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Objective3

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  • 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)Agdon, Berverlyn B. September 20, 2011Signal Spectra and Signal Processing/BSECE 41A1 Score: Engr. Grace Ramones Instructor
  • 2. Objectives: Demonstrate PCM encoding using an analog-to-digital converter (ADC). Demonstrate PCM encoding using an digital-to-analog converter (DAC) Demonstrate how the ADC sampling rate is related to the analog signal frequency. Demonstrate the effect of low-pass filtering on the decoder (DAC) output.
  • 3. Sample Computation Step2 Step 6 Step 9 Step 12 Step 14 Step 16 Step 18
  • 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 acommunications channel using electronic circuits. Information is defined as knowledge orintelligence such as audio voice or music, video, or digital data. Often the information idunsuitable for transmission in its original form and must be converted to a form that issuitable for the communications system. When the communications system is digital,analog signals must be converted into digital form prior to transmission.The most widely used technique for digitizing is the analog information signals fortransmission on a digital communications system is pulse-code modulation (PCM), which wewill be studied in this experiment. Pulse-code modulation (PCM) consists of the conversionof a series of sampled analog voltage levels into a sequence of binary codes, with eachbinary number that is proportional to the magnitude of the voltage level sampled.Translating analog voltages into binary codes is called A/D conversion, digitizing, orencoding. The device used to perform this conversion process called an A/D converter, orADC.An ADC requires a conversion time, in which is the time required to convert each analogvoltage into its binary code. During the ADC conversion time, the analog input voltagemust remain constant. The conversion time for most modern A/D converters is shortenough so that the analog input voltage will not change during the conversion time. Forhigh-frequency information signals, the analog voltage will change during the conversiontime, introducing an error called an aperture error. In this case a sample and holdamplifier (S/H amplifier) will be required at the input of the ADC. The S/H amplifieraccepts the input and passes it through to the ADC input unchanged during the sample
  • 5. mode. During the hold mode, the sampled analog voltage is stored at the instant ofsampling, making the output of the S/H amplifier a fixed dc voltage level. Therefore, theADC input will be a fixed 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 readuntil the previous conversion time is complete. The sampling rate is important because itdetermines the highest analog signal frequency that can be sampled. In order to retainthe high-frequency information in the analog signal acting sampled, a sufficient number ofsamples must be taken so that all of the voltage changes in the waveform are adequatelyrepresented. Because a modern ADC has a very short conversion time, a high samplingrate is possible resulting in better reproduction of high0frequency analog signals. Nyquistfrequency is equal to twice the highest analog signal frequency component. Althoughtheoretically analog signal can be sampled at the Nyquist frequency, in practice thesampling rate is usually higher, depending on the application and other factors such aschannel bandwidth and cost limitations.In a PCM system, the binary codes generated by the ADC are converted into serial pulsesand transmitted over the communications medium, or channel, to the PCM receiver one bitat a time. At the receiver, the serial pulses are converted back to the original sequenceof parallel binary codes. This sequence of binary codes is reconverted into a series ofanalog voltage levels in a D/A converter (DAC), often called a decoder. In a properlydesigned system, these analog voltage levels should be close to the analog voltage levelssampled at the transmitter. Because the sequence of binary codes applied to the DACinput represent a series of dc voltage levels, the output of the DAC has a staircase (step)characteristic. Therefore, the resulting DAC output voltage waveshape is only anapproximation to the original analog voltage waveshape at the transmitter. These stepscan be smoothed out into an analog voltage variation by passing the DAC output through alow-pass filter with a cutoff frequency that is higher than the highest-frequencycomponent in the analog information signal. The low-pass filter changes the steps into asmooth curve by eliminating many of the harmonic frequency. If the sampling rate at thetransmitter is high enough, the low-pass filter output should be a good representation ofthe 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 ADCwill convert each of the sampled analog voltages into 8-bit binary code as that represent
  • 6. binary numbers proportional to the magnitude of the sampled analog voltages. Thesampling frequency generator, connected to the start-of conversion (SOC) terminal onthe ADC, will start conversion at the beginning of each sampling pulse. Therefore, thefrequency of the sampling frequency generator will determine the sampling frequency(sampling rate) of the ADC. The 5 volts connected to the VREF+ terminal of the ADC setsthe voltage range to 0-5 V. The 5 volts connected to the output (OE) terminal on the ADCwill keep the digital output connected to the digital bus. The DAC will convert thesedigital codes back to the sampled analog voltage levels. This will result in a staircaseoutput, which will follow the original analog voltage variations. The staircase output of theDAC feeds of a low-pass filter, which will produce a smooth output curve that should be aclose approximation to the original analog input curve. The 5 volts connected to the +terminal of the DAC sets the voltage range 0-5 V. The values of resistor R and capacitorC determine the cutoff frequency (fC) of the low-pass filter, which is determined fromthe equationFigure 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 atransmission line to the receiver and converted back to parallel format before beingapplied to the DAC input. In Figure 23-1, the ADC output is connected to the DAC inputby the digital bus for demonstration purposes only.
  • 7. 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 kHzQuestion: How did the measure sampling frequency compare with the frequency of thesampling frequency generator? They have a difference of 50 kHz.How did the sampling frequency compare with the analog input frequency? Was it morethan twice the analog input frequency? It is 20 times the analog input frequency. Yes it is more than twice the analog inputfrequency.How did the sampling frequency compare with the Nyquist frequency? The Nyquist is 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 DACstaircase output? The staircase output and the sampling generator output are both in digitalformStep 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 oscilloscope
  • 8. Questions: What happened to the DAC output after filtering? Is the filter outputwaveshape a close representation of the analog input waveshape? The DAC output became analog after it was being filtered. Yes.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 completionagain.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 = 9.5µsQuestion: How does the time between the samples in Step 8 compare with the time between the samples in Step 1? The time between the samples in Step 8 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.
  • 9. 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 = 66.5µsQuestion: How does the time between samples in Step 15 compare with the time between samples in Step 11? It is 3.5 times higher than the time in Step 11.
  • 10. 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? It is 6.28 times smaller than the Nyquist 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? The filter output frequency is the same with the analog input. Yes, it is expected based on the sampling frequency
  • 11. CONCLUSION: After conducting the experiment, I conclude PCM coding can be demonstratedthrough ADC while DAC is used for PCM decoding. Also, based on the output graph displayedby the oscilloscope, the staircase output frequency is almost the same with the samplingfrequency generator, and is several times higher than the input analog frequency. Thesampling time is inversely proportional to the sampling frequency. The output of the filter islike the input analog signal. Its frequency is also the same with the input analog frequency.The cutoff frequency remains constant even if the sampling generator frequency changes butis inversely proportional to the capacitance and resistance. Finally, the Nyquist frequency isalways 6.28 times higher than the sampling frequency generated by the ADC.

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