BIRZEITUNIVERSITY ELECTRICAL ENGINEERING DEPARTMENT ANALOG AND DIGITAL COMMUNICATION LAB (ENEE 411) Last Update: January -2013 Table of Contents Experiment 1 AM Modulation and Detection .............................................................. 4 Experiment 2 DSB-SC and SSB ................................................................................. 11 Experiment 3 FM Modulation and Demodulation .................................................... 19 Experiment 4 FDM ..................................................................................................... 26 Experiment 5 ADC ...................................................................................................... 34 Experiment 6 DAC ...................................................................................................... 42 Experiment 7 PCM ...................................................................................................... 52 Experiment 8 TDM ..................................................................................................... 57 Experiment 9 ASK (Amplitude Shift Keying) ............................................................ 63 Experiment 10 FSK (Frequency Shift Keying) .......................................................... 70 Experiment 11 BPSK(Binary Phase Shift Keying) .................................................... 75 Experiment 12 QPSK(Quadri- Phase Shift Keying) .................................................. 80 Experiment 13 Delta Modulation and Demodulation ............................................... 83 2 EXPERIMENT.1 AM Modulation and Demodulation Objectives: To understand the theory of amplitude modulation and demodulation. To design and implement the two types of AM modulator: transistor and balanced modulator. To design and implement the two types of AM demodulator: the diode detection and the product detection. To understand the measurements and adjustments of AM modulator and demodulator. PrelaB Work: Use MATLAB command and M files to draw the demodulated signal after the envelope detector given that: S AM (t ) Ac [1 cos(m t )] cos(c t ) 1. Write the mathematical expression for the demodulated signal. 2. Use MATLAB command and M files to draw the demodulated signal for the following three cases: a. Ac=16v, modulation index=0.22, modulating signal frequency=800Hz b. Ac=16v, modulation index=1, modulating signal frequency=800Hz c. Ac=16v, modulation index=1.85, modulating signal frequency=800Hz 3. Discuss your result in each part .you must write the commands which are used in the Pre-lab. Equipment Required: 2 AC Function Generators DC Power Supply ETEK ACS-3000-02 Module Connection wires 3 Theory: Modulation: the process by which some characteristic (like: amplitude, frequency or phase) of a carrier signal is varied in accordance with the modulating signal (message signal).The signal modulation is used in order to transmit messages over long distances and also to transmit signals from various sources simultaneously over a common channel. Amplitude Modulation (AM): The process in which the amplitude of the carrier varies linearly with message signal. The general formula for the modulated AM signal: A S AM (t ) Ac 1 m cos(2f m t ) cos(2fct )............(1 1) Ac A : M odulation index whic h is equal to m . Ac Am : Amplitude of the message signal with volt unit. f m : Frequency of the message signal. fc : Frequency of the carrier signal. Ac : Amplitude of the carrier signal with volt unit. From the above formula we find that in order to generate an AM signal we just need to add a DC signal with the message signal then multiply the added signal with the carrier signal. The analog multiplier is the basic modulator that is used to generate AM signal as shown in fig1.1: Fig(1.1): Analog Multiplier. 4 Since the audio signal is hidden in the double side bands and the carrier signal does have no data. the AM modulation is lower efficiency than double side band suppressed carrier (DSB-SC) modulation but its demodulation circuit is much simpler. In electronics circuits a multiplier is implemented by the nonlinear characteristics of active elements. From equation (1-2) we find that the double side bands are proportional to µ so larger µ is getting better efficiency.Product detector (synchronous detector) 5 .Envelope detector (asynchronous detector) 2. So it is unable to recover the signal at the receiver using the envelope detector (as we will see later). which means that the variation of the carrier is no longer sinusoidal and the signal is distorted . AM modulator practical circuits: As we said previously we can implement the AM modulators by using a multiplier.Balanced Modulator (MC1496) Demodulation: is the process of restoring the message signal at the receiver side.The Modulation Index µ: There is an important parameter in the AM modulation which is called the modulation index (µ) which is equal to Am/Ac .While the second term represents the carrier signal. The transmission power efficiency η: ( ) The modulation index is smaller or equal to . If the double side bands get stronger then the transmission efficiency is getting better. The Frequency Spectrum for the AM modulated signal: Equation (1-1) can be written as : ( ) [ ( ( ) ) ( ( ) )] ( ) ( ) The first term of equation (1-2) represents the double side band signals . In this experiment we will show two types of AM demodulators: 1.Transistor AM modulator 2.So if µ <1 an over modulation will happen for the AM signal . There are many circuits’ works as a multiplier but in this experiment we will deal with two types: 1. fc<<10 fm 2. Fig(1. RC is the time constant of the RC low pass filter .Detection Using Envelope detector: Fig(1-2) shows the envelope detector operation. The envelope detector is able to recover the message signal if the following conditions are achieved: 1.Tc>>RC>>Tm . If there is an over modulation we can use the product detector in order to recover the signal. 6 .µ=[0.2): Envelope detector.1] 3. Tm=1/fm.Where Tc=1/fc. After the diode rectifies the AM signal (removing the negative part) then the RC low pass filter obtains the AM envelope which is the message signal. However. Procedure: Transistor AM modulator: 1. For the envelope detector which is asynchronous detector. The output of the product as shown in fig (1-3): ( ) ( ) [ ( )] ( ) ( ) The first term of eq(1-4) is a DC signal while the second is the message signal .use the function generator to input a sine wave 600mV amplitude and 1KHz frequency.Refer to ACS3-1 on ETEK kit ACS-3000-02 module. At the carrier signal port (Carrier I/P) .At the audio input port (Audio I/P) .By using the oscilloscope. The two types of the detectors have its own advantages and disadvantages.Adjust VR1 so that the AM signal is maximum without distortion (VR1 is used to change the operation point of the transistor and it also controls the magnitude of the carrier) .Observe the signals at TP1. 2.input a sine wave 1. its circuit is simple but its performance is not better as the product detector .7 amplitude and 500 KHz frequency. the third term is the second harmonic of the AM signal which is rejected by the low pass filer.Detection Using Product detector: Fig(1. the product detector’s circuit is more complicated and requires synchronous for both carrier signal and AM signal (same phase and same frequency). otherwise the quality of the output will be affected. observe the AM modulated signal at the modulator output port (AM O/P).3): Product detector. 7 . 3.TP2 and TP3. 4. Keep the connection for the balanced modulator the same as in the previous part.record your results.input a sine wave 2V amplitude and 500 KHz frequency. Balanced AM modulator: 1. Product Detector: 1.Adjust VR2 so that the AM signal is maximum without distortion (VR2 controls the gain of the modulator) .At the audio input port (Audio I/P).Let J1 short.8KΩ (R10 determines the magnitude of the bias current for the modulator). 2.By using the oscilloscope. Adjust VR1 so that the value of µ is less than 1 (VR1 controls the value of µ). At the carrier signal port (Carrier I/P) . Let µ<1 (over modulation) observe the AM signal.Change the value of VR1 until µ=1 (100% modulation). observe the AM modulated signal at the modulator output port (AM O/P). 3. 6. use the function generator to input a sine wave 350mV amplitude and 3KHz frequency. 5. 8 .Adjust VR1 so that µ is maximum and VR2 so that AM O/P1 approximate between 150350mVp-p. (Keep the connection as it is for the second part ).Observe the signal at TP1. Diode Detector: 1.Connect AM O/P1 to the input AM I/P of the product (Coherent) detector (ACS 4-2 on ETEK ACS-3000-02 module). 2. 2. 4.TP3 and TP4.Refer to ACS3-2 on ETEK kit ACS-3000-02 module. 6. 4.input a sine wave 2V amplitude and 500 KHz frequency.5.Try to change the frequency and the amplitude of the message signal.Connect AM O/P1 to the input AM I/P of the diode detector (ACS 4-1 on ETEK ACS3000-02 module). record your results.J2 open so that R10=6.Try to change the amplitude of the message signal and its frequency.input a sine wave 2V amplitude and 500 KHz frequency.use the function generator to input a sine wave 500mV amplitude and 1KHz frequency.Change the value of VR1 so that µ is less than 1.At the audio input port (Audio I/P).Adjust VR1 so that µ is less than 1.At the audio input port (Audio I/P) . notice the effects on the AM signal. 4. At the carrier signal port (Carrier I/P) . 5. Record your results. At the carrier signal port (Carrier I/P) .TP2. 3.Change the message signal amplitude and frequency and notice their effects on the signal at the detector output. use the function generator to input a sine wave 350mV amplitude and 3KHz frequency. Record your results. 3. Observe the signal at the detector output. 7.5sin(2 l000t)sin(2 455000t) Explain what is this wave and the meaning of the parameters: 5.5.Adjust the value of VR1 (controls the amplitude of the carrier). µ =0.5.75.5. 455000 What is the modulation coefficient of the above wave and what is the relation between the modulated wave amplitude and the' carrier wave amplitude? What is the bandwidth of this AM modulated wave? 9 . Calculate the power efficiency for different modulation index µ=0. 6. µ =0.Try to change the frequency of the message signal and the frequency of the carrier.VR2(controls the amplitude of the message signal) and VR3(controls the gain of the detector) so that the signal at the output of the detector is maximum without distortion. An AM detector gets a wave with the following mathematical expression: V(t) = 5(1 + 0.25.Connect the carrier signal input of the product detector (Carrier I/P) with the same carrier signal in AM modulator (for synchronization).0. µ =1 2. 1000. Questions: 1. To understand the waveforms and frequency spectrums of DSB-SC and SSB signals. Theory: DSB-SC and SSB modulation: Recall that the AM modulated signal is given by: ( ) [ ( ( ) ) ( ( ) )] ( ) ( ) 10 .EXPERIMENT. to show graphically the time domain of SSB-SC modulated Signal. Equipment Required: 2 AC Function Generators DC Power Supply ETEK ACS-3000-03 Module Connection wires.2: DSB-SC and SSB Objectives: To understand the theory of DSB-SC and SSB modulation and demodulation. PreLab work: Using Matlab software and Simulink. To design and implement the DSB-SC and SSB modulators and demodulators. To understand the measurements and adjustments of DSB-SC and SSB modulators and demodulators. Taking the modulating signal m(t ) cos 2 (1500)t and the carrier signal c(t ) 4 cos 2 (100000)t . It is not necessary to transmit both side-bands. 11 . The information represented by the modulating signal is contained in both the upper and the lower sidebands. to make the carrier amplitude equals to zero. This will explain that the AM modulation has a low transmission efficiency (it could be only 33% in the best case). Fig(2-2): Frequency spectrum of DSB-SC.Fig(2-1): Frequency spectrum of AM. So the idea from Double Side Band Suppressed Carrier modulation (DSBSC) is to suppress the carrier or in other words. We can use DSB-SC to obtain SSB modulation. In SSB modulation we eliminate the carrier and one sideband. Either one can be suppressed at the transmitter without any loss of information. therefore. the bandwidth required for SSB is theoretically one-half that required when both sidebands are transmitted. Since the message signal is hidden in the double side bands and the carrier does not contain any signal. a power savings of over 83 percent is realized. the power is consumed in the carrier during the transmission of AM signal. This technique will improve the power efficiency. Additionally. We utilize two DSB-SC modulators and let the phase difference between the two audio signals and the two carriers to become 90 degree. (2-2) (DSB-SC)Q = cos2π(fc-fm)t .Fig(2-3):Frequency spectrum of SSB ( Lower sideband). Fig(2-4):Frequency spectrum of SSB ( Upper sideband). We can also use the DSB-SC modulation to obtain SSB modulation. (2-3) 12 .quadrature component and (DSB-SC)I-in phase component where: (DSB-SC)I = cos2π(fc-fm)t+ cos2π(fc+fm)t ……………………………………. i.cos2π(fc+fm)t ……………………………………. Fig(2-5):block diagram of DSB-SC modulation.e: (DSB-SC)Q. the power consumption of SSB modulation is less than DSB-SC modulation so the sequence of power consumption for these different types of modulation is as follows: AM <DSB-SC < SSB.5) During transmission. In this experiment we will utilize balanced modulator MC1496 to design DSB-SC modulated signal. LSSB= (DSB-SC)I+ (DSB-SC)Q= cos2π(fc-fm)t ……………………………….Equations (2-2) and (2-3) show that both (DSB-SC)I and (DSB-SC)Q connect to an adder to obtain USSB or LSSB at the output port. Implementation of SSB modulator: From equations (2-4) and (2-5).(2. Fig(2-6):Circuit diagram of DSB-SC modulation by utilizing MC1496. we know that the SSB modulator is the combination of two DSB-SC modulators. Figure (2-6) is the block diagram of SSB modulator. Implementation of DSB-SC modulator: DSB-SC modulation is a kind of AM modulation so we can use the structure of AM modulator to implement DSB-SC.(DSB-SC)Q= cos2π(fc + fm)t ………………………………. 13 . (2.4) USSB= (DSB-SC)I. Fig(2-8):Circuit diagram of phase shifter. 14 .Fig(2-7):Circuit diagram of SSB modulator. 15 .(2.5M(f). DSB-SC and SSB demodulation: As we know: ( ) ( ) ( ( ) )…………………..…. The block diagram for the DSB-SC demodulator is shown in the figure below: Fig(2-10):Block diagram of DSB-SC demodulator (Coherent).Fig(2-9):Circuit diagram of linear adder.6) Multiply equation (2-6) by 2cos(2πfct): ( ) ( ) ( )[ ( ( ) ) ] ( ) )] ( ) By using Fourier Transform on equation (2-7) we get: ( ) ( ) [ ( ) ( When ( ) passes through a low pass filter which its bandwidth equals or larger than the bandwidth of the message signal but smaller than 2fc then the only term left in equation (2-8) is 0. Adjust the variable resistor “Phase Adjust” so that the phase shift between TP3 and TP4 is 90º. observe TP5 (DSB-SC(Q)) then adjust VR1(gain adjustment) so that the output amplitude is maximum without distortion. 4. 2.At the audio input port (Audio I/P) put a sine wave with 400mV amplitude and 1KHz frequency. adjust VR4 (modulation index µ) so that µ=1. Then at the carrier input port (Carrier I/P) put a sine wave with also 400 mV amplitude and 200KHz frequency. Also. Implementation of DSB-SC and SSB demodulator: DSB-SC and SSB demodulator which is called a coherent product detector will be implemented in this circuit using MC1496 as shown in the diagram. adjust VR3 (modulation index µ) so that µ=1. observe TP6 (DSB-SC(I)) then adjust VR2(gain adjustment) so that the output amplitude is maximum without distortion. 3.Record your result.By using oscilloscope. Also. 5. 16 .From equations (2-4) and (2-5) we conclude that we can use the demodulator above figure (210)as a SSB demodulator.Record your result.By using oscilloscope. Adjust the variable resistor “QPS” so that the phase shift between TP1 and TP2 is 90º. 6.By using oscilloscope. Procedure: Part 1: DSB-SC and SSB modulators: 1. Fig(2-11):Circuit diagram of the coherent product detector. observe the signals at the audio output ports TP1 and TP2 at the same time.Refer to module ACS5-1 on ETEK ACS-3000-03 Kit. observe the signals at the carrier output ports TP3 and TP4 at the same time.By using oscilloscope. Adjust VR1 and VR2 so that the amplitude at (Audio O/P) is maximum without distortion then record the waves of the product detector at TP1 and TP2 . input the same carrier signal in ACS5-1 to the carrier signal input port (Carrier I/P) in ACS6-1. 2. 8. Part 2: DSB-SC demodulator: 1.Connect the modulated DSB-SC(I) signal in module ACS5-1 to the input terminal (DSB-SC/SSB I/P) of the product detector in module ACS6-1.At the same time.Try to change the amplitude of the message signal then its frequency and observe the effects. 4.Let J1 is open and J2 is short and repeat the steps above. 3.Keep all connections as they are.Keep only the connections at the modulator side. 9. Record your results. Part 3: SSB demodulator: 1. 17 .By using Oscilloscope.Adjust VR1 and VR2 so that the amplitude at (Audio O/P) is maximum without distortion then record the waves of the product detector at TP1 .Try to change the amplitude of the carrier signal then its frequency. 2. TP2 and (Audio O/P) . observe the output signal of the product detector (Audio O/P) in ACS6-1. input the same carrier signal in ACS5-1 to the carrier signal input port (Carrier I/P) in ACS6-1.Let J1 is open and J2 is short and repeat the steps above. Let J1 be short circuit and J2 open circuit.At the same time.Connect the modulated SSB signal (SSB O/P) in module ACS5-1 to the input terminal (DSB-SC/SSB I/P) of the product detector in module ACS6-1. observe the output signal of the product detector (Audio O/P) in ACS6-1.To implement the product detector of DSB-SC refer to ACS6-1 on ETEK ACS3000-03 module. 4. 6. 3. 5.7.By using Oscilloscope. c. subtract the dc term and show that the detector’s output is linearly proportional to m(t). Calculation of the frequency deviation and the bandwidth. Investigating the influence of changing the frequency of the signal at the modulator voltage output. b. Find the message signal m(t). Differentiate s(t) with respect to t and plot ds(t)/dt for -1 ≤ t ≤ 1. Modulation and detection of the FM signal using MC4046 and LM565 detectors. 18 . d. Calculation of the modulation coefficient. Plot s(t) versus t for -1 ≤ t ≤ 1.EXPERIMENT. Apply ds(t)/dt to an ideal envelope detector. Equipment Required: ETEK ACS-3000-04(MC4046 and LM566 Modules) Power supply Oscilloscope Audio signal generator Banana wires PreLab work: Consider the frequency modulated signal: S (t ) cos[2 (17)t 4 sin(2t )] a.3 FM Modulation and Demodulation Objectives: Creating a modulated FM wave using MC4046 and LM566 Modulators. Notice how this operation transforms an FM waveform into an AM waveform. The General Formula of FM Modulated Wave t S FM (t ) Ac cos 2fct 2Kf m( )d ………………………………………….Theory: Frequency Modulation: the process by which frequency of the carrier must be varied with respect to the message signal...2) When message signal m(t ) Am cos(2fmt ) What is the formula of FM modulated wave? The modulation coefficient KfAm f ……………………………………… (3. The instantaneous frequency given by: fi (t ) fc Kfm(t ) ……………………………………………………………….. (3... FM modulation is implemented by controlling the instantaneous frequency of a voltage-controlled oscillator (VCO).. FM modulation (Direct Method): In practice.1) Where Kf: is the proportionality constant with unit (HZ/volt) A typical characteristics of VCO looks like these: Fig(3-1): A typical characteristics of VCO. The amplitude of the input signal voltage controls the oscillation frequency of the VCO output signal.3) fm fm Where f : is the peak frequency deviation. Fm: message signal frequency 19 . (3. .........2)...Band width of FM Modulated signal: The band width of fm modulated signal given by Carson’s rule as the following: BWFM 2(f fm) ...................6) 0 Where: m(t ) = modulating signal f c = carrier frequency Ac = carrier amplitude 20 ......2): FM modulator and demodulator..... t s(t ) Ac cos (2f c t 2k f m( )d ) …………………………………………….... The differentiator basically produces an AM-like signal that is then demodulated by the envelope detector block.. Power efficiency: the total transmitted power is constant and is independent of the message signal...... Constant power 2..... which uses a cascaded differentiator with an envelope detector circuit as illustrated in Fig (3....... (3....4) BWFM 2 fm(1 ) …………………………………………………………. (3.…... Fig (3......... Our focus in this experiment will be on the Slope Method......5) Advantages of FM Modulation : 1.................. (3.... FM DeModulation : FM signals can be demodulated using different techniques..... Better noise immunity [good quality] 3..... Connect the signal generator to Audio signal input (Audio I/P) and set the amplitude of the generator 10vp-p and 1 KHz frequency of the sine wave. (3. Differentiating s(t) in (3.7) fm Where k f Am f = frequency deviation = f = Modulation index (Deviation ration) fm There are two types of FM signals depending on the value of .8) dt 0 Note that equation (3.9) You must add capacitor to do dc blocking.8) similar to AM modulated signal. What is the output after the capacitor? Procedure: Part One: Modulation and Demodulation using MC4046 and PLLMC4046Modules fig (3.7)with respect to t we get: t ds(t ) 2f c 2k f m(t ) Ac sin (2f c t 2k f m( )d ) …………………. 3.1) becomes k f Am s(t ) Ac cos (2f c t sin 2f m t Ac cos(2f c t sin 2f m t …………… (3. Connect the oscilloscope to the output of the modulator (FM O/P) and observe the output signal and take at least three measurements of frequency variations..adjust the variable resistor VR1 so that the output signal 20 KHz square wave. Repeat step 2 and step 3 for triangular input signal and square input signal and draw the Modulated signal in each case. The output after the envelope detector given as the following: Ac 2f c 2k f m(t )………………………………………………………………. Connect the oscilloscope to (FM O/P) output in CD4046 module and observe the output signal . then equation (1.4): 1.…. NBFM ( 1 ) and WBFM (all ). (3. 21 . 4.3) and fig (3.. 2.k f = sensitivity factor If we let the modulating signal be a pure sinusoid m(t ) Am cos 2f mt . Fig. 7. 6. Connect the output port (FM/OP) of the VCO MC4046 to the input port (FM I/P) of the PLLMC4046. Part Two: Modulation using LM566 Modules: 22 . Fig (3.4): FM Modulation and demodulation using PLL MC4046 module.3): FM Modulation and demodulation using MC4046 module. Adjust the free running frequency (fo) of the VCO output port TP1 to 20KHz in PLLCD4046 Module. At the Audio input port (Audio I/P) of the VCO connect the function generator choose sine wave with frequency 1 KHz and choose a suitable value of amplitude to recover the output.(3.5. Connect the oscilloscope to the output of the FM modulator (FM O/P) port then measure at least three variations in frequency and draw the FM modulated signal. 5. 2. Connect the oscilloscope to (VCO O/P) port adjust the variable resistor VR1 so that the free running frequency fo equal to 20 KHz. Connect the signal generator to Audio signal input port (Audio I/P) and select sinusoidal signal with amplitude 10vp-p and frequency 1 KHz.5): FM modulator and demodulator using LM566 module. At the Audio input port (Audio I/P) Connect the function generator and select sinusoidal signal with maximum amplitude and 1 KHz frequency. Connect the oscilloscope to (FM O/P) port adjust the variable resistor VR1 so that the frequency of the (FM O/P) output equal to 20 KHz 3. the desired bandwidth for the square modulating signal.1. 4. Fig (3. Set J3 short circuit and J1 and J2 open circuit to choose C5=10nf in LM565 module. Part Four: Voltage and Frequency conversion using LM565 module: 23 . Connect the oscilloscope to the output of the FM detector (Audio O/P) and draw the output. the modulation coefficient. Connect the output port of (FM O/P) of the VCO LM566 module to the input port (FM I/P) of the PLL LM565. 3. Set J1 and J3 short circuit and J2 open circuit the selected capacitor C4=10nf in the FM modulator LM566 Module 2. 4. Repeat step 3 and step 4 for triangular and square signal and calculate the maximum frequency deviation. 5. Part Three: Modulation and Demodulation using PLL LM565 Module: 1. 5KHz 2KHz 2. Set J2 short circuit and J3 and J1 open circuit C2=100nf 2. Set J3 short circuit and J1 and J2 open circuit C5=10nf 7.5KHz 1KHz 1. 24 .5KHz 4KHz 0. 6.5KHz 3KHz 3.6): Voltage and frequency conversion using LM565 module. Then change the input frequency as shown in the table (2-2) and measure the amplitude of (Audio O/P) at each frequency. Draw the frequency vs the voltage output (Characteristic of the VCO).5KHz Amplitude Table (3-1) Fig (3. Then change the input frequency as shown in the table (2-1) and measure the amplitude of (Audio O/P) at each frequency. Adjust the variable VR1 so that the free running frequency fo of the (VCO O/P) equal 20 KHz. Frequency 0. Adjust the variable VR1 so that the free running frequency fo of the (VCO O/P) equal 2 KHz. At the demodulated FM input port (FM I/P) connect the signal generator and choose square wave with 5vp-p and 20 KHz frequency. 5. Set J1 open circuit this means that SW1 is open 4. At the demodulated FM input port (FM I/P) connect the signal generator and choose square wave with 5vp-p and 2 KHz frequency.1. 3. 8. Set J1 open circuit this means that SW1 is open 9. 10. 25 . Frequency 8KHz 12KHz 14KHz 15KHz 18KHz 22KHz 23KHz 24KHz 25KHz Amplitude Table (3-2) Questions: Question#1: 1) What are the significant frequencies and the power in each harmonies of a FM modulated wave for: β=0.5(refer to Bessel function table) and for: β=2(refer to Bessel function table) When the carrier wave is: Vc(t) =8 cos(2π50000t) What is the 99% power bandwidth for the each β? Refer to the table of Bessel functions. Draw the frequency Vs the voltage output (Characteristic of the VCO). J0 (β) power. c. Carrier wave in fc frequency and Vc . β appears in Brackets because the Bessel functions are dependent on β. b. Waves in f c ± n· f m frequencies and Vc . Jn (β) are Bessel functions of the first kind of order n and argument β. Jn (β)) power for n>0. 26 .Bessel function table Hint to solve the question: a. we utilize the simplest technique which is the AM modulation to implement the modulator. FDM is used to transmit multiple signals over the same communication channel simultaneously. The conversion of the frequency is controlled by the carrier signal.EXPERIMENT. There are two types of signal division Time Division Multiplexing (TDM) and Frequency Division Multiplexing (FDM). then the usage of the channel is very low and the efficiency is also not good. To design and implement the FDM multiplexer and Demultiplexer. Connection wires. Then the audio signals will be sent into the modulator so that the frequency range of the signals will shift to different region. However. Theory: If the transmission channel consists only of one modulated signal. DC Power Supply.1) assumes that all the input audio signals are low pass pattern and after each input signal. the channel must be able to transmit multiple signals. As you know the frequency range of the sound is 300Hz to 3 KHz so in order to transmit this kind of signal via a single channel. Then the modulated signal will pass through a band pass filter which can limit the signal bandwidth to prevent the interference between each signal. Finally. there will be a low pass filter to remove all the unwanted signals except the audio signals. unlike TDM. Figure (4. Like TDM. FDM does not use pulse modulation.4 FDM Multiplexer and Demultiplexer Objectives: To understand the operation theory of frequency Division Multiplexing FDM and Demultiplexing. Equipment Required: ACS11-1 and ACS12-1 of ETEK ACS-3000-06 module.1) is the system block diagram of FDM. 27 . As compare to TDM. FDM Multiplexing: Figure (4. in order to comfort with the economic benefit. Therefore. the signals will be added by a linear adder. Therefore. we utilize AM modulation to implement FDM system and sampling to implement TDM system. such as in the telephone system. we must divide the signal into several slots to prevent the interference then we can obtain the signal at the receiver. the output from each balanced modulator is a DSBSC signal. As you know.1):Block diagram of FDM Multiplexer.2):Circuit diagram of DSB-SC modulation by utilizing MC1496.(4. Then. we build each balanced modulator by utilizing MC1496 and use different carriers for each modulator.(4. the DSB-SC signals will be added by a linear adder in order to produce the FDM signal. 28 . In this experiment. Fig.Fig. Fig.5) shows the second way to implement the FDM demultiplexer which is called synchronous product detection. this filter will remove the signal which its frequency is larger and lower than f0 and only left a single DSB-SC modulated signal. After the signal passes through the synchronous product detector. we will add a LPF to remove all the unwanted signals and recover the original audio signal.Fig.(4. 29 . After that.3):Circuit diagram of the linear adder.4):Block diagram of FDM demultiplexer (first method).(4. Let the FDM signals pass through a band pass filter. FDM Demultiplexing: There are two ways to implement FDM demultiplexer.4). The first way is shown in figure (4. While Figure (4. this signal will pass through the LPF which recover the modulated signal and obtain the original audio signal. Fig. Fig.(4.5):Block diagram of synchronous product detector.7):Circuit diagram of the LPF.(4.Fig.6):Circuit diagram of synchronous product detector. 30 .(4. Adjust the variable resistor “Modulator Adjust 3” so that the output is DSB-SC modulated signal. 8.Using the oscilloscope to observe output signal of the balanced modulator 3(TP9). 5. 7. 6. Adjust the variable resistor “Modulator Adjust 2” so that the output is DSB-SC modulated signal.Using the oscilloscope to observe the carrier signal from the carrier signal generator 1 output (TP2) . 2.Refer to the audio signal generator in ACS11-1 of ETEK ACS-3000-06 module.Adjust the variable resistor “Carrier Gain Adjust3” so that the output amplitude of the carrier is 620mV.Adjust the variable resistor “Modulator Adjust 1” so that the output is DSB-SC modulated signal. 31 .Using the oscilloscope to observe output signal of the balanced modulator 1(TP5) .Using the oscilloscope to observe the audio signal from the signal generator 2 output (TP3) .Adjust the variable resistors “Audio Frequency Adjust3” and “Audio Gain Adjust3” to obtain an output audio signal with 1.Using the oscilloscope to observe the audio signal from the signal generator 1 output (TP1) .Using the oscilloscope to observe output signal of the balanced modulator 2(TP6).Adjust the variable resistor “Carrier Gain Adjust2” so that the output amplitude of the carrier is 620mV.2 kHz frequency and 620mV amplitude.Using the oscilloscope to observe the audio signal from the signal generator 3 output (TP7) . 3.Adjust the variable resistors “Audio Frequency Adjust1” and “Audio Gain Adjust1” to obtain an output audio signal with 500 Hz frequency and 620mV amplitude. 10.Procedure: FDM Multiplexing: 1.Using the oscilloscope to observe the carrier signal from the carrier signal generator 2 output (TP4) .Adjust the variable resistor “Carrier Gain Adjust1” so that the output amplitude of the carrier is 620mV.Using the oscilloscope to observe the carrier signal from the carrier signal generator 3 output (TP8) .Adjust the variable resistors “Audio Frequency Adjust2” and “Audio Gain Adjust2” to obtain an output audio signal with 800 Hz frequency and 620mV amplitude. 11. 4. 9.Refer to the carrier signal generator in ACS11-1 of ETEK ACS-3000-06 module. 5)) and the low pass filter (shown in fig.(4.12. 4.(4.To implement a product detector (shown in fig. Record your results.Using the oscilloscope to observe output signal waveform of FDM output port (FDM O/P). 2. “Gain Adjust 2”. FDM Demultiplexing: 1. “Carrier Adjust 3” and “Gain Adjust 3” so that the output waveforms are maximum without distortion.Connect (FDM O/P) in ACS11-1 to (FDM I/P) in ACS12-1. refer to figure ACS12-1 on ETEK ACS-3000-06 module. (Audio O/P2) and (Audio O/P3). 32 . then adjust the variable resistors “Carrier Adjust 1”. “Gain Adjust 1”. (TP4) to (Carrier I/P2) and (TP8) to (Carrier I/P3).Using oscilloscope to observe the output signal waveforms of (Audio O/P1).6)). “Carrier Adjust 2”.Connect the carrier signal (TP2) in ACS11-1 to (Carrier I/P1) in ACS12-1. 3. and this binary code is corresponding with the mid-value. then all of the errors comprise the error value of ADC. Equipment Required: ETEK-DCS-3000-07 module Signal Generator Oscilloscope DC-Power supply Theory: Figure 5-1 is the characteristic curve of an ideal 3-bit analog to digital converter. the control 33 . therefore. and the analog input range is from 0 V to 5 V. successive approximation ADC. This is because the ± 1/2 LSB becomes small. In this chapter. sample-and-hold. during the processing of converter. only the successive approximation ADC is discussed.5 Analog to Digital Conversion (ADC) Objectives: To understand the operation theory of analog to digital converter. the quantization error will reduce. therefore. flash ADC and tracking ADC. the required input voltage value also changes The methods of conversion for analog to digital converter are various. S&H circuit will capture the input signal Vin to avoid normally can be divided as AID conversion methods are digital-ramp ADC. and also includes the previous converter that has the analog error. which is provided with 8-bit resolution. at each range all the analog values use the same binary code to represent. One of the methods to reduce the quantization error is to increase the number of bits of the converter. we will discuss on the operation theory of successive approximation ADC. Quantization value (Q) means when the digital output changes LSB. Therefore. The more the numbers of bits. Figure 5-2 is the block diagram of successive approximation ADC. sample-and-hold. it consists of ± 1/2 least significant bit (LSB) quantization uncertainty or quantization error. We can divide the input signal into 8 (2^3 = 8) ranges. At this moment . To implement the analog to digital converter by using ADC0804 and ADC0809. the more the numbers of ranges and the data signal will be more detail.EXPERIMENT. When we input the analog signal. S&H circuit will capture the input signal Yin to avoid any signal change during con version period. then D0 to D7 remains at " 1 ". make second bit D6 as " 1 ". If the input voltage Vin is higher than V( D). follow by the most significant bit. MSB D7 is set to " 1 ". after 34 . otherwise alters to " 0 ".logic will store all the bit s and reset to" 0 ". Next. Thus. the output voltage of DAC is: Fig(5-1) Fig(5-2) This voltage is half of the reference voltage Vref. ) is connected to ground. if Yin is higher than V(D). 00000000 (OOH) represents 0. ordinarily used single input terminal and Vin ( . The analog input voltage range is from 0 V to 5 V with single 5 V power supply. then D0 to D7 are in floating condition. the frequency range starts from 100 kHz to 800 kHz. If any pins of CS and RD are high.01953 V. one is analog ground (A GND) and another one is digital ground (D GND). ADC0804 Analog to Digital Converter ADC0804 is a 20-pin DIP package with an 8-bit resolution single channel IC. ADC will start the conversion. when CS and WR are Low. INTR will alter to low. then D6 remains at " 1 " otherwise alters to " 0 ". Similarly for the others until the comparison of D0 to D7 have been completed. then we can generate the ADC operating time. If we add a resistor and capacitor at CLK R (pin 19) and CLK IN (pin 4). then we can obtain the complete D7 to D0 digital output. In figure 5-3. As a result of this IC contains of 8-bit resolution.passing through a DAC then obtain an output voltage V(D). Fig(5-3) 35 . The unadjusted error of ADC0804 is ±1 LSB. each step will be 5/256 = 0. then the 1/2 reference voltage equals to power supply voltage Vee. which is 0. Pin6 Vin (+) and pin7 Vin (-) are differential analog signal inputs.01953 V . CLK IN (Pin 4) is the clock input . INTR is at high level and then after the conversion completed.9805 V. if pin 9 is floating . 15 mW power consumption and 100 us conversion time . the D0 to D 7 of ADC0804 is the 8-bit output pins . offset error and non-linearity error. so it has 2^8=256 quantization steps. WR is the write control signal. when CS and RD are low. if the reference voltage is 5 V . which includes full-scale error. Figure 5-3 shows the pins diagram of ADC0804.00 V and 1111111 (FFH) represents 4. the digital data will be sent to the output pins. During the conversion period. ADC0804 will do the clear action. Pin 9 (Vref /2) is 1/2 of the reference voltage. at this moment comparing the new V(D) and Vin. when WR backs to high. ADC0804 has a built-in Schmitt trigger as shown in figure 5-4. ADC0804 has two ground terminals. and record the measured results in table 5-1. Adjust VR2 so that the input voltage of the analog signal input port (TP3) is 0 V. CS and RD are short circuit. At this moment. 36 . 3. Adjust VR1 so that the voltage of TP1 is 2. ADC0804 analog voltage input range is 0 V to 5 V. Use the digital voltage meter to measure the reference voltage input port (TP1). the Vin ( -) is short circuit. C1 and R3 is used to control the clock of the circuit. Fig(5-4) Procedure: ADC0804 analog to digital converter 1. then let WR and INTR connect to SW1 in order to simulate the control signal. R2 and VR1. Vref / 2 is provided by R1 . Refer to the circuit diagram in figure 5-4 on ETEK ACS-3000-07 module. 2. We can determine the clock signal by the external R and C via pin 4 and pin 19. Figure 5-5 is the circuit diagram of ADC0804 analog to digital converter. By using oscilloscope. so that the IC is enable.Therefore. the analog signal input range is controlled by VR2 and input through the Vin ( +) terminal and at the same time. we need not input an external clock signal to CLK IN terminal .5 V. Set J1 be open circuit. observe on the TP2. 4. 5. Set J1 be short circuit to maintain the output digital signal. Observe on the changes of LED, LED "on" represents " I", LED "off' represents "0". ADC0809 analog to digital converter 1. Refer to the circuit diagram in figure 5- 5 on ETEK ACS-3000-07 module. 2. At the CLK input port (CLK I/P), input 120 kHz frequency and a TTL signal with 5 V offset. 3. Let SW3, SW2 and SW1 switch to GND (push down the slide switch), at this moment, the multiplexer selects to channel 0 and the analog signal is inputted from the IN0 input. 4. Use the digital voltage meter to measure the TP1 of channel 0. Adjust VR1 so that the input voltage of TP1 is similar to the values in table 5-2. Observe on the changes of LED, LED "on" represents “1’’. LED "off' represents "0", then record the measured result s in table 5-2. 5. Use the digital voltage meter to measure the TP2 of channel 1 until TP7 of channel 6, and then record the measured results in table 5-3. Fig(5-5) 37 Table(5-1) Table(5-2) 38 Table(5-3) 39 EXPERIMENT.6 Digital to Analog Conversion (DAC) Objectives: To understand the operation theory of digital to Analog converter. To implement the digital to analogue converter by using ADC0804 and ADC0809. Equipment Required: ETEK-DCS-3000-07 module Signal Generator Oscilloscope DC-Power supply Theory: Digital to analog converter (DAC) is a device, which converts the digital signal to analog signal. We normally store a digital signal in a media or transmission line. Then a DAC changes the digital signal to an analog signal in order to control data display or further analog signal processing. For example, from a digital communication system, when a receiver receives the digital modulation signal , then after via a demodulator and decoder, we can obtain the digital signal, and follow by using DAC to convert this digital signal to the analog signal. Next we will discuss the basic operation theory of DAC. Basically, DAC is a digital code that represents digital value converted to analog voltage or current. Figure 6-1(a) is a genera 14-bit DAC binary codes , the digit al input terminal [D3 D2 D1 D0] are manipulated by the register in a digital system . The 4-bit code represent s 2^4= 16 groups of 2 . binary state value, as shown in figure 6-1(b). For every binary code input, DAC will output a voltage (Vout), which is double or other order of the binary value. According to this, analog output voltage Vout and the digital input binary values are the equivalent. If the DAC output is current, l out, the theory is similarly. Figure 6-2 is the basic block diagram of DAC. The reference voltage (Vref) is used to provide the reference voltage during conversion, Then due to the magnitude of the input binary code , the digital control switch will output different binary codes to the resistors network. Normally, the DAC analog output is represented by current, if we want to obtain the voltage output, we need to connect an operational amplifier, which can convert the current to voltage level. 40 Fig(6-1a) Fig(6-1b) Fig(6-2) Resolution and Step Size: The resolution of DAC illustrates that when the digital input terminal changes a unit. Refer to 41 . which is normally the LSB levels. it will produce a small change at the analog output terminal. The output waveform of DAC is every step with 1V change. We call this situation as full-scale output. Figure 6-3 shows 16 types digital inputs corresponding to the 16 levels of output steps waveform. when the digital input value changes a unit Vout will change at least 1 V. which is 15 V.figure 6-1(b). there are only 15 steps size. When the counter generates 0000. Resolution is also called step size because Vout will change. the DAC output is the maximum value. so the resolution is 1V. Fig(6-3) DAC 0800 Digital to Analog Converter: DAC 0800 is a cheap and commonly used 8-bit DAC. From 0 V to 15 V (full-scale) . Generally. R-2R ladder resistors network and transistor switch. When the counter generates 1111. Figure 6-3 shows a 4-bit binary counter as DAC digital input signal. For example. Figure 6-4 is the pins diagram 42 . the DAC output is 0 v. under the ± 5 V condition. when the digital input step varies from one to another. if the step size is 1 V then the difference between the steps is 1V. the counter has a clock input. so it can output 16 types of statuses continuously in cycle . N bits of DAC will produce 2N different levels and 2N-1 steps size. the power loss is approximately 33mWand the settling time is approximately 85 ns.5 V to ± 18 V. The voltage power supply range is between ± 4. Resolution or step size is to indicate the difference between two steps. the internal circuit consists of reference voltage power supply. of DAC0800. Figure 6-5 is the circuit diagram of DAC0800 single polarity voltage output. The negative reference voltage is GND and passes through R2 to connect to Vref (-) (pin 15). which D7 ~ Do are the 8-bit digital input s. The positive reference voltage is + 5V and passes through R1 to connect to Vref(+) (pin14). the output current as in equation(6-2) below I out is: Fig(6-4) 43 . The reference current Iref that passes through R1 can be expressed as Iref in the following equation (6-1): At the current output terminal (pin4). TP4. Refer to the circuit diagram in figure 6-7 or figure ACS 14-2 on ETEK ACS-3000-07 module. 44 . SW3 and SW4 in table 6-l. record the measured results in table 6-1. SW2. TP2 .Fig(6-5) Procedure: Part1: R-2 R network DAC: 1. According to the switching of SW 1. 2. TP5 of R-2Rnetwork and output port of D/A converter (Vout) . TP3. By using voltage meter to measure TP1 . SW2. Let SW I. 3. Fig(6-6) Part 2: DAC0800 unipolar voltage output 1. " 1" represents as "+5 V"). Refer to the circuit diagram in figure 6-6(b) or figure ACS14-1 on ETEK ACS-3000-07 module. SW3 and SW4 switch to 1 ("0" represents as GND. 4. Using equation (6-2) and equation Vout=I out *R3 to calculate the theoretical values of the output current I out and output voltage Vout (Rr = 4. Then record the measured results in table 6-2. In table 6-3. 6. the binary values are used as the digital inputs. which 0 represents GND And 1 represents +5 V. Calculate the step size and record the calculation in table 6-2. J3 be open circuit. 5. I out = Ifs) 45 . Using digital voltage meter to measure the output voltage (O/P). J2 and J3 be short circuit. the binary values are used as the digital inputs. In table 6-2. then record in calculation in table 6-2. 4. Remove the current meter and let J1 be short circuit. then substitute I out and Ifs into equation Vout=2*I out*R4. Fig(6-7) Part three: DAC0800 bipolar voltage output 1. Calculate the step values and record the calculation in table 6-3.7 k). (note: when D0 to D7 is 1. Using equation (6-2) to calculate I out and Ifs . which" 0 “represents GND. Set Jl . Refer to the circuit diagram in figure 6-7 or figure ACS 14-2 on ETEK ACS-3000-07 module. finally record the measured results in table 6-3 .Ifs*R4 Find the theoretical value of output voltage Vout. 3. Finally record the measured results in table 6-2.” 1 “represents +5V. Set J1 and J2 be short circuit. then connect the digital current meter to 11 for measuring the output current Iout. Let J1 be open circuit. 3. 2.2. 7. J1 be short circuit. Using digital voltage meter to measure the output voltage Vout .5. Table(6-1) 46 . Let J1 and J3 be open circuit. then record the measured results in table 6-3 . Finally record the measured results in table 6-3. 6. J2 be short circuit. Connect the digital current meter to J1. Let J1 and J2 be short circuit. J3 be open circuit. Let J2 and 13 be open circuit. Finally record the measured results in table 6-3 . Calculate I out + I~out and record the measured results in table 6-3. then measure the output current Iout . Connect the digital current meter on J2 to measure the output current Iout. 8. Table(6-2) 47 . Table(6-3) 48 . 49 . Find the quantized values of x(t) for 0 t 1 d. To consider reasons for using digital signal transmission of analog signals. a. one that assigns 000 to the first level and 111 to the eight levels).the quantized samples are then applied to a natural binary encoder. Plot x(t) for 0 t 1 b. Find the values of the sampled signal over 0 t 1 c.EXPERIMENT.e.7 Pulse Code Modulation (PCM) Objectives: T o study the operation of a PCM encoder. +1). Find the sequence of binary digits observed at the encoder output for 0 t 1 Theory: Pulse-code modulation (PCM) is a digital representation of an analog signal where the magnitude of the signal is sampled regularly at uniform intervals. then quantized to a series of symbols in a digital (usually binary) code. Equipment Required: ETEK-DCS-6000-03 module Signal Generator Oscilloscope DC-Power supply Prelab works: Consider the sinusoidal test signal x(t ) cos(2t ) this signal is applied to a sampler operating at 10 sample per second followed by a 8 level quantizer with a range of (-1. (i. To study the operation of a PCM decoder. PCM has been used in digital telephone systems and is also the standard form for digital audio in computers. As a result of these instantaneous transitions. provided a sampling frequency that is sufficiently greater than that of the input signal. This output would then generally be filtered and amplified. and operate similarly to ADCs. can operate without introducing significant distortions within their designed frequency bands. The electronics involved in producing an accurate analog signal from the discrete data are similar to those used for generating the digital signal. no explicit filtering is done at all.e: greater than. The sampling theorem suggests that practical PCM devices. inherent losses in the system compensate for the artifacts — or the system simply does not require much precision. They produce on their output a voltage or current (depending on type) that represents the value presented on their inputs. the discrete signal will have a significant amount of inherent high frequency energy. In some systems. the procedure of modulation is applied in reverse. the signal would be passed through analog filters that suppress artifacts outside the expected frequency range (i. There are many conditions for any PCM system: Choosing a discrete value near the analogue signal for each sample (quantization error must be minimum) Between samples no measurement of the signal is made. the next value is read and the output of the system is shifted instantaneously (in an idealized system) to the new value. the maximum resolvable frequency). as it's impossible for any system to reproduce a signal with infinite bandwidth. These devices are DACs (digital-to-analog converters).Block diagram of PCM modulator: Demodulation of PCM: To produce output from the sampled data. mostly harmonics of the sampling frequency To smooth out the signal and remove these undesirable harmonics. After each sampling period has passed. due to the sampling theorem 50 . Some systems use digital filtering to remove the lowest and largest harmonics. Q=8. Sampler should operate at a rate of fs 2w Absolute error less than where is the step of the quantizer. 8 1 8 51 . If 2Ais the peak to peak variation of the message signal and Q is the number of quantization levels. then 2A Q where Q 2 n The above fig represent a uniform quantizer with A=4v. and n is the number of 2 bits per sample. then Draw the output of each terminal and determine the amplitude and the frequency of each output. Fig (7. 2.Procedure: Consider the circuit diagram in figure 7-1 in ETEK DCS-6000-03 module. Set J1 of DCS6-1 short circuit and connects the output terminal (PCMO/P) of modulated PCM signal of DCS5-1 to the input terminal (PCM I/P) of demodulation PCM signal of DCS6. Set J1 short circuit and from the signal input terminal (I/P). Connect the oscilloscope to observe on the output terminal of low-pass filter (Tl).1.1) 1. connect the signal generator and Set the amplitude of the sinusoidal signal 5 vp-p and 500Hz frequency. feedback point of output signal (T3) and output signal Terminal of PCM (OP). After that connect the output terminal (T4) with2048 kHz square wave to the CH1 of the oscilloscope and output terminal (T6) of modulated signal to CH2 of the oscilloscope. Consider the circuit diagram in figure 7-2 in ETEK DCS-6000-03 module. By 52 .2) 3. input terminal of audio signal (T2). Fig (7. 2048 kHz square wave generator (T2). 53 . feedback point of output signal (T3) and output signal Terminal of PCM (OP). Set J2 short circuit and J1 open and from the signal input terminal (I/P). observe on the output terminal of buffer (TI). Connect the oscilloscope to observe on the output terminal of low-pass filter (Tl). Set J2 of DCS6-1 short circuit and J1 open and connects the output terminal (PCM O/P) of modulated PCM signal of DCS5-1 to the input terminal (PCM I/P) of demodulation PCM signal of DCS6. then record the Measured results and draw each output and determine the frequency and the amplitude for each output. demodulated PCM signal output Terminal (T4) and signal output terminal (Audio O/P). Repeat step (5-8) when the frequency of the function generator change to 1 KHz. 9. connect the signal generator and Set the amplitude of the sinusoidal signal 5 vp-p and 500Hz frequency. 2048 kHz square wave generator(T2). 7. connect the oscilloscope. observe on the output terminal of buffer (TI).using oscilloscope. 8 kHz square wave generator (T3). then Draw the output of each terminal and determine the amplitude and the frequency of each output. 5. input terminal of audio signal (T2). After that connect the output terminal (T4) with 2048 kHz square wave to the CH1 of the oscilloscope and output terminal (T6) of modulated signal to CH2 of the oscilloscope. 4. 8 kHz square wave generator (T3). 8. Repeat step (1-3) when the frequency of the function generator change to 1 KHz.1. then record the measured results and draw each output and determine the frequency and the amplitude for each output. 6. demodulated PCM signal output Terminal (T4) and signal output terminal (Audio O/P). DC Power Supply. these slots are also called as sampling values. Time division indicates the signal is divided into several slots in time domain.EXPERIMENT.8 TDM Multiplexer and Demultiplexer Objectives: To understand the operation theory of Time Division Multiplexing TDM and Demultiplexing. Equipment Required: ACS9-1 and ACS10-1 of ETEK ACS-3000-05 module.2 below: Fig(8. Therefore. The basic structure of TDM system is shown in figures 8. then this method can achieve the function of multiplexing.1): TDM with two signals m1(t) and m2(t). Theory: Time Division Multiplexing TDM: TDM means multiple signals can be transmitted over the same transmission channel. Connection wires.1 and 8. and then these slots will transmitted to the receiver by following a fixed time slots. If the fixed time slot is large enough for other sampling value of other signal to fill in. 54 . To design and implement the TDM multiplexer and Demultiplexer. 3) which can generate a fixed timing as the switching circuit. in this experiment we utilize sinusoidal. square and triangle waves as the several groups of signals to achieve the TDM modulation.Fig(8. Fig(8.3): Circuit diagram of time generator.Therefore. The implementation of the TDM Multiplexer: As a result of TDM uses the same channel to transmit several group of signals . 55 .2): Circuit structure of TDM system. Since the TDM uses the time slots to transmit signal so we need to produce a time generator circuit (shown in fig 8. Fig(8.5): Circuit diagram of TDM multiplexer. On the other hand. we can utilize the pulse at a certain period to process different number of channels. The example of a simple TDM system is shown below in fig (8.5) shows the circuit diagram of TDM multiplexer. the receiver can also separate the signals of different channels accurately. TDM Demultiplexing: After we divided the time of transmission channel into several time slots.6). When t1 is high the triangular wave will occur at the output port while when t2 is high then the square wave will be at the output port and when t3 is high the output will be the sinusoidal wave. Therefore. there will be a small gap between each time slots which is known as guard time and it is used to prevent the interference between the symbol and jitter of the multiplexer. 56 . Fig(8. The fig (8. according to the synchronous signal at the transmitter.4): Time sequence of time generator. The most important thing is the synchronization between both time generators in the transmitter and receiver so the system will be able to recover the original signal. Fig (8.Fig(8.7): Circuit diagram of TDM demultiplexer.6): Transmission diagram of TDM system. Procedure: TDM Multiplexing: Refer to figure ACS9-1 of ETEK ACS-3000-05 module. Fig(8. The implementation of the TDM Demultiplexer: Fig (8. then we can obtain the input sequences which are the triangle.3) is also used as a synchronous signal generator for Demultiplexing.7) is the circuit diagram of TDM demultiplexer. When TDM signal inputs by matching with the synchronous signal generator. square and sinusoidal waveforms. 57 . 13.Observe again (TP4) and the output signal of square wave output port (O/P2) of the TDM demultiplexer. Record your result. Record your result. 4.Use CH2 to observe TDM output port (TDM O/P). By using CH1 of the oscilloscope.Observe the output signal of (TP1) of TDM demultiplexer.By using oscilloscope. 7. 58 . adjust the variable resistor VR1 so that the amplitude of TP2 is maximum without distortion.By using CH1 of the oscilloscope.Connect the sine wave output port (TP6) of the TDM multiplexer to the sine wave input port (TP4) of the TDM demultiplexer.Use CH2 to observe TDM output port (TDM O/P).Connect the triangular wave output port (TP4) of TDM multiplexer to (TP2) of TDM demultiplexer. 9. 8. Record your result. 2. 1.Again use CH1 and CH2 of the oscilloscope to observe (TP2) and the square wave output port (O/P2) of the TDM demultiplexer.Use the oscilloscope to observe (TP4) and the output signal of triangular wave output port (O/P1) of the TDM demultiplexer. 11.Use CH2 to observe TDM output port (TDM O/P).By using CH1 of the oscilloscope. Record your result. 12.Observe again (TP3) and the output signal of sine wave output port (O/P3) of the TDM demultiplexer. at this moment the counter of the clock is slow. observe the output signal of the square wave at port (TP5) . 2. observe the output signal of the sine wave at port (TP6) .Using the oscilloscope to observe the signals at (TP2) and (O/P1) of the TDM demultiplexer. Record your result.Turn the variable resistor “Clock Adj.Observe again (TP3) and the output signal of square wave output port (O/P2) of the TDM demultiplexer.Also observe both (TP2) and the sine wave output port (O/P3) of the TDM demultiplexer.Connect the square wave output port (TP5) of the TDM multiplexer to the square wave input port (TP3) of the TDM demultiplexer. adjust the variable resistor VR3 so that the amplitude of TP3 is maximum without distortion. observe the output signal of triangular wave output port (TP1).Connect the output port (TDM O/P) of TDM multiplexer in ACS9-1 to the input port (TDM I/P) of TDM demultiplexer in ACS10-1. 10.” left to the end. TDM Demultiplexing: Refer to figure ACS10-1 of ETEK ACS-3000-05 module. adjust the variable resistor VR3 so that the amplitude of TP1 is maximum without distortion. Record your result.Use the oscilloscope to observe (TP3) and the output signal of triangular wave output port (O/P1) of the TDM demultiplexer.1. Record your result. 6.Observe the output signal of square wave output port (TP2). 3. 6. 5. 5. 3. observe the output signal of the triangular wave at port (TP4) . 4.Observe the output signal of sine wave output port (TP3). 59 .Observe again (TP4) and the output signal of sine wave output port (O/P3) of the TDM demultiplexer.14. EXPERIMENT.9 Amplitude Shift Keying (ASK) Objectives: Creation of ASK an modulated signal Detection of the modulated signal using envelope detector Detection of the modulated signal using bandpass filter followed by an envelope detector Equipment Required: TPS3-3431 Power supply Banana wires Theory: The general block diagram that represent the generation of ASK signal given bellow The bit stream consists of a sequence of binary digits as demonstrated below for the sequence (10110100): The data after converting it to uni-polar non-return to zero [u-NRZ] is as the following: 60 . The data after multiplication with a carrier [vcos(2pifct)] is an ASK signal as the following: The detection of an ASK signal is done using envelope detection as the follows: Refer to envelope detection of AM signal Try to find the output after each stage of envelope detector. The output after the two Schmitt trigger must be as the following (in the case of perfect data recovery): 61 . Note it is not enough to do detection using envelope detector you must use two Schmitt trigger to convert the analogue output to digital output. Demodulation process using non-coherent demodulator: Disadvantage of ASK modulated signal 1. Usable only for Low data rate . Amplitude not constant this causes the detection process to be very difficult. Connect the data transmitter output to the ASK modulator input Set the BIN/QUAD switch to the BIN position. 2. 62 . Procedure: Connect the trainer to the power supply. Remarks: The spectrum of ASK signal resembles that of Normal AM modulation. Connect the power supply to the Mains and turn it ON. In your notebook. Connect the modulator output to the envelope detector input. Because of the low sampling rate. Move the CH2 scope probe from the modulator output to the detector's Output. Connect the CH1 scope probe to the modulator input. This frequency should be approximately 12 KHz. Connect the output of the detector to the upper Schmitt trigger amplifier 63 . In your notebook draw the shape of the signals-the modulated signal (at the detector input) and the output of the detector. You should see the transmitted data on channel CH1. Connect the CH2 scope probe to test point TP 1. Measure or calculate the frequency of the carrier wave. Move the output of the CH2 scope probe to the modulator output. Set the switches to the binary number 0101010 1 and watch the transmitted Data signals.You should see the Fl carrier wave. draw the shape of the signals .the modulator (at the Modulator input) and the signal at the modulator output. you can notice only some of the F1Cycles. Decrease the time base a little in order to see more FI cycles during Transmission. Connect the output of the ASK modulator to the input of the bandpass filter(BPF) Connect channel 2 of the scope probe to the filter input. Move channel 2 of the scope probe from the detector output to the amplifier output.. In your notebook draw the shape of the signals-the modulated signal (at the detector input) and the signal at the amplifier output. Draw the filter input Connect channel 2 of the scope probe to the output of the filter Draw the output of the filter Connect the filter output to the envelope detector input Connect the detector output to the upper Schmitt amplifier input Connect the Schmitt trigger1 output to the Schmitt trigger2 input Connect the Schmitt trigger2 output to the data receiver input 64 . 00001111. Draw the signal described in the mathematical expression . What will the shape of the wave at the next filter output be? What will the shape of the wave at the next filter output be? 65 .The carrier wave is at an intensity of 20 vpp and a frequency of 15 KHZ.00111100 Check whether the frequency of the modulating signal (in the data transmitted) has an effect on the filtration and detection signals. Repeat all the steps for different data sequence:00110011. The modulating signal is a symmetrical square wave at a frequency of 1000HZ . Make sure that the binary number indicated on the data transmitter switches appears on the lights of the data receiver. Questions: An ASK modulated signal is transmitted to an AM detector. What will the shape of the wave at the next filter output be? What is the desired time-constant of the detector for the desired signal? 66 . …. 2.10 Frequency Shift Keying (FSK) Objectives: Creation of FSK modulated signal Detection of the FSK signal using non-coherent detector Equipment Required: TPS3-3431 Power supply Banana wires Theory: In binary FSK symbols (1) and (0) are distinguished from each other by transmitting one of the Two sinusoidal signals that differ in frequency. Generation of FSK signal 1..for binary (0) s2 (t ) A cos(2f 2t …………. Direct method of generation using VCO [voltage controlled oscillator]. Coherent FSK 2.EXPERIMENT. Non-coherent FSK What is the major difference between the two types of FSK signal? The formula of FSK modulated signal s1 (t ) A cos(2f1t ………………. In-direct method of generation Two type of FSK modulated signal 1.for binary (1) 67 . Voltage control oscillator 68 . Two oscillator method 2.The general shape of the FSK modulated signal The detection of FSK modulated signal done using: BPF[band-pass filter] Envelope detector Two Schmitt trigger What is the operation of each detection parts? Direct Methods of FSK generation: 1. 69 . Connect the CH2 scope probe to test point TP 2. This frequency should be approximately 25 KHz. Measure or calculate the frequency of the carrier wave.Demodulation using non-coherent demodulator: Procedure: Connect the trainer to the power supply. Connect the CH1 scope probe to the modulator input. You should see the transmitted data on channel CH1. Connect the data transmitter output to the FSK modulator input Set the BIN/QUAD switch to the BIN position. Set the switches to the binary number 01010101 and watch the transmitted Data signals. You should see the F2 carrier wave. Move the output of the CH2 scope probe to the modulator output. Connect the power supply to the Mains and turn it ON. Connect the modulator output to the Bandpass filter input(BPF). 70 .the modulator (at the Modulator input) and the signal at the modulator output. Repeat all the steps for different data sequence:00110011. In your notebook. draw the shape of the signals . Draw the filter input Connect channel 2 of the scope probe to the output of the filter Draw the output of the filter Connect the filter output to the envelope detector input Connect the detector output to the upper Schmitt amplifier input Connect the Schmitt trigger1 output to the Schmitt trigger2 input Connect the Schmitt trigger2 output to the data receiver input Make sure that the binary number indicated on the data transmitter switches appears on the lights of the data receiver.00001111.00111100 Check whether the frequency of the modulating signal (in the data transmitted) has an effect on the filtration and detection signals. Connect channel 2 of the scope probe to the filter input. . 71 .11 Binary Phase Shift Keying (BPSK) Objectives: Creation of a BPSK modulated signal Detection of the signal using a BPSK detector Equipment Required: TPS3-3431 Power supply Banana wires Theory: General block diagram for BPSK generation: Prove that the output of the following diagram is given by: s1 (t ) A cos(2f c t 0 ……………….for binary (1) The following fig represents the BPSK modulated signal.EXPERIMENT.for binary (0) s2 (t ) A cos(2f c t ……………….... XOR-gate 2.Note that bits 1 and 0 have the same amplitude and the same frequency but they differ in phase. Remarks: The spectrum of the binary BSK resembles that of double sideband suppressed carrier modulation. The following table represents Null to Null band width for each digital modulation Digital modulation technique Null to Null band width ASK 2rb BPSK 2rb FSK 3rb QPSK 2rs= rb technique: 72 . LPF [low pass filter] Exercise: draw the block diagram and explain its operation Detection of BPSK using coherent demodulator: Exercise: Analyze the operation of each block in the above block diagram. BPSK Detection: The detection of BPSK is achieved by: 1. 73 . draw the shape of the signals .Where: rb: represent the data rate rs: symbol rate (One symbol consists of two binary digits) Procedure: Connect the trainer to the power supply. Connect the data transmitter output to the modulator input Set the BIN/QUAD switch to the BIN position. Connect the BPSK modulator output to the BPSK detector input. Connect the power supply to the Mains and turn it ON. You should see the carrier wave F1 in the phase shift 180 Move the output of the CH2 scope probe to the modulator output and return probe of CH1 to the modulator input.the modulator (at the Modulator input) and the signal at the modulator output.You should see the F1 carrier wave. Connect the BPSK detector output to the data receiver input. You should see the transmitted data on channel CH1. Connect the CH2 scope probe to test point TP4. Connect the CH1 scope probe to the modulator input. Set the switches to the binary number 01010101 and watch the transmitted Data signals. Connect the CH1 scope probe to test point TP1. In your notebook. 00001111.00111100 Check whether the frequency of the modulating signal (in the data transmitted) has an effect on the filtration and detection signals. Make sure that the binary number indicated on the data transmitter switches appears on the lights of the data receiver. Repeat all the steps for different data sequence:00110011. 74 . the odd-numbered bits have been assigned to the in-phase component and the evennumbered bits to the quadrature component (taking the first bit as number 1).120.90.60. The two carrier waves are a cosine wave and a sine wave.270]. Equipment Required: TPS-3431 Power supply Banana wires Theory: Quadri-means four messages. Because we use four messages then we need at least two bits to represent each message.180. QPSK signal in the time domain The modulated signal is shown below for a short segment of a random binary data-stream.180] or full cycle[0.12 Quadri Phase Shift Keying (QPSK) Objectives: Creation of a QPSK modulated wave. PSK-means the variation happens on phase only. Here. The total signals 75 . as indicated by the signal-space analysis above.EXPERIMENT. Detection of a signal using a QPSK detector. M1 00 M2 01 M3 10 M4 11 The phase varied along half the cycle[0. the sum of the two components. The general block diagram of the modulator: Generation of QPSK signal using two Where: Basis function1= 2 cos(2f c t ) Ts 76 . The topmost waveform alone matches the description given for BPSK as shown below. Basis function2= 2 sin( 2f c t ) Ts Detection of QPSK using the above generator: Space representation of QPSK: General block diagram of the QPSK receiver: 77 . 78 . Connect the scope probe to test point TP5. Draw the signals.You should see the carrier wave F1 with a phase shift of 120°. Draw the signals. It is very difficult to see the signals in an ordinary scope because the signals are not stable due to the phase changes. Draw the signals.You should see the carrier wave Fl. Connect the scope probe to test point TP4. Set the BIN/QUAD switch to the QUAD position. F1 180° and Sout Set the switches to the binary numbers 00011011. F1 120°. You should see the transmitted data on channel CH2 according to the clock pulses. which appear on the scope's screen. respectively. Connect the CH2 scope probe to test point TP6. Connect the CH2 scope probe to the Din0 modulator input. Connect the data transmitter Dout0 and Doutl outputs to the Din0 and Din1 Modulator inputs.You should see the carrier wave Fl with a phase shift of 180°.You should see the carrier wave F I with a phase shift of -60° (300°). Draw the shape of signals F1 0°. Connect the power supply to the Mains and turn it ON. Fl 60°. Connect the CH1 scope probe to test point TP3 (the transmitter clock). which appear on the scope's screen. which appear on the scope's screen. Connect the CH1 scope probe to test point TP1.Procedure: Connect the trainer to the power supply. Connect the QPSK modulator output to the QPSK detector input. Connect the CH2 scope probe to the Din1 receiver input. You should see the received data on channel CH2 according to the clock pulses. You should see the received data on channel CH2 according to the clock pulses. Connect the CH1 scope probe to test point TP9 (the receiver clock). It is very difficult to see the phase changes because the data rate is much lower than the signal frequency. Draw on a graph the clock pulses. Check that the binary number indicated by the switches on the data Transmitter appears in the lights on the data receiver. the Dout0 and the Doutl channels Signals. Connect the detector outputs to the corresponding data receiver inputs. 79 . Repeat the above steps for the binary numbers: 00110011 and 11100100. Connect the CH2 scope probe to the Din0 receiver input. Connect the CH2 scope probe to the Dinl modulator input. the Dout0 and the Doutl channels Signals. You should see the transmitted data on channel CH2 according to the clock pulses. Draw on a graph the clock pulses. Move the CH2 probe output to the modulator output. EXPERIMENT. the subtraction between the low frequency signal x(t) and the signal xs(t) will produce a difference signal d(t) . this is because digital signal can recover the original signal easily by the comparator. To understand the signal waveforms of delta modulation. Furthermore.13 Delta Modulation and Detection Objectives: To understand the delta modulation. Equipment Required: ETEK-DCS-6000-04 module Signal Generator Oscilloscope DC-Power supply Theory: The Operation Theory of Delta Modulation: Delta modulation is a kind of source coding which can convert the analog signal to digital signal. Design and implementation of delta modulator. To understand the operation of the delta demodulation. filtering the unwanted signal and so on. From figure 4-1. then we can obtain a signal given as: ∆(t)={+ ∆ if ∆(t)>0} 80 . Therefore the expression of the difference signal d(t) is given as: d(t)= x(t)-xs(t) However the difference signal d(t) will be converted by a limiter. the transmission quality of digital signal is better than analog signal . which is the former sampling value. The block diagram of delta modulator is shown in figure 4-1. After that we can deal with the digital signal easily such as encoding. where Xs(t) is a reference signal. we know that if the difference signal d(t) is larger than zero. which is ∆ during the transmission. if the PCM signal after encoded is 8 bits . After that the output signal will feedback to integrator for integration and the output signal of integrator will again compare to the input signal to obtain the value of +∆ or -∆ . then we only need to transmit a variation value which is -∆. this means that the reference signal X s (t) is lower than the low frequency signal x(t). Figure 13-2 is the basic circuit diagram of delta modulation. which can prevent the interference from noise. if the difference signal d(t) is lower than zero. The comparator is to compare the audio signal and the output signal of integrator. which is 8 times more than the original bandwidth. As for PCM modulation. we can estimate the next sampling value base on the former sampling value. Although PCM modulation increases the quality of transmission. thus. 81 . so this situation can greatly save the transmission bandwidth. For example. the estimated value is too small and we need to increase the next estimated value by ∆on the other words. then the difference will be sampled by the D-type flip -flop and the output signal is a TTL digital signal. Thus it can be seen that every sampling value is related to the former sampling value. However delta modulation can reduce the transmission bandwidth and achieve the quality of transmission as PCM modulation. it also increases the transmission bandwidth. Then we only need to concentrate on the transmission of the estimated value and need not care about the quantification encode. this situation will increase transmission bandwidth. we only need to send a variation value. every sampling value is independently and need to be quantified for encoding. The audio signal will pass through a low -pass filter to remove the unwanted signals.∆(t)={-∆ if ∆(t)<0} Fig(13-1) Hence. then the transmission bandwidth will be BT >8fs /2= 4fs = 8 W . therefore. Nevertheless. when AB=11 . Analog switch is a structure of multiplexer. U2 is the conversion of unipolar to bipolar circuit. This is because the gain of the integrator will affect the slope of the output signal of integrator. When AB=OO. therefore. we need to convert the unipolar signal to bipolar signal. which can compare the audio signal and the output signal of integrator. then we get :Av=vo/vin=(R16/R15)/(1+S/WH) Where :WH=1/R16*C1 Fig(13-2) 82 . U3 is an inverse integrator. the signal will pass through R14 to integrator. R11 and send into integrator. we can improve the low frequency response of the integrator. R13 . this method can prevent the occurrence from slope overload. Assume that R16 and C1 are equivalent impedance. the signal will pass through R14 . then the output square wave sign al will be sampled by a D-type flip -flop and finally the output signal is the delta modulation signal.We modified the circuit diagram of delta modulation in figure 13-2 to figure 13-3. therefore. Tile purpose of the analog switch is the selection of the amplified gain of integrator. From figure 13-3 we add a multiplexer to control the gain of the integrator. Since there is no output signal from integrator by inputting the unipolar square wave signal. The expression without R16 is given as: Vo=-Vc= ∫ By adding a shunt resistor R16 between integrator U3 and capacitor C1. R12 . U1is the comparator. input a 5 Vp-p and 32 kHz TTL signal. 83 . Next at the CLK input port (I/P2). the output port of the conversion from unipolar to bipolar (TP3). To implement a delta modulator circuit as shown in figure (1) or refer to figure DeS7-1 on ETEK DCS-6000-04 module.Fig(13-3) Procedure: 1. Then observe the input signal (TP1). the output port of tunable gain (TP4). At the signal input port (I/Pl). the output port of integrator (TP5) and the output port of delta modulation signal (O/P) by using oscilloscope. Finally record the measured results. input a 5 Vp-p and 500 Hz sine wave frequency. Set J2 and J3 be short circuit the connection between Xo and X is on. Fig (1) 2. the output port of comparator (TP2). input a 5 Vp-p and 500Hz sine wave frequency. input a 5Vp-p and 64 kHz TTL signa1. To implement a delta demodulator circuit as shown in figure (2) or refer to figure DCS8-l on ETEK DCS-6000-04 module. low-pass filter (TP4) . integrator (TP5) and signal output port (O/P). 8. Next at the CLK input port (I/P2). observe on the output signal waveforms of sampling signal output port (TPl). 5. At the CLK input port (I/P2) of the delta demodulator. Change the input signal to 5 Vp-p and 2KHz frequency and same input CLK. TP5 and O/P signal. TP4. 7. unipolar-tobipolar (TP2). Set J2 and J4 be short circuit. Change the input CLK TTL signal to 5 Vp-p and 128KHz frequency and input signal 5 Vp-p and 2KHz frequency repeat steps 2-4 and record the measured results. At the CLK input port (UP2) of the delta demodulator.3. Then by using oscilloscope. record the measured results. Finally record the measured results. Connect the modulated delta signal (O/P) in figure DCS7-l to the input terminal (UP1) of the delta demodulator in figure DCS8-1. tunable gain (D). Fig (2) 4. observe on the output signal waveforms of TP1. input a 5 Vp-p and 32 kHz TTL signal. Then by using oscilloscope. Finally. the connection between XJ and X is on for both modulator and demodulator circuit. TP3. Then by using oscilloscope. repeat steps 2-4 and record the measured results. Connect the modulated delta signal (O/P) in figure DCS7-l to the input terminal (I/P l) of the delta demodulator in figure DCS8-1. 84 . TP2. input a 5 Vp-p and 64 kHz TTL signal. 10. Change the input CLK TLL signal to 5 Vp-p and 128KHz frequency and input signal 5 Vp-p and 500Hz frequency repeat steps 2-4 and record the measured results. At the signal input port (I/P l). 9. 6. TP3. TP2.5KHz frequency repeat steps 9-10 and record the measured results. Change the input CLK TTL signal to 5 Vp-p and 128KHz frequency and input signal 5 Vp-p and 1. 11. Change the input CLK TLL signal to 5 Vp-p and 128KHz frequency and input signal 5 Vp-p and 500Hz frequency repeat steps 9-10 and record the measured results. TP5 and O/P signal. 12. TP4. repeat step 9-10 with same CLK and record the measured results.5KHz frequency.observe on the output signal waveforms of TP1. 85 . Change the input signal to 5 Vp-p and 1. Finally record the measured results. 13.
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