ANALOG COMMUNICATION+LIC LAB MANUAL

Transcription

1 ANALOG COMMUNICATION+LIC LAB MANUAL As per the revised syllabus of VTU 2010 FIFTH SEM AUTHOR Mr.Yashodhara OUR SINCERE THANKS TO Mr. K.Ezhilarasan FIRST EDITION 2013 S A M B H R A M I N S T I T U T E OF T E C H N O L O G Y

2 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 2 ANALOG COMMUNICATION LAB + LIC LAB Subject Code: 10ECL58 IA Marks: 25 No. of Practical Hrs/Week: 03 Exam Hours: 03 Total no. of Practical Hrs. 42 Exam Marks: Second order active LPF and HPF 2. Second order active BPF and BRF 3. Schmitt Trigger Design and test a Schmitt trigger circuit for the given values of UTP and LTP 4. Frequency synthesis using PLL. 5. Design and test R-2R DAC using op-amp 6. Design and test the following circuits using IC 555 a. Astable multivibrator for given frequency and duty cycle b. Monostable multivibrator for given pulse width W 7. IF amplifier design 8. Amplitude modulation using transistor/fet (Generation and detection) 9. Pulse amplitude modulation and detection 10. PWM and PPM 11. Frequency modulation using 8038/ Precision rectifiers both Full Wave and Half Wave.

3 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 3 CONTENTS SL.NO. EXPERIENTS [1]. Second order low pass filter (LPF) [2]. Second order high pass filter (HPF) [3]. Second order Band pass filter (BPF) [4]. Second order Band rejection filter (BRF) [5]. Schmitt trigger [6]. Precision rectifier [A]. Half wave [B]. Full wave PAGE NO [7]. Astable multi-vibrator [8]. Monostable multi-vibrator [9]. R-2R ladder [10]. Pulse width modulation [11]. Pulse position modulation [12]. Frequency modulation [13]. Pulse amplitude modulation [14]. Amplitude modulation

4 ANALOG COMMUNICATION+LIC LAB MANUAL 4 Experiment-1 Second Order Low Pass Filter AIM:- Design a Second order active low pass filter for a given cut-off frequency fc=1 khz. And conduct an experiment to draw frequency response and verify roll-off. COMPONENTS REQUIRED: - Op-amp 741 IC, Resistors, Capacitors, Connecting wires, Connecting board, CRO-probes. THEORY:-A low pass filters is the filter that passes low frequency signals but attenuates Signals with higher frequencies than the cut-off frequency. It is a Second order low pass filter which means that noise above a certain preset cut-off frequency is weakened by 40db/decade. Op-amp stage is unity gain amplifier. They are often rated for general audio/video, automotive, avionics, commercial, computers and many industrial, medical and military applications. Circuit for the LPF uses two stage passive RC filters connect to the input of Noninverting op-amp. The frequency response of op-amp will be same as that of passive RC filter, except that the amplitude of output signal is increased by pass band voltage gain (AF) of the amplifier and non-inverting amplifier as this is given as 1+Rf/R1.For noninverting amplifier circuit the voltage gain of the filter is generally expressed in decibels and is a function of feedback (Rf) divided by corresponding input resistor (R1) value and is given by Voltage gain=20log (Vout/Vin) = Where Af= pass band gain of a filter,(1+rf/r1) f=frequency of input signal in hertz fc=cut-off frequency in hertz CIRCUIT DIAGRAM:-

5 ANALOG COMMUNICATION+LIC LAB MANUAL 5 Design:Choose a cut off frequency fc=1 khz Assume pass band gain A=2 then Rf=R1 Choose Rf=R1=10KΩ Assume C1=C2=0.1µf Using the formula Calculate R when C=0.1µf we get R=1.59KΩ Choose R2=R3=1.5KΩ Procedure:[1].Before wiring the circuit checks all the components using multimeter and IC tester. [2].Design the filter for gain and make the connection as shown in the circuit diagram. [3].Set the signal generator amplitude (input voltage) say 2Vp-p and observe the input Vin and output Vout signals of the circuit simultaneously on CRO screen. [4].By varying the frequency of the input from 100Hz to higher KHz range note the frequency of signal and corresponding output voltage across pin 6 of IC with respect to the ground. [5].The output voltage remains constant at lower frequency range and drops its amplitude by 40db/decade after designed cutoff frequency. [6].Tabulate the reading in tabular column and plot the graph with frequency along X-axis and gain in db along Y-axis. To find the Roll-off factor for LPF:- keep the input signal amplitude constant, adjust the input frequency at 10fc,note the output signal amplitude. The difference in the gain of the filter at fc and 10fc gives the roll-off factor. Tabular column:input frequency in Hz Vout in volts [Vout/Vin] Gain(db)=20log[Vout/Vin] K 2K 3K 4K 5K 6K 7K 8K 10k

6 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 6 Result:-The circuit has been verified for a cut off frequency 1 KHz for Roll-off= -40db/decade.

7 ANALOG COMMUNICATION+LIC LAB MANUAL 7 Experiment-2 Second Order High Pass Filter AIM:- Design a Second order active High pass filter for a given cut-off frequency fc=1 khz. And conduct an experiment to draw frequency response and verify roll-off. COMPONENTS REQUIRED:- Op-amp 741 IC, Resistors, Capacitors, Connecting wires, Connecting board, CRO-probes. THEORY:-High pass filters are used in applications requiring the rejection of lowfrequency signals. One such application is in high-fidelity loudspeaker systems. Music contains significant energy in the frequency range from around 100Hz to 2 KHz, but highfrequency drivers can be damaged if low frequency audio signals of sufficient energy appear at their input terminals. A high-pass filter between the broadband audio signal and the tweeter. In conjunction with a LPF for the low frequency driver. Circuit diagram:- Design:Choose a cut off frequency fc=1 khz Assume pass band gain A=2 then Rf=R1 Choose Rf=R1=10KΩ Assume C1=C2=0.1µf Using the formula Calculate R when C=0.1µf we get R=1.59KΩ Choose R2=R3=1.5KΩ

8 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 8 Procedure:- [1].Before wiring the circuit checks all the components using multimeter and IC tester. [2].Design the filter for gain and make the connection as shown in the circuit diagram. [3].Set the signal generator amplitude (input voltage) say 2Vp-p and observe the input Vin and output Vout signals of the circuit simultaneously on CRO screen. [4].By varying the frequency of the input from 100Hz to higher KHz range note the frequency of signal and corresponding output voltage across pin 6 of IC with respect to the ground. [5].The output voltage remains constant at lower frequency range and drops its amplitude by 40db/decade after designed cutoff frequency. [6].Tabulate the reading in tabular column and plot the graph with frequency along X-axis and gain in db along Y-axis. To find the Roll-off factor for HPF:- keep the input signal amplitude constant, adjust the input frequency at 0.1fc,note the output signal amplitude. The difference in the gain of the filter at fc and 0.1fc gives the roll-off factor. Tabular column:- Input frequency in Hz Vout in volts [Vout/Vin] Gain(db)=20log[Vout/Vin] K 2K 3K 4K 5K 6K 7K 8K 10k

9 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 9 Result:-The circuit has been verified for a cut off frequency 1 KHz for Roll-off= +40db/decade.

10 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 10 Experiment-3 Second Order Band Pass Filter AIM:- Design a Second order active Band pass filter for a given frequency fc1=1 khz and fc2=6khz.and conduct an experiment to draw frequency response and verify roll-off. COMPONENTS REQUIRED:- Op-amp 741 IC, Resistors, Capacitors, Connecting wires, Connecting board, CRO-probes.

11 Circuit Diagram:- A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 11

12 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 12 Design:-Specifications f1=1 KHz f2=6 KHz Bandwidth=5 KHz = 2.45 KHz Quality factor Q=fc/BW=0.49 hence it is wide band pass filter Choose R1=R6=10K then therefore Rf=5.86KΩ since Af=1.586 Choose C=0.1µF then R=1.5KΩ=R2=R3 Choose C=0.1µF then R=265Ω=R4=R5 Procedure: - [1]. Before wiring the circuit, check all the components. [2]. Design the two filters for the desired cut off frequencies and make the connections as shown in the circuit diagram. [3]. Set the signal generator amplitude to 10V peak to peak and observe the input voltage and output voltage on the CRO. [4]. By varying the frequency of input from Hz range to KHz range, note the frequency and the corresponding output voltage across pin 6 of the op amp with respect to the gnd. [5]-.The output voltage (VO) remains constant at lower frequency range. [6]. Tabulate the readings in the tabular column. [7]. Plot the graph with f on X-axis and gain in db on Y axis. Tabular column:- Input frequency in Hz Vout in volts [Vout/Vin] Gain(db)=20log[Vout/Vin] K 2K 3K 4K 5K 6K 7K 8K 10k

13 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 13 Result:-The circuit is designed and verified for given cut-off frequencies.

14 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 14 Experiment-4 Second Order Band Elimination Filter AIM:- Design a Second order active Band elimination filter for a given frequency fc1=5khz and fc2=9khz.and conduct an experiment to draw frequency response and verify roll-off. COMPONENTS REQUIRED:- Op-amp 741 IC, Resistors, Capacitors, Connecting wires, Connecting board, CRO-probes.

15 Circuit Diagram:- A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 15

16 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 16

17 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 17 Tabular column:- Input frequency in Hz Vout in volts [Vout/Vin] Gain(db)=20log[Vout/Vin] K 2K 3K 4K 5K 6K 7K 8K 10k Roll-off factor:- For LPF = [Gain at 10fc2 Gain at fc2] For HPF=[Gain at fc1 - Gain at 0.1fc1] Result:- The circuit is verified for a given cut-off frequencies and Roll-off factor is calculated.

18 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 18 Experiment-5 Schmitt Trigger AIM:- To design and test Schmitt trigger for the given values of Case[a] Case[b] UTP 4V -1V LTP 2V -3V COMPONENTS REQUIRED: - Op-amp 741 IC, Resistors, Connecting wires, Connecting board, CRO-probes.

19 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 19

20 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 20 Design:- Assume Vsat=10V Case[a].UTP=4V LTP=2V Vref=3.3V Case[b].UTP=-1V LTP=-3V Vref=-2.2V Waveform:- Result:-Schmitt trigger is designed and verified for the given values of UTP and LTP.

21 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 21 Experiment-6[a] Half wave precision rectifier Aim:-To design and obtain half wave precision rectification using OP-Amp and to observe the transfer characteristics. Components required:-op-amp, resistors, diode, connecting board, connecting wires. Theory:-Rectifiers are often called into action to measure signal strength. A rectifier is an electrical device that converts AC to DC the process is known as rectification. Although the series diode is the classic rectifier, it can t rectify signals than its own forward voltage 0.7V threshold which must be overcome before appreciable conduction occurs, therefore above this threshold we have Due to this, output of conventional rectifier is distorted, as shown in the figure below Circuit Diagram:-

22 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 22 Procedure:- [1].Check all the components,make connections as shown in the figure. [2].Set the signal generator amplitude 1vp-p sine wave. [3].Observe the input and rectified output signal waveform on the CRO. [5].Plot the graph. Result:-Half wave precision rectifier is designed using an op-amp and the transfer charecteristics are plotted.

23 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 23 Experiment-6[b] Full wave precision rectifier Aim:-To design and obtain full wave precision rectification using OP-Amp and to observe the transfer characteristics. Components required:-op-amp, resistors, diode, connecting board, connecting wires. Theory:-The full wave rectifier depends on the fact that both the half-wave and summing amplifiers are precision circuits. It operates by producing an inverted half-wave rectified signal and then adding that signal at double amplitude to the original signal in the summing amplifier. Vo=-Vi+2Vi=Vi The result is a reversal of the selected polarity of the input signal. The resistor values shown are reasonable, the resistors themselves must be of high precision in order to keep the rectification process accurate. Circuit diagram:- Procedure:- [1].Before wiring the circuit, check all the components using multimeter. [2].Make the connections as shown in the figure. [3].Set the signal generator amplitude 1Vp-p sine wave. [4].Observe the input and rectified output signal waveform on CRO. [5].Plot the waveform.

24 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 24 Result:-Full wave precision rectifier id designed using an op-amp and the transfer characteristics is plotted.

25 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 25 Experiment-7 Astable multivibrator Aim:-To design an Astable multivibrator using IC 555 to generate a clock signal of [a].frequency 1KHz with 0.50 duty cycle.(symmetrical) [b].frequency 1KHz with 0.75 duty cycle.(asymmetrical) Components required:-ic 555, capacitors, resistors, connecting boards and wires. Theory:-An astable multivibrator often called a free running multivibrator, is a rectangular wave generating circuit unlike the monostable multivibrator, this circuit does not require an external trigger to change the state of the output hence the name free running. However the time during which the output is either high or low is determined by the two resistors and capacitors, which are connected externally. In the circuit initially when the circuit output is high capacitor equals 2/3 Vcc comparator- 1 triggers the flip flop and the output goes high, then the cycle repeats. The time during which capacitor charges from 1/3 Vcc is equal to the time the ouput is high and given by tc=0.69(ra+rb)c. Similarly the time during the capacitor discharge from 2/3 Vcc to 1/3 Vcc is equal to the output is low and is given by td=0.69(ra+2rb)c Thus the total period of the output waveform is T=ta+td.and the frequency of oscillation is F=1.44/(Ra+2Rb)C. To calculate Duty cycle:- Note Ton and Toff from CRO o/p waveform and D=Ton/(Ton+Toff).

26 Circuit Diagram:- A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 26

27 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 27

28 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 28 Result:- The astable multivibrator circuit has been verified for both symmetric and unsymmetric designs.

29 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 29 Experiment-8 Monostable multivibrator Aim:-To design a Monostable multivibrator using IC 555 for a given pulse width. Components required:-ic 555, capacitors, resistors, connecting boards and wires. Theory:-The monostable multivibrator often called a one shot multivibrator,is a pulse generator circuit in which the duration of the pulse is determined by the R-C network connected externally to the 555 timer. In such a vibrator, one state of output is stable while the other is quasi-stable (unstable).for auto triggering of the output from quasistable to stable energy is stored by an externally connected capacitor C to a reference level. The time taken in storage determines the pulse width. The transition of the output from stable state to quasi-stable state is accomplished by external triggering. Circuit Diagram:- Design:-Tc=1msec,Tc=1.1RC let C=0.1µF hence R=10KΩ Time period of the trigger input R C =0.0016T1 let T1=3msec and C =0.01µF Hence R =470Ω

30 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 30 Procedure:- [1].Connections are made as shown in circuits. [2].Apply suitable inputs to monostable vibrator. [3].Observe the waveform across timing capacitor in one channel and output in other channel. [4].Connect o/p between capacitor and verify waveform. Result: - Observed monostable output with pulse width=2/3vcc.

31 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 31 Experiment-9 R-2R ladder network Aim:-[a].To design 4 bit DAC using R-2R ladder network. [b].to generate a staircase waveform using the DAC circuit. Components required:-op-amp, resistors, diode, connecting board, connecting wires. Theory:-A ladder is a series/parallel resistor network, a R-2R ladder requires only two resistor value R and 2R. Nowadays digital systems are used in many applications because of their increasingly efficient, reliable and economical operation. Since digital systems such as microcomputers use a binary system of ones and zeros, the data to be put into the microcomputer have to be converted from analog form to digital form. The circuit that performs this conversion and reverse conversion are called A/D and D/A converters respectively. D/A converter in its simplest form use an op-amp and resistors either in the binary weighted form or R-2R form. The fig. below shows D/A converter with resistors connected in R-2R form. It is so called as the resistors used here are R and 2R. The binary inputs are simulated by switches b0 to b3 and the output is proportional to the binary inputs. Binary inputs are either in high (+5V) or low (0V) state. The analysis can be carried out with the help of Thevenin s theorem. The output voltage corresponding to all possible combinations of binary inputs can be calculated as below. V0 = - RF [ (b3/2r) + (b2/4r) + (b1/8r) + (b0/16r) ] Where each inputs b3, b2, b1 and b0 may be high (+5V) or low (0V). The great advantage of D/A converter of R-2R type is that it requires only two sets of precision resistance values. In weighted resistor type more resistors are required and the circuit is complex. As the number of binary inputs is increased beyond 4 even D/A converter circuits get complex and their accuracy degenerates. Therefore in critical applications IC D/A converter is used. Some of the parameters must be known with reference to converters. They re resolution, linearity error, settling time etc. Resolution = 0.5V / 28 = 5 / 256 = Procedure:- [1].Verify all the components and patch chords whether they are in good condition. [2].Make the connections as shown in circuit diagram. [3].Give the supply to trainer kit. [4].For different digital inputs measure the output voltage using multi-meter. [5].Verify whether the theoretical value is matching with practical values and observes the outputs.

32 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 32 D3 D2 D1 D0 Theoretical value Practical value Procedure (TWO):- [1].Connect DAC circuit using R-2R ladder network and Op-amp voltage follower. [2].Construct a modulo-16 counter using a suitable digital IC like [3].Apply the clock and observe the staircase O/P waveform on the CRO,sketch the waveform.

33 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 33 Result:-The analog signals are converted to digital and the analog data is recovered.

34 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 34 Experiment-10 Pulse width modulation Aim:-To conduct an experiment to generate PWM for a given frequency. Components required:-op-amp, resistors, diode, connecting board, connecting wires. Theory:- Pulse width Modulation (PWM) is also known as Pulse duration Modulation (PDM). Three variations of PWM are possible. In One variation, the leading edge of the pulse is held constant and change in the pulse width with signal is measured with respect to the leading edge. In other Variable, the tail edge is held in constant and w.r.t to it, the pulse width is measured in the third variation, the center of the pulse is held constant and pulse width changes on either side of the center of the pulse. The PWM has the disadvantage when compared to PDM that its pulses are of varying width and therefore of varying power content, this means the transmitter must be powerful enough to handle the max width pulses. Circuit Diagram:- Procedure:- [1]. Make the connections as shown in the circuit diagram, [2]. Set the carrier amplitude to 4Vp-p and frequency 4 KHz. [3]. Set the signal amplitude to 2Vp-p and frequency 1 khz. [4]. Observe the o/p signal at pin 6of 2nd op-amp and observe the variation in pulse width by varying the modulating signal amplitude. [5]. Draw PWM Waveform [6]. Now connect the output to the demodulate circuit and observe the signal it matches with m (t).

35 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 35 Waveform:- Result:- The circuit to generate a PWM signal is designed and the output waveforms are observed.

36 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 36 Experiment-11 Pulse Position modulation Aim:-To conduct an experiment to generate PPM for a given frequency. Components required:-op-amp, resistors, diode, connecting board, connecting wires. Theory: - In this type of modulation, the amplifier and width of the pulses is kept constant while the position of each pulse with reference to the position of a reference pulse is changed according to the instantaneous sampled value of the modulating signal. Pulse position modulation is observed from pulse width modulation. Any pulse has a leading edge and trailing edge in this system the leading edge is held in fixed position while the trailing edge varies towards or away from the leading edge in accordance to the instantaneous value of sampled signal. Circuit Diagram:- Design:- Pulse width=200µsec Tp=1.1RC assume C=0.01µf hence R=18KΩ

37 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 37 Procedure:- [1]. Make the connections as shown in the circuit diagram. [2]. Set the carrier amplitude to around 4v (p-p) and frequency = 4 KHz. [3]. se the signal amplitude to around 2v (p-p) and frequency around 1KHz [4]. Observe the output signal at pin no: 3 of the 555 timer and also observe the variation in pulse position by varying the modulating signal amplitude [5]. Draw the PPM waveform. Result:- The circuit to generate a PPM signal of pulse width 200 µs is designed and the output waveform of PPM was observed.

38 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 38 Experiment-12 Frequency modulation Aim:-To design and conduct a suitable experiment to generate an FM wave using IC Components required:-op-amp, resistors, diode, connected board, connecting wires. Theory: - Frequency modulation: FM is that form of angle modulation in which the instantaneous frequency is varied linearly with the message signal. The IC 8038 waveform generator is a monolithic integrated circuit capable of producing high accuracy sine square, triangular, saw tooth and pulse waveforms with a minimum number of external components. Basic principle of IC 8038 The operation of IC 8038 is based on charging and discharging of a grounded capacitor C, whose charging and discharging rates are controlled by programmable current generators Ia and Ib. When switch is at position A, the capacitor charges at a rate determined by current source Ia. Once the capacitor voltage reaches Vut, the upper comparator (CMP 1) triggers and reset the flip-flop out put. This causes a switch position to change from position A to B. Now, capacitor charge discharging at the rate determined by the current sink Ib. Once the capacitor reaches lower threshold voltage, the lower comparator (CMP 2) triggers and set the flip-flop output. This causes the switch position to change from position B to A. And this cycle repeats. As a result, we get square wave at the output of Flip flop and triangular wave across capacitor. The triangular wave is then passed through the on chip wave shaper to generate sign wave. To allow automatic frequency controls, currents Ia and Ib are made programmable through an external control voltage Bi. For equal magnitudes of Ia and Ib, output waveforms are symmetrical conversely, when two currents are unequal, output waveforms are asymmetrical. By making, one of the currents much larger than other we can get saw tooth waveform across capacitor and rectangular waveform at the output of flip-flop.

39 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 39 Circuit Diagram:- Design:- Let R=Ra=Rb Let f=33 khz f= 3*(2*Ra-Rb)/10*Rac*Ra Substituting for R&C in above equation, we get f=0.3/rc Let R=10k ohms Therefore C =0.001µF Procedure:- [1]. Rig up the circuit as shown in the figure. [2]. Apply +12,-12V from the supply. [3]. Observe the sinusoidal waveform at pin 2.It should be same as design carrier frequency. [4]. Switch on signal generator and apply the signal amplitude of 8Vp-p and frequency of 5 khz. [5]. Observe the output between pin 2 and ground. [6]. Sketch the waveforms. Show the graph of message carrier and modulation signal.

40 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 40 Calculation:- To calculate modulation index β Δf=(fmax-fmin)/2= Hz For fm=5khz β=δf/fm Transmission Bandwidth Bt=2(Δf+ fm) Result:- The frequency modulation is seen and the transmission bandwidth was found to be khz.

41 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 41 Experiment-13 Pulse Amplitude modulation Aim:-To conduct an experiment to generate PAM signal and also design a circuit to demodulate the PAM signal. Components required:-op-amp, resistors, diode, connecting board, connecting wires. Theory:- In PAM the amplitude of the pulses are varied in accordance with the modulating signal. (Denoting the modulating signal as m (t). PAM is achieved simply by multiplying the carrier with the m (t) signal. The balanced modulators are frequency used as multipliers for this purpose. The Output is a series of pulses, the amplitude of which vary in proportion to the modulating signal. The form of pulse Amplitude modulation shown in the circuit diagram is referred to as natural PAM because the tops of the pulses follow the shape of the modulating signal. As shown in fig, the samples are taken at regular interval of time. If enough samples are taken, a reasonable approximation of the signal being sampled can be constructed at the receiving end. This is known as PAM. Circuit Diagram:-

42 DESIGN: - Fc>> 1/RC i.e., R>1/Fc*C Let Fc =15 khz and C=0.1μF Therefore R~680Ω A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 42 PROCEDURE: [1]. Make the Connections as shown in circuit diagram. [2]. Set the carrier amplitude to 2 Vpp and in the frequency of 5 khz to 15 khz. [3]. Set the i/p Signal amplitude to around 1V (p-p) and frequency to 2 khz. [4]. Connect the CRO at the emitter of the transistor and observe the Pam waveform. [5]. Now connect the O/p(i.e. PAM) signal to the demodulation circuit and observe the signal if it matched plot the waveform RESULT: The circuit to generate PAM signal and to demodulate the PAM signal were Designed and the waveform were observed.

43 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 43 Experiment-14 Amplitude modulation Aim: - To generate AM signal, information signal given the collector. Also, demodulate it. Measure the modulation index using two different methods. Components Required:- IFT, AFT, SL 100/BF 194 transistor, resistors, capacitors, diode 0A79, connecting board, connecting wires and CRO. Theory: - The modulator is a linear power amplifier that takes the low-level modulating signal and amplifies it to a high power level. The modulating output signal is coupled through a modulating transformer to the Class C amplifier. The secondary winding of the modulation transformer is connected in series with collector supply voltage Vcc of the Class C amplifier. This means that modulating signal is applied in series with the collector power supply voltage of the Class C amplifier applying collector modulation. In the absence of the modulating input signal, there will be zero modulation voltage across the secondary of the transformer. Therefore, the collector supply voltage will be applied directly to the Class C amplifier generating current pulses of equal amplitude and output of the tuned circuit will be a steady sine wave. When the modulating signal occurs, the a.c. voltage across the secondary of the modulating transformer will be added to and subtracted from the collector supply voltage. This varying supply voltage is then applied to the Class C amplifier resulting in variation in the amplitude of the carrier sine wave in accordance with the modulating signal. The tuned circuit then converts the current pulses into an amplitude-modulated wave. Design:- Let fm= khz m= RC>>tc or RC (1/mωm) Or RC/3= (1/mωm) ωm=2πfm Assuming value of C=0.01μF Substituting value of C and fm=1 khz, we get R=9.5kΩ 10kΩ m= (Vmax-Vmin)/ (Vmax+Vmin) Vm= (Vmax-Vmin)/2 Procedure:- 1. Design the collector modulator circuit assuming fm=1 khz and m=0.5 take C=0.01μF. 2. Before wiring, check all components using multimeter. 3. Make connections as shown in figure. 4. Set the carrier frequency to 2v and 455 khz. 5. Set the modulating signal to 5v and 1 khz. 6. Keep carrier amplitude constant and vary the modulating voltage in steps and measure Vmax and Vmin, and calculate modulation index.

44 A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L Tabulate the reading taken. 8. Feed AM output to Y-plates and modulation signal yo X-plates of CRO. Obtain trapezoidal pattern. 9. Plot the graph of modulating signal versus modulation index. Circuit Diagram:- Vmax in volts Vmin in volts μ(mod index) Vm in volts

45 Result:- AM is observed for different modulation index. A N A L O G C O M M U N I C A T I O N + L I C L A B M A N U A L 45