Low-pass filter regarding Laplace.

In summary, low-pass and high-pass filters allow signals of specific frequencies to pass while attenuating others. These signals can refer to voltage or current in a circuit. To solve a problem involving these filters, one must use Laplace transform and know how to express components such as R, C, and L in terms of their Laplace representations. Differentiation is not needed. A common mistake is to make an algebraic error, resulting in incorrect units in the final answer.
  • #1
aryise
2
0

Homework Statement



Low-pass filter allows signals of low frequency to pass while attenuating high frequency signals; similarly, a high-pass filter allows signals of high frequency to pass while attenuating low frequency signals. The signals here can refer to voltage or current in the circuit. There is a link to a picture. I don't quite understand how to solve it.

Link: http://img198.imageshack.us/img198/5677/mathb.png

Homework Equations



I understand that Vin = Vc + Vr.
How do I use Laplace transform on Vin = Vc + Vr. AnD if Vout is the voltage across the resistor R.

The Attempt at a Solution



I am totally stuck.
 
Last edited by a moderator:
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  • #2
Do you know how to represent circuits and basic components R, C, L in s domain?
This following link may help.
http://tintoretto.ucsd.edu/jorge/teaching/mae140/lectures/6sdomain.pdf


aryise said:

Homework Statement



Low-pass filter allows signals of low frequency to pass while attenuating high frequency signals; similarly, a high-pass filter allows signals of high frequency to pass while attenuating low frequency signals. The signals here can refer to voltage or current in the circuit. There is a link to a picture. I don't quite understand how to solve it.

Link: http://img198.imageshack.us/img198/5677/mathb.png

Homework Equations



I understand that Vin = Vc + Vr.
How do I use Laplace transform on Vin = Vc + Vr. AnD if Vout is the voltage across the resistor R.

The Attempt at a Solution



I am totally stuck.
 
Last edited by a moderator:
  • #3
You will need the Laplace transform for the excitation voltage U(t)sin(wt). You also need to know how to express L, C and R in terms of their Laplace representations, as klondike points out.
 
  • #4
Thank you! I finally solve it... :)
 
  • #5
aryise said:
Thank you! I finally solve it... :)

Hey how did you solve it? i have a simillar question. Help would be greatly appreciated
 
  • #6
Cfty said:
Hey how did you solve it? i have a simillar question. Help would be greatly appreciated

See my post of June 24.
 
  • #7
rude man said:
See my post of June 24.

I can't understand. Argh.kNot so good in this.

I understand SinWt = iR + iC

Then i applied laplace : W/(s^2+w^2) = IR + IC.

Is this correct?
 
  • #8
Cfty said:
I can't understand. Argh.kNot so good in this.

I understand SinWt = iR + iC

Then i applied laplace : W/(s^2+w^2) = IR + IC.

Is this correct?

sin(wt) = iR + iC is not correct. R and C are not the same type of impedance!
But you got the right transform for sin(wt).
 
  • #9
rude man said:
sin(wt) = iR + iC is not correct. R and C are not the same type of impedance!
But you got the right transform for sin(wt).

I tried it with W/(S^2 + W^2) = IR + I/CS

Do i need to diffrentiate anywhere in this question?. I got the first two parts for the showing right. I can't seem to get the final part. I followed the laplace transform on top by making I the subject and doing partial fractions. Then when i found I i laplace inversed it to Find i. And Since Vout is VR : i took the i*R.
 
  • #10
Cfty said:
I tried it with W/(S^2 + W^2) = IR + I/CS
This is correct

Do i need to diffrentiate anywhere in this question?.
No.
I got the first two parts for the showing right. I can't seem to get the final part. I followed the laplace transform on top by making I the subject and doing partial fractions. Then when i found I i laplace inversed it to Find i.
That is correct.
And Since Vout is VR : i took the i*R.
Yes, if I understand you correctly, Vout= iR.
Show your math in detail.
 
  • #11
Hey This is my written solutions. If you could tell me where i went wrong I would greatly appreciate it.
 

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  • #13
rude man said:
This is correct



Show your math in detail.


Hope It helps
 
  • #14
Your equation at the top of page 1 is correct. And your answer is almost correct, obviously.

You just made an algebraic mistake somewhere. You can see that the last term in your answer is incorrect since it's dimensionally incorrect. Each term must be dimensionless
(actually it's Volts but you didn't include volts in your excitation function).

I can't follow your math from the pictures. I will leave it to you to find the mistake you made.

Good luck and good work!
 

Related to Low-pass filter regarding Laplace.

1. What is a low-pass filter?

A low-pass filter is a type of electronic circuit that allows low-frequency signals to pass through while attenuating high-frequency signals. It is commonly used to remove noise or unwanted frequencies from a signal.

2. How does a low-pass filter work?

A low-pass filter works by using a combination of capacitors and resistors to create a frequency-dependent impedance. This impedance causes high-frequency signals to be attenuated while low-frequency signals are passed through with minimal loss.

3. What is the Laplace transform?

The Laplace transform is a mathematical tool that converts a function of time into a function of complex frequency. It is commonly used in signal processing and control systems to analyze the behavior of systems over time.

4. What is the relationship between a low-pass filter and the Laplace transform?

The Laplace transform is commonly used to analyze the frequency response of a low-pass filter. By taking the Laplace transform of the filter's transfer function, we can determine how the filter will affect different frequencies in a given input signal.

5. What are some applications of low-pass filters in real-world systems?

Low-pass filters have a wide range of applications in various systems, including audio processing, telecommunications, and control systems. They are used to remove noise from audio signals, reduce interference in communication systems, and smooth out control signals in feedback loops.

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