# Complex Filter Function - 3D graph

Discussion in 'General Electronics Chat' started by kdillinger, Nov 29, 2009.

1. ### kdillinger Thread Starter Active Member

Jul 26, 2009
141
3
I have been reviewing Maxim's Filter Primer application note for an idea to create my own 3D graphical utility. This goes in conjunction with the MFB derivation I posted a few weeks ago.
This work has certainly stressed the limits of my knowledge and understanding, so I need another set of brains to clarify a few things for me.
I want to start with a simple low pass filter per Maxim's app note, referring to figure 2b. I understand why they highlighted the jω axis in red as it is the frequency response of the filter and the 2D version as I am extremely familiar with regarding Bode plots (figure 2c). What I am confused about is the σ axis. Where did this comes from and how it is graphed especially since it is >0.
I can calculate the pole location of a simple RC filter (-1/RC), I know the cutoff frequency, time constant, and in my mind σ=0 during sinusoidal steady state.
What simple thing am I missing?

Thoughts and ideas will be greatly appreciated...especially at this hour.

2. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
In a complex exponential function, σ is the factor which causes the envelope to grow without limit (σ > 0), or decay (σ < 0). The root locus will determine if the transfer function is stable or unstable and under what conditions of gain or other paramerters.

We often joke that : "The positive real roots disappear or the system does."

The roots of a polynomial equation are either real, or occur in complex conjugate pairs. Each complex root leads to an exponential that looks like:

e^(σ ± jω) = (e^σ)*(e^±jω)

From that expression you can see the effect of the σ

3. ### kdillinger Thread Starter Active Member

Jul 26, 2009
141
3
Hello Papabravo,

Thank you for the response. I remember the root locus, but I am still not sure what their figure 2b graph is telling me. Perhaps I am being dense.

If I were to pick values for the filter, say R=1kΩ and C=1μF then the pole for this transfer function is [1/(sRC+1)] and is located at -1000 on the negative real axis. There are no imaginary roots for this transfer function. If I remember correct the zeros are located at -∞ and +∞.

Referring back to 2b, the response is highlighted at σ=0 so the magnitude starts at 0dB and rolls off with increasing frequency as expected.

Now what I am still having difficulty with is that at some value of σ on their graph, the magnitude response is large. I am trying to correlate this large value or peak in the response to the negative real part (σ) of the pole at -1000.

Like I said before, I must be dense to not see the obvious.

4. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
The 3D plot in the app note has negative values along the σ-axis. For positive σ the response grows without bound. To compute the magnitude for a complex point s = σ ± jω, just do the substitution and compute the magnitude.

As indicated in the app note σ is related to the transient behavior of the system rather than the steady state behavior.