect adder input C to FG2, set as sine 9: Pulse Modulation 96 Connect adder output D to the sample-and-hold input, TP4 Notice that the SYNC signal and FG1 have the same frequency To obtain a stable oscilloscope of the adder output, the FG2 frequency must be an integral multiple of FG1, an adjustment that is somewhat delicate 2: Adjust the FG2 signal to 4.000 kHz, just twice FG1 Adjust both amplitudes so FG2 is about half FG1, and their sum not exceeding 4.0 VP–P For N=16 compare the sampleand-hold input with the low-pass filter output Sketch the display 3: What N value gives the best resemblance between the original analog signal, fm, and the de-modulated PAM signal? Observation: A glance at Fig shows that in PAM the bandwidth of the transmitted signal far exceeds the base band, fm Recall that the reason for modulation methods as BPSK, QPSK and 16–QAM is to reduce the bandwidth of the transmitted signal Those methods are used for wireless transmission through free space, where bandwidth if a main concern PAM, as wire transmission, has exclusive use of all available bandwidth What is of concern for wire transmission is the ease of signal regeneration of the attenuated signal at various points along the signal path The rectangular pulse shape must be restored, specifically, the pulse amplitude since pulse width is a common property The difference between AM and FM broadcast quality suggests that random interference has a greater effect on amplitude than on frequency or timing For this reason various methods have been devised to convert the information carried by the amplitude of constant-width pulses into the width of constant-amplitude pulses, just for transmission At the receiving end (perhaps after many regenerations while in transit) is then converted back to PAM-form for demodulation In the following activity we examine one such method, Pulse Width Modulation (PWM) Pulse Width Modulation (PWM) Pulse Width Modulation, (or Pulse Duration Modulation) is a modulation technique which varies the duty cycle or the “on time” of constant-amplitude pulses The PWM pulse width changes in direct proportion with the PAM pulse height on which it is based Pulses or Levels? ? ? Fig shows pulses, in which the voltage drops to zero during a portion of each period Fig shows change in signal levels with no return to zero Can these changes in signal levels really be called pulses (perhaps a pulse train with 100% duty cycle!)? This is not really a problem since the sampling that produces the flat level shown in Fig in practice are not transmitted, but serve as in intermediate step toward other modulation methods Fig Comparing modulation 9: Pulse Modulation 97 Fig PAM to PWM circuit As seen in Fig 8, pulse width is greatest at maximum signal voltage, and least at minimum signal voltage For zero signal, the pulses have a 50% duty cycle A ramp generator is the heart of this conversion from pulse height to width (and back again, too) Fig 10 Proper ramp adjustment The ramp voltage is adjusted to rise from 0.0 v to 5.0 v in a time 1/fs, the period of the sampling frequency As shown in Fig 10 the PWM signal is the output of a comparator, which remains high as long as the PAM signal is less than the ramp voltage The higher the PAM pulse, the longer the comparator remains high, and the greater the width of the PWM pulse Activity 6: PWM Modulation 1: Prepare the PAM signal: Set FG1 as (1.000 kHz, sine, 4.0 VP–P) Set xN = 16 Connect FG1 to TP4, Connect A to TP5 Connect TP6 to TP9 2: Adjust Ramp Connect fs from A to TP10 , Connect Channel #1 to TP11 Adjust Ramp output, as in Fig Repeat ramp adjustment for xN = 8, 4, Return adjustment for xN = 16 Note: each time xN is changed, ramp must be re-adjusted ! 3: Connect Channel #1 to TP9 to view PAM input Connect Channel #2 to TP12 to view PWM output Adjust TIME/DIV to display one full cycle Sketch the screen display Describe the screen display for xN = 8, 4, 9: Pulse Modulation 98 Demodulation of the PWM signal We generated the PWM signal in two steps: Sampling to PAM and PAM to PWM This suggests that PWM demodulation use in reverse the same two-step process The same ramp voltage generator that converted voltage to time can also convert time back to voltage Also used are two sample-and-hold circuits, as in Activity 2a As shown in Fig 3, there are two inputs; the signal to be sampled and the narrow trigger pulse that captures and holds the input value The start of each ramp cycle and the rising edge of each PWM pulse coincide with each narrow fs pulse The PWM duration is the time between the rising and falling edges of each pulse In modulation, the time for the ramp voltage to rise to the PAM level determined the pulse duration; in demodulation the level of the ramp voltage at the falling edge of the PWM pulse determines the PAM pulse height A block diagram of the demodulation Fig 11 PWM back to PAM process is shown in Fig 11 The differentiator block converts the falling edge of each PWM pulse into a positive-going narrow spike, suitable to latch the ramp voltage level into the first sampleand-hold unit The next following fs pulse latches this level into the second sample-andhold unit which provides the desired PAM output pulse, with height directly proportional to the PWM pulse width This can then be demodulated by the circut described in Activity 2b to regain the original analog signal Activity 7: PWM demodulation 1: Connect the PWM Demodulator panel as shown in Fig 11 Set xN=16 The same Ramp adjustment is used for both modulation and demodulation 2: View the Differentiator output at TP 15 Describe the waveform: _ 3: View and compare the waveforms at TP16 and TP18: _ 4: View and compare the original PAM waveform at TP9 and the recovered PAM waveform at TP18: 5: Connect the recovered PAM, TP18, to the low-pass filter input, TP7 View and compare the original analog waveform at TP4 and the recovered analog waveform at TP 8: _ 9: Pulse Modulation 99 Pulse Position Modulation In the above activities we have been converting analog signals into pulse trains to be transmitted by wire The power transmitted by each pulse ( product of pulse width and square of pulse height) is not constant, but is determined by the flat-top sampling level Pulse position modulation, PPM, is a method to maintain a constant (and minimum) power level independent of the original analog signal A diagram is shown in Fig 12 A constant height extremely narrow pulse or spike is generated at some instant during each sampling period The time delay between the moment of sampling and the spike is directly proportional to the sample level This method conveniently starts from a PWM signal, so that the spike coincides with the falling edge of each PWM pulse The required hardware for this process is extremely simple, as shown in Fig 13 Since each PPM spike occurs at the falling edge of a Fig 12 PPM generated from PWM PWM pulse, simply invert the PWM signal and feed this to a one-shot multivibrator, triggered by a rising edge of a pulse Activity 8: PPM Modulation 1: Since PPM is based on PWM, configure the module as in Activity 3a, PWM Modulation Connect TP12 to TP19 2: Connect the PPM modulator, TP19, to Channel #1, and Fig 13 PWM to PPM the output, TP20, to Channel #2 Describe the display: _ _ 3: What is the approximate PPM pulse height and width: _ 4: Does each PPM pulse rise at a high-to-low transition of the PWM pulse? Pulse Position Demodulation For PWM demodulation at the receiving end, we need the sample pulse train, fs, which is generated at the transmitting end Since the rising edge of each received pulse coincides with fs there is no problem It is different with PPM, since fs cannot be extracted from the received signal Other methods must be devised However, on our module we work around this difficulty by using the same fs source for both modulation and demodulation From the diagram in Fig 11, we note that the PWM pulse duration is the time interval between the fs and PPM pulses Recall that in a D-latch the edge-triggered CLK transfers the D input to the Q output A negative CLR signal at any time brings Q low Fig 14 Shows the D input permanently high Each fs brings Q high and each following inverted PPM pulse brings Q low The Q output is the recovered PWM Fig 14 PPM to PWM 9: Pulse Modulation 100 signal, which may then converted back to PAM as in Activity 3b, which in turn is converted to analog by the low-pass filter The process is straightforward, provided the fs signal is available Activity 9: PPM Demodulation 1: Use the connections of Activity 4a above Also add the connections shown in Fig 14 2: Connect the PPM at TP22 to Channel #1 and the demodulated output at TP23 to Channel #2 How does this display compare with Fig 12? _ _ 3: Connect the demodulated PPM signal at TP23 to the input of the PWM demodulator at TP 14 (as in Fig 11) Compare with the oscilloscope the demodulated PWM signal at TP18 with the original PAM signal at TP6 How faithful is the recovery? _ 4: Finally connect the recovered PAM signal at TP18 to the input of the low-pass filter at TP7 Compare the low-pass filter output at TP8 with the original analog signal at TP4: _ _ 5: After all these successive modulations and demodulations, how faithful to the original is the final reconstruction? 6: What effect might there be if the sampling rate, fs, were 32 or 64 times the analog frequency, fm ? _ Pulse Code Modulation For the three modulation methods already considered, pulse height, width and position (each based on the sample levels of the analog signal) we modulated and demodulated on adjacent panels on the same physical module In reality, the connection between panels would not be a simple jumper Instead, the modulation and demodulation units would be joined by a rather lengthy transmission path with distributed capacitance and inductance and contaminated by noise, factors that distort the pulse shape The pulse may be ideal when it leaves the transmitter but on arrival at the receiver, its real shape may be similar to that suggested by Fig 15 Compared with PWM, PAM is more subject to errors, and PPM is more difficult to demodulate In PWM, the accuracy of the Fig 15 Pulse shapes 9: Pulse Modulation 101 sampled analog value depends on the accuracy of the ratio of pulse width to sample period The sample period is the same for every pulse, so there are ways to recover accurately this value at the receiver Ramps are used at both ends, and it is not too difficult to make them identical The problem is the accurate determination at the receiver of the pulse width Here is where pulse code modulation, PCM, comes in Instead of sending the actual pulse whose width must be measured at the receiver, PCM measures the pulse width at the transmitter and sends only the measured value of this width in binary form We recall the steps followed in PWM: 1) sample the analog signal at a rate, fs; 2) compare this sample level with a fixed rate-of-rise ramp voltage; 3) set the pulse width equal to the time needed for the ramp voltage to rise to the sample level All of this can be done by an Analog Digital Converter chip ADC0804, which accepts an analog signal that it samples at a user-determined sampling frequency, provides an adjustable ramp and comparator, and outputs an eight-bit binary measure of the pulse-width to sample-period ratio Of course we can‟t send an eight-bit byte in parallel over a single transmission path, so a parallel-in / serial-out shift register is inserted into the signal path before transmission At the receiver-end the process is reversed A companion serial-in / parallel-out shift register recovers the eight-bit byte and passes it along to a Digital Analog Converter chip, DAC0832 gives an analog voltage just equal to the original sampling level, essentially PAM The original analog signal is recovered by adding a low-pass filter Fig 16 Pulse Code Modulation / Demodulation Activity 10: ADC Counting The eight-bit ADC chip used in the module converts the analog input to one of 256 divisions of the nominal 5.00 reference level To examine this behavior connect the ADC input at TP24 to the variable 0.0 to 5.0 volt source at E Also connect a DC voltmeter between TP24 and ground The LEDs display the ADC chip output Determine the input DC voltage for each display (Note: a single LED is lit for the 1, 2, , 128 displays) LED Volts 16 32 64 128 255 Is there a linear relation between voltage and display? 9: Pulse Modulation 102 Activity 11: ADC and Shift Timing Each sampling cycle of the ADC is started by a high-to-low transition, measured at TP28, which occur at a frequency, fs, provided by the system clock unit on the PCM panel The frequency is approximately 15 kHz (This is close to the fs signal at A, for FG1 = 1.00 kHz and xN = 16.) In what follows this is referred to as the LOAD signal 1: With a frequency counter, measure the exact LOAD frequency, fs: _ 2: Connect oscilloscope Channel #2 to LOAD, TP28 TRIGGER the display from Ch Set TIME/DIV = 10 s Describe it: _ _ 3: The eight LEDs show the byte output of the ADC which is also fed to the parallel-in /serial-out shift register The action of this register is controlled by two signals: the LOAD, fs, and a square-wave signal of frequency 8.000 x fs, referred to as the SHIFT signal Measure the SHIFT frequency at TP29: _ Calculate the ratio of the measured SHIFT to LOAD frequencies: _ With which edge of the SHIFT signal does the LOAD pulse coincide? The parallel-in / serial-out shift register contains a chain of eight 1-bit cells On a LOAD signal, these cells are filled with the data at the parallel-input ports On a SHIFT signal, the data of each cell is shifted to the cell following ( the first cell is emptied, and the last cell data is moved to the output terminal, TP25 ) If more than eight SHIFT signals occur without a LOAD, only zero bits appear at the serial output So in our module a LOAD signal should occur after every eighth SHIFT pulses 4: Connect the ADC analog input, TP24, to FG1 set as 1.00 kHz, sine, 4.0 V P–P Connect the shift register output, TP25 to Channel #1 Maintain the Channel #2 at LOAD Compare the channel displays: _ _ 5: Vary the ADC input signal from FG1 and compare Channel #1 with the LED display: _ _ 6: Often one or more LEDs appear dim, neither off or full on Explain this in terms of the Channel #1 display and possible noise at the input terminal: 9: Pulse Modulation 103 Activity 12: At the Receiver At the receiver end, the process is reversed The serial-in / parallel-out shift register (input at TP26) also contains a chain of eight registers Each low-to-high SHIFT pulse moves each cell data to the following cell, accepts one bit of input data ( at TP26) into the first cell, while the former data of last cell is lost Cell data is continuously available at each of the eight parallel output terminals ( as well as at the following eight DAC ports) and may change with each SHIFT pulse However only on a LOAD pulse is the data, present at the DAC parallel inputs, accepted into the chip for conversion to a DC level (at TP27) Once accepted, the conversion to a DC level is practically instantaneous A LOAD pulse starts the sampling of input analog data by the ADC (at TP24), and a LOAD pulse also moves to the DAC output (TP27) the DC level of the previously sampled data Notice that the input to the ADC (TP24) is analog, but the output of the DAC (TP27) is in PAM form The ADC internally samples the analog data before conversion from a DC level to a byte form 1: Connect the Transmitter to the Receiver, TP25 to TP26 Connect FG1 to B, FG2 to C and the Transmitter input, TP24, to D Set FG1 to 1.000 kHz, sine, 4.0 VP–P Set FG2 amplitude to 0.0 Connect Channel #1 to PCM Transmitter input at TP24 and Channel #2 to PCM output at TP27 Describe the display: _ 2: Connect the PCM output, TP27, to the low-pass filter, TP7, and connect Channel #2 to the filter output at TP8 Compare the two screen traces: _ 3: Obtain a compound input signal by setting FG2 to 3.00 kHz, sine Adjust FG1 and FG2 amplitudes so their sum does not exceed 4.0VP–P Make a fine adjustment of FG2 frequency to obtain a steady scope pattern: Describe the display: _ _ _ In most previous activities we used FG1 as the starting analog signal and xN as the sampling signal, fs This guaranteed that the Nyquist relation, Eq 5, is satisfied, and the sampling frequency is an integral multiple of the analog frequency, factors that promote a stable oscilloscope display Recall that the sampling frequency, fs or LOAD, used on the PCM panels is approximately but not exactly 15.0 kHz If this is an integral multiple of FG1, the PAM signal at TP27 and low-pass output at TP8 will appear steady on the oscilloscope display 4: Set the amplitude of FG2 back to 0.0 Set FG1 frequency near to 500 Hz ( 1/30 fs) Connect the PCM analog input, TP24, to FG1 and Channel #1 Connect the PAM output, TP27, to the low-pass filter input, TP7 Connect Channel #2 alternately to the input and output of the low-pass filter Describe the display as FG1 is varied near 1/30 fs: _ _ 9: Pulse Modulation 104 5: Repeat for FG1 1/20 fs and 1/10 fs and describe the displays: 6: As the analog input frequency is a smaller fraction of fs, the PAM output, before and after filtering, becomes less stable, but the wave shape is easily discernable However, as FG1 approaches ½ fs the Nyquist requirement becomes important Experiment for FG1 approaching and going beyond ½ fs and describe the results: _ _ _ In PAM PWM and PPM the amplitude, width or position vary continuously, not in discrete steps This proportional nature of modulation does not qualify as a fully digital modulation scheme Pulse Code Modulation, (PCM) is truly a digital scheme The term modulation is perhaps a misnomer because the samples are not modulated, but rather are quantized and coded Quantizing simply means rounding off a sample in discrete steps or quantizing levels The number of quantizing levels available depends on the number of bits used in each code Therefore, the more bits used for coding, the better the representation we can have of our analog signal since the round off errors are minimized However more bits in-parallel means more bits to be sent in-serial, increasing the SHIFT to LOAD ratio A distinction was made above between wire and wireless transmission PCM is suitable for both In wireless transmission, an increased bit rate is always accompanied by an undesirable frequency spread Observation In most of the experiments of this series we had sent various pulse patterns from a transmitter to a receiver and evaluated the process on the basis of similarity of the received and transmitted waveforms However, more is required Suppose the transmitted signal is a continuous stream of identical bytes, for example, 7‟s (=00000111) or 14‟s (=00001110) or 28‟s (=00011100) or 56‟s (=00111000) and so on For each or these transmissions, the end-to-end byte stream would appear identical if viewed at random Somehow, we need to “package” the bytes or other patterns so we know where one package ends and the next package starts Various methods exist, which we have not considered in the present set of modulation experiments 9: Pulse Modulation 105 9: Pulse Modulation 106 9: Pulse Modulation 107 Pulse Modulation Module Parts List Part C1, C2, C13, C22 C3, C4 C5, C6 C7, C8, C11, C15 C9, C10, C17, C28 C12, C16, C25, C30 C14 C18 C19 C20 C21 C23 C24, C26, C29 C27 C31 D1, D2, D8 D3-D7 R1-R4, R13-R14,R60-R62 R5-R6, R21, R34, R38, R7-R8, R11-R12, R19 R9 R10, R64 R12 R16, R20, R44 R17 R18 R22, R35, R40 R23, R26, R28, R36, R41, R46, R51 R24 R25 R27 R29 R30 R31, R52-R59 R32 R33, R45 R37, R42 R39, R47, R49-R50 R43 R48 R63 POT1 POT2 IC1, IC8, IC14, IC16 IC2 IC3 IC4-IC5, IC11-IC12, IC17, IC28 IC6 IC7, IC15, IC18 IC9 IC10 IC13 IC19 IC20 IC21 IC22 IC23 IC24 IC25 IC26 IC27 T1-T4 C9013 9: Pulse Modulation Value Description 0.1µF 0.33µF 330µF/16V 10µF/16V 0.01µF 0.001µF 3.9nF 22nF 18nF 6.8nF 4.7nF 10nF 0.005µF 1.2nF 100pF 1N4733A 1N4148 10kΩ 3.3kΩ 100kΩ 9.1kΩ 2.2kΩ 12kΩ 2.7kΩ 5.1kΩ 3.3MΩ 220Ω 2kΩ 75Ω 1.2kΩ 1.5kΩ 22kΩ 1.8kΩ 330Ω 3kΩ 1kΩ 100Ω 470Ω 560Ω 270Ω 33kΩ 10kΩ 50kΩ LM393N 7805T 7905T LF353N CD40106N CD4053N CD4046N CD4040N LM318N 74LS123N 74LS74N NE555N ADC0804 74LS93N 74LS14N 74LS165N 74LS164N DAC0832 ceramic capacitor mylar capacitor electrolytic capacitor electrolytic capacitor ceramic capacitor ceramic capacitor mylar capacitor mylar capacitor mylar capacitor mylar capacitor mylar capacitor mylar capacitor ceramic capacitor mylar capacitor ceramic capacitor 5.1V zener diode signal diode ¼ W resistor ¼ W resistor ¼ W resistor ¼ W resistor ¼ W resistor ¼ W resistor ¼ W resistor ¼ W resistor ¼ W resistor ¼ W resistor ¼ W resistor ¼ W resistor ¼ W resistor ¼ W resistor ¼ W resistor ¼ W resistor ¼ W resistor ¼ W resistor ¼ W resistor ¼ W resistor ¼ W resistor ¼ W resistor ¼ W resistor ¼ W resistor 10-turn Bourns potentiometer mono potentiometer dual comparator +5V voltage regulator -5V voltage regulator dual FET-input op-amp CMOS hex inverting Schmitt trigger triple 2-channel analog multiplexer micropower PLL 12-stage binary/ripple counter high-speed op-amp retriggerable monostable multivibrator Dual D-type positive edge flip-flop timer 8-bit A/D converter decade, divide by 12, binary counter hex inverting Schmitt trigger 8-bit parallel load shift register 8-bit parallel out shift register 8-bit D/A converter general purpose NPN transistor 108 Ateneo de Davao University Electronic Communication Series Electric Circuits I: Direct Current Electric Circuits II: Alternating Current Electronics I: Basic Components Electronics II: Amplifiers and Oscillators Electronics III: Operational Amplifiers Communications I: AM and FM Communications II: Digital Communications Digital Logic Circuits, with Verilog HDL Industrial Electronics LOGO! PLC: Learning a Programmable Logic Controller We are a university in a Third World country, the Philippines We believe that more than chalk and whiteboard pens are needed to train a communication engineer for today‟s world “Hands on” is a must for every student Excellent student laboratory equipment is readily available on the world markets Yet the funding necessary for us to purchase such equipment, and in the quantity we desired, was completely unavailable Our only viable option was to design and fabricate locally the materials of which before we only dreamt For each item of laboratory equipment student instructional material had to be prepared, as shown in the above listing With a view to share with other institutions the fruit of our own endeavors, we are making these student manuals freely available Permission is given to copy this material, and to suitably modify it to the needs of a particular institution 109