Low Pass Filters and Laplace Transform

Click For Summary
The discussion centers on the impracticality of realizing an ideal "brick-wall" lowpass filter in practice. Participants note that while r(t) can represent any periodic signal, it must be analyzed as a sum of harmonics, particularly focusing on lower frequencies for lowpass filters. The transfer function H(s) must be converted to H(jω) to examine frequency response, revealing that the abrupt changes in amplification at specific frequencies are unachievable. The challenge of proving this using the Laplace transform is highlighted, as it inherently assumes no pre-existing activity before t = 0, making realization impossible for causal networks. Ultimately, the Fourier transform provides insights into the issue, indicating that the ideal filter's impulse response cannot exist in a practical setting.
jendrix
Messages
120
Reaction score
4

Homework Statement



Given that r(t) = L^-1 (Inverse laplace) *H(S) and by making the link between the time-domain and frequency-domain responses of a network, explain in detail why the ideal “brick-wall” lowpass filter is not realisable in practice. [/B]

Homework Equations

The Attempt at a Solution



Is r(t) any input signal or is it a specific type? I am completely stumped by this.

Regards
 
Physics news on Phys.org
jendrix said:
Is r(t) any input signal or is it a specific type?
In principle r(t) could be any type of periodical signal, but speaking of a "low pass filter", lower frequencies are meant. So r(t) must be regarded as a sum of its harmonics, ( Fourier transform ).

Having a transfer function, H(s), you must substitute the "s", so that it becomes H(jω). Then you asked to explain why the amplification in the filter cannot be changed in steps, varying ω.
 
jendrix said:
Is r(t) any input signal or is it a specific type?
From the context of your post, I'd assume H(s) is the transfer function of the brick-wall lowpass filter and so r(t) is its time-domain response (impulse response).

If you determine r(t), it should be pretty clear why it's unrealizable.
 
Hesch said:
In principle r(t) could be any type of periodical signal, but speaking of a "low pass filter", lower frequencies are meant. So r(t) must be regarded as a sum of its harmonics, ( Fourier transform ).

Having a transfer function, H(s), you must substitute the "s", so that it becomes H(jω). Then you asked to explain why the amplification in the filter cannot be changed in steps, varying ω.
Is it because you can't vary w as it is a filter and therefore w will stop at a certain value?
 
ω is the frequency of the input. You may set it to any value.

I cannot explain why a filter that makes a "brick-wall" step in the amplification from say ω = 9.99999 to 10.00001 is unrealizable, since english is not my mothers tongue.
 
Find the impulse response of whatever brick-wall filter you're considering to see why it's unrealizable.

The Wikipedia page on Sinc filters should be a big help.
 
I don't think this can be proven via the Laplace transform, at least not the commonly understood single-sided (s-s) laplace transform. That's because the s-s laplace a priori assumes no activity before t = 0 (the double-sided (d-s) transform allows this, as does the Fourier transform.)

With the Fourier transform it can readily be shown that the inverse transform of the brick filter has response for t < 0 but this is impossible for a causal network output where the input is zero for t < 0, hence the filter realization is impossible too. But the problem asked to use the laplace to prove this & I don't think that's doable (unless as I say the d-s laplace is used).
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 4 ·
Replies
4
Views
765
  • · Replies 13 ·
Replies
13
Views
5K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
4K