# pole of a low-pass filter

#### anik321

Joined Oct 25, 2008
22
In a simple RC low pass filter, the simplified transfer function comes out to:

1/(1+jwRC)

Solving for the root in the denominator, we see a pole exists at S=j/RC

This is also the w3db point where the magnitude of the transfer function starts falling.

Question: How is the w3db point also a pole? When you are making the bode plot, why does the magnitude not blow up at this point?

#### The Electrician

Joined Oct 9, 2007
2,914
Since s = jω, the pole is at s = -1/RC, not j/RC.

The Bode plot is along the positive imaginary axis, so you don't pass through the pole, which is on the negative real axis.

#### anik321

Joined Oct 25, 2008
22
Since s = jω, the pole is at s = -1/RC, not j/RC.

The Bode plot is along the positive imaginary axis, so you don't pass through the pole, which is on the negative real axis.
Electrician, thank you for the quick response. But i dont think I still understand

How is the bode plot along the positive imaginary axis (specially the magnitude plot)? The 3db point (and all other values on the magnitude plot) is an observable physical quantity - how can it be along the imaginary axis?

I feel as though my fundamental understanding may not be correct.

#### steveb

Joined Jul 3, 2008
2,436
How is the bode plot along the positive imaginary axis (specially the magnitude plot)? The 3db point (and all other values on the magnitude plot) is an observable physical quantity - how can it be along the imaginary axis?
The Fourier Transform is a special case of the Laplace Transform with $$s=j\;\omega$$. The bode plot is just a representation of the Magnitude and Phase of the Fourier Transform.

#### StayatHomeElectronics

Joined Sep 25, 2008
1,073
The magnitude does not blow up because the denominator is not going to zero at that point. The three db point is determined by w(3db) = 1/RC. Substitute that back into the equation and you get

1/(1+jw/w(3db))

To get the magnitude plot, you need to take the magnitude of the complex number.

So, at w = w(3db), the denominator is not zero.

#### The Electrician

Joined Oct 9, 2007
2,914
Electrician, thank you for the quick response. But i dont think I still understand

How is the bode plot along the positive imaginary axis (specially the magnitude plot)? The 3db point (and all other values on the magnitude plot) is an observable physical quantity - how can it be along the imaginary axis?

I feel as though my fundamental understanding may not be correct.
A Bode plot is simply the magnitude of the transfer function as the jω variable is allowed to vary from some low frequency to some higher frequency.

A typical pole-zero plot is on the complex plane, and the jω axis is vertical. Your pole is on the left horizontal (negative real) axis, and the jω variable never encounters the pole. If your pole were complex with a zero real part, then your plot would have a singularity, but that's not what you have.