Analog Communication Lab Manual 01, 2013 · 1 Kanara College Society (R) Shree Vidyadhiraj Polytechnic Kumta (N.K), 581343 Department of Electronics and Communication Engg. Analog

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Kanara College Society (R)

Shree Vidyadhiraj Polytechnic Kumta (N.K), 581343

Department of Electronics and Communication Engg.

Analog Communication Lab ManualAnalog Communication Lab ManualAnalog Communication Lab ManualAnalog Communication Lab Manual For

III Semester Diploma (E&C)

Omkar V. Bhat (B.E.) Lecturer, E&C Dept. S.V.Polytechnic, Kumta

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CONTENTCONTENTCONTENTCONTENT

Syllabus……………………………………………………………………………………. 3

Scheme of evaluation……………………………………………………………………… 4

Conducting experiments

1 Verification of Maximum power Transfer Theorem……………………………………… 5

2 Series resonance…………………………………………………………………………… 7

3 Parallel resonance…………………………………………………………………………. 10 4 a) T type Low Pass Filter………………………………………………………………….. 13 b) Pi type Low Pass Filter…………………………………………………………………. 16

5 a) T type High Pass filter………………………………………………………………….. 18 b) Pi type Low Pass Filter…………………………………………………………………. 21

6 Active Low Pass Filter…………………………………………………………………….. 23 7 Active High Pass Filter……………………………………………………………………. 26 8 Active Band Pass Filter…………………………………………………………………… 29 9 Active Band Elimination Filter……………………………………………………………. 32 10 a) T type Attenuator……………………………………………………………………….. 35 b) Pi type Attenuator………………………………………………………………………. 38

11 AM modulator…………………………………………………………………………….. 41 12 AM detector……………………………………………………………………………….. 44 13 FM modulator……………………………………………………………………………... 47 14 FM Detector……………………………………………………………………………….. 49 15 Pre-Emphasis network…………………………………………………………………….. 51 16 De-Emphasis network……………………………………………………………………... 54 17 a) Clipper circuit…………………………………………………………………………... 57 b) +ve and –ve clampers…………………………………………………………………... 59

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SYLLABUSSYLLABUSSYLLABUSSYLLABUS GENERAL EDUCATIONAL OBJECTIVES:

After the completion of this laboratory work, the student should be able to

1 Draw the circuit diagrams for resonant circuits, filters, attenuators, Equalizers, AM and FM modulators and

demodulators.

2 Assemble the above circuits.

3 Test the above circuits for their performance.

4 Analyze the above circuits. .

GRADED EXERCISES

1 Verification of Maximum power Transfer Theorem.

2 Plot the frequency response characteristics of series resonant Circuits and find their Bandwidth and Q factor.

3 Plot the frequency response characteristics of parallel resonant Circuits and find their Bandwidth and Q

factor.

4 Design, construct and test Constant K and PI type T section Low pass filter for the given cutoff frequency and

Ro.

5 Design, construct and test Constant K and PI type T section High Pass filter for the given cutoff frequency

and Ro.

6 Design, construct and test Active Low pass filter.

7 Design, construct and test Active High pass filter.

8 Design, construct and test Active Band Pass Filter (Using Dual or quad OPAMP)

9 Design, construct and test Active Band Reject Filter.

10 Design and construct T & PI type Attenuator for the given attenuation & Ro.

11 Construct AM modulator and determine modulation index, observe the AM signal.

12 Construct envelope AM detector and observe the effect of time constant RC.

13 Construct FM modulator and observe FM signal.

14 Construct FM Detector circuit using PLL 565 and observe the waveform.

15 Construct and test the operation of Pre-Emphasis network by plotting frequency response using OPAMP.

16 Construct and test the operation of De-Emphasis network by plotting frequency response. (Using OPAMP)

17 Construction and testing of following wave shaping circuits

i) Combinational clipper ii) +ve and –ve clampers

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SCHEME OF EVALUATIONSCHEME OF EVALUATIONSCHEME OF EVALUATIONSCHEME OF EVALUATION

1 Lab record 05

2 Writing Two circuit diagrams. 30

3 Conduction of one Experiment 30

4 Result. 15

5 Viva-voce 20

Total 100

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Expt. No: 1 Date:

MAXIMUM POWER TRANSFER THEOREM Aim: To verify maximum power transfer theorem. Apparatus: Power supply, multi meters, Resistor 1KΩ, Decade resistance box (DRB). Circuit diagram:

Fig (1.1) Symmetrical network

Tabular column:

Sl. No.

Load resistance RL, in Ohms

Current through load resistor, IL in m.A.

Voltage across load resistor, VL in volts

Output power Pout=VL.IL in m.W.

1 2 3 4 5 6 7 8 9 10 11 12

100 300 500 700 900 1k 2k 3k 4k 5k 7k 10k

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Theory: Statement: “In any symmetrical network, the power transferred from the source to the load will be maximum, when the source impedance is complex conjugate of load impedance.” Let Source impedance

Load impedance

For maximum power transfer, conditions are

Ri= RL Xi= -XL

We conclude that for maximum power transfer, load resistance must be equal to source resistance and load reactance must be opposite in nature i.e. if one is the positive reactance the other one should be negative (if one is inductive other one should be capacitive).

For dc circuits there will not be any reactive components. Therefore source impedance Zi= Ri and load impedance ZL= RL. Hence condition for maximum power transfer in dc circuits is Ri= RL. In the circuit diagram source resistance Ri is 1 K Ohm. Therefore power transferred from the source to the load is maximum when load resistance RL is equal to 1 K Ohm. Procedure:

1. Make the connections as shown in the circuit diagram. 2. Keep the input voltage Vin=10V 3. Set the DRB (RL) to 100 ohms. 4. Measure the current (IL) through load resistor and the voltage (VL) across load

resistor. Record these data in the tabular column. 5. Repeat step 4 for different values of RL. 6. For each reading, calculate output power using the formula Pout=VL.IL

Result:

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Expt. No: 2 Date:

SERIES RESONANCE Aim: To design the series resonant circuit for the given resonant frequency and to plot the frequency response characteristic of that series resonant circuit. Apparatus: Resistor 100 Ohms, Inductor 1m.H. (DIB), Capacitor 1µ.F. (DCB), Signal generator, Multi meter. Circuit diagram:

Fig (2.1) Series resonant circuit

Design: Given, Series resonant frequency= 5 KHz We know that, series resonant frequency, fs= 1/2π√LC Assume C= 1 µ.F.

Therefore, L= _________ m.H.

Nature of graph:

Band width = f2-f1

= _______Hz Fig (2.2) Universal resonance curve (Frequency response characteristic)

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Tabular column: Vin= 5V

Sl. No.

Frequency in Hz

Current I, in m.A.

1 2 3 4 5 6 7 8 9 10 11 12

100 200 300 500 700 900 1K 2K 3K 5K 7K 10K

Theory: The resonant circuit is a combination of R, L and C elements. The interesting characteristic of such circuit is that, they exhibit maximum or minimum current output at a particular frequency. This property is called as resonance. The frequency at wich resonance occurs is called as resonant frequency. The R, L and C components may be connected in series or in parallel. Accordingly, there are two types of resonant circuits. They are

1. Series resonant circuit 2. Parallel resonant circuit (Anti resonance)

Series resonance: Fig (2.1) shows series resonant circuit. Series resonance is a phenomenon in a series RLC circuit. At series resonance inductive reactance is equal to capacitive reactance. Hence inductive and capacitive reactances cancel out each other. The two reactances act as a short circuit since no voltage drops across them. All the applied voltage drops across resistor R. At resonance the net impedance of the circuit is purely resistive. Fig (2.2) shows frequency response characteristic of series resonant circuit. This characteristic is also called as universal resonance curve. It is the plot of current flowing through the resonant circuit v/s frequency. This curve is bell shaped curve, with its peak value at f=fs. For all other values of frequency current flowing through the circuit is less than Io. This is because for a frequency less than fs, capacitive reactance will be more and for a frequency more than fs inductive reactance will be more. In either case, the net impedance of the circuit is more than its resistance R. Hence the current is less. From the graph it is clear that current reduces on either side of fs gradually. Mark a point on current axis, where the current is equal to 0.707 X Ip. Extend

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that point on the plot. The extended line cuts the plot at two points. Extend both the points on to the frequency axis. We will get two frequencies f1 and f2. Observe that between f1 and f2 current flowing through the circuit is more, whereas below f1 and above f2 current is less. We conclude that the circuit is frequency selective. It allows more current for a range of frequencies and less current for all other frequencies. The range of frequency in which current is more is called pass band (between f1 & f2) and range of frequency in which current is less is called stop band. Frequencies f1 and f2 are called cut off frequencies because they separate pass band and stop band. Procedure:

1. Design the series resonant circuit for the given resonant frequency (Assume the value of C & calculate the value of L using appropriate formula).

2. Make the connections as shown in the circuit diagram. 3. Keep the input voltage Vin= 5V by varying the amplitude knob of signal generator

and maintain it constant throughout the experiment. 4. Vary the input frequency in steps and note down the corresponding current. 5. Plot the graph of current vs. frequency. 6. Find out the band width from the graph.

Result:

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Expt. No: 3 Date:

PARALLEL RESONANCE Aim: To design the parallel resonant circuit for the given resonant frequency and to plot the frequency response characteristic of that parallel resonant circuit. Apparatus: Resistor 100 Ohms, Inductor 1m.H. (DIB), Capacitor 1µ.F. (DCB), Signal generator, Multi meter. Circuit diagram:

Fig (3.1) Parallel resonant circuit

Design: Given, Parallel resonant frequency, fp= 5 KHz We know that, series resonant frequency, fp= 1/2π√LC Assume C= 1 µ.F.

Therefore, L= _________ m.H.

Nature of graph:

Fig (3.2) Frequency response characteristic of parallel resonance circuit

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Sl. No.

Frequency in Hz

Current I, in m.A.

1 2 3 4 5 6 7 8 9 10 11 12

100 200 300 500 700 900 1K 2K 3K 5K 7K 10K

Theory: Fig (3.1) shows parallel resonant circuit. Parallel resonance is a phenomenon in a parallel LC circuit. At parallel resonance inductive susceptance is equal to capacitive susceptance. Hence inductive and capacitive susceptances cancel out each other. At parallel resonance net admittance of the circuit is given by

Y= G+jbc-jbl

Where, ‘Y’ is admittance ‘G’ is conductance ‘bc’ is capacitive susceptance ‘bl’ is inductive susceptance.

At parallel resonance bl = bc. Therefore Y = G. We conclude that at parallel resonance net admittance of the circuit is equal to conductance only and is minimum. Since admittance is minimum at resonance impedance is maximum. Hence current at parallel resonance is minimum.

Fig (3.2) shows frequency response characteristic of parallel resonant circuit. It is also called as universal anti resonance curve. It is having inverted bell shape with its minimum current value at f = fp. At all other frequencies current flowing through the circuit is more. We conclude that parallel resonant circuit is also frequency selective. It allows less current at frequencies surrounding fp. i.e. parallel resonant circuit rejects these frequencies. Hence circuit is called rejecter circuit.

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Procedure: 1. Design the parallel resonant circuit for the given resonant frequency (Assume the

value of C & calculate the value of L using appropriate formula). 2. Make the connections as shown in the circuit diagram. 3. Keep the input voltage Vin= 5V by varying the amplitude knob of signal generator

and maintain it constant throughout the experiment. 4. Vary the input frequency in steps and note down the corresponding current. 5. Plot the graph of current vs. frequency.

Result:

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Expt. No: 4a Date: T-TYPE LOW PASS FILTER

Aim: To design, construct and test Constant K, T type Low Pass filter for the given cutoff frequency and Ro. Apparatus: Decade Inductance Box (DIB), Decade Capacitance Box (DCB), 600 Ohms resistor, spring board or tag board, CRO or Multi meter & wires. Circuit diagram:

Fig (4.1) T type LPF

Design: Given: Cut off frequency fc=2 KHz Characteristic impedance, Ro=600 Ohms We know that for HPF L= Ro/ Π fc C= 1/ Π fcRo There for, L= ____________ m H

L/2= ____________ m H

C= ____________ µ F

Nature of Graph:

Fig (4.2) Gain vs frequency and attenuation vs frequency characteristic

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Sl. No.

Frequency in Hz

Out put voltage, Vo in Volts

Gain= Vo/Vin Attenuation= Vin/Vo

1 2 3 4 5 6 7 8 9 10 11 12

100 200 300 500 700 900 1k 2K 3K 5K 7K 10K

Theory: Filter is an electronics circuit, which passes a specified band of frequencies and blocks unwanted frequencies. Pass band: Pass band is a band of frequencies passed by the filter. Stop band: Stop band is a band of frequencies blocked by the filter. Cut off frequency: It is a frequency which separates pass band and stop band.

Classifications of filters Based on the type of components used to construct the filter, they can be classified as

1. Passive Filter 2. Active Filter

Based on the nature, filters can be classified as 1. Low pass filter 2. High pass filter 3. Band pass filter 4. Band rejection or elimination filter

Based on the shape, filters can be classified as 1. T-Type filter 2. Π- type filter

Based on the relationship between series and shunt arm impedances, filters can be classified as

1. Constant K filter or prototype filter 2. m derived filter

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Low pass filter is a filter having one pass band and one stop band with one cut off frequency. It allows all the frequencies between zero and cut off frequency fc. On the other hand it blocks all the frequencies above the cut off frequency. Procedure:

1. Design the T- type low pass filter for the given cut off frequency and characteristic impedance.

2. Connections are made as shown in the circuit diagram choosing appropriate components.

3. Connect CRO to the signal generator and keep 5 V (p-p) input. 4. Vary the frequency of input signal in steps and note down the out put voltage. 5. Calculate gain and attenuation for each reading. 6. Plot the graph of gain vs. frequency and attenuation vs. frequency.

Result:

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Expt. No: 4b Date: Π-TYPE LOW PASS FILTER

Aim: To design, construct and test Constant K, Π- type low Pass filter for the given cutoff frequency and Ro. Apparatus: Decade Inductance Box (DIB), Decade Capacitance Box (DCB), 600 Ohms resistor, spring board or tag board, CRO or Multi meter & wires. Circuit diagram:

Fig (4.3) Π- type low Pass filter

Design: Given: Cut off frequency fc=2 KHz Characteristic impedance, Ro=600 Ohms We know that for HPF L= Ro/Π fc C= 1/ Π fcRo There for, L= ____________ m H C= ____________ µ F

C/2=____________ µ F Nature of Graph:


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Sl. No.

Frequency in Hz



1 2 3 4 5 6 7 8 9 10 11 12

100 200 300 500 700 900 1k 2K 3K 5K 7K 10K

Procedure:

1. Design the Π- type low pass filter for the given cut off frequency and characteristic impedance.



Result:

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Expt. No: 5a Date: T-TYPE HIGH PASS FILTER

Aim: To design, construct and test Constant K, T type High Pass filter for the given cutoff frequency and Ro. Apparatus: Decade Inductance Box (DIB), Decade Capacitance Box (DCB), 600 Ohms resistor, spring board or tag board, CRO or Multi meter & wires. Circuit diagram:

Fig (5.1) T type high pass filter

Design: Given: Cut off frequency fc=2 KHz Characteristic impedance, Ro=600 Ohms We know that for HPF L= Ro/4 Π fc C= 1/4 Π fc Ro There for, L= ____________ mH C= ____________ µF

Nature of Graph:


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Sl. No.

Frequency in Hz



1 2 3 4 5 6 7 8 9 10 11 12 13

100 200 300 500 700 900 1k 2K 3K 5K 7K 10K 20K

Theory: High pass filter is a filter having one pass band and one stop band with one cut off frequency. It allows all the frequencies lying above the cut off frequency fc. On the other hand it blocks all the frequencies below the cut off frequency fc. Gain and attenuation characteristics of high pass filter are shown in fig (5.2). We can observe that high pass filter has one stop band and one pass band. Cut off frequency fc separates pass band and stop bands.

T type passive high pass filter circuit is shown in fig (5.1). This circuit is designed for the cut off frequency 2 KHz. The circuit is terminated by its characteristic impedance Ro. The value of Ro is 600 Ohms. Here series arm contains capacitance and the shunt arm contains inductance. The Values of L and C are calculated by using the formulas

L= Ro/4 Π fc

C= 1/4 Π fc Ro

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Procedure: 1. Design the T- type high pass filter for the given cut off frequency and

characteristic impedance. 2. Connections are made as shown in the circuit diagram choosing appropriate

components. 3. Connect CRO to the signal generator and keep 5 V (p-p) input. 4. Vary the frequency of input signal in steps and note down the out put voltage. 5. Calculate gain and attenuation for each reading. 6. Plot the graph of gain vs. frequency and attenuation vs. frequency.

Result:

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Expt. No: 5b Date: Π-TYPE HIGH PASS FILTER

Aim: To design, construct and test Constant K, Π- type High Pass filter for the given cutoff frequency and Ro. Apparatus: Decade Inductance Box (DIB), Decade Capacitance Box (DCB), 600 Ohms resistor, spring board or tag board, CRO or Multi meter & wires. Circuit diagram:

Fig (5.3) Π- type High Pass filter

Design: Given: Cut off frequency fc=2 KHz Characteristic impedance, Ro=600 Ohms We know that for HPF L= Ro/4 Π fc C= 1/4 Π fcRo There for, L= ____________ mH C= ____________ µF

Nature of Graph:


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Sl. No.

Frequency in Hz



1 2 3 4 5 6 7 8 9 10 11 12 13

100 200 300 500 700 900 1k 2K 3K 5K 7K 10K 20K

Procedure:

7. Design the Π- type high pass filter for the given cut off frequency and characteristic impedance.



Result:

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Expt. No: 6 Date: ACTIVE LOW PASS FILTER

Aim: To design, construct and test active low pass filter using op-amp. Apparatus: op-amp µA741, Resistors- 1K, 10K, 10K pot or DRB, Capacitor-0.1µF, Signal generator, CRO or Multi meter, wires and bread board Circuit diagram:

Fig (6.1) Active low pass filter

Design: Amplifier gain Af= 1+(Rf/R1)

= _________

Given: Cut off frequency fc= 1KHz

We know that, fc= 1/2ΠRC

By assuming the value of capacitor, C=0.1µF

Therefore, R= _________KΩ

Nature of graph:


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Sl. No.

Frequency in Hz

Output voltage, Vo in Volts


1 2 3 4 5 6 7 8 9 10 11 12 13

100 200 300 500 700 900 1k 2K 3K 5K 7K 10K 20K

Theory: An electric filter is often frequency selective circuit that passes a specified band of frequencies and blocks or attenuates signals of frequencies outside this band. Filters can be either active or passive. Elements used in passive filters are resistors, capacitors and inductors. On the other hand active filters employ op-amps in addition to resistors and capacitors. An active filter offers the following advantages over a passive filter:

a) Gain and frequency adjustment flexibility: Since the op-amp is capable of providing a gain, the input signal is not attenuated as it is in passive filter. In addition active filter is easier to adjust.

b) No loading problem: Because of the high input resistance and low out put resistance of the op-amp, the active filter does not cause loading of the source or load.

c) Cost: Typically, active filters are more economical than passive filters. This is because variety of cheaper op-amps and the absence of inductor. Fig. shows the active low pass filter that uses an RC network for filtering. Note

that the op-amp is used in the non-inverting configuration. Resistors R1 and Rf determine gain of the filter. Procedure:

1. Design the active low pass filter for the given cut off frequency (assume the value of C and find out the value of R using the formula fc= 1/2ΠRC).

2. Make the connections as shown in the circuit diagram.

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3. Keep the input voltage Vin= 5V by varying the amplitude knob of signal generator. 4. Connect CRO or Multi meter across load resistor RL 5. Vary the input frequency in steps and note down the output voltage, Vo for each

frequency. 6. Calculate gain and attenuation for each reading. 7. Plot the graph of gain vs. frequency and attenuation vs. frequency.

Result:

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Expt. No: 7 Date: ACTIVE HIGH PASS FILTER

Aim: To design, construct and test active low pass filter using op-amp. Apparatus: op-amp µA741, Resistors- 1K, 10K, 10K pot or DRB, Capacitor-0.1µF, Signal generator, CRO or Multi meter, wires and bread board Circuit diagram:

Fig (7.1) Active high pass filter

Design: Amplifier gain Af= 1+(Rf/R1)

= _________

Given: Cut off frequency fc= 1KHz

We know that, fc= 1/2ΠRC

By assuming the value of capacitor, C=0.1µF


Nature of graph:


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Sl. No.

Frequency in Hz



1 2 3 4 5 6 7 8 9 10 11 12 13

100 200 300 500 700 900 1k 2K 3K 5K 7K 10K 20K

Theory: The basic operation of an Active High Pass Filter (HPF) is exactly the same as that for its equivalent RC passive filter circuit, except that this type of circuit has an operational amplifier or op-amp included within its design for amplification and gain control. Like the active Low Pass filter circuit, the simplest form of an active high pass filter is to connect a standard inverting or non-inverting operational amplifier to the basic RC high pass passive filter circuit. However, unlike Passive High Pass Filters which have an "infinite" frequency response, the maximum pass band frequency response of an active High Pass Filter is limited to the characteristics or bandwidth of the op-amp being used within the circuit design. The maximum frequency response of an op-amp is limited to the Gain/Bandwidth product or open loop voltage gain (A V) of the operational amplifier being used.

Fig (7.1) shows the active high pass filter that uses an RC network for filtering.

Note that the op-amp is used in the non-inverting configuration. Resistors R1 and Rf determine gain of the filter. Procedure:

1. Design the active high pass filter for the given cut off frequency (assume the value of C and find out the value of R using the formula fc= 1/2ΠRC).

2. Make the connections as shown in the circuit diagram. 3. Keep the input voltage Vin= 5V by varying the amplitude knob of signal generator. 4. Connect CRO or Multi meter across load resistor RL.

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5. Vary the input frequency in steps and note down the output voltage (Vo) for each frequency.

6. Calculate gain and attenuation for each reading. 7. Plot the graph of gain vs. frequency and attenuation vs. frequency.

Result:

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Expt. No: 8 Date:

ACTIVE BAND PASS FILTER Aim: To design, construct and test active band pass filter. Apparatus: : op-amp µA741, Resistors- 1K, 10K, 22K pot or DRB, Capacitor-0.1µF, Signal generator, CRO or Multi meter, wires and bread board. Circuit diagram:

Fig (8.1) Active band pass filter

Design: Gain of A1= 1+(Rf/R1)

= _________

Gain of A2= 1+(R'f/R1')

= _________

Gain of BPF = (Gain of A1) X (Gain of A2)

= _________

Given: Lower cut off frequency, fL= 200Hz

Upper cut off frequency, fH= 1KHz

We know that, lower cut off frequency, fL= 1/2ΠRC

Upper cut off frequency, fH= 1/2ΠR'C'

By assuming the value of capacitors, C= C'= 0.1µF


R'= _________KΩ

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Nature of graph:

Fig (8.2) Gain vs frequency characteristic


Sl. No.

Frequency in Hz


Gain= Vo/Vin

1 2 3 4 5 6 7 8 9 10 11 12 13 14 14 14 17

10 30 50 70 100 200 300 500 700 900 1K 2K 3K 5K 7K 10K 20K

Theory: Principal characteristic of any filter is its ability to pass frequencies relatively unattenuated over a specified band or spread of frequencies called the "Pass band". For a low pass filter this pass band starts from 0Hz and continues up to the specified cut-off frequency. Equally, for a high pass filter the pass band starts from this cut-off frequency and continues up to infinity or the maximum open loop gain for an active filter.

However, the Active Band Pass Filter is slightly different in that it will only pass

frequencies or signals within a certain "Band" or range of frequencies that are set

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between two cut-off or corner points labeled "lower cut off frequency" (ƒL) and "higher cut off frequency" (ƒH) while attenuating any signals outside of these two points.

A simple Active Band Pass Filter can be easily made by cascading together a

single Low Pass Filter with a single High Pass Filter. The cut-off or corner frequency of the low pass filter (LPF) is higher than the cut-off frequency of the high pass filter (HPF) and the difference between the cut off frequencies will determine the "Bandwidth" of the filter.

The high or upper cut of frequency (ƒH) and the lower cut-off frequency (ƒL) are

calculated as same as before in the standard low and high pass filter circuits. The amplifier defines the overall voltage gain of the circuit. The Bandwidth of the circuit is the difference between the upper and lower cut off frequencies. For example, if the cut-off frequencies are at 200Hz and 1KHz then the bandwidth of the filter would be given as: Bandwidth (BW) = 1KHz – 200Hz = 800Hz.

Procedure:

6. Design the band pass filter for the given cut off frequencies (Assume the values of C & C' and calculate the values of R & R' using appropriate formulas).

7. Make the connections as shown in the circuit diagram. 8. Keep the input voltage Vin= 5V by varying the amplitude knob of signal generator. 9. Connect CRO or Multi meter across load resistor RL. 10. Vary the input frequency in steps and note down the output voltage (Vo) for each

frequency. 11. Calculate the gain for each reading. 12. Plot the graph of gain vs. frequency.

Result:

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Expt. No: 9 Date:

ACTIVE BAND ELIMINATION FILTER Aim: To design, construct and test active band elimination filter. Apparatus: : op-amp µA741, Resistors- 1K, 10K, 3.3K, 22K pot or DRB, Capacitor-0.1µF, Signal generator, CRO or Multi meter, wires and bread board. Circuit diagram:

Fig (9.1) Active band elimination filter Design: Gain of A1= 1+(Rf/R1)

= _________

Gain of A2= 1+(R'f/R1')

= _________

Gain of BPF = (Gain of A1) X (Gain of A2)

= _________

Given: Lower cut off frequency, fL= 200Hz

Upper cut off frequency, fH= 1KHz

We know that, lower cut off frequency, fL= 1/2ΠR'C'

Upper cut off frequency, fH= 1/2ΠRC

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By assuming the value of capacitors, C= C’= 0.1µF


R’= _________KΩ

Nature of graph:

Fig (9.2) Gain vs frequency characteristic

Tabular column:

Vin= 5V Sl. No.

Frequency in Hz


Gain= Vo/Vin

1 2 3 4 5 6 7 8 9 10 11 12 13 14 14 14 17

10 30 50 70 100 200 300 500 700 900 1K 2K 3K 5K 7K 10K 20K

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Theory: A Band-Stop Filter is a circuit that allows most frequencies to pass, but blocks or attenuates a certain range or band of frequencies. It is also known as a ‘band-elimination’ filter or a ‘band-rejection filter’. The band-stop filter is the opposite of the band-pass filter. The range of frequencies that a band-stop filter blocks is known as the ‘stopband’, which is bound by a lower cut-off frequency fL and a higher cut-off frequency fH.

An ideal band-stop filter is one whose stopband is completely rejected by it, while

allowing all other frequencies to pass unchanged. In an ideal band-stop filter, the transition of the response from outside the stopband to within the stopband and vice versa is instantaneous. Of course, an ideal band stop filter doesn't exist in the real world, i.e., complete attenuation within the stopband can not be achieved while frequencies outside the stopband undergo some level of attenuation.

Fig (1) shows a band reject filter using a low pass filter, a high pass filter and a summing amplifier. To realize the band reject filter the cut off frequency of the high pass filter must be larger than the cut off frequency of the low pass filter. In addition, gain of both the high pass and low pass sections must be equal. The frequency response of the band reject filter is also shown in fig (2). Procedure:

1. Design the band elimination filter for the given cut off frequencies (Assume the values of C & C' and calculate the values of R & R' using appropriate formulas).

2. Make the connections as shown in the circuit diagram. 3. Keep the input voltage Vin= 5V by varying the amplitude knob of signal generator. 4. Connect CRO or Multi meter across load resistor RL. 5. Vary the input frequency in steps and note down the output voltage (Vo) for each

frequency. 6. Calculate the gain for each reading. 7. Plot the graph of gain vs. frequency.

Result:

35

Expt. No: 10a Date:

T TYPE ATTENUATOR

Aim: To design, construct and test T type attenuator for the given attenuation of 10dB and Ro 600Ω. Apparatus: Signal generator, CRO or Multi meter, Resistors 330Ω, 390Ω & 600Ω Design: Given, Ro= 600Ω Attenuation = 10dB We know that, Attenuation in dB= 20 log N 10dB= 20 log N Therefore, log N= 10/20 N= Antilog [10/20] N= 3.16 We know that, R1= [(N-1)/ (N+1)] Ro

R1= [(3.16-1)/ (3.16+1)] X 600

Therefore, R1 = 311.53Ω (Choose 330Ω resistor)

We know that, R2= [(2N)/ (N²-1)] Ro

R2= [(2X3.16)/ (3.16²-1)] X 600

Therefore, R1 = 422Ω (Choose 390Ω resistor)

Circuit diagram:

Fig (10.1) T type attenuator

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Tabular column: Vin= ______V

Sl. No.

Frequency in Hz


Attenuation in dB= 20 log( Vin/Vo)

1 2 3 4 5 6 7 8 9 10 11 12

100 200 300 500 700 900 1K 2K 3K 5K 7K 10K

Nature of graph:

Fig (10.2) Attenuation vs frequency characteristic

Theory: Attenuator is an electrical network used to reduce the signal level by a given amount. The function of attenuator is exactly opposite to that of an amplifier. An amplifier is used to increase the signal level. Thus the attenuation is reverse of amplification.

Attenuation = Vin/Vout or Iin/Iout or Pin/Pout

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For an attenuator attenuation is always greater than 1. (i.e. Vin/Vout > 1, Vout/Vin < 1). But for amplifier attenuation is always less than 1 and gain is always greater than 1. (i.e. Vin/Vout < 1, Vout/Vin > 1).

There are three units for attenuation. They are Bel, DeciBel (dB) and Neper.

A symmetrical T-type resistive attenuator is as shown in the fig (10.1). Note that this circuit is terminated by Ro. If the input resistance Rin is equal to the terminated resistance (i.e. Rin=Ro), then Ro becomes characteristic resistance. Procedure:

1. Calculate the values of R1 and R0 by using appropriate formulas. 2. Make the connections as shown in the circuit diagram choosing appropriate

components. 3. Keep the input voltage Vin= 2V and maintain it constant throughout the

experiment. 4. Vary the frequency in steps and note down the output voltage Vo, for each

frequency. 5. Calculate the attenuation for each reading by using the formula, Attenuation in

dB= 20 log( Vin/Vo) 6. Plot the graph of frequency v/s attenuation.

Result:

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Expt. No: 10b Date:

π TYPE ATTENUATOR Aim: To design, construct and test π type attenuator for the given attenuation of 10dB and Ro 600Ω. Apparatus: Signal generator, CRO or Multi meter, Resistors 330Ω, 390Ω & 600Ω Design: Given, Ro= 600Ω Attenuation = 10dB We know that, Attenuation in dB= 20 log N 10dB= 20 log N Therefore, log N= 10/20 N= Antilog [10/20] N= 3.16 We know that, R1= [((N²-1))/2N] Ro

R1= [(3.16²-1)/ (2X3.16)] X 600

Therefore, R1 = 853Ω (Choose 820Ω resistor)

We know that, R2= [(N+1)/ (N-1)] Ro

R2= [(3.16+1)/ (3.16-1)] X 600

Therefore, R1 = 1.155KΩ (Choose 1.2KΩ resistor)

Circuit diagram:

Fig (10.3) Π- type attenuator

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Tabular column:

Vin= ______V

Sl. No.

Frequency in Hz


Attenuation in dB= 20 log( Vin/Vo)

1 2 3 4 5 6 7 8 9 10 11 12

100 200 300 500 700 900 1K 2K 3K 5K 7K 10K

Nature of graph:

Fig (10.4) Attenuation vs frequency characteristic

Theory:

Types of attenuators 1. Resistive and Capacitive attenuators 2. Fixed and Variable attenuators 3. Symmetrical and asymmetrical attenuators

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4. T-type and π-type attenuators A π-type symmetrical resistive attenuator is as shown in the fig (10.3). R1 is

series arm resistance and R2 is shunt arm resistance. Note that the circuit is terminated by the resistance Ro. If the input resistance of the circuit is also equal to Ro, then Ro becomes characteristic resistance.

Procedure:

1. Calculate the values of R1 and R0 by using appropriate formulas. 2. Make the connections as shown in the circuit diagram choosing appropriate

components. 3. Keep the input voltage Vin= 2V and maintain it constant throughout the

experiment. 4. Vary the frequency in steps and note down the output voltage Vo, for each

frequency. 5. Calculate the attenuation for each reading by using the formula, Attenuation in

dB= 20 log( Vin/Vo) 6. Plot the graph of frequency v/s attenuation.

Result:

41

Expt. No: 11 Date: AM MODULATOR

Aim: To construct AM modulator and to determine modulation index ‘m’. Apparatus: Resistors 1KΩ, Capacitors 0.01µF, Diode, Decade Inductance Box, Signal generators, CRO, tag board, wires. Circuit diagram:

Fig (11.1) AM modulator circuit

Tank circuit design: Frequency of oscillation, fc= 1/2π√L1C1 Assume C1= 0.01uF We know that carrier frequency, fc = 10 KHz Therefore L1 = _________ m.H. Tabular column:

Sl No. Vmax, in Volts Vmin, in Volts Modulation index, m in percentage

1

2

3

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Waveforms:

Fig (11.2) Wave forms

a) Carrier signal b) Modulating signal c) AM signal

Theory: Amplitude modulation: “Amplitude modulation is a process in which the amplitude of high frequency carrier is varied in accordance with the amplitude of a low frequency modulating signal.” In amplitude modulation, the information signal varies the amplitude of the carrier sine wave. In other words, the instantaneous value of the carrier amplitude changes in accordance with the amplitude of the modulating signal. Fig (11.1) shows amplitude modulator circuit. It is a circuit which generates amplitude modulation. Fig (11.2) shows a single frequency sine wave modulating a high frequency carrier signal.

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Note that the carrier frequency remains constant during the modulation process but its amplitude varies in accordance with the modulating signal. An increase in the modulating signal amplitude causes the amplitude of the carrier to increase. An increase or decrease in the amplitude of the modulating signal causes a corresponding increase or decrease in both the positive and negative peaks of the carrier amplitude. Modulation index: In order for proper AM to occur, the modulating signal voltage must be less than carrier voltage. Therefore, the relationship between the amplitudes of the modulating signal and the carrier is important. This relationship is expressed in terms of a ratio known as the modulation index, m.

Modulation index is the ratio of the modulating signal voltage to the carrier voltage.

Modulation index, m=Vm/Vc, or

The modulation index should be a number between 0 and 1. If the amplitude of the modulating voltage is higher than the carrier voltage, m will be greater than 1. This will cause severe distortion of the modulated wave form. This condition is called over modulation. When m=1 the condition is called full modulation. Whenever the modulation index is multiplied by 100, the degree of modulation is expressed as a percentage. In this case modulation index is called percentage modulation.

%m= (Vm/Vc) x100 Procedure:


2. Apply 20V/1KHz modulating signal and 20V/10KHz carrier signal. 3. Connect the CRO across the tank circuit and observe the AM waveform. 4. Measure Vmax and Vmin from the AM waveform and calculate the

modulation index, m. 5. Repeat the above step for different values of modulating signal and carrier

signal voltages. 6. Sketch modulating signal, Carrier signal and AM wave forms.

Result:

44

Expt. No: 12 Date: AM DETECTOR

Aim: To construct envelope detector and to observe the demodulated waveform. Apparatus: Resistors 1KΩ, Capacitors 0.01µF & 0.1µF, Diodes, Decade Inductance Box, Decade resistance box, Signal generators, CRO, tag board, wires. Circuit diagram:

Fig (12.1) AM Modulator

Fig (12.2) AM Detector (Envelope detector)

Filter design: Filter cut off frequency, fm = 1/2πR4C2 Assume C2 = 0.1uF We know that fm = 1 KHz Therefore, R4 = _________ K Ohms.

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Waveforms:


a) AM signal b) Demodulated signal Theory: Fig (12.1) shows the amplitude modulator circuit and fig (12.2) shows AM demodulator (detector) circuit. A demodulator is a circuit that accepts a modulated signal and recovers the original modulating information. A demodulator circuit is the key circuit in the radio receiver. Diode detector: The simplest and most widely used amplitude demodulator is the diode detector shown in fig (12.2). The AM signal is applied to the rectifier circuit consisting of diode, capacitors and resistors. The diode conducts when the negative half cycles of the AM signals occur. During the positive half cycles, the diode is reverse biased and no current flows through it.

To recover the original modulating signal filter is connected after the diode. The filter is designed such that capacitors have very low impedance at the carrier frequency. At the frequency of the modulating signal, they have much higher impedance. The result

46

is that capacitors effectively short or filter out the carrier, thereby leaving the original modulating signal. The fig 12.3b shows the demodulated signal. Procedure:


2. Connect the out put of AM modulator to the in put of envelope detector. 3. Connect the CRO to the output of envelope detector. 4. Observe the demodulated wave form, measure the frequency of this

waveform and compare it with the original modulating signal. 5. Sketch AM wave and demodulated waveform.

Result:

47

Expt. No: 13 Date: FM MODULATOR

Aim: To study FM modulator and to observe the FM wave form. Apparatus: Silicon micro system’s ‘Frequency modulation and Demodulation’ kit, and CRO. Block diagram:

FM signal Modulating (CH2 of CRO) Signal (CH1 of CRO) Fig (13.1) Block diagram of FM modulator

Waveforms:


a) Carrier signal b) Modulating signal c) FM signal

Theory: Frequency modulation: “Frequency modulation is a process in which the frequency of a high frequency carrier is varied in accordance with the amplitude of a low frequency modulating signal.” In FM, the carrier amplitude remains constant, while the carrier frequency is changed by the modulating signal. As the amplitude of the information

Modulating signal

generator

FM modulator

48

signal varies, the carrier frequency shifts in proportion. As the modulating signal amplitude increases, the carrier frequency increases. If the amplitude of the modulating signal decreases, the carrier frequency decreases. The reverse relationship can also be implemented. A decreasing modulating signal will increase the carrier frequency above its center value; whereas an increase in modulating signal will decrease the carrier frequency below its center value. As the modulating signal amplitude varies, the carrier frequency varies above and below its normal center frequency with no modulation. The amount of change in carrier frequency produced by the modulating signal is known as the frequency deviation. Maximum frequency deviation occurs at the maximum amplitude of the modulating signal.

The frequency of the modulating signal determines how many times per second the carrier frequency deviates above and below its nominal center frequency 100 times per second. This is called the frequency deviation rate.

An FM signal is illustrated in fig (13.2c). With no modulating signal

applied, the carrier frequency is a constant amplitude sine wave at its normal constant center frequency. The modulating information signal [Fig (13.2a)] is a low frequency sine wave. As the sine wave goes positive, the frequency of the carrier increases proportionately. The highest frequency occurs at the peak amplitude of the modulating signal. As the modulating signal amplitude decreases, the carrier frequency decreases. When the modulating signal is at zero amplitude, the carrier will be at its center frequency point.

When the modulating signal goes negative, the carrier frequency will

decrease. The carrier frequency will continue to decrease until the peak of the negative half cycle of the modulating sine wave is reached. Then, as the modulating signal increases towards zero, the frequency will again increase. Procedure:

1. Connect the trainer kit to the mains supply and switch ON. 2. Observe the modulating signal at ‘modulator out put’ by varying ‘frequency

control’ & ‘amplitude control’ knob. 3. Observe the unmodulated carrier signal at ‘FM out’. 4. Connect the ‘modulator O/P’ to the ‘modulating I/P’. 5. Connect CH1 of the CRO to the modulating I/P & CH2 of CRO to the ‘FM

out’ as shown in the block diagram. Observe FM wave & compare it with modulating signal.

Result:

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Expt. No: 14 Date: FM DEMODULATOR

Aim: To study FM demodulator and to observe the demodulated wave form. Apparatus: Silicon micro system’s ‘Frequency modulation and Demodulation’ kit, and CRO. Block diagram:

Modulating Demodulated O/P

Signal (CH1 of CRO) (CH2 of CRO)

Fig (14.1) Block diagram of FM demodulator Waveforms:

Fig (14.2) Wave forms a) Modulating signal (CH1 of CRO) b) FM signal c) Demodulated signal (CH2 of CRO)

Theory: Fig (14.1) shows the block diagram of FM demodulator. Demodulation is the process of recovering the low frequency modulating signal. Here in FM demodulator the frequency modulated signal is inputted. The output of the demodulator is the original low frequency modulating signal. Fig (14.2) shows original modulating signal, FM signal and demodulated signal.

Modulating signal

generator

FM

modulator

FM demodulator

50

There are literally dozens of circuits used to demodulate or detect FM signals. The well-known Foster-Seeley discriminator and the ratio detector were among the most widely used frequency demodulators at one time, but today these circuits have been replaced with more sophisticated IC demodulators. Anyhow, they are still found in older equipment. The most widely used detectors today include the pulse-averaging discriminator, the quadrature detector and the phase locked loop (PLL). The PLL is the best of all frequency demodulators in use. Its ability to provide frequency selectivity and filtering give it a signal to noise ratio superior to any other type of FM detector. The linearity of the VCO ensures the highly accurate reproduction of the original modulating signal. Although PLLs are complex, they are easy to apply because they are readily available in low cost IC form. Procedure:

1. Connect the trainer kit to the mains supply and switch ON. 2. Observe the modulating signal at ‘modulator out put’ by varying ‘frequency

control’ & ‘amplitude control’ knob. 3. Observe the unmodulated carrier signal at ‘FM out’. 4. Connect the ‘modulator O/P’ to the ‘modulating I/P’. 5. Connect ‘FM out’ to ‘FM in’ of demodulator. 6. Connect CH1 of CRO to the ‘modulating I/P’ & CH2 to the ‘demodulator

O/P’ as shown in the block diagram. Observe the demodulated wave form and compare it with the modulating signal.

Result: Expt. No: 15 Date:

PRE-EMPHASIS

51

Aim: To construct and test the operation of Pre-Emphasis network by plotting frequency response using OPAMP. Apparatus: Op-amp, power supply, Resistors 820 Ω, 100 Ω, 2.2 K Ω & 15K Ω Capacitor 0.1µF, Signal generator, wires. Circuit diagram:

Fig (15.1) Circuit diagram and frequency response of pre-emphasis

Design: Given, Lower break frequency, fl=2.1 KHz, Upper break frequency fu=15 KHz We know that, fl=1/2пrC and fu=1/2пRC Choose C=0.1µF Therefore, r=757.88 Ω (choose 820 Ω resistor) R=106.10 Ω (choose 100 Ω resistor) Also, r/R = Rf/R1 Therefore, 757.88/106.10 = Rf/R1 7.14= Rf/R1 7.14R1= Rf Choose R1=2.2 K Ω Therefore Rf= 15.71 K Ω (choose 15K Ω resistor)

52

Tabular column: Input voltage Vi= 2V

Frequency in Hz


Gain= Vo/Vi

Gain in dB= 20log(Vo/Vi)

100 200 300 500 700 1k 2k 3k 5k 7k 10k 13k 15k 17k 20k

Theory: Even though FM is immune to noise, noise still interferes with an FM signal. This is particularly true for the high frequency components in the modulating signal. To overcome this problem, most FM systems use a technic known as pre-emphasis. At the transmitter, the modulating signal is passed through a simple network which amplifies the high frequency components more than the low frequency components. The simplest form of such a circuit is the high pass filter. This circuit is shown in fig (15.1). The pre-emphasis curve is also shown. Here fl is lower break frequency and fu is upper break frequency. The pre-emphasis circuit linearly enhances frequencies higher than fl. After the frequency fu the signal enhancement flattens out. The frequency fu is computed with the expression fu= 1/2пRC This frequency is usually set at high value beyond the audio range. Procedure:

1. Design the pre-emphasis network for the given lower and upper break frequencies. 2. Apply input sine wave of 2V peak to peak.

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3. Vary the frequency in steps and note down the output voltage, Vo 4. Calculate the gain and gain in dB for each reading. 5. Plot the graph of gain in dB v/s frequency.

Result: Expt. No: 16 Date:

DE-EMPHASIS

54

Aim: To construct and test the operation of De-Emphasis network by plotting frequency response using OPAMP. Apparatus: Op-amp, power supply, Resistors 820Ω & 1KΩ, Capacitor 0.1µF, Signal generator, wires. Circuit diagram:

Fig (16.1) Circuit diagram and frequency response of de-emphasis

Design: Cutoff frequency, fc=2.1 KHz We know that, fc=1/2пRC Choose C=0.1µF Therefore, R=757.88 Ω (choose 820 Ω resistor) Choose Rf= R1= 1KΩ Tabular column:

Input voltage Vi= 2V

55

Frequency in Hz


Gain= Vo/Vi

Gain in dB= 20log(Vo/Vi)

100 200 300 500 700 1k 2k 3k 5k 7k 10k 13k 15k 17k 20k

Theory: In telecommunication, de-emphasis is the complement of pre-emphasis, in the anti-noise system called emphasis. Emphasis is a system process designed to decrease, (within a band of frequencies), the magnitude of some (usually higher) frequencies with respect to the magnitude of other (usually lower) frequencies in order to improve the overall signal-to-noise ratio by minimizing the adverse effects of noise.

In serial data transmission, de-emphasis has a different meaning, which is to reduce the level of all bits except the first one after a transition. That causes the high frequency content due to the transition to be emphasized compared to the low frequency content which is de-emphasized. This is a form of transmitter equalization; it compensates for losses over the channel which is larger at higher frequencies.

De-emphasis is commonly used in audio digital recording, record cutting and FM radio transmission.

Procedure:

1. Design the de-emphasis network for the cut off frequency of 2.1 KHz.

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2. Apply input sine wave of 2V peak to peak. 3. Vary the frequency in steps and note down the output voltage, Vo 4. Calculate the gain and gain in dB for each reading. 5. Plot the graph of gain in dB v/s frequency.

Result: Expt. No: 17a Date:

CLIPPER CIRCUIT

57

Aim: To construct and test combinational clipper circuit. Apparatus: Signal generator, diode OA79, resistor 1kΩ, power supply and CRO. Circuit diagram:

Fig (17.1) Combinational clipper circuit

Waveforms:


a) Input waveform b) Output waveform Theory:

A clipper circuit clips a fraction of its input signal keeping the remaining part of the signal unchanged. The circuit consists of resistor, diodes and external dc sources. A positive clipper clips a fraction of positive half cycles where as negative clipper clips a fraction of negative half cycles. A combinational clipper clips fraction of both positive and negative half cycles. Combinational clipper is a most general form of

58

diode clipper. The circuit diagram of this clipper is shown in fig (17.1). Here the clipping occurs at the level of the clipping voltage, V1 and V2. Classification: Practical clippers may be classified into two types: (a) Shunt Clippers, and (b) Series Clippers. Application: It is used in television sets and FM receivers. It is also used for amplifier and different types of op-amps through which we can do some mathematical operations. Procedure:

1. Make the connections as shown in the circuit diagram using appropriate components.

2. Apply 1 KHz, 10V sinusoidal signal using signal generator. 3. Connect the CRO as shown and observe the output waveform.

Result: Expt. No: 17b Date:

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CLAMPER CIRCUIT Aim: To construct and test positive and negative clampers. Apparatus: Signal generator, diode OA79, 0.1µF capacitor and CRO. Circuit diagram:

Fig (17.3) Positive clamper Fig (17.4) Negative clamper Waveforms:


Theory:

Clamper is a circuit that "clamps" a signal to a different dc level. The different types of clampers are positive, negative and biased clampers. A clamping network must have a capacitor and a diode.

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Positive Clamper: The circuit for a positive clamper is shown in the fig (17.3). During the negative half cycle of the input signal, the diode conducts and acts like a short circuit. The capacitor is charged to the peak value of input voltage and it behaves like a battery. During the positive half of the input signal, the diode does not conduct and acts as an open circuit. This gives a positively clamped voltage. Negative Clamper: The circuit for a negative clamper is shown in the fig (17.4). During the positive half cycle the diode conducts and acts like a short circuit. The capacitor charges to peak value of input voltage. During this interval the output Vo which is taken across the short circuit will be zero. During the negative half cycle, the diode is open. Procedure:

1. Make the connections as shown in the circuit diagram using appropriate components.

2. Apply 1 KHz, 10V sinusoidal signal using signal generator. 3. Connect the CRO across diode and observe the output waveform.

Result:

Analog Communication Lab Manual 01, 2013 · 1 Kanara College Society (R) Shree Vidyadhiraj Polytechnic Kumta (N.K), 581343 Department of Electronics and Communication Engg. Analog

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