EXPERIMENT – 1

OBJECT:

To study active low pass filter and to plot its frequency response.

THEORY:

The figure shows a LPF that uses an RC network for filtering. The op-
amp is used in the non inverting mode, hence it does not load the RC network.
Resistors R1 and Rf determine the gain of the filter.
The voltage at the non inverting terminal is

 1 
 j 
 C  .V
V1 =
 1 
in
R  j 
 C 

Vin
or V1 =
1  j 2fRC

and the output voltage

 Rf 
V0  1  V 1  A f V1
 R1 

Vo Af 1
  , fH 
Vin  f  2RC
1  j  
 fH 

Vo Af
the gain magnitude  2
Vin  
f
1  
 fH 
TRANSFER FUNCTION:

1
Vin
Vin
V1 = CS 
1 1  SRC
R
CS
  Rf 
Vo  1  V1  AV1
 R1 

Vo 1  R  A
  1  f  
Vin 1  SRC  R1  1  SRC

A
T 
1  (RC ) 2
PROCEDURE:

1. Connect function generator, CRO and  15V supply at respective

terminals.
2. Keep input voltage at 1V peak to peak.
3. Measure output at various signal frequencies.
4. Calculate gain magnitude and gain in dB.
5. Plot the frequency response and find bandwidth.
6. Plot on semilog graph paper.
OBSERVATION:

Input Voltage = 1 V (peak to peak)

Input Frequency Gain Magnitude Gain in dB

10 Hz onwards
 f H = 900 Hz

Theoretically
1
fH   795.8 Hz
2RC

RESULT:

The frequency response of the active low pass filter has been plotted.
The higher cutoff frequency came out to be ……….. Hz.
EXPERIMENT – 2

OBJECT:

To study active high pass filter and to plot its frequency response.

THEORY:

The high pass filter is formed simply by interchanging the frequency

determining resistors and capacitor. Figure shows a high pass filter. All the
frequencies higher than the lower cut off frequency fL are pass band
frequencies with the highest frequency determined by the closed loop band
width of the op-amp.

For the high pass filter.

V0 = AfRCS /(1+SRC)

 f 
Af  
V0
  fL 
2
Vin  f 
1   
 fL 

Transfer function
T(S) = AfRCS /(1+SRC)

PROCEDURE:
1. Connect function generator, CRO and  15V supply at respective
terminals.
2. Fix the input voltage at 1V Peak to Peak.
3. Measure output at various signal frequencies.
4. Calculate the gain magnitude and gain in dB.
5. Plot the frequency response and find band width on Semilog graph
paper.
OBSERVATIONS:
Input Voltage = 1 V (peak-peak)
Input Frequency Gain Magnitude Gain in dB
10 Hz onwards

f H = ……… Hz
RESULT:
The frequency response for the active high pass filter is plotted. The lower
cut off frequency came out to be
f L =………………… Hz
Theoretically
1
fL   400 Hz
2RC
EXPERIMENT – 3

OBJECT:

To study active band pass filter and to plot its frequency response
curve.

THEORY:

A band pass filter (BPF) has a pass band between two cut off frequencies
f H and f L  f H  f L  . Any input frequency outside this band is attenuated.
There are two types of band pass filters:
1. Wide band pass and 2. Narrow band pass.

We define a filter as wide band pass if figure of merit or quality factor Q

< 10. On the other hand if Q > 10 we call the filter as narrow band pass filter.
This Q is a measure of selectivity. The higher is the value of Q, more selective
is the filter or narrower is its bandwidth. The relationship between Q, the 3
dB BW and the centre frequency f 0 f is given by

fC fC
Q 
BW fH  fL

For the wide band pass filter the centre frequency f C can be defined as

fC  fH  fL

In a narrow band pass filter the output voltage peaks at the centre
frequency.
A wide BPF can be formed by simply cascading HPF and LPF sections which
is generally the choice for simplicity of design and performance. To obtain a
40 dB/decade slope, two 1st order filters are cascaded while to obtain a 80
dB/decade slope, two IInd order filter are cascaded.
Figure shows a  20 dB decade wide BPF composed of 1st order HPF and

LPF. To realize a band pass response, however, f H must be greater than f L .

PROCEDURE:
1. Connect function generator and CRO at the input and output terminals
respectively.
2. Keep input at 1V peak to peak.
3. Connect  15 V power supply to the respective terminals.
4. Measure the output voltage.
5. Vary the input signals frequency and measure the corresponding
output.
V0
6. Calculate the gain magnitude and in dB.
Vin
7. Plot the frequency response and find the bandwidth.
OBSERVATIONS:
Input Voltage = 1 V (peak-peak)
Input Frequency Gain Magnitude Gain in dB
10 Hz onwards

CALCULATIONS:

Lower cut off frequency

1
Theoretically fL   160 Hz
2RC

1
Upper cut off frequency fH   1591 Hz
2RC
Practically

fH fL
Upper cut off frequency f H =……….Hz Q  0.31
fH  fL

Lower cut off frequency f L= …………… Hz

RESULT:

The frequency response of active band pass filter is as shown. The upper
and lower cutoff frequencies are …….. Hz & …….. Hz respectively. Bandwidth
= ………..Hz.

EXPERIMENT – 4
OBJECT:

To study All Pass Filter (APF) and the plot its frequency and phase
response.

THEORY:

An All Pass Filter (APF) passes all frequency components of the input
signal without attenuation, while providing predictable phase shift for
different frequencies of input signal. When signals are transmitted over
transmission lines such as telephone wires, they undergo change in phase.
To compensate for these phase changes all pass filters are required. They are
also called equalizers or phase connectors.

Transfer function:
By applying superposition theorem o/p voltage can be expressed as
V 0  V0 1  V0 2

where,
V0 1 =  Vin

 Vin 
V0 2 = 2 
 1  SRC 
2Vin
V0  Vin 
1  SRC
V0 1  SRC
 
Vin 1  SRC

T S   1

 2fRC   2fRC 
and phase angle   tan 1    tan  1 
 1   1 
 2 tan 1 2fRC 
For fix values of R and C, the phase angle changes from 0 – 180 as
frequency varies from 0 -  minimum sign indicate that V0 logs Vin.
If the position of R and C are inter changed then the phase shift between
input and output becomes positive.
For
 1
f 
2RC
2RC
  2 tan 1  2  45  90 0
2RC
PROCEDURE:
1. Connect function generator, CRO and dc supply ±15 at respective
terminals.
2. Fix the input voltage to 1V (p-p).
3. Check the phase shift between input and output.
4. Calculate the phase shift expected using the given formula.
5. Compare the theoretical & practical values of phase shift.

OBSERVTIONS:
Gain = 1
Theoretical phase shift   2 tan  12fRC 
R = 10 K
C = 0.01  F

Th   Pr
% error =  100
Th

Frequency Phase Shift % Error

S.No.
(KHz) Th  Pr

1 10 Hz onwards

6
RESULT:

The phase response of the All Pass Filter is as shown in the graph.
Maximum percentage error between theoretical and practical values is ……%.
EXPERIMENT – 5

OBJECT:

To study second order Butterworth high pass filter & to plot its
frequency response.

THEORY:

A second order high pass filter can be formed from a second order low
pass filter by simply interchanging the frequency determining resistor and
capacitor figure shows a second order Butterworth high pass filter.
The voltage gain magnitude of the second order high pass filter is given
by:
V0 Af
 4
Vin f 
1   0 
 f 
where,

A f = Pass band gain

f = frequency of input signal

f C = lower cut off frequency

Transfer function:

Rf
Le the gain be  . Then  = 1+
R1
The DAM for the circuit will be

1 2 3 4
1  sC  sC 0 0 

2  sC G2  2 sC  sC  G2 
3 0  sC G3  sC  G3 
 
4 0  G2 0 G2 

Generating constrained matrix by C3 +  C4 and then deleting 4th row

and 4th column.
 1 2 3
1  sC  sC 0 

2  sC G2  2 sC  sC  G2 
3  0  sC  sC  (1   )G3 

Now,
V4 V3 y 31
T   1
V1 V1 y1

s 2 C 2
T 
G2  G2  2sC sC  G3   ( s 2 C 2 )

s 2 C 2

s 2 C 2  sC (G2  G3 )  G2 G3 (1   )

PROCEDURE:

1. Connect function generation, CRO and  15 V supply at respective

terminals.
2. Fix the input voltage at 1V.
3. Measure output voltage at various signal frequencies.
4. Calculate the gain magnitude & gain in dB.
5. Plot the frequency response & find bandwidth.

OBSERVATIONS:

Input Voltage = 1 V (peak-peak)

Input Frequency Gain Magnitude Gain in dB
10 Hz onwards
RESULT:

The frequency response for the second order Butterworth high pass filter is
plotted. The lower cut off frequency f L = …………. Hz.
EXPERIMENT – 6

OBJECT:

To study Butterworth second order low pass filter.

THEORY:

A stop band response having 40 dB/decade roll off is obtained with IInd
order LPF. A Ist order LPF can be converted to a second order type simply
using an additional RC network as in circuit diagram.
Second order filters are important because higher order can be designed
using them. The gain of the filter is set by R1 and Rf while the higher cutoff
frequency is determined by R1, R3, C2 & C3.

TRANSFER FUNCTION:
By applying KCL at nodes (1) & (2).
V1  Vin V1  V2 V1  V0
  0
R2 R3 1
C2 s
V1  Vin R3  V1  V2 R2  V1  V0 R2 R3 C 2 s  0
=> V1 R3  R2  R2 R3 C 2 s   V2 R2  Vin R3  V0 R2 R3 C 2 s  0 ...(1)

V2 V  V1
 2 0 ...(2)
1 R3
C2 s

=> V2 R3 C 3 s  V2  V1   0

V2 1  R3 C 3 s   V1

Substituting the value of V1 in (1)

=> V2 R3  R2 R3 C 2 s   R32 C 3  R2 R3 C 3 s  R1 R32 C 2 C 3 s 2

= Vin R3  V0 R2 R3 C 2 s

K
V0 R2 C 2 C 3 R3
=>  T (s) 
Vin sR2 C 2  R3 C 3  R2 C 3  KR2 R3 C 2  1
s2  
R2 R3 C 2 C 3 R2 R3 C 2 C 3
 Rf
where, K = 1+
R1
This transfer function is of the form.

K 02
T (s) 
s 2   0 Q s   02

1
0 
R2 R3 C 2 C 3

T  j  
K
Now,
1    0 
2n

n=2

K
 T ( j )  4

1    
 0 

PROCEDURE:

1. Connect function generation, CRO and  15V supply at respective

terminals.
2. Fix the input voltage at 1 V.
3. Measure output voltage at various signal frequencies.
4. Calculate the gain magnitude and gain in decibel.
5. Plot the frequency response and find bandwidth.

OBSERVATIONS:

Input Voltage = 1 V (peak-peak)

Input Frequency Gain Magnitude Gain in dB
10 Hz onwards
RESULT:

The frequency response for the second order Butterworth high pass
filter is plotted. The higher cut off frequency f H = …….Hz = …….. KHz.

1
Theoretically f H 
2 R2 R3 C 2 C 3

f H=……………………… Hz
EXPERIMENT – 7
OBJECT:
To study a notch filter.

THEORY:
The band reject filter, also called band stop or band elimination filter,
attenuates the frequencies in the stop band while passes all other frequencies.
The band reject filter can be classified as:
(a) Wide band reject filter.
(b) Narrow band reject filter.
The narrow band reject filter is commonly called Notch filter because of
its high quality factor Q (>10). The bandwidth of the notch filter is much
smaller than the bandwidth of wide band reject filter. The relationship
between Q and the bandwidth is given by.

fC fC
Q 
BW fH  fL

where, f H = higher cut off frequency.

f L = lower cut off frequency.

For a notch filter, f H and f L are situated very close to each other and

the centre frequency f C is the one where output voltage peaks down wards.

PROCEDURE:
1. Connect function generator, CRO and  15 V supply at respective
terminals.
2. Fix the input voltage.
3. Measure output at various signal frequencies.
4. Calculate the gain magnitude & gain in dB.
5. Plot the frequency response on semilog graph paper.
 OBSERVATIONS:
Input Voltage = 1 V (peak-peak)
Input Frequency Gain Magnitude Gain in dB
10 Hz onwards

Notch frequency f N = …………..Hz

RESULT:
The frequency response of the notch filter is as shown in the graph. The
notch frequency f N = ………….. Hz.
EXPERIMENT – 8

OBJECT:
To study a band reject filter.

THEORY:
The band reject filter is also called a band stop or band elimination
filter. In this filter, frequencies are attenuated in the stop band while they are
passed outside this band.
The quality factor Q is a measure of selectivity of BRF, Greater is the
value of Q smaller is the band width.
fC fC
Q 
BW fH  fL

Centre frequency f C  fH fL

Band reject filter can be realized by a low pass filter a high pass filter
and a summing amplifier. The low cutoff of HPF must be greater than high
cutoff of LPF. Also the pass band gain of both the HPF and LPF must be
equal.

PROCEDURE:
1. Connect function generator and CRO at the input and output terminals
respectively.
2. Keep input at 1V peak to peak.
3. Connect  15 V power supply to the respective terminals.
4. Measure the output voltage.
5. Vary the input signals frequency and measure the corresponding
output.
V0
6. Calculate the gain magnitude and in dB.
Vin
7. Plot the frequency response and find the bandwidth.
OBSERVATIONS:
Input Voltage = 1 V (peak-peak)
Input Frequency Gain Magnitude Gain in dB
10 Hz onwards
RESULT:
The frequency band which is lie between f L= ……. And f H =…….. rejected.
EXPERIMENT – 9

OBJECT:

To study Chebyshev second order low pass filter.

THEORY:

Chebyshev filters are also called as “Equal-ripple filters”. The transfer

function has the appearance :

1
T
1  C n2 ( x)
2

where, C n (x) is known as the Chebyshev polynomial and is given as:

C n ( x) cos(n cos 1 x) for x  1

 cos(n cosh 1 x) for x  1

Compared to the Butterworth approximation the Chebyshev

approximation, yield increased stop-band attenuation but at the expense of
allowing some ripples in the pass band. This approximation distributes the
error evenly throughout the pass band in an oscillatory shape.

TRANSFER FUNCTION:

T(S)=-G1 (G4 G5)/(S C1 + G2)* [G6 (G4 + G5)+ S C3 G5]

The transfer function can be calculated by the IAM method.

The DAM for the circuit is

1 2 3 4 5 6
1  G1  G1 .0 .0 0 0 
2  G1 G1  sC 2  sC1  G2  G3  sC1  G2 0 0 0 

3 0 sC1  G2 sC1  G2  G4  G4 0 0 
 
4 0 0  G4 G 4  G5  G5  sC 3 
5 0 0 0  G5 G5  G 6  G 7  G6 
 
6  0 0 0  sC 3 sC 4 G6  sC 3 
PROCEDURE:
1. Connect function generator, CRO and  15 V supply at respective
terminals.
2. Fix the input voltage.
3. Measure output at various signal frequencies.
4. Calculate the gain magnitude & gain in dB.
5. Plot the frequency response on semilog graph paper

RESULT: The Magnitude response is plot and the ripple have find for the
value of n =…………