Ee-315 Electric Filter _2011

8/3/2019 Ee-315 Electric Filter _2011

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PRACTICAL WORK BOOKFor Academic Session 2011

ELECTRIC FILTERS (EE-315)For

T.E (EE) & T.E (TC)

Name:

Roll Number:

Class:

Batch: Semester/Term:

Department :

Department of Electrical EngineeringNED University of Engineering & Technology


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Electric Filters ContentsNED University of Engineering and Technology Department of Electrical Engineering

Revised 2011MMAB

CONTENTS

Lab.

No.

Da te d

List of Experiments

Pa ge

No .

Re ma rk s

1Introduction to filters

1

2To investigate inverting and non-

inverting configurations of OP-AMP7

3To investigate the Frequency response of

741 OP-AMP in close-loop11

4To implement First order low pass active

filter17

5 To implement First order High passactive filter

23

6 To implement higher order filters usingcascade connection.

29

7 To obtain low pass response usingBiquad circuit

35

8 To obtain band pass response using

Biquad circuit

41

9 To obtain high pass response usingBiquad circuit

47

10 To obtain band stop response usingBiquad circuit

53

11 To implement Generalized impedanceconverter circuit.

59

12 Demonstration of a voice processingMATLAB based program.

65

13LAB PROJECT(a)

To design and implement BASS control

circuit.67

14LAB PROJECT(b)

To design and implement MID control

circuit.70

15LAB PROJECT(c)

To design and implement TREBLE

control circuit.

73


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Electric Filters Introduction to FiltersNED University of Engineering and Technology Department of Electrical Engineering

1

LAB SESSION 01

Introduction to Filters

(Design Terminologies)OBJECTTo familiarize with the basic filter terminologies and understand the behavior of basic

filters circuits.

APPARATUSBread-board, 1K resistor, 0.1F capacitor, Oscilloscope and a function generator.

THEORY

Electric filters as the name implies are circuits capable to allow or reject signals. Thedecision of accepting or rejecting a signal is based on its frequency. The basic idea

behind filter theory is that in time domain a single non-sinusoidal periodic signal iscomposed of an infinite number of sinusoids and using filter circuits we may select a

desired frequency signal. The idea is illustrated below

Consider a signal shown in figure (1).

The signal is composed of threedifferent frequencies each of different

magnitude. The component signals are

shown subsequently in figure (2).

Normally the signal we encounter is intime domain, which is the sum of

other signals of different frequencies.

The component signals cant be

extracted directly, since they all add

up at a single instant of time. Whereasin frequency domain these signals are treated as separate signals , so we design

circuits that could handle frequency based characteristic of the signals . Processing the

resultant signal can lead us to any of the desired component.

Fig 1

Fig 2


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Signal Source

R

L

Filter design PhilosophyAs mentioned above we design our system based on the information of signal

frequencies. Design approach is simple; we have certain known ranges of input signals

that contain the required signal. In short, we know system input and output. Based on

the frequency information we select our system components. Its obvious to expect that

our circuit have frequency dependent elements.Note: we dont cut other signals from the input signal in filters. We just suppress them

and amplify the desired frequency.

Based on this information lets discuss frequency response of a simple RL circuit.

Consider that we have a signal source that contains number of frequency components.

As mentioned previously, every frequency is treated separately in frequency domain. If

we take the output across the inductor, inductor will behave short circuited for the low

frequency signals, hence lower frequency will be by passed by the inductor. Whereas

inductor wont bypass high frequency signals (open circuited for high frequencies) andtheir magnitude wont decrease at the output terminal. Since the circuit wont let low

frequencies to propagate at the output and allows only high frequencies, it is a HIGHPASS filter circuit. The output signal will be a processed high frequency signal with

little low frequency components.

The frequency response of the typical HIGH PASS circuit is shown below. Themagnitude of low frequency signal is smaller than the high frequency signal.

Lets define certain terminologies to describe frequency response of the system.


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Design Terminologies

Pass Band:

The range of frequencies at which the output is little attenuated is called as pass band.

We shall treat these signals as the accepted signals.

Stop Band:The range of frequencies at which the output is significantly attenuated is called as stop

band. We shall treat these signals as rejected signals or unwanted signals.

Cutoff Frequency:The frequency associated with the boundary between stop band and an adjacent pass

band is called as the cutoff frequency. It is the frequency at which the output is 0.707

times the maximum value in the pass band.

Lets review the plot in terms of these definitions

Besides HIGH PASS filters, we have LOW PASS, BAND PASS and BAND STOP

filters. We shall study each response in detail. Right now lets consider another filter

circuit based on RC circuit . You are required to plot the frequency response for the

circuit, and determine the cutoff frequency of your circuit.

Stop

Band PASS BAND

Cutofffrequency


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CIRCUIT DIAGRAM

PROCEDURE Set the function generator to sinusoidal function with peak value of 1 volt

Connect the circuit elements as shown in the above diagram

Vary the frequency of the function generator from 100Hz to 2000Hz and

measure the peak values of the output signal form oscilloscope.

OBSERVATIONSBased on the observations, plot the magnitude of output signal against their

corresponding frequencies. Also predict the cutoff frequency of the response. Can youname this filter?

S. No Frequency(Hz) Output Voltage(Peak)

01 100

02 200

03 400

04 600

05 800

06 1000

07 120008 1400

09 1500

10 1600

11 1800

12 2000

XFG1

C1

0.1uF

R1

1k

2

XSC1

A B

Ext Trig+

+

_

_ + _

1

0

0


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Plot

1.00

0.95

0.90

0.85

0.80

0.75

0.70

0.65

0.60

0.55

0.50

0.45

0.40

0.35

0.30

0.25

0.20

0.15

0.10

0.05

0.00 0 200 400 600 800 1000 1200 1400 1600 1800 2000

Gain

Frequency in Hz


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CONCLUSION:

The above characteristics shows that the circuit is a , with a

cutoff frequency of Hz.


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Electric Filters Operational AmplifiersNED University of Engineering and Technology Department of Electrical Engineering

7

LAB SESSION 02

Operational Amplifiers

OBJECT

To investigate the inverting and non-inverting mode of Op-amp(741-IC)

APPARATUS

Bread-board, 10 K resistors, Dual Power supply, multi-meter and a 714 Op-amp IC.

THEORY

Operational amplifiers are voltage controlled voltage sources. They differ from ordinary

amplifiers by having two inputs. The operation is such that the output voltage is thedifference of the two input voltages multiplied by an overall gain factor. A typical circuit

diagram of an ideal Op-amp is shown below

)( vvAvo

v-

v+

A


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The 741 Operational amplifier exhibits ideal behavior for D.C signals, the behavior

deviates as the frequency of the signal exceeds audio range. The specifications along

with the pin diagram of a typical 741 are shown below

SYMBOL VARIABLE VALUE UNIT

A Open loop gain 210 5

Ri Input Resistance 2106

Ro Output Resistance 75

Vos Input offset voltage 0.001 V

Ibs Input bias current 8108 A

SR Slew Rate 5105

V/s

Cc Compensation Capacitance 31011 F

PIN CONFIGURATION OF 741-IC

Pin number 2 and 3 are the two input terminals, where as pin number 6 is the output pin.

We connect positive and negative supply voltages to pin number 7 and 4 respectively.


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CIRCUIT DIAGRAMS

Inverting Mode:In inverting mode input is connected to the inverting terminal of op-amp, as shown in

the circuit diagram. The gain for the inverting mode is

i

f

R

RGain

OBSERVATIONS

Rf R i Vin Vo(expected) Vo(measured) Error20K 10K 2V

U1

741

3

2

4

7

6

51

PositiveSupply

10V

NegativeSupply

-10V

V1

2 V

Ri

10kOhm

Rf

20kOhm

XMM1


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Non-inverting Mode:

In non-inverting mode input is connected to the non-inverting terminal of op-amp, as

shown in the circuit diagram. The gain for the non-inverting mode is

i

f

R

RGain 1

This shows that the gain of the non-inverting configuration can never be less than 1.

OBSERVATIONS

Rf R i Vin Vo(expected) Vo(measured) Error

20K 10K 2V

CONCLUSION:

U1

741

3

2

4

7

6

51

V1

2 VRi

10kOhm

PositiveSupply

10V

NegativeSupply

-10V

U2

20kOhm

XMM1


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Electric Filters Frequency Response of Op-ampNED University of Engineering and Technology Department of Electrical Engineering

11

LAB SESSION 03

Frequency Response of 741 Op-amp

OBJECTTo investigate the Frequency response of 741 OP-AMP in close-loop

APPARATUSBread-board, 10K resistor, 100K resistor, dual power supply, multi-meter and a 714

Op-amp IC.

THEORY

Most of the times, for the sake of simplicity we design filter circuits using ideal model ofOp-amp. The ideal model suggests that Op-amp has an infinite open loop gain for all

frequencies i.e. op-amp gain is independent of frequency. In the real devices such as 741

we have a small integrated capacitor, for the reason of stability, which introduces a low

frequency dominant pole in the open loop transfer function. A typical expression for the

open loop gain is shown below. The expression and the plot show that the gain isfrequency dependent. As we increase the frequency of the signal the gain of op-amp

reduces.

)1028.17)(4.31(

1028.17)(

6

12

sssA


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Using op-amp in closed loop, we may reduce the gain of the amplifier and can achieve

more bandwidth over which the gain remains constant. Its a kind of trade off between

gain and bandwidth since the product of gain bandwidth for any device remains

constant. For lower values of gain i.e. from 1 to 10 741 ICcan be used for audio range

signals. If you want more gain, you have to cascade lower gain stages.

CIRCUIT DIAGRAMS

Low Gain, Higher Bandwidth(a)

741IC

Ri

10kOhm

Rf

10kOhm

XFG1

XSC1

A B

G

T

High Gain, Lower Bandwidth(b)

741IC

Ri

10kOhm

Rf

10kOhm

XFG1

XSC1

A B

G

T

100


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OBSERVATIONSFor circuit (a)

For circuit (b)

S. No Frequency Hz Gain = Vout/ Vin

01 20002 1000

03 2000

04 4000

05 6000

06 8000

07 10,000

08 12,000

09 14,000

10 16,000

11 18,000

12 20,000

13 24,000

14 30,000

15 50,000

16 100,000

S. No Frequency Hz Gain = Vout/ Vin200

1000

2000

4000

6000

8000

10,000

12,000

14,000

16,00018,000

20,000

24,000

30,000

50,000

100,000


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Plot (a)

Frequency in Hz

Gain


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Plot (b)

Frequency in Hz

Gain


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CONCLUSION:


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ectric Filters First order low passNED University of Engineering and Technology Department of Electrical Engineering

17

LAB SESSION 04

First order Low pass FilterOBJECTTo design and investigate the response of 1

storder low pass filter (active)

APPARATUSBread-board, pair of resistances and a pair of capacitances, Oscilloscope, function

generator, dual power supply and 741 op-amp IC

THEORY

A typical bilinear transfer function contains single pole and zero.

)(

)()(

ps

zsKsT

Where K is the over-all gain of the system and z and p represents pole and zero. In order

to achieve a low pass responsez > p

A typical response of the system can be shown as


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Design Activity

With the help of admittance or impedance model of inverting configuration of op-amp

we can achieve the above form of transfer function

The circuit is shown below

The transfer function obtained through this circuit is given as

)(

)()(

22

11

2

1

GsC

GsC

Y

YsT

Show all the design steps on the next page, for the following design parameters and

implement the circuit.

K =

z =

=


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Calculations:

Based on the design parameters the elements values obtained are

C1 =C2 =

R1 =

R =


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OBSERVATIONS

Now based on your circuit design, select suitable range of frequencies for the input

signal and obtain the gain for every frequency. Also plot the gain against frequency.

S. No Frequency Hz Gain = Vout/ Vin db Gain

01

02

03

04

05

06

07

08

09

10

11

12

13

14

15

16


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Plot

Frequency in Hz

Gain


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CONCLUSION:


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Electric Filters First order High passNED University of Engineering and Technology Department of Electrical Engineering

23

LAB SESSION 05

First order High pass FilterOBJECTTo design and investigate the response of 1

storder High pass filter (active)

APPARATUSBread-board, pair of resistances and a pair of capacitances, Oscilloscope, function

generator, dual power supply and 741 op-amp IC

THEORY

A typical bilinear transfer function contains single pole and zero.

)(

)()(

ps

zsKsT

Where K is the over-all gain of the system and z and p represents pole and zero. In order

to achieve a High pass responsep > z

A typical response of the system can be shown as


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Design Activity

With the help of admittance or impedance model of inverting configuration of op-amp

we can achieve the above form of transfer function

The circuit is shown below

The transfer function obtained through this circuit is given as

)(

)()(

22

11

2

1

GsC

GsC

Y

YsT

Show all the design steps on the next page, for the following design parameters and

implement the circuit.

K =

z =

=


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CALCULATIONS:

Element values obtained

C1 =

C2 =

R1 =

R2 =


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OBSERVATIONS




01

02

03

04

05

06

07

08

09

10

11

12

13

14

15

16


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Plot

Frequency in Hz

Gain


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CONCLUSION:


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Electric Filters Cascade designNED University of Engineering and Technology Department of Electrical Engineering

29

LAB SESSION 06

Cascade DesignOBJECTTo implement higher order filters using cascade connection.

APPARATUSBread-board, pair of resistances and a pair of capacitances, Oscilloscope, function-

generator, dual power supply and two 741 op-amp ICs

THEORY

Higher order filter circuits can be implemented using 1st order circuits connected in a

chain, so called cascade connection. A cascaded system is a system that consists ofsmall subsystems such that the output of one subsystem is the input of the other . A

simple cascade connection block diagram is shown below

Where T(s) is the system overall transfer functionUsing cascade connection, we can achieve band pass and band stop response since they

cant be implemented form a 1st order system. Cascading a low pass and a high pass

can provide these responses. However there are some conditions which we have to

satisfy before obtaining the over-all response i.e. no subsystem in the cascade

connection is going to load other subsystem . We shall implement a band pass responsewith the help of 1st order filter circuits.

T2T1 T3

T(s) = (T1)(T2)(T )


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CIRCUIT DIAGRAM

The circuit diagram shown above consists of two cascaded bilinear systems with the

overall system function of

)(

)(

)(

)()(

44

33

22

11

GsC

GsC

GsC

GsCsT

Now select the element values to achieve the following system function

)22000(

)28000(

)2500()(

s

s

s

ssT


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CALCULATIONS:


C1 = C3 =C2 = C4 =

R1 = R3 =

R2 = R4 =


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OBSERVATIONS




01

02

03

04

05

06

07

08

09

10

11

12

13

14

15

16


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Plot

Frequency in Hz

Gain


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CONCLUSION:


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Electric Filters Biquad Circuit(L.P)NED University of Engineering and Technology Department of Electrical Engineering

35

LAB SESSION 07

Biquad Circuit (Low pass response)OBJECTTo design and investigate low pass response using Biquad circuit

APPARATUSBread-board, resistors, capacitors, Oscilloscope, function-generator, dual power supply

and three 741 op-amp ICs

THEORY

Biquad circuit is also known as Universal filter. We can implement low pass, band

pass, high pass and band stop, 2nd order filters using biquad circuit. Dealing with thesecond order circuits we normally express system response in terms of gain (H),

quality factor (Q) and resonance frequency (o).

While designing low pass filter we keep the quality of the system less than 1, so that

the response curve wont be peaky. This ensures that the system treat all the pass band

signals equally.

Following are some example response for Q = 7 (Fig 1) and Q = 0.8 (Fig 2).

Fig 1 Fig 2


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CIRCUIT DIAGRAM

Since the output at the last stage is the product of the transfer functions of the sub-

stages, the output equation can be given as

5

6

2411

2

11

3

)()( R

R

CR

s

GsC

GV

GsC

GVV

oin

o

Starting from the above equation reduce the equation to a simplified Low pass function

and compare it to the standard form i.e.

22

2

)/()(

oo

o

sQs

HsT

Also determine the element values for the following given system parameters

Given data

o =H =

Q =


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CALCULATIONS:


C1 = R3 =C2 = R4 =

R1 = R5 =

R2 = R6=


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OBSERVATIONS




01

02

03

04

05

06

07

08

09

10

11

12

13

14

15

16


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Plot

Frequency in Hz

Gain


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CONCLUSION:


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Electric Filters Biquad Circuit(B.P)NED University of Engineering and Technology Department of Electrical Engineering

41

LAB SESSION 08

Biquad Circuit (Band pass response)OBJECTTo design and investigate band pass response using Biquad circuit


and three 741 op-amp ICs

THEORY

We shall now implement band pass response using biquad circuit. Unlike low pass

response, here, we shall keep the quality factor high. Increasing the quality factor forband pass response will increase the selectivity of a particular frequency.


Fig 1 Fig 2


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CALCULATIONS:


C1 = R3 =C2 = R4 =

R1 = R5 =

R2 = R6=


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OBSERVATIONS




01

02

03

04

05

06

07

08

09

10

11

12

13

14

15

16


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Plot

Frequency in Hz

Gain


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CONCLUSION:


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Electric Filters Biquad Circuit(B.S)NED University of Engineering and Technology Department of Electrical Engineering

47

LAB SESSION 09

Biquad Circuit (Band Stop response)OBJECTTo design and investigate Band stop response using Biquad circuit


and four 741 op-amp ICs

THEORY

We shall now implement band stop response using biquad circuit. Unlike low pass

response, here, we shall keep the quality factor high. Increasing the quality factor forband stop response will increase the selectivity of a particular frequency.


Fig 1 Fig 2


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CIRCUIT DIAGRAM

For band stop response we have added another op-amp to the previous circuit which isadding the input and the band pass response and equated the two resistances i.e. R1=R3

)( .PBino VVV

Starting from the above equation reduce the equation to a simplified Band stop function

and compare it to the standard form of the band stop response i.e.

22

22

)/()(

oo

o

sQs

ssT


Given data

o =

H = 1/Q

Q =


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CALCULATIONS:


C1 = R4 =C2 = R5 =

R1 = R6=

R2 =R3= R7=R8 = R9=


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OBSERVATIONS




01

02

03

04

05

06

07

08

09

10

11

12

13

14

15

16


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Plot

Frequency in Hz

Gain


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CONCLUSION:


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Electric Filters Biquad Circuit(H.P)NED University of Engineering and Technology Department of Electrical Engineering

53

LAB SESSION 10

Biquad Circuit (High Pass response)OBJECT

To design and investigate High pass response using Biquad circuit

APPARATUS

Bread-board, resistors, capacitors, Oscilloscope, function-generator, dual power supplyand four 741 op-amp ICs

THEORY

We shall now implement High pass response using biquad circuit. In designing High

pass filter we keep the quality of the system less than 1, so that the response curve

wont be peaky. This ensures that the system treat all the pass band signals equally.


Fig 1 Fig 2


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CIRCUIT DIAGRAM

For High pass response we have added another op-amp to the three op-amp biquad

circuit which is adding the input, the band pass response and the low pass response. Wehave also equated the three resistances i.e. R3=R1=R2

)( .. PLPBino VVVV

Starting from the above equation reduce the equation to a simplified High pass function

and compare it to the standard form of the High pass response i.e.

22

2

)/()( oo sQs

s

sT


Given data

o =

Q =


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CALCULATIONS:


C1 = R4 =C2 = R5 =

R1 = R2 =R3= R6=

R7=R8 = R9=R10=


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OBSERVATIONS




01

02

03

04

05

06

07

08

09

10

11

12

13

14

15

16


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Plot

Frequency in Hz

Gain


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CONCLUSION:


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Electric Filters GIC circuitNED University of Engineering and Technology Department of Electrical Engineering

59

LAB SESSION 11

Generalized Impedance ConverterOBJECT

To simulate inductance using a generalized impedance converter circuit

APPARATUS

Bread-board, resistors, capacitors, Oscilloscope, function-generator, dual power supplyand a pair of 741 op-amp ICs

THEORY

The elements we discussed so far for the implementat ion of active filters were

capacitors, resistors and op-amps. We havent discussed any approach which includes

inductor in active circuit. The reason of not using inductors is that they require more

space, since the dimension of an inductor is proportional to its inductance and for lower

frequencies we need higher inductance values to achieve considerable impedance. Theother reason is that due to flux linkages, they could induce voltages to the adjacent

elements.

We can simulate inductance through alternate method, since the design approach for

the passive network is simpler than the active approach.

Consider a passive circuit, shown below. The response across the resistor will be a

band pass. To simulate inductance we shall use a GIC circuit, which will convert a

resistance into an inductance

Signal Source

R

L

C


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CIRCUIT DIAGRAM OF GIC

The relationship between Zin and Z can be given as

ZZZ

ZZZ

in

42

31

With the help of GIC circuit, implement a series RLC circuit with the following values

and obtain response curve across the resistor.

1 Z3Z2

ZZin

Element values of the passive circuit

C = 1FR = 10K

L = 1mH


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CALCULATIONS:


Z =Z1 =

Z2 =

Z3 =

Z4 =


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OBSERVATIONS




01

02

03

04

05

06

07

08

09

10

11

12

13

14

15

16


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Plot

Frequency in Hz

Gain


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CONCLUSION:


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Electric Filters Voice Processing Using MatlabNED University of Engineering and Technology Department of Electrical Engineering

65

LAB SESSION 12

Voice Processing Using MatlabOBJECT

To implement a transfer function over voice signal and observe its output generated byMATLAB.

APPARATUS

MATLAB software

OVERVIEW OF MATLAB AND RELATED TOOL BOXES

It will be covered by the instructor during the Lab session.

MAIN PROGRAM

Our main simulation can be divided into three main blocks

Input/output block Transfer function DAC/ADC

BASS CONTROL MODULE

Speaker

OutputInput

10

Gain1From Mic

In1 Out1

DAC

300*2*pi

s+300*2*pi

BASS

In1 Out1

ADC


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Electric Filters Voice Processing Using MatlabNED University of Engineering and Technology Department of Electrical Engineering

66

The input/output blocks are MIC and speaker taken from Signal Processing Block Set.

These modules allow us to obtain voice signals directly from the MIC and to reproduce it

through speakers after processing the signal. Our processing will be a real-time

processing. These modules work for digital signals only. The output of MIC and the

input of speaker are in digital form, where as s-domain transfer function requirescontinuous-time signal. In order to achieve compatibility among the blocks we have

added DAC (digital to analog) and ADC (analog to digital) converters. You may create

any transfer function for your voice signal of range (20 to 5000 Hz). You need a faster

processor for the signal beyond this range.In our case, a low pass filter is implemented. When you simulate the program you can

feel the difference in the input and output voice. The low pass will reduce the sharp

contents of voice. Mean while the scopes connected to the input and output sides also

provides you an opportunity to view the modification that took place during the

processing in the voice signal. High frequency components will be reduced and theoutput wave form will be a smoother one. A sample input and its output is shown below


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Electric Filters Design of BASS control NED University of Engineering and Technology Department of Electrical Engineering

67

LAB SESSION 13

Design Bass Control CircuitOBJECT

Design and implement bass control circuit

EXPLANATION

BASS CONTROL provides boost to low frequency notes of audio signal such as beat. It

is a low pass filter and process audio signals of range 20 to 300 Hz (approx). Most of thelow frequency sound lies in this region. Low frequency notes are power hunger and need

more power than other range of audio signals. At the output stage of BASS circuit an

especially designed transducer Woofer is connected.

You need an audio input that must contain sound of lower notes to process this signal.The Bass control circuit wont allow or significantly suppress signals ranging form 301

to 20KHz.

Based on the necessary design procedures, select your circuit elements and mention all

the calculations and assumptions below.

CALCUALTIONS

BODE PLOT


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CIRCUIT DIAGRAM


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Electric Filters Design of MID control NED University of Engineering and Technology Department of Electrical Engineering

70

LAB SESSION 14

Design MID Control CircuitOBJECT

Design and implement Mid control circuit

EXPLANATION

MID CONTROL provides boost to voice frequency notes of audio signal. It is a Band

pass filter and process audio signals of range 300 to 4000 Hz (approx).

You need an audio input that must contain voice of human to process. The Mid control

circuit wont allow or significantly suppress signals ranging form 20 to 300 Hz and from4001 to 20KHz.



CALCUALTIONS

BODE PLOT


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CIRCUIT DIAGRAM


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Electric Filters Design of TREBLE control NED University of Engineering and Technology Department of Electrical Engineering

73

LAB SESSION 15

Design Treble Control CircuitOBJECT

Design and implement Treble control circuit

EXPLANATION

TREBLE CONTROL provides boost to high frequency notes of audio signal. It is a high

pass filter and process audio signals of range 4001 to 20000 Hz (approx).

You need an audio input that must contain sharp sounds. The Mid control circuit wont

allow or significantly suppress signals ranging form 20 to 4000 Hz. The output isconnected to an especially designed transducer named as TWEETER



CALCUALTIONS

BODE PLOT


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CIRCUIT DIAGRAM

Ee-315 Electric Filter _2011

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