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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!

A screenshot showing the pole frequency(Wp of a low pass filter)

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  • \$\begingroup\$ Can you check your definitions in bold type, particularly the word 'frequency' \$\endgroup\$ – Chu Jun 24 '17 at 14:14
  • \$\begingroup\$ ... a pole is a value of s that makes the transfer function infinite. \$\endgroup\$ – Chu Jun 24 '17 at 14:18
  • \$\begingroup\$ @Chu I don't understand. What exactly is 'S' here? Isn't it the frequency itself? I understand that S is complex here but that doesn't clear my doubt. \$\endgroup\$ – Sumanth Jun 24 '17 at 15:20
  • \$\begingroup\$ s is sometimes referred to as frequency because of the close relationship between the Laplace transform and the Fourier transform. It gets confusing because the substitution \$\small s\rightarrow j\omega\$ gives the steady state frequency response of a TF. In your example, poles at s=-2 and -3 do not mean that the TF has infinite gain at any particular frequency. In fact there is no frequency where the gain is infinite for that TF. Better to stick with the definition of a pole as the value of s where the TF is infinite. \$\endgroup\$ – Chu Jun 24 '17 at 15:30
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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: -

enter image description here

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.

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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.

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  • \$\begingroup\$ so can we safely say that once we convert the TF from S domain to Frequency​ Domain, the concept of TF becoming infinity at a pole(frequency not 's') does not hold true anymore? \$\endgroup\$ – Sumanth Jun 25 '17 at 12:36
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    \$\begingroup\$ Exactly, we can. In the frequency domain at the pole's frequency the TF lessens of a 3dB quantity and continues to decrese as long as the frequency increases from the pole's frequency (called cut-off frequency) to higher values, as you can evaluate by plotting the corresponding bode diagram. upload.wikimedia.org/wikipedia/commons/thumb/6/60/… In the frequency domain the magnitude of a TF can tend to infinity only if you have a 2nd order system (at least) and the frequency is set at its resonance frequency. \$\endgroup\$ – cyberdyne Jun 25 '17 at 14:06

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