Designing first order Butterworth low pass filter

Thread Starter

naspek

Joined Feb 17, 2010
45
Hey there.. This is my homework..
i really need someone who can check whether i'm doing
it right or not.. =)

Question:
Design a first order Butterworth low pass filter with cut-off frequency fo = 20 kHz and the gain K = 1

fo = 20kHz
K = 1

When K = 1, we can ignore R1 and Rf
formula :
ωo = 1/RC
2∏fo = 1/RC

Set R = 1kΩ

2∏fo = 1 / (1000) C
C = 1 /
(1000)(2∏fo)
C = 7.96nF

Answer: R = 1k
Ω ; C = 7.96nF

My answer is correct? :)

 

Thread Starter

naspek

Joined Feb 17, 2010
45
ok! done with the first part..
now is the complete question..

Design a first order Butterworth low pass filter with cut-off frequency fo = 20 kHz and the gain K = 1. You will need to simulate the circuit using Multisim and shows that the circuit meet the required specification.

i)Show all calculations steps and assumption made in your circuit design.
(DONE!)

ii)Show the circuit schematic with all the resistor and capacitor values labelled.(attachment)

iii)Perform frequency response analysis on the circuit and obtain the bode diagram. The bode diagram should show the voltage gain (Vout / Vin) in dB vs the signal frequency (0 to 100 MHz). Does the -3dB cut-off frequency match the desired cut-off frequency fo = 20 kHz ? Explain. From the graph, estimate the roll-off rate. (attachment)


I inserted my schematics and graph.. am i right?
:)
 

Attachments

Audioguru

Joined Dec 20, 2007
11,249
what do u mean? :confused:
The simple first-order lowpass filter is just the resistor and capacitor. It must have a fairly high resistance load. An opamp has a very high input resistance and can be used as an output buffer if needed to drive a fairly low resistance.

A second-order Butterworth filter must have an active amplifier like a transistor or opamp.
 

Audioguru

Joined Dec 20, 2007
11,249
The Rl in your circuit is whatever the opamp is driving. The load of your RC lowpass filter is the extremely high input resistance of the opamp.

Your circuit is correct. It is an RC lowpass filter with an opamp buffering its output.
 

Thread Starter

naspek

Joined Feb 17, 2010
45
The Rl in your circuit is whatever the opamp is driving. The load of your RC lowpass filter is the extremely high input resistance of the opamp.

Your circuit is correct. It is an RC lowpass filter with an opamp buffering its output.
ok.. so.. the value for RL , can i choose it randomly
or, there is formula to get the perfect value for RL?
 

Audioguru

Joined Dec 20, 2007
11,249
The value of Rl in your circuit has nothing to do with the simple RC lowpass filter. It is the value of the resistance that the opamp is able to drive.

There is no perfect value for Rl of an opamp. It is 2k ohms or more for ordinary opamps.
 

Thread Starter

naspek

Joined Feb 17, 2010
45
The value of Rl in your circuit has nothing to do with the simple RC lowpass filter. It is the value of the resistance that the opamp is able to drive.

There is no perfect value for Rl of an opamp. It is 2k ohms or more for ordinary opamps.
is my answer is wrong, if i didn't include the load resistor?
 

Audioguru

Joined Dec 20, 2007
11,249
It is phoney baloney school work, not a real circuit.
Nobody knows what is the value of the load. It could be anything.
The quiz should have listed the load resistance if it is important to the calculations of whether the circuit will work or not.
 

Audioguru

Joined Dec 20, 2007
11,249
You zipped the file instead of posting the schematic and graphs. I don't want to unzip it again so I don't know if your schematic is correct.
 

Audioguru

Joined Dec 20, 2007
11,249
Your schematic shows an opamp without a power supply so it won't work.
I don't know what you are asking about it because it is extremely simple.
 
Top