NATIONAL COLLEGE OF SCIENCE & TECHNOLOGY                      Amafel Bldg. Aguinaldo Highway Dasmariñas City, Cavite      ...
Objectives:   1.   Demonstrate PCM encoding using an analog-to-digital converter (ADC).   2.   Demonstrate PCM encoding us...
Sample ComputationStep2Step 6Step 9Step 12Step 14Step 16Step 18
Data Sheet:MaterialsOne ac signal generatorOne pulse generatorOne dual-trace oscilloscopeOne dc power supplyOne ADC0801 A/...
previous conversion time is complete. The sampling rate is important because it determines thehighest analog signal freque...
Figure 23–1 Pulse-Code Modulation (PCM)                                                                                   ...
Question: How did the measure sampling frequency compare with the frequency of the samplingfrequency generator?    It is a...
Question: How does the time between the samples in Step 8 compare with the time between the           samples in Step 1?  ...
Question: How does the time between samples in Step 15 compare with the time between samples           in Step 11?        ...
CONCLUSION:        I conclude that ADC and DAC can be use for Pulse Code Modulation. The output waveformproduced was a sta...
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  1. 1. NATIONAL COLLEGE OF SCIENCE & TECHNOLOGY Amafel Bldg. Aguinaldo Highway Dasmariñas City, Cavite EXPERIMENT 2 Digital Communication of Analog Data Using Pulse-Code Modulation (PCM)Balane, Maycen M. September 20, 2011Signal Spectra and Signal Processing/BSECE 41A1 Score: Engr. Grace Ramones Instructor
  2. 2. Objectives: 1. Demonstrate PCM encoding using an analog-to-digital converter (ADC). 2. Demonstrate PCM encoding using an digital-to-analog converter (DAC) 3. Demonstrate how the ADC sampling rate is related to the analog signal frequency. 4. Demonstrate the effect of low-pass filtering on the decoder (DAC) output.
  3. 3. Sample ComputationStep2Step 6Step 9Step 12Step 14Step 16Step 18
  4. 4. Data Sheet:MaterialsOne ac signal generatorOne pulse generatorOne dual-trace oscilloscopeOne dc power supplyOne ADC0801 A/D converter (ADC)One DAC0808 (1401) D/A converter (DAC)Two SPDT switchesOne 100 nF capacitorResistors: 100 Ω, 10 kΩTheoryElectronic communications is the transmission and reception of information over a communicationschannel using electronic circuits. Information is defined as knowledge or intelligence such as audiovoice or music, video, or digital data. Often the information id unsuitable for transmission in itsoriginal form and must be converted to a form that is suitable for the communications system.When the communications system is digital, analog signals must be converted into digital form priorto transmission.The most widely used technique for digitizing is the analog information signals for transmission on adigital communications system is pulse-code modulation (PCM), which we will be studied in thisexperiment. Pulse-code modulation (PCM) consists of the conversion of a series of sampled analogvoltage levels into a sequence of binary codes, with each binary number that is proportional to themagnitude of the voltage level sampled. Translating analog voltages into binary codes is called A/Dconversion, digitizing, or encoding. The device used to perform this conversion process called an A/Dconverter, or ADC.An ADC requires a conversion time, in which is the time required to convert each analog voltage intoits binary code. During the ADC conversion time, the analog input voltage must remain constant.The conversion time for most modern A/D converters is short enough so that the analog inputvoltage will not change during the conversion time. For high-frequency information signals, theanalog voltage will change during the conversion time, introducing an error called an aperture error.In this case a sample and hold amplifier (S/H amplifier) will be required at the input of the ADC. TheS/H amplifier accepts the input and passes it through to the ADC input unchanged during the samplemode. During the hold mode, the sampled analog voltage is stored at the instant of sampling,making the output of the S/H amplifier a fixed dc voltage level. Therefore, the ADC input will be afixed dc voltage during the ADC conversion time.The rate at which the analog input voltage is sampled is called the sampling rate. The ADCconversion time puts a limit on the sampling rate because the next sample cannot be read until the
  5. 5. previous conversion time is complete. The sampling rate is important because it determines thehighest analog signal frequency that can be sampled. In order to retain the high-frequencyinformation in the analog signal acting sampled, a sufficient number of samples must be taken sothat all of the voltage changes in the waveform are adequately represented. Because a modern ADChas a very short conversion time, a high sampling rate is possible resulting in better reproduction ofhigh0frequency analog signals. Nyquist frequency is equal to twice the highest analog signalfrequency component. Although theoretically analog signal can be sampled at the Nyquistfrequency, in practice the sampling rate is usually higher, depending on the application and otherfactors such as channel bandwidth and cost limitations.In a PCM system, the binary codes generated by the ADC are converted into serial pulses andtransmitted over the communications medium, or channel, to the PCM receiver one bit at a time. Atthe receiver, the serial pulses are converted back to the original sequence of parallel binary codes.This sequence of binary codes is reconverted into a series of analog voltage levels in a D/A converter(DAC), often called a decoder. In a properly designed system, these analog voltage levels should beclose to the analog voltage levels sampled at the transmitter. Because the sequence of binary codesapplied to the DAC input represent a series of dc voltage levels, the output of the DAC has astaircase (step) characteristic. Therefore, the resulting DAC output voltage waveshape is only anapproximation to the original analog voltage waveshape at the transmitter. These steps can besmoothed out into an analog voltage variation by passing the DAC output through a low-pass filterwith a cutoff frequency that is higher than the highest-frequency component in the analoginformation signal. The low-pass filter changes the steps into a smooth curve by eliminating many ofthe harmonic frequency. If the sampling rate at the transmitter is high enough, the low-pass filteroutput should be a good representation of the original analog signal.In this experiment, pulse code modulation (encoding) and demodulation (decoding) will bedemonstrated using an 8-bit ADC feeding an 8-bit DAC, as shown in Figure 2-1. This ADC will converteach of the sampled analog voltages into 8-bit binary code as that represent binary numbersproportional to the magnitude of the sampled analog voltages. The sampling frequency generator,connected to the start-of conversion (SOC) terminal on the ADC, will start conversion at thebeginning of each sampling pulse. Therefore, the frequency of the sampling frequency generator willdetermine the sampling frequency (sampling rate) of the ADC. The 5 volts connected to the VREF+terminal of the ADC sets the voltage range to 0-5 V. The 5 volts connected to the output (OE)terminal on the ADC will keep the digital output connected to the digital bus. The DAC will convertthese digital codes back to the sampled analog voltage levels. This will result in a staircase output,which will follow the original analog voltage variations. The staircase output of the DAC feeds of alow-pass filter, which will produce a smooth output curve that should be a close approximation tothe original analog input curve. The 5 volts connected to the + terminal of the DAC sets the voltagerange 0-5 V. The values of resistor R and capacitor C determine the cutoff frequency (fC) of the low-pass filter, which is determined from the equation
  6. 6. Figure 23–1 Pulse-Code Modulation (PCM) XSC2 G T A B C D S1 VCC Key = A 5V U1 Vin D0 S2 D1 V2 D2 D3 Key = B 2 Vpk D4 10kHz D5 0° Vref+ D6 Vref- D7 SOC VCC OE EOC 5V D0 D1 D2 D3 D4 D5 D6 D7 ADC V1 Vref+ R1 VDAC8 Output 5V -0V Vref- 100Ω 200kHz U2 R2 10kΩ C1 100nFIn an actual PCM system, the ADC output would be transmitted to serial format over a transmissionline to the receiver and converted back to parallel format before being applied to the DAC input. InFigure 23-1, the ADC output is connected to the DAC input by the digital bus for demonstrationpurposes only.PROCEDURE:Step 1 Open circuit file FIG 23-1. Bring down the oscilloscope enlargement. Make sure that the following settings are selected. Time base (Scale = 20 µs/Div, Xpos = 0 Y/T), Ch A(Scale 2 V/Div, Ypos = 0, DC) Ch B (Scale = 2 V/Div, Ypos = 0, DC), Trigger (Pos edge, Level = 0, Auto). Run the simulation to completion. (Wait for the simulation to begin). You have plotted the analog input signal (red) and the DAC output (blue) on the oscilloscope. Measure the time between samples (TS) on the DAC output curve plot. TS = 4 µsStep 2 Calculate the sampling frequency (fS) based on the time between samples (TS) fS = 250 kHz
  7. 7. Question: How did the measure sampling frequency compare with the frequency of the samplingfrequency generator? It is almost equal. The difference is 50 kHz.How did the sampling frequency compare with the analog input frequency? Was it more than twicethe analog input frequency? The sampling frequency is more than 20 times of the analog input frequency. Yes it is more thantwice the analog input frequency.How did the sampling frequency compare with the Nyquist frequency? It is 2π or 6.28 times more than the sampling frequency.Step 3 Click the arrow in the circuit window and press the A key to change Switch A to the sampling generator output. Change the oscilloscope time base to 10 µs/Div. Run the simulation for one oscilloscope screen display, and then pause the simulation. You are plotting the sampling generator (red) and the DAC output (blue).Question: What is the relationship between the sampling generator output and the DAC staircaseoutput? They are both digital.Step 4 Change the oscilloscope time base scale to 20 µs/Div. Click the arrow in the circuit window and press the A key to change Switch A to the analog input. Press the B key to change the Switch B to Filter Output. Bring down the oscilloscope enlargement and run the simulation to completion. You are plotting the analog input (red) and the low-pass filter output (blue) on the oscilloscopeQuestions: What happened to the DAC output after filtering? Is the filter output waveshape a closerepresentation of the analog input waveshape? The DAC output became analog. Yes, it is a close representation of the analog input. TheDAC lags the input waveshape.Step 5 Calculate the cutoff frequency (fC) of the low-pass filter. fC = 15.915 kHzQuestion: How does the filter cutoff frequency compare with the analog input frequency? They have difference of approximately 6 kHz.Step 6 Change the filter capacitor (C) to 20 nF and calculate the new cutoff frequency (fC). fC = 79.577 kHzStep 7 Bring down the oscilloscope enlargement and run the simulation to completion again.Question: How did the new filter output compare with the previous filter output? Explain. It is almost the same.Step 8 Change the filter capacitor (C) back to 100 nF. Change the Switch B back to the DAC output. Change the frequency of the sampling frequency generator to 100 kHz. Bring down the oscilloscope enlargement and run the simulation to completion. You are plotting the analog input (red) and the DAC output (blue) on the oscilloscope screen. Measure the time between the samples (TS) on the DAC output curve plot (blue) TS = 9.5µs
  8. 8. Question: How does the time between the samples in Step 8 compare with the time between the samples in Step 1? It doubles.Step 9 Calculate the new sampling frequency (fS) based on the time between the samples (TS) in Step 8? fS=105.26HzQuestion: How does the new sampling frequency compare with the analog input frequency? It is 10 times the analog input frequency.Step 10 Click the arrow in the circuit window and change the Switch B to the filter output. Bring down the oscilloscope enlargement and run the simulation again.Question: How does the curve plot in Step 10 compare with the curve plot in Step 4 at the higher sampling frequency? Is the curve as smooth as in Step 4? Explain why. Yes, they are the same. It is as smooth as in Step 4. Nothing changed. It does not affect the filter.Step 11 Change the frequency of the sampling frequency generator to 50 kHz and change Switch B back to the DAC output. Bring down the oscilloscope enlargement and run the simulation to completion. Measure the time between samples (TS) on the DAC output curve plot (blue). TS = 19µsQuestion: How does the time between samples in Step 11 compare with the time between the samples in Step 8? It doubles.Step 12 Calculate the new sampling frequency (fS) based on the time between samples (TS) in Step 11. fS=52.631 kHzQuestion: How does the new sampling frequency compare with the analog input frequency? It is 5 times the analog input.Step 13 Click the arrow in the circuit window and change the Switch B to the filter output. Bring down the oscilloscope enlargement and run the simulation to completion again.Question: How does the curve plot in Step 13 compare with the curve plot in Step 10 at the higher sampling frequency? Is the curve as smooth as in Step 10? Explain why. Yes, nothing changed. The frequency of the sampling generator does not affect the filter.Step 14 Calculate the frequency of the filter output (f) based on the period for one cycle (T). T=10kHzQuestion: How does the frequency of the filter output compare with the frequency of the analog input? Was this expected based on the sampling frequency? Explain why. It is the same. Yes, it is expected.Step 15 Change the frequency of the sampling frequency generator to 15 kHz and change Switch B back to the DAC output. Bring down the oscilloscope enlargement and run the simulation to completion. Measure the time between samples (TS) on the DAC output curve plot (blue) TS = 66.5µs
  9. 9. Question: How does the time between samples in Step 15 compare with the time between samples in Step 11? It is 3.5 times more than the time in Step 11.Step 16 Calculate the new sampling frequency (fS) based on the time between samples (TS) in Step 15. fS=15.037 kHzQuestion: How does the new sampling frequency compare with the analog input frequency? It is 5 kHz greater than the analog input frequency.How does the new sampling frequency compare with the Nyquist frequency? The Nyquist frequency is 6.28 times larger than the sampling frequency.Step 17 Click the arrow in the circuit window and change the Switch B to the filter output. Bring down the oscilloscope enlargement and run the simulation to completion again.Question: How does the curve plot in Step 17 compare with the curve plot in Step 13 at the higher sampling frequency? They are the same.Step 18 Calculate the frequency of the filter output (f) based on the time period for one cycle (T). f=10kHzQuestion: How does the frequency of the filter output compare with the frequency of the analog input? Was this expected based on the sampling frequency? It is the same. Yes, it is expected.
  10. 10. CONCLUSION: I conclude that ADC and DAC can be use for Pulse Code Modulation. The output waveformproduced was a staircase wave. However, the low-pass filter output is like the input analog signal. TheADC sampling rate affects the frequency of the sampling signal. As the ADC sampling rate increases, thefrequency of the sampling signal also increases. On the other hand, the filter frequency was not affectedby the rate of the sampling generator from the ADC. The analog frequency is the same as the frequencyof the filter. The filter’s cutoff frequency is inversely proportional to the capacitor, as the capacitorincreases, the cutoff frequency decreases. The Nyquist frequency is always 6.28 times larger than thesampling frequency.

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