Experiement 8

• Objectives:•

• • • •

Familiarization with the low-pass, high-pass ,band-pass and band-stop filters. Find the frequency response of a seriese RC curciut. Plot the magnitude and Phase response of a seriese RC curciut. Design, build and test a low pass filter. Design, build and test a high pass filter.

• Equipments:• Digital multimeter (DMM). • Breadboard. • Audio signal generator. • Oscilloscope. • Wires.

• Contents:• •

Series RC circuit analysis(Low pass filter). Series RC circuit analysis(High pass filter).

• Theory:A network which functions to permit circuit frequencies to pass, and not to permit other frequancies,is called filter. • If the filter passes all frequencies up to certain frequency , and blocks all frequencies above that frequency is called a low-pass filter as shown in fig1. •

• If the filter passes all frequencies above a certain frequency , and blocks all frequencies less that frequency is called a high-pass filter as shown in fig 2. A more complicated filter has three frequency ranges; two versions are the band –pass filter and the band –stop filter (bandrigiction) as shown in fig 3 . • The frequency at which the power equals half of its maximum value and when the voltage drop to 0.707 times of its maximum value it is refers to as break frequency or corner frequency. ** Vload = Vin / 2 1/2 •

• Every filter has at least one frequency that permit to pass frequencies above and under that frequency it is called the cut off frequency and it is given in the following expressions:** fc = 1/ 2*3.14 RC

• Results and analysis:For the low–pass filter we construct the following circuit:-

•

Input Freq

Input amplitude Voltage(V)

1 2

2Vp 2Vp

2.3

Output amplitude

Phase Shift (ø)

Gane=Vo/Vin

1.8 1.6

a=0.9,b=2.1,ø=38.97 a=1.1,b=1.8,ø=39.29

1.9/2=0.95 1.5/2=0.75

2Vp

1.5

a=1.1,b=1.7,ø=36.34

1.2/2=0.56

5

2Vp

0.9

a=0.9,b=1.1,ø=44.9

1.2/2=0.6

10

2Vp

0.5

a=0.6,b=0.8,ø=57.14

0.436/2=0.218

20

2Vp

0.3

a=0.6,b=0.7,ø=56.44

0.26/2=0.13

50

2Vp

0.2

a=0.3,b=0.4,ø=36.8

0.14/2=0.07

100

2Vp

0.1

a=0.2,b=0.3,ø=32.29

0.11/2=0.055

200

2Vp

0.1

a=0.2,b=0.3,ø=27.6

0.125/2=0.0625

500

2Vp

0.1

a=0.1,b=0.2,ø=21.8

0.11/2=0.055

1000

2Vp

0.08

a=0.3,b=0.3,ø=21.7

0.105/2=0.0525

•

Voltage(Vp)

For the value of the Phase shift we use the equation: ø=sin-1a/b.

The characteristics for the low-pass filter:

Vo-f in characterstic

Vo

2 1.5 1

Series1

0.5 0 0

500

1000

1500

f in

*fc=1/2*3.14*68*1*10-6=2.3KHz

Phase shift

•

70 60 50 40 30 20 10 0

Series1

0

500

1000 f in

1500

• For the high-pass filter we construct the following curcuit:-

Input Freq

Input amplitude Voltage(V)

Output amplitude Voltage(Vp)

Phase Shift (ø)

Gane=Vo/Vin

100Hz 500Hz

2Vp 2Vp

0.4*0.2=0.08 1.9*0.2=0.38

a=0.4,b=0.4,ø=90 a=1.9,b=2,ø=71.8

0.08/2=0.04 0.38/2=0.19

1K

2Vp

1.4*0.5=0.7

a=1.3,b=1.6,ø=54.3

0.7/2=0.35

2K

2Vp

1.1*1=1.1

a=0.9,b=1.3,ø=43.8

1.1/2=0.55

2.3K

2Vp

1.2*1=1.2

a=0.9,b=1.3,ø=44

1.2/2=0.54

5K

2Vp

1.6*1=1.6

a=0.7,b=1.7,ø=24.31

1.6/2=0.8

10K

2Vp

1.75*1=1.75

a=0.4,b=1.9,ø=12.15

1.75/2=0.875

50K

2Vp

1.9*1=1.9

a=0.1,b=2,ø=2.36

1.9/2=0.95

100K

2Vp

1.95*1=1.95

a=0.1,b=2,ø=2.3

1.95/2=0.975

500K

2Vp

1.96*1=1.96

a=0.8,b=2,ø=23.57

1.96/2=0.98

1000K

2Vp

1.98*1=1.98

a=1.2,b=1.6,ø=48.6

1.98/2=0.99

The characteristics for the high-pass filter :

Vo

Vo-f in characterstic 2.5 2 1.5 1 0.5 0

Series1

0

500

1000

1500

f in

2.5 Phase shift

•

2 1.5

Series1

1 0.5 0 0

200

400

600 f in

800

1000

1200

• Conclusion:We can design the filters by constructing the series RC circuit. If we need low-pass filter we take the output voltage from the capacitor. • If we need high-pass filter we take the output voltage from the resistor. • We can calculate the break down frequency by using the equation: fc=1/2 *3.14 RC. • •