Analog comm lab manual

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Analog comm lab manual

  1. 1. Analog Communications Lab ______________________________________________________GRIET/ECE 1 of 86
  2. 2. Analog Communications Lab ______________________________________________________ ContentsContents.........................................................................................................................................21. Amplitude Modulation and Demodulation.............................................................................42.Frequency Modulation and Demodulation............................................................................133.Balanced Modulator................................................................................................................214.Pre-emphasis and De-emphasis..............................................................................................275.Characteristics of mixer..........................................................................................................346.Phase detection and Measurement using PLL......................................................................397.Synchronous Detector..............................................................................................................458.SSB System...............................................................................................................................49MATLAB PROGRAMS.............................................................................................................521.Amplitude Modulation............................................................................................................532.Demodulation of AM wave using Hilbert transform............................................................573.Demodulation using Diode detector.......................................................................................594.DSBSC Modulation..................................................................................................................615.DSBSC Demodulation.............................................................................................................636.Generation DSBSC using Balanced Modulator....................................................................657.SSBSC Modulation..................................................................................................................678.Demodulation of SSBSC..........................................................................................................699.Frequency Spectrum of Amplitude Modulated Wave..........................................................7110.Frequency Spectrum of Amplitude Modulated Wave........................................................73Output Waveform.......................................................................................................................7311.Frequency Spectrum of SSBSC............................................................................................7512.Performing Pre-emphasis and De-emphasis........................................................................7713.Frequency Modulation..........................................................................................................7914.Demodulation of FM wave....................................................................................................8115.Frequency Spectrum of FM..................................................................................................8316.Construction of SQUELCH circuit......................................................................................85GRIET/ECE 2 of 86
  3. 3. Analog Communications Lab ______________________________________________________GRIET/ECE 3 of 86
  4. 4. Analog Communications Lab ______________________________________________________ 1. Amplitude Modulation and DemodulationAim: To study amplitude modulation and demodulation and to calculate modulation index bychanging the modulating signal’s amplitude.Equipment required: Analog Communication Kit, CRO, Connecting wires.Components required: 1. Transistors: BC107 (2 nos.) 2. Diodes: A79 (1 no.) 3. Resistors: 8 K-1 No, 10K-(2nos), 100 K-(4nos) 4. Capacitors:0.1µF-(2nos),-0.0015µ-(1no), 1KpF-(2nos),22µF-(1no),0.01 µF-(1no) 5. Inductors -2.5mH-(1no)Theory: In Amplitude modulation of a carrier signal is varied by the modulating voltage whosefrequency is invariably, lower than that of the carrier frequency. In practice, the carrierfrequency may be high frequency (HF), while the modulating frequency is audio frequency.Formally AM is defined as a system of modulation in which the amplitude of the carrier signalis made proportional to the instantaneous amplitude of the modulating voltage.Let the carrier voltage and the modulating voltage, Vc and Vm respectively be represented by Vc = VcSinωct Vm= VmSinωmt Note that phase angle has been ignored in both expressions since it is unchanged by theamplitude modulation process. Its conclusion here would merely complicate the precedingwithout affecting the result. However, it will certainly not be possible to ignore phase anglewhen we deal with frequency and phase modulation. From the definition of AM, it follows that the (max amplitude V c of the unmodulatedcarrier will have to be made proportional to the instantaneous modulating voltage V mSinωmtwhen the carrier is amplitude modulated.Frequency spectrum of the AM wave: We shall show mathematically that the frequencies present in the AM wave are thecarrier frequency and the first pair of side band frequencies, where a sideband frequency isdefined as fSB = fc ±n fm and in first pair n=1When a carrier signal is amplitude modulated, the proportionality constant is made to unity, andthe instantaneous modulating voltage variations are superimposed onto carrier amplitude. Thus,when there is temporarily no modulation, the amplitude of the carrier is equal to its unmodulatedvalue. When modulation is present, the amplitude of the carrier is varied by its instantaneousvalue. This situation is illustrated in fig 1, which shows how the max amplitude of theamplitude- modulated voltage is made to vary in accordance with modulating voltagechanges.Fig1 also shows that something unusual (distortion, as it happens) will occur if Vm isGRIET/ECE 4 of 86
  5. 5. Analog Communications Lab ______________________________________________________greater than Vc.This and the fact that the ratio Vm/Vc often occurs, leads to the followingdefinition of modulation index. m=Vm/Vc → Eq. 1GRIET/ECE 5 of 86
  6. 6. Analog Communications Lab ______________________________________________________The modulation index is a number lying between 0 and 1 it is very often expressed as apercentage and called the percentage modulation.It is possible to write an equation for the amplitude of the amplitude-modulated voltages, thuswe have A = Vc+Vm = Vc+VmSinωmt+Vc+ VcSinωct A=Vo(1+m Sinωmt)The instantaneous value of the resulting amplitude modulated wave is V =A Sinωct = Vc(1+m Sinωmt) Sinωct →Eq. 2Eq 2 may be expanded by means of the trignometrical relation Sin x Sin y = ½ {Cos(x-y) – Cos(x+y)} to give V = VcSinωct + mVc/2Cos (ωct-ωmt) - mVc/2Cos (ωct-ωmt) → Eq. 3 It has been shown that the eq of an amplitude modulated wave contains three terms. Thefirst term is identical to eq1 and represents the unmodulated carrier. It is thus apparent that theprocess of amplitude modulation has the effect of loading to the unmodulated wave , rather thanchanging it .The two additional terms produced are the two side bands out end . The frequencyof the lower side band (LSB) is fc-fm and the frequency of the upper side band (USB) is f c+fm .The very important conclusion to be made at this stage is that the band width required foramplitude modulation is twice the frequency of the modulating signal. In modulation by severalsine waves simultaneously, as in the AM broadcasting service, the band width required is twicethe highest modulated frequency.Representation of AMAmplitude modulation may be represented in any of three waves depending on the point ofview . Fig 2 shows the frequency spectrum and so illustrates equation 3 . AM is shown simply asconsisting three discrete frequencies , of these the central frequency , i.e. the carrier has thehighest amplitude , and other two are disposed symmetrically about it , having amplitude whichare equal to each other , but which can never exceed half the carrier amplitude.The appearance of the amplitude modulated wave is of great interest, and it is shown in fig2 forone cycle of the modulating sine wave. It is derived from fig1 , which showed an amplitude , orwhat may now be called the top envelope , is given by the relation A = (V c + VmSinωmt ).Similarly , the maximum negative amplitude , or bottom envelope , is given by - A = - (V c +VmSinωmt ). The modulated wave extends between these two limiting envelopes and has arepretion rate equal to the unmodulated carrier frequency.It will be recalled that Vm= mVc and it is now possible to use this relation to calculate the index(or depth ) of modulation from the wave form of fig 2 as follows. Vm = (Vmax – Vmin)/2 Vc = Vmax – Vm = Vmax – ( (Vmax – Vmin)/2 ) = (Vmax – Vmin) / 2 M = Vm / Vc = ( (Vmax – Vmin) / 2) / ( (Vmax – Vmin) / 2) M=(Vmax – Vmin) / (Vmax + Vmin)GRIET/ECE 6 of 86
  7. 7. Analog Communications Lab ______________________________________________________Circuit Diagram : AM Modulator AM DemodulatorProcedure:1) Connect the circuit as shown in the circuit diagram.2) Apply the 100 KHz carrier signal and amplitude of 6V(p-p) to the input of AM modulator at100 KΩ pot and 1 KHz of modulating signal to the AM modulator at 100 KΩ pot as shown inthe circuit diagram.3) Apply the power supply of 12V as shown in the circuit diagram.4) Observe the amplitude modulated wave synchronization with the modulating signal on adual trace CRO following figure shown the connections.GRIET/ECE 7 of 86
  8. 8. Analog Communications Lab ______________________________________________________5) Adjust the 10 KΩ linear pot for carrier suppression and 100KΩ linear pot for propermodulation.i.e. 100%.6) Now by varying the amplitude of the modulating signal , the depth of modulation varies.Calculate the maxima and minima points of modulated wave on the CRO and calculate thedepth of modulation using formula.GRIET/ECE 8 of 86
  9. 9. Analog Communications Lab ______________________________________________________Waveforms:Result: Amplitude modulated signal is generated and original signal is demodulated from AMsignal Depth of modulation is calculated for various amplitude levels of modulating signals.GRIET/ECE 9 of 86
  10. 10. Analog Communications Lab ______________________________________________________GRIET/ECE 10 of 86
  11. 11. Analog Communications Lab ______________________________________________________Attach one Graph Sheet hereGRIET/ECE 11 of 86
  12. 12. Analog Communications Lab ______________________________________________________GRIET/ECE 12 of 86
  13. 13. Analog Communications Lab ______________________________________________________ 2.Frequency Modulation and DemodulationAim: To study the process of frequency modulation and demodulation and to calculate thedepth of modulation by varying the modulating voltage.Equipment Required: Analog communication kit , CRO, connecting wires.Components Required: 1. IC’s:8083,LM-565,TL08IC(OR)TL084 2. Resistors:4.7kΩ-(2nos),10kΩ-(5nos),15kΩ-(1nos),30kΩ-(1nos),300kΩ-(2nos) 3. Capacitors:0.1µF-(2nos),0.001µF-(3nos),0.0022µF-(1no)Theory:The general equation of an unmodulated wave, or carrier, may be written as X=A sin (ωt+ф)Where X = instantaneous value of voltage or current. A = (maximum) amplitude ω = angular velocity (rad/s) ф = phase angle, radNote that we represent an angle in radians.If any one of these three parameters is varied in accordance with another signal, normally of alower frequency, then the second signal is called the modulation and the first is said tomodulated by the second. In the frequency modulation, frequency of the carrier is made to vary.For simplicity, it is again assumed that the modulation, signal is sinusoidal. This signal has twoimportant parameters which must be represented by the modulation process without distortion:namely, its amplitude and frequency. It is assumed that the phase relations of a complexmodulation signal will be preserved. By the definition of frequency modulation, the amount bywhich the carrier frequency is varied from its unmodulated value, called the deviation, is madeproportional to the instantaneous value of the modulating voltage. The rate at which thisfrequency variations or oscillations takes place is naturally equal to the modulating frequency.The situation is illustrated in fig 1, which shows the modulating voltage and the resultingfrequency-modulate wave. Fig -1 also shows the frequency variation with time which is seen tobe identical to the variation with time of the modulating voltage. As an example of FM, allsignals having the same amplitude will deviate the frequency by the same amount, says 45 kHz,no matters what their frequencies. Similarly, all signals of the same frequency, says 2 kHz, willdeviate the carrier at the same rate of 2000 times per second, no matter what their individualamplitudes. The amplitudes of the frequency modulated wave remains constants at all times, thisis , in fact the greatest signal advantage of FM.GRIET/ECE 13 of 86
  14. 14. Analog Communications Lab ______________________________________________________Fig 1:Mathematical Representation of FM:The instantaneous frequency of the frequency modulated wave is given by f = fc (1+k Vmcosωmt)Where fc = unmodulated (or average carrier frequency) K = proportional constant Vmcosωmt = instantaneous modulating voltage (cosine being preferred for simplicity in calculations)The maximum deviation for this particular signal will occur when the cosine term has itsmaximum value, that is +(or)-1. under these conditions, the instantaneous frequency will be f =fc(1+KVm) so that the maximum deviation δ=KVmfcThe instantaneous amplitude of the FM signal will be given by a formula of the form V = A sin {F ωc ωm} = A sin ф equation1Where F (ωc ωm), is some function, as yet undetermined, of the carrier and modulatingfrequencies. This function represents an angle and will be called for convenience. The problemnow is to determine the instantaneous value for this angle. Fig -2GRIET/ECE 14 of 86
  15. 15. Analog Communications Lab ______________________________________________________ As fig-2 shows, ф is the angle traced out by the vector A in time t.If A were rotating witha constant angular velocity, say p.This angle ф would be given by P t in radians.In thisinstance,however the angular velocity is anything but constant. It is, in fact, governed by theformula for ω=ωc(1+kVmcos ωmt).in order to find f, ω must be integrated with respect to time.Thus θ = ∫ωdt= ωc (1+KVmcosωmt)dt θ = ωc∫(1+KVmcosωmt)dt θ = ωct+ KVmsinωmt/ωm θ = ωct+ KVmsinωmt/fm θ = ωct+ δ /fmsinωmtThe deviation utilized, in turn the fact that ωc is constanst,the formula. ∫cosnx dx = (sinnx/n), which had shown that K Vm f c = δ.Equation 1 may now be subdivided to give the instantaneous value of the FM voltage, thus V = A Sin(ωct+δ/fm Sinωct) Mf= max.frequency deviation / modulating frequency= δ/fm V = A Sin (ωct+mf Sinωct)It is important to note as the modulated frequency decrease and the modulating voltageamplitude i.e. δ remains constant, the modulation index increase. This will be basis fordistinguishing frequency modulation from phase modulation. Note also that m f,which is the ratioof two frequencies is measured in radians.Circuit description: In this kit, frequency modulation is generated by using IC8038.The frequency of the waveform generator is a direct function of the DC voltage at terminal8 (measured from V+).By altering this voltage, frequency modulation is performed. For smalldeviations (e.g.+ (or) – 10%) the modulating signal can be applied directly to pin 8,merelyproviding DC decoupling with a capacitor as in Fig 3.An external resistor between pins 7 and 8 is not necessary, but it can be used to increase inputimpedance from about 8KΩ (pin 7 and 8 connected together) to about (R=8KΩ).GRIET/ECE 15 of 86
  16. 16. Analog Communications Lab ______________________________________________________For larger FM deviations or for frequency sweeping, the modulating signal is applied betweenthe positive supply voltage at pin 8.During the demodulation,FM output is given to a phase lock loop(565).We have seen that,during lock, the average dc level of the phase comparator output is directly proportional to thefrequency of the input signal.As the frequency shifts, it is the output which causes the VCO to shift and key tracking. In otherwords,the phase comparator output is an exact replica of the original modulating audio signal.Fig-4 shows connections of 565 as FM Demodulator. The component values shown are for acarrierfrequency of 82KHZ approx. The demodulated output is followed by a three stage filter toremove Rf component. A small capacitor of 0.01µF as connected between pins 7 and 8 toeliminate possible oscillations in the current source.Circuit Diagram:GRIET/ECE 16 of 86
  17. 17. Analog Communications Lab ______________________________________________________Procedure:Step 1: connect the circuit as shown in the circuit diagram.Step 2: check the circuit properly and apply the power supply to the circuitStep 3: observe the carrier signal from the FM modulator at pin 2 of the IC 8083, which is 82KHZ.Step 4: apply the modulating signal frequency of 4KHZ, 6Volts(p-p) from the function generatorto the FM input at pin 8 as shown in the figure below.Fig:Step 5: Trigger CRO with respect to CH1.adjust amplitude of the modulating signal until we getundistorted FM output. It is difficult to trigger FM on analog CRO.That is why you adjustmodulating signal amplitude until small distortions notified in Fm output.GRIET/ECE 17 of 86
  18. 18. Analog Communications Lab ______________________________________________________Step 6: Calculate maximum frequency and minimum frequency from the FM output andcalculate modulating index using table shown in belowDemodulation fig: FM DemodulationSTEP7:Connect the FM output to the demodulator input as shown in figure.Step8 :In this condition decrease the amplitude if the modulating signal generator until we getundistorted demodulated output(1 Vp-p). Observations:S.NO AMP fmax fmin(Hz) ∆f(Hz) (p-pV) fc+∆f fc+∆f β=Result:The frequency modulated signal is generated and original signal is demodulated and modulationindex is calculated for Fm signalGRIET/ECE 18 of 86
  19. 19. Analog Communications Lab ______________________________________________________Attach one normal graph sheet hereGRIET/ECE 19 of 86
  20. 20. Analog Communications Lab ______________________________________________________GRIET/ECE 20 of 86
  21. 21. Analog Communications Lab ______________________________________________________ 3.Balanced ModulatorAim: 1. To construct and properly adjust a balanced modulator and study its operation. 2. To observe that the output is a double sideband with a suppressed carrier signal. 3. To adjust it for optimum carrier suppression. 4. To verify that the output audio level directly affects the double side-band output amplitude. 5. To observe that the output is maximum with zero audio input. 6. To measure carrier only output and the peak side- band output and to calculate the carrier suppression.Equipment Required:Analog communication kit, CRO connecting wires.Component Required:1. IC’S:1496- (1no)2. Resistors:100Ω-(1no),820Ω-(1no),1kΩ-(1no)1.2kΩ-(1no),2.7kΩ-(1no),10kΩ-(3nos),47kΩ- (1nos),5OkΩ-POT3. Capacitors: 1µF-(1no),22µF-(1no),100µF-(2nos)Theory:One circuit that lends itself extremely well to balanced modulator application is the differentialamplifier. A simplified diagram of a differential amplifier is shown in figure 1.FIG 1: A Differential amplifier used as a balanced modulatorGRIET/ECE 21 of 86
  22. 22. Analog Communications Lab ______________________________________________________Q3 acts as the current source for Q1 and Q2 .If the RF input is applied to the bases of Q1 and Q2in phase, current through both transistors will be identical and the voltage difference across theoutput will be zero. This is the common mode rejection of the differential amplifier and it hasbalanced modulator out the carrier. The audio input is applied to the base of Q3.This upsets the circuits balance. As a result, theaudio mixing and RF signals are mixed across Q1 and Q2 .This is non linear mixing and,therefore, side bands appear at the output. However, the carrier or RF input does not. Since it isa common mode signal, it is rejected.A differential amplifier must be constructed of transistors whose characteristics are very closelymatched. Forming the transistors on a single chip of silicon as done with ICs ensure thisnecessary match.Therefore, the differential amplifier is ideally suited to integrated circuit construction.Fig 2IC Balanced Modulators:Fig-2 shows IC that has been specifically designed for use as balanced modulators. Fig-2 is the1496 balanced modulator, which is manufactured by Motorola, National and Signetics. Thisdevice uses a differential amplifier configuration similar to what was previously described. Itscarrier suppression is rated at a minimum of -5db with a typical value of -65db at 500 kHz.GRIET/ECE 22 of 86
  23. 23. Analog Communications Lab ______________________________________________________Step-By–Step Procedure:A balanced modulator using a 1496 integrated circuit .You will verify that it does suppress thecarrier and also adjust it for optimum carrier suppression.1. Balanced modulation circuit is designed using 1496 integrated circuit, and connect balancedmodulator circuit as per the circuit diagram.2. Apply the power supply =12 to the pin No.8 of IC 1496.3. Give the Audio frequency of 200 Hz sine wave from Function Generator 1 and 100KHZ Sinewave form from the Function Generator 2 which are provided on the Analog CommunicationLaboratory Kit as a inputs. Adjust R1 (1k Ω linear pot) and R2 (50k Ω linear pot) for mid –range which are provided on the Analog Communication Kit. Connect your oscilloscope to theoutput and set the vertical input control to 1v/cm and the sweep to 1ms.cm.4. Check the circuit diagram properly and switch on the Analog Communication Kit. AdjustOscilloscope’s trigger control for a stable display. You may use square as carrier wave andobserve the output waveform.5. Vary R1 (1K) both clockwise, counter clockwise, what effect does it have on the output=_____________6.Disconnect the modulating wave input R1 (1KΩ) .The output should now be close to zero. Setyour Oscilloscope vertical input to 0.2 v/cm. Now adjust R2 for minimum output. If possible,increase the Oscilloscope’s vertical input sensitivity to measure the output voltage.EOUT CARRIER ONLY =___________________7. Set the vertical input control to 1v/cm. Connect the sine input to R1(1K) and adjust R1 formaximum output without producing clipping. Measure the peak sideband output voltage.EPK SIDEBANDS =____________________8.Calculate the carrier suppression in DB.DB =20LOGDB =______________________9.Turn off your experiment and disconnected your circuit.OBSERVATIONS:1. AF signal frequency= 200 Hz.2. RF signal frequency = 100 KHZ.3. Varying R1 ↑, DSBC amplitude (p-p) ↑ proportionally.4. After disconnecting Sine input to R1.5. Eout carrier only = 20 mV (p-p).6. Epk sidebands = 2.4V (p-p).7. Carrier suppression in db = 20 log EPK Sideband / Eout carrier only =41.5GRIET/ECE 23 of 86
  24. 24. Analog Communications Lab ______________________________________________________Waveforms:Input and output waveforms of balanced modulatorResult:1. The Double side band suppressed carrier signal is obtained (Balanced output i.e.,100%modulation is obtained).2. Carrier Suppression in DB is calculatedi.e.,41.5dbGRIET/ECE 24 of 86
  25. 25. Analog Communications Lab ______________________________________________________Attach normal graph sheet hereGRIET/ECE 25 of 86
  26. 26. Analog Communications Lab ______________________________________________________GRIET/ECE 26 of 86
  27. 27. Analog Communications Lab ______________________________________________________ 4.Pre-emphasis and De-emphasisAim: To study the frequency response of the Pre-emphasis and De-emphasis circuits and drawthe graphs.Equipment Required:Analog communication kit, C.R.O, connecting wiresComponents Required:1. Resistors: 47Ω, 100Ω, 1kΩ, 50k Ω-pot2. Capacitors: 0.1µf, 0.01µf, 0.001µf, 0.047µfTheory: Frequency modulation is much more to noise than amplitude modulation and issignificantly more immune than phase modulation. A single noise frequency will effect the output of a receiver only if it falls with in its pass band .The carrier and noise voltages will mix andif the difference is audible, it will naturally interfere with the reception of wanted signals. The noise has a greater effect on the higher modulating frequencies than on lower ones.Thus, if the higher frequencies were artificially boosted at the transmitter and correspondinglycut at the receiver, improvement in noise immunity could be expected. This boosting of highermodulating frequencies, in accordance with a prearranged curve, is termed pre-emphasis, andthe compensation at the receiver is called de-emphasis. If two modulating signals have the same initial amplitude, and one of them is pre-emphasized to (say) twice this amplitude, where as the other is unaffected (being at a muchlower frequency), then the receiver will naturally have to do de-emphasize the first signal by afactor of 2, to ensure that both signals have the same amplitude in the output of the receiver.Before demodulation, i.e., while susceptible to noise interference the emphasized signal hadtwice the deviation it would have had without pre-emphasis, and was thus more immune tonoise. Side band voltages are de-emphasized with it, and therefore have correspondingly loweramplitude than they would have had without emphasis again their effect on the output isreduced. The amount of pre emphasis U.S FM broadcasting, and in the sound transmissionaccompanying television, has been standardized at 75 micro seconds, whereas a number of theother services, notably CCIR and Australian TV sound transmission, use 50 micro seconds. Theusage of microseconds for defining emphasis is standard. A 75 microsecond’s de-emphasis Corresponds to a frequency response curve that is 3 db down at the frequency whosetime constant RC is 75 microseconds. This frequency is given by f=1/2ΠRC and is therefore2120HZ: with 50-micro seconds de-emphasis it would have been 3180HZ.If emphasis is applied to amplitude modulation, some improvement will also result, but it is notas great as in FM because the highest modulating frequencies in AM are no more affected bynoise than any others. Apart from that, it would be difficult to introduce pre-emphasis and de-emphasis in existing AM services since extensive modifications would be needed, particularly inview of the huge numbers of receivers in use.GRIET/ECE 27 of 86
  28. 28. Analog Communications Lab ______________________________________________________Circuit Diagrams:Procedure: 1. Construct the circuit as shown in the circuit diagram. 2. Observe the 3. I/P waveform on the CRO in channel 1. 4. Adjust the amplitude of the sine wave using the amplitude knob to a particular voltage, say 4V or 6V or 10V etc. 5. Measure the O/P voltage in the CRO and note down in the observation table. 6. Calculate the attenuation and Log f values as shown in the observation table. 7. Draw the graph frequency (X-axis) and attenuation in db (Y-axis) to show the emphasis curves on a semi log graph. 8. Various values of R and C are available so that the time constant in suitably selected depending upon the application.Pre-Emphasisi/p voltage=4volts Frequency Output Log f Attenuation (db) (Hz) (volts) 20 log eo/eiGRIET/ECE 28 of 86
  29. 29. Analog Communications Lab ______________________________________________________De-Emphasis I/p Voltage=10voltsFrequency in Hz Output in volts Logf Attenuation in db 20 log eo/eiFigure 1Frequency response of Pre-emphasis &De-emphasisResult: The frequency response of Pre-emphasis and De-emphasis circuits obtained.GRIET/ECE 29 of 86
  30. 30. Analog Communications Lab ______________________________________________________GRIET/ECE 30 of 86
  31. 31. Analog Communications Lab ______________________________________________________Attach one semilog graph Sheet hereGRIET/ECE 31 of 86
  32. 32. Analog Communications Lab ______________________________________________________GRIET/ECE 32 of 86
  33. 33. Analog Communications Lab ______________________________________________________GRIET/ECE 33 of 86
  34. 34. Analog Communications Lab ______________________________________________________ 5.Characteristics of mixerAim:To measure the following parameters of mixer of the signal band (Medium wave: 550 kHz to1.5MHz) Radio Receiver.1 Conversion gain2 Image suppression (or) rejection3 Sensitivity of the mixerEquipment Required:1 AM-FM signal generator (AP-LAB)2 Audio signal generator3 Oscilloscope4 Radio Receiver Dynamic demonstrator5 Wires for subsystems interconnection6 Spectrum analyzers.Theory: A Frequency changer(mixer or converter) is a nonlinear resistance having two sets ofinput frequency (FR and LO) and one set of output (IF)a mixer is used to shift the input receivedRF signal (Frequency band of 550kHZ to 1.5KHz) to low frequency Intermediate Frequency(IF) of 455kHz.The block diagram of the mixer is as follows.Fig:Image frequency and its rejection: The frequency of the local oscillator f0 in majority of receivers, is kept higher than thesignal frequency fs and is given by f0 = fs + fi (where fi is the intermediate frequency) of fs = f0-fi.When fs and f0 are mixed together in the mixer, they produce the difference frequency f i and it isthe only frequency that is amplified by the IF stage. If still higher frequency fsi=f0+fi manages to reach the mixer, then this will also produce f iwhen mixer with f0. This spurious If signal will also be amplified by the IF stage, and willtherefore produce interference. This results in two stations being received simultaneously, whichis undesirable fsi is called the image frequency and is defined by fsi=fs+2fiThe rejection of this image frequency by a single tuned circuit is measured by ‘α’ the rejectionratio. it is the ratio of gain at the signal frequency to the gain at the image frequency and givenby α =√1+q²ρ²Where ρ=(fsi/fs)-(fs/fsi)GRIET/ECE 34 of 86
  35. 35. Analog Communications Lab ______________________________________________________If the receiver has an RF stage, then there will be two tuned circuits tuned to fs,the rejection willbe calculated by the same formula and the total rejection will be the product of the two.image rejection depends on the front end selectivity of the receiver and must be achieved beforethe IF stage, because after the unwanted frequency enters the IF amplifier, it is impossible toremove it from the unwanted signal. For short wave range ,an RF stage is essential for frequencyrejection.RadioReceiverDynamicDemonstratorGRIET/ECE 35 of 86
  36. 36. Analog Communications Lab ______________________________________________________Procedure:(i)Conversion Gain:(a) set the RF signal level to about 100mv at a frequency of 100KHZ from an AM-FMsignal generator and noted as v1 volts.(b) Apply the RF signal at the antenna input of the antenna stage.(c) As transistor T1 acts both as LO stage with sufficient output power and mixer, the IFoutput is selected by IF-T in IF amplifier stage.(d) Tune the gang condenser to tune antenna stage and also tune the coiling mixer-Localoscillator stage for maximum output at IF-T1 of IF stages.(e) Measure the IF level(V2 volts) at the point of IF-T1 without connection to the base ofT2.If the level of IF is low at this point, measure the IF level at the output coil of IF-T3(withoutconnecting detector diode OA79) and subtract the IF amplifier gain to get the IF detectedlevel(V2 volts).(f) The conversion gain/loss is calculated as 20logV2/V1db (if V2 is less than v1then it isconversion loss).(g) The level of V1 and V2 can be measured either with oscilloscope or spectrum analyzer.(ii) Image Suppression/Rejection: (a) As L.O frequency in radio receiver is above the RF signal by IF frequency (455 KHz), the image frequency is above L.O frequency by IF frequency. Image frequency = L.O frequency + IF frequency. (b) Apply the RF signal of 1000KHz,from AM-FM signal generator at a level of about 50 or 100mV at the antenna terminal of the antenna stag and note the level as V1. (c) Measure the IF output level at IF-T1 of IF amplifier as V2. (d) Calculate RF(or desired input) gain as 20logV2/V1db. (e) Select the image signal (RF+2IF =1000+2*455=1910KHZ)from AM-FM signal generator and repeat steps (b) and (c) above and measure v2(Im) (f) Calculate Receiver gain/loss at image frequency, as 20log V2 (Im)/V1(Im) is a level of input signal of 1910KHz to the receiver at image frequency. This value should be less than RF gain for better performance of the receiver(iii) Sensitivity of The Receiver:(a) Apply 1000 KHz RF signal at level of 100mV at the input of the receiver(either at antennainput or base input of T1)(b) Measure the (S/N)ratio of the audio output across the loud speaker(c) Decrease the RF input level of AM-FM signal generator till (S/N)ratio of step(b) above,decrease by 3db or till the unacceptable level of audio tone is heard in the speaker. The RF levelat this point is the sensitivity of the mixture and measured as db µV or dbmV.Observations:A. Conversion gain: RF (100 KHz) input = 150Mv(a)If output (at IF – T3) = 9.2V = A(b)If input level (at IF – T1) at 455 KHz required to get the same IF output of 9.2 V = 150 mV= B. If B ≥A then conversion gain = 20 log (A/B) If B <A then conversion gain = 20 log (A/B)GRIET/ECE 36 of 86
  37. 37. Analog Communications Lab ______________________________________________________As same levels are obtained in case (a) and (b) above, the conversion gain = 0 db.B.Image suppression (or) rejection:RF input = 50mV (at 1000KHz)IF output voltage = 9V.Image = 1000 + 910 = 1910KHzIF output (at input of 910 KHz) = 0.2V.g1 = RF to IF gain = 20 log (9V/50mV) = 45.1dbg2 = Image to IF gain = 20 log (0.2V/50mV) = 12db Image rejection = (g1 – g2)db = 33.1dbC.Sensitivity of the mixer:RF input level at 1000 KHz = 50mV.Measure AF output level impedance switch position of AFpower meter at 6Ω position.AF output level = 4vp-p (or) 0.35watts.RF (100KHz) input level required for half power level of AF p-p measured above (2Vp-p or0.175mW)= 2.5mV.Sensitivity = 20log (2.5mV/1µV) = 67.95dbµV.Result:Characteristics of mixer are studied. Conversion gain, Image suppression and Sensitivity ofmixer are obtained.GRIET/ECE 37 of 86
  38. 38. Analog Communications Lab ______________________________________________________GRIET/ECE 38 of 86
  39. 39. Analog Communications Lab ______________________________________________________ 6.Phase detection and Measurement using PLLAim: To determine the phase difference in capture range using PLL.Equipment Required:Analog communication kit, CRO, connecting wires.Components Required:I.C’S:4046-(1no).Resistors:560Ω-(1no),4k7Ω-(1no),100kΩ-(1no),27kΩ-(1no)Capacitors: 0.1µ F-(1no), 0.01µF-(1no), 0.001µF-(1no), 4.7µF-1No.Theory:As same as the theory written in voltage Controlled Oscillator using PLL. (I.E., Experiment no:2)Circuit Diagram :GRIET/ECE 39 of 86
  40. 40. Analog Communications Lab ______________________________________________________Procedure:Step 1:Connect the circuit as per the circuit diagram as shown in figure give the input at pin 14,i.e. sine wave of 1KHZ and peak to peak of 6 volts . And observe the input sine wave in channel1 of CRO.Step 2:Now connect the pin 3 and pin 4 to channel 2 of CRO which is the output of the IC 4046.And measure the output frequency on CRO (channel 2) which is equal to the input frequency(i.e. 1KHZ).Step 3:Now connect the pin -9 to ground of the 4046 with piece of wire, and record the resultantoutput frequency of the phase locked loop.F1 =____________HzThis output frequency is the lower range of VCO which is determined by the 0.1µF capacitorconnected between pins 6 and 7 and 100KW resistor connected between pin 12 and ground.Step 4:Now with same wire connect the pin 9 to +5 volts supply, record the output which ishigher than the one you measured in step 3.Fh=_____________HzThis frequency is the upper range of VCO which is determined by the 0.1µF capacitorconnected between pin 6 and 7 and the 560Ω resistor connected pins 11 and ground.Step 5:Now remove the connection between pin 9 and the +5 volts supply and measure theoutput which is the same as the input.GRIET/ECE 40 of 86
  41. 41. Analog Communications Lab ______________________________________________________Step 6:Now slowly increase the input frequency; you should observe that, the outputfrequencies also increase. In fact, the output frequencies follow the changes of the inputfrequency and should be exactly equal. Check the input frequency to confirm this.Step 7:Continue to slowly increase the input frequency of PLL and stop when the outputfrequency of the PLL does not continue to increase. Measure the input frequency and recordresult Fin(H)=___________Hz. This is the upper range of VCO which is same as measured instep4 .The phase locked loop then follows input frequency changes for frequencies below thisupper range.Step 8:Now decrease the input frequency, At some point the output frequency will remainconstant. Measure the input frequency and record the result.Fin (L) =____________Hz This frequency is lower range of VCO and which is same as measured in step3.Consequqntly the PLL circuit follow changes in the input frequency for any frequencybetween the lower and upper range of the VCO. There for the loop is locked.Fin(H)- Fin(L). Which is lock range, =_____________HzStep 9:Measure the frequency just below the lowest range of the lock range, which is termed ascapture range.Step 10:In capture range observe and measure the phase difference between the i/p and o/pwaveforms at the same frequencies. Δф =t1-t2/T *360°=_______OBSERVATIONS:RESULT:Lock range=_________________Capture range=_______________Phase detection=______________GRIET/ECE 41 of 86
  42. 42. Analog Communications Lab ______________________________________________________GRIET/ECE 42 of 86
  43. 43. Analog Communications Lab ______________________________________________________Attach normal graph sheet hereGRIET/ECE 43 of 86
  44. 44. Analog Communications Lab ______________________________________________________GRIET/ECE 44 of 86
  45. 45. Analog Communications Lab ______________________________________________________ 7.Synchronous DetectorAim:To obtain the demodulating (or) message signal using DSB or SSB synchronous orcoherent detector.Equipment Required:Analog communication kit , CRO , Connecting wires.Components Required:1. IC’s:1496–2No.2.Resistors:100Ω-(2nos),10Ω-(1no),820Ω-(2nos)47KΩ-(1nos),1KΩ-(2nos),1K-Pot(2nos),1.2KΩ(3nos),50KΩPot-(2nos),2.7KΩ-(2nos),7.3296KΩ-(1nos)3. Capacitors: 0.047µF, 0.1mF, 22mF, 100mF.Block Diagram:Procedure:1. A connection was made as per circuit diagram.2. A balanced modulator circuit was connected for both Transmitter and Receiver.3. Carrier signal was same and was applied to two balanced modulator circuits.4. Apply the power +12V to pin 8 of both the circuits.5. The output of the Receiver balanced modulator circuit was connected to low pass filter.6. Low pass filter circuit was designed according to the requirement.7. And the modulating signal was obtained at the LPF.GRIET/ECE 45 of 86
  46. 46. Analog Communications Lab ______________________________________________________Observations:Transmitter Receiverfc=15 KHz fc=15 KHzfm=462Hz T=2.3msec.After Filter:T=2.3msec.fm = 434.78 Hz.Designing LPFfm = 462HZ = 1/2∏RCLet c = 0.047µF.R= 1/( 2∏ ×0.047×10 -6× 462)R = 7.3296K ΩResult:Thus the modulating signal obtained from the DSB detector.GRIET/ECE 46 of 86
  47. 47. Analog Communications Lab ______________________________________________________Attach normal graph sheet hereGRIET/ECE 47 of 86
  48. 48. Analog Communications Lab ______________________________________________________GRIET/ECE 48 of 86
  49. 49. Analog Communications Lab ______________________________________________________ 8.SSB SystemAim:To generate a SSB signal using Balanced Modulator.Equipment Required:Analog communication kit, CRO, Connecting wires.Components Required:1. IC’s 1496–1No.2. Resistors: 100Ω-1, 10Ω-1820Ω-1, 47KΩ-1, 1KΩ-1, 1KΩ-Pot-1, 2.7KΩ-1.3. Capacitors: 0.047µF-1, 0.1mF-1, 22mF-1,100mF–1.Block Diagram:Procedure:1. Obtain a DSB-SC signal using balanced modulator2. Design a Band pass filter (BPF) for a given specifications to eliminate one of the side bandfrom DSB signal.3. Connect the output of balanced modulator to the input of BPF.4. Obtain the output from BPF to get SSB signal which is either LSB or USB.5. Note down its frequency.Wide Band BPF Circuit Diagram :GRIET/ECE 49 of 86
  50. 50. Analog Communications Lab ______________________________________________________Observations:Design of BPFWe consider wide band BPF, Which is a cascade of HPF & LPFfc=6.3 KHz fm = 200HzLet us consider USB/SSB = fc + fm = 6.5 Hz.fL=6.3 KHz fO=6.5KHz fH=6.7 KHzHPF LPFfL= 1/2∏RC=6.3 kHz Fh= 1/2∏RC=6.7 KHzChoose C=.047uf Let C=.01ufRL=537 ohms Rh=2.37K ohmsResult:Thus the SSB signal was generated using balanced modulatorGRIET/ECE 50 of 86
  51. 51. Analog Communications Lab ______________________________________________________Attach semilog graph sheet hereGRIET/ECE 51 of 86
  52. 52. Analog Communications Lab ______________________________________________________ MATLAB PROGRAMSGRIET/ECE 52 of 86
  53. 53. Analog Communications Lab ______________________________________________________ 1.Amplitude Modulationt=0:0.0001:0.02msg=10*cos(2*pi*100*t)carr=20*cos(2*pi*1000*t)amw=(20+msg).*cos(2*pi*1000*t)subplot(2,1,1)plot(t,msg)title(Message signal)subplot(2,1,2)plot(t,carr)title(Carrier signal)figure;subplot(3,1,1)plot(t,amw)title(Under Modulation)msg=30*cos(2*pi*100*t)amw=(20+msg).*cos(2*pi*1000*t)subplot(3,1,2)plot(t,amw)title(over modulation)msg=20*cos(2*pi*100*t)amw=(20+msg).*cos(2*pi*1000*t)subplot(3,1,3)plot(t,amw)title(100 % modulation)GRIET/ECE 53 of 86
  54. 54. Analog Communications Lab ______________________________________________________ OUTPUT WAVEFORMSGRIET/ECE 54 of 86
  55. 55. Analog Communications Lab ______________________________________________________GRIET/ECE 55 of 86
  56. 56. Analog Communications Lab ______________________________________________________GRIET/ECE 56 of 86
  57. 57. Analog Communications Lab ______________________________________________________ 2.Demodulation of AM wave using Hilbert transformt=0:0.0001:0.02fc=1000Ec=7Carr=Ec*sin(2*pi*fc*t)fm=100Em=7Mod=Em*sin(2*pi*fm*t)Am=(Ec+Mod).*(sin(2*pi*fc*t))disp(Performing Amplitude Demodulation using Hilbert transform);Am_hil=hilbert(Am)Am_abs=abs(Am_hil)Am_Demod=Am_abs-mean(Am_abs)disp(plotting the results);figure;subplot(4,1,1);plot(t,Mod);title(Message Waveform);%xlabel(Time(sec));ylabel(Amplitude);subplot(4,1,2);plot(t,Carr);title(carrier waveform);%xlabel(Time(Sec));ylabel(Amplitude);subplot(4,1,3);plot(t,Am);title(amplitude modulated wave form);%xlabel(Time(sec));ylabel(Amplitude);subplot(4,1,4);plot(t,Am_Demod);title(Amplitude demodulated waveform);%xlabel(Time(sec));ylabel(Amplitude);GRIET/ECE 57 of 86

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