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  • NATIONAL COLLEGE OF SCIENCE & TECHNOLOGY Amafel Bldg. Aguinaldo Highway Dasmariñas City, Cavite EXPERIMENT NO. 2 Digital Communication of Analog Data Using Pulse-Code Modulation (PCM)Lopera, Jericho James L. September 20, 2011Signal Spectra and Signal Processing/BSECE 41A1 Score: Engr. Grace Ramones Instructor
  • 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.
  • Sample Computation Step2 Step 6 Step 9 Step 12 Step 14 Step 16 Step 18
  • 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 is suitablefor the communications system. When the communications system is digital, analog signalsmust be converted into digital form prior to transmission.The most widely used technique for digitizing is the analog information signals for transmissionon a digital communications system is pulse-code modulation (PCM), which we will be studiedin this experiment. Pulse-code modulation (PCM) consists of the conversion of a series ofsampled analog voltage levels into a sequence of binary codes, with each binary number thatis proportional to the magnitude of the voltage level sampled. Translating analog voltages intobinary codes is called A/D conversion, digitizing, or encoding. The device used to perform thisconversion process called an A/D converter, or ADC.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 voltage mustremain constant. The conversion time for most modern A/D converters is short enough so thatthe analog input voltage will not change during the conversion time. For high-frequencyinformation signals, the analog voltage will change during the conversion time, introducing anerror called an aperture error. In this case a sample and hold amplifier (S/H amplifier) will berequired at the input of the ADC. The S/H amplifier accepts the input and passes it through tothe ADC input unchanged during the sample mode. During the hold mode, the sampled analogvoltage is stored at the instant of sampling, making the output of the S/H amplifier a fixed dcvoltage level. Therefore, the ADC input will be a fixed dc voltage during the ADC conversiontime.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 untilthe previous conversion time is complete. The sampling rate is important because it determinesthe highest 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
  • so that all of the voltage changes in the waveform are adequately represented. Because amodern ADC has a very short conversion time, a high sampling rate is possible resulting inbetter reproduction of high0frequency analog signals. Nyquist frequency is equal to twice thehighest analog signal frequency component. Although theoretically analog signal can besampled at the Nyquist frequency, in practice the sampling rate is usually higher, depending onthe application and other factors 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 atime. At the receiver, the serial pulses are converted back to the original sequence of parallelbinary codes. This sequence of binary codes is reconverted into a series of analog voltagelevels in a D/A converter (DAC), often called a decoder. In a properly designed system, theseanalog voltage levels should be close to the analog voltage levels sampled at the transmitter.Because the sequence of binary codes applied to the DAC input represent a series of dcvoltage levels, the output of the DAC has a staircase (step) characteristic. Therefore, theresulting DAC output voltage waveshape is only an approximation to the original analogvoltage waveshape at the transmitter. These steps can be smoothed out into an analog voltagevariation by passing the DAC output through a low-pass filter with a cutoff frequency that ishigher than the highest-frequency component in the analog information signal. The low-passfilter changes the steps into a smooth curve by eliminating many of the harmonic frequency. Ifthe sampling rate at the transmitter is high enough, the low-pass filter output should be a goodrepresentation 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 willconvert each of the sampled analog voltages into 8-bit binary code as that represent binarynumbers proportional to the magnitude of the sampled analog voltages. The samplingfrequency generator, connected to the start-of conversion (SOC) terminal on the ADC, will startconversion at the beginning of each sampling pulse. Therefore, the frequency of the samplingfrequency generator will determine the sampling frequency (sampling rate) of the ADC. The 5volts connected to the VREF+ terminal of the ADC sets the voltage range to 0-5 V. The 5 voltsconnected to the output (OE) terminal on the ADC will keep the digital output connected to thedigital bus. The DAC will convert these digital codes back to the sampled analog voltage levels.This will result in a staircase output, which will follow the original analog voltage variations. Thestaircase output of the DAC feeds of a low-pass filter, which will produce a smooth output curvethat should be a close approximation to the original analog input curve. The 5 volts connectedto the + terminal of the DAC sets the voltage range 0-5 V. The values of resistor R andcapacitor C 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 being applied tothe DAC input. In Figure 23-1, the ADC output is connected to the DAC input by the digital busfor demonstration purposes 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 kHzQuestion: How did the measure sampling frequency compare with the frequency of thesampling frequency generator? The sampling time is almost equal, however, the frequencies have a differrence of 50 kHz.How did the sampling frequency compare with the analog input frequency? Was it more thantwice the analog input frequency? It is much higher; in fact, it is 20 times larger than the input frequency. Yes, it is more than twice the analog input frequency.How did the sampling frequency compare with the Nyquist frequency?
  • The Nyquist frequency is higher. 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? Both outputs are both in digitalStep 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 aclose representation of the analog input waveshape? It became an analog signal that lags the input analog signal. Yes, it is a close representation of 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 (f C). 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? The sampling frequency is much higher than the input frequency. It is 10 times the 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 (T S) 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? The sampling time 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? The new sampling frequency 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.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 much lesser 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? They are the same. Yes, it should be the same for this sampling frequency.
  • CONCLUSION: Based on the input and the output displayed by the oscilloscope, I conclude that the inputanalog signal can be converted into digital through PCM. An ADC is use for PCM encoding whileDAC is use for PCM decoding. The staircase signal is the DAC output and its frequency isgenerated by the signal frequency generator connected to the ADC. The sampling time is inverselyproportional to the sampling frequency. Add to that, the filter output is analog like the input analogsignal. Also, the frequency of the filter output is the same as the input analog signal. The cutofffrequency is inversely proportional to the capacitor, as the capacitor increases, the cutofffrequency decreases. Lastly, the frequency of the sampling signal is much lesser than the Nyquistfrequency.