What exactly is Pole Frequency in a Filter and how does it affect the frequency response?

Question

I'm having great trouble understanding the concept of a pole in a real world electric circuit. I do understand what a 'pole' is and what a 'zero' is, in the point of view of a 'Transfer Function' but when I'm study bode plots, the definition seems to differ.

WHAT I ALREADY KNOW: (Assuming a Voltage Transfer Function i.e Vout/Vin)

"A pole frequency is that frequency at which the transfer function of a system approaches infinity"

And similarly "A Zero frequency is that frequency at which the transfer function of a system approaches Zero"

THE QUESTION:

1) Why does the magnitude Bode Plot of the response of a filter NOT approach Infinity at a pole? (and why is the -3dB point at the pole frequency?)

2) In the attached image, why is Wp (Omega subscript:P) called the pole frequency when the denominator clearly does not become zero at that frequency?

3) Dealing in the S domain if a transfer function turns out to be 1/(s+2)(s+3) how can the Negative pole frequencies i.e s=-2,s=-3 be produced physically?What are the poles in this circuit?

I feel I'm missing something very significant here. Please help!

Answer 1

1) Why does the magnitude Bode Plot of the response of a filter NOT approach Infinity at a pole?

Try looking at this picture and recognize that poles may exist as infinities in the bode plot but more usually they are "behind" it: -

2) In the attached image, why is Wp (Omega subscript:P) called the pole frequency when the denominator clearly does not become zero at that frequency?

The denominator becomes zero "behind" the bode plot. See above for relationship between bode and pole-zero diagrams.

3) Dealing in the S domain if a transfer function turns out to be 1/(s+2)(s+3) how can the Negative pole frequencies i.e s=-2,s=-3 be produced physically?What are the poles in this circuit?

They aren't physical at all - they don't exist except as a mathematical model to explain things. The only thing that exists on the 3D image above (bottom left) is the bode plot.

Answer 2

All depends on the variable you're using to represent your transfer function (TF). The definition of pole you know is based on using the complex variable s. Indeed, in that case you have that poles are the solutions of the characteristic polynomial at the denominator of the TF. In your case s=-(1/RC) per given RC:

$$A)\ \lim_{s\to-(1/RC)} \frac{V_{out}}{V_{in}} (s)=\lim_{s\to-(1/RC)} \frac{1}{1+sRC}=+\infty $$

Since this is a matematical approach different from using the frequency domain, calling s=-(1/RC) as frequency is incorrect (clarifying in part your 3rd question).

But, if you find more familiar talking about frequencies, we can move from the sdomain to the frequency domain. The two domains are linked by the relationship: $$s=j\omega$$ (let's remember that s is a complex variable) where omega represents any generic frequency value from 0 to infinity. Therefore in the frequency domain your TF remains:

$$B)\ \frac{V_{out}}{V_{in}} (j\omega)= \frac{1}{1+j\frac{\omega}{\hat{\omega}}}$$ with $$\hat{\omega}=1/RC$$ and can be dimensionally considered as a frequency per chosen R and C.

Therefore:

1) Look at the different representaion of the TF between A) and B). The -3dB is because if you chose the representation in the frequency domain and calculate the module in dB, at the frequency associated to the pole you will have:

$$|\frac{V_{out}}{V_{in}}|_{dB} (\omega=\hat{\omega})= 20\ \log\ |{\frac{1}{1+j}}| = 20\ \log\frac{1}{\sqrt{2}} = -3 [dB] $$

2) Explained with A);

3) Poles are s=-2, s=-3 and they are not frequencies: they are poles in the s domain associated with the frequencies $$\omega=2\ rad/s$$ and $$\omega=3\ rad/s$$ in the frequency domain and can be set properly by the RCs values.