0% found this document useful (0 votes)
6 views8 pages

Experiment 1

Uploaded by

sksaikrishna1018
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
0% found this document useful (0 votes)
6 views8 pages

Experiment 1

Uploaded by

sksaikrishna1018
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
You are on page 1/ 8

Dept.

of Electronics and Electrical Communication Engineering


Indian Institute of Technology Kharagpur

ANALOG COMMUNICATION LAB


(EC39001)

EXPERIMENT 1: Basic Filter Design

Name: P Vamshi Krishna


Roll Number: 23EC10052
Group Number: 5 (Even Batch)
Date of Experiment: 31-07-2025
Date of submission: 05-08-2025
Introduction:
In this experiment, we designed passive low pass, high pass and band pass filters with
specific cutoff frequencies and observed their frequency response. Extending, we also
designed an active low pass filter with a specific gain using an operational amplifier. Also,
we designed a voltage adder circuit using an inverting operational amplifier.
High pass and low pass filters have wide range of applications in digital signal processing,
interpolation etc.
The cutoff frequencies were as follows:
Low Pass Filter = 10kHz.
High Pass Filter = 4kHz.
Band Pass Filter = 4kHz & 10kHz.
Active Low Pass Filter = 4kHz (gain=2)
Adder Circuit = V1+V2 = Vout

Key Objectives:
Task 1: Design 1st order passive filter with following corner frequencies: {G =5}
 Low pass filter with cut-off frequency FH = {4 x [(G - 1) % 4] + 10} = 10 kHz.
 High pass filter with cut-off frequency FL = {4 x [(G - 1) % 4] + 4} = 4 kHz.
 Band pass filter with above fH and FL.
 Find out the 3dB frequency from both time and frequency domain representation of
signal on DSO.
 Find out the amplitude of signal at fL/2, 2fH.

Task 2: Implement 1st order active LPF using Op-Amp with above specifications and a gain
of 2. Observe the output on DSO.

Task 3: Take two voltage signals x1(t) and x2(t) and add them using Op-Amp using the
formula: (-1) G{x1(t) + [(G - 1) %4+1] x2(t)}

So, Vout = x1(t) + x2(t) {G =5}

Components Used:
1) Digital Signal Oscilloscope
2) Function Generator
3) Resistors- 16Ω, 40Ω, 1kΩ
4) Capacitor-1uF
5) Operational Amplifier
6) Breadboard
7) Connecting wires
8) 12V power Supply
Theory:
In the field of signal processing, a filter is a system that isolates desired parts of a signal by
suppressing or eliminating unwanted features. This is most often achieved by blocking or
passing specific frequency ranges.
Passive Filters
Passive filters are circuits composed entirely of passive electronic components:
 Resistors (R)
 Inductors (L)
 Capacitors (C)
The defining characteristic of passive filters is that they function without an external power
source. They do not contain any amplifying components (e.g., transistors), which makes their
design simpler but can limit their performance.
Active Filters
Active filters are implemented using a combination of passive components and active
(amplifying) components that require an external power source.
 Components: Typically use op-amps in conjunction with resistors and capacitors.
 Advantages: They can achieve a high Q factor (a measure of selectivity) and can be
designed to resonate without the use of inductors, which can be bulky and expensive.
 Limitations: The primary drawback of active filters is that their effective frequency
range is limited by the bandwidth of the active components, such as the op-amp.

Circuit Design and Circuit Diagram


Task 1:

Circuit 1: LPF Passive

⇒ RC ~ 1.59 x 10-5
FH = 10 KHz = 1/(2πRC)

We chose R = 16Ω & C = 1uF

Circuit 2: HPF Passive


FL = 4 KHz = 1/(2πRC)
⇒ RC ~ 3.97 x 10-5
We chose R = 40Ω & C = 1uF

For Band Pass Filter, we will use the above components in Band Pass Filter Configuration, as
we require FL = 4KHz & FH = 10KHz.

Task 2:
Circuit 3: LPF Active

To obtain a Gain(K) of 2, we
choose Rf = R2 = 1kΩ

⇒ RC ~ 1.59 x 10-5
Fc = 10kHz. = 1/2πRC

We chose R = 16Ω & C = 1uF

Task 3:

Circuit 4: Adder for G = 5

To obtain Vo = V1 + V2
We chose R1 = R2 = Rf = 1kΩ
Observations:
LPF at cutoff frequency, fH = 10kHz

HPF at cutoff frequency, fL = 4kHz


Band-Pass Filter (without buffer) at cutoff frequency, fH = 10kHz

Band-Pass Filter (without buffer) at cutoff frequency, fL = 4kHz


Active Low Pass Filter at cutoff frequency, fH = 10kHz

Adder circuit using op-amp at 1kHz with V1=5V and V2=5V


Discussion:
We implemented the filters mentioned in the task using breadboard and components in
the lab. For each configuration, we varied the frequency and measured the output peak-to-
peak voltage value for a given input voltage. We derived compared the theoretical cut-off
frequencies with the ones observed.

 For task 1, we designed low pass, high pass and band pass filters with passive
components; as per the specified cut-off frequencies. For task 2, we designed a LPF
with an op-amp. (Active Component). For task 3, we designed a adder circuit using
op-amp. that adds two signals.
 While designing the band pass filter, when the high pass and low pass stages were
cascaded without a buffer, we faced great attenuation in the output due to loading
effects. But we could preserve the nature of the waveform.
 To overcome this attenuation, we insert a buffer between the stages. When an op-amp
buffer was used the attenuation was mitigated and could obtain outputs close to
theoretical values but the waveform was slightly distorted.
 We compared the theoretical value of the amplitude being 0.707 times the value at
that of flat band frequencies to judge our positions of cutoff frequency.
 We detected noise in the DSO for the output signals when used an op-amp while
performing bandpass filtering.
 The operational amplifier must be provided with an appropriate voltage source, as it
could result in problems with the op-amp circuitry and could go into saturation if the
input voltages are greater than the supply giving wrong outputs on DSO.
 It was observed that, since adder circuit made using an inverting amplifier, it’s output
response was 180 degrees phase shifted with respect to input (inverted)

You might also like