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I was reading this document and I found something that I did not understand ...

https://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202016.pdf

Suppose the following transfer function :

$$\frac{Vout(s)}{Vin(s)} = \frac{1+\frac{s}{w_z}}{1+\frac{s}{w_p}}$$

When : $$s = -w_z <=> Vout(-w_z) = 0 $$

The question is what is s ? In the Laplace domain s is defined as : $$s = \sigma + jw$$ In the Fourier domaine sigma is equal to 0. So Fourier domaine is included in the Laplace domaine. Suppose s is equal to : $$s = jw$$ Then $$-w_z = jw$$ or $$w = jw_z$$

And now how I can apply a signal at a frequency : $$f = 2\pi \cdot jw_z$$ ?

Thank you very much and have a nice day !

Carl
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Jess
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1 Answers1

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Why the output voltage of a transfer function is not equal to zero when the exciting signal is at the frequency of the zero

Setting the denominator to 1 (because you are talking about the zero conditions rather than poles), we get: -

$$H(s) = 1+\dfrac{s}{\omega_z}$$

And, if the imaginary part of s is chosen to be \$j\omega_z\$, the transfer function is: -

$$H(s) = 1+j$$

And clearly that isn't zero but, how to find a value of s that produces a zero?

To make this happen means we have to set the TF to zero and solve for s in its general form: -

$$0 = 1+\dfrac{\alpha+j\omega}{\omega_z}$$ $$-1 = \dfrac{\alpha+j\omega}{\omega_z}$$ $$-\omega_z = \alpha+j\omega$$ $$j\omega = -\alpha - \omega_z$$

But, in the real world of signal generators, oscilloscopes and bode plots, we cannot make our signal like that; the "zero" does occur, but not along any axis we can see on a bode plot. If you plotted the pole zero diagram (this of course includes an axis for \$\sigma\$ as well as \$j\omega\$) you would see it. But, the pole zero diagram is really an abstraction from the real world and, is mathematically useful but only if you grasp it.

Fourier is along the \$j\omega\$ axis and, Laplace is the fuller story to put it simply.

For instance, an example diagram I have shows the bode plot of a 2nd order low pass filter and the pole-zero diagram combined: -

enter image description here

If you looked at the bode plot part (our real world of oscilloscopes and generators) you would see this: -

enter image description here

And behind the bode plot (into the page so to speak) you can see a pole. In other words, what we see in the world of oscilloscopes and signal generators is just a thin slice of a bigger mathematical picture called the pole zero diagram.

Andy aka
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  • Thank you for your answer :) As you said the answer is $$jω=−α−ωz$$ what is $$α$$ It does not depends on the system ? It probably depends on the input signal. So why not set it to 0 ? Otherwise if you trace the magnitude of the transfer function that I wrote, It cannot be equal to 0. – Jess Jan 08 '21 at 10:40
  • It depends on the system - it is inside the transfer function. We can't alter it by changing the signal. – Andy aka Jan 08 '21 at 10:41
  • I do not get it ^^ If $$H(s) = 1+\frac{s}{w_z}$$ , if you set $$s=jw$$ the transfer does not depends on $alpha. Did I do an error ? – Jess Jan 08 '21 at 10:48
  • Well as Chu said if wz is real I can't define it as an imaginary ... So I should say $sigma$ is equal to -wz – Jess Jan 08 '21 at 10:52
  • If you set s = jw then that is perfectly valid but, you only see the value of the transfer function along the jw axis (the bode plot). The TF (in its entirety) is a 3 dimensional thing with sigma along another axis. Look at the pictures. wz is fixed by the circuit like 1/CR or R/L - they are real numbers. – Andy aka Jan 08 '21 at 10:56
  • And as you said, if we define s to be equal to sigma( I do not know how you do to write it as the greek leter), when sigma is equal to -wz at that moment H(s) is null – Jess Jan 08 '21 at 10:57
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    Thank you for your analysis ;) – Jess Jan 08 '21 at 10:57
  • when sigma is equal to -wz at that moment H(s) is null - correct but, of course, you cannot see that on a bode plot or by using an oscilloscope! – Andy aka Jan 08 '21 at 11:00
  • I think that it is possible to see it ! But you must apply a waveform which is not sinusoidal as it correspond to sigma equal 0. But if you apply a decay exponential something like this you shoud be able to see Vout equal 0 ? Isn't it? It is not possible to see it on a Bode plot – Jess Jan 08 '21 at 11:02