0
\$\begingroup\$

I'm reading a manual about op. amps. (op amps for everyone) which, at a certain point, speaks about poles and zeros. I'm trying to understand the following example:

Given the following band reject filter's transfer function

enter image description here

the poles and zeros are supposed to be the following:

enter image description here

enter image description here

One of the poles is supposed to be 0.44/tau, however, if I apply it to the transfer function, specifically to 1 + tau*s/0.44 = 1 + tau*j*w/0.44, my understanding is that I get 1 + tau*j*(0.44/tau)/0.44. So the taus multiplying and dividing go away, and the same goes for the 0.44s. That leaves 1 + j. This doesn't make the denominator 0 but it should in order to be a pole. I have the same problem with the other poles and zeros. What am I missing?

\$\endgroup\$
0
1
\$\begingroup\$

Corner frequencies are not the poles of the transfer function and hence you wouldn't get value of denominator equal to zero at those frequencies ,these frequencies in some books refer as pole or zeroes but actually they are not !

Poles and Zeroes of a transfer function is decided by real part of 'S' not by imaginary parts of S(i.e frequency part). But for bode plot we already assumed that $$Re[S] =0$$ and so concept of pole Zero doesn't make sense

\$\endgroup\$
0
\$\begingroup\$

There is a pole at \$s=\frac{-0.44}{\tau}\$

If you plug that into the formula for G, in the denominator you will have

\$(1 + \frac{-0.44}{\tau}\frac{\tau}{0.44}) = 1-1 = 0\$

\$\endgroup\$
5
  • \$\begingroup\$ Then there is a typo in the book? \$\endgroup\$ – Martel Feb 11 at 0:17
  • \$\begingroup\$ If the book is missing the negative sign, then yes. \$\endgroup\$ – Math Keeps Me Busy Feb 11 at 0:19
  • \$\begingroup\$ But how do you get rid of the j? Because s=omega*j \$\endgroup\$ – Martel Feb 11 at 0:19
  • \$\begingroup\$ I solved the equation \$(1+\frac{\tau s}{0.44}) = 0\$ for \$s\$. If you want you can say that \$s=j\omega = \frac{-0.44}{\tau}\$ or \$\omega = \frac{0.44j}{\tau}\$ (by mulitplying both sides by \$-j\$) \$\endgroup\$ – Math Keeps Me Busy Feb 11 at 0:23
  • \$\begingroup\$ the j is supposed to be there. it just means your transfer function at that frequency has a phase shift. the poles with real s, aka imaginary omega, correspond to exponential chatacteristic. imaginary s, aka real omega, means sinusoidal characteristic \$\endgroup\$ – Pete W Feb 11 at 0:23
0
\$\begingroup\$

I have been looking at it an seen that, in a previous example of the same book, the 'poles' are calculated using the modulus of the gain. If that is the case, if

$$\omega=\frac{0.44}\tau$$

that leaves a \$(1 + j)\$ factor in the denominator (because the \$\tau\$ and \$0.44\$ cancel out). Applying the modulus, which is $$\sqrt{1^2 + j^2} $$ and taking into account that \$j^2 = -1\$,the whole denominator goes to 0, so it is a pole as far as I understand.

EDIT

The above is wrong as some comments have pointed out. First, I have realized that the text I'm reading speaks about breakpoints and they seem to be called zeros/poles as well. Breakpoints are not supposed to be the roots of numerator and denominator respectively. Instead, they seem to be points where the slope of the gain changes its sign, see breakpoints. Therefore, all what I have done above is wrong because I was looking for roots when that is not what I should find.

Now, as far as I understand from the text, these breakpoints are located when each of the factors of numerator/denominator become \$\sqrt{2}\$ (not sure why) and for some reason this indicates when the slope changes sign.

In addition, it seems that when a breakpoint in the denominator changes changes the slope to negative, while one in the numerator changes it to positive.

However, if a breakpoint in the numerator is hit and the previous slope was negative, it changes the slope to flat, not directly to possitive. This is unless the actual breakpoint is double, as the example of the post, in which the numerator is \$(1 + \tau s)(1 + \tau s)\$, so \$\omega=1/\tau\$ is there two times. See below:

enter image description here

In addition, a breakpoint in the denominator changes the slope to flat if it was positive before, or to negative if it was flat before.

\$\endgroup\$
4
  • 1
    \$\begingroup\$ forget the omega. just write the transfer function out in "s", and then the poles are literally just the roots of the denominator polynomial, and the zeros are the roots of the numerator polymomial. they can very well be be complex. \$\endgroup\$ – Pete W Feb 11 at 16:02
  • \$\begingroup\$ Doing that would mean that the root in this case is \$-\frac{0.44}\tau\$ (minus), doesn't it? \$\endgroup\$ – Martel Feb 11 at 16:05
  • 1
    \$\begingroup\$ that's right. the fact that the real component is negative (in the s plane) means it is stable (it would have to be the case for all the poles) \$\endgroup\$ – Pete W Feb 11 at 16:06
  • 1
    \$\begingroup\$ re: your update ... the expression in your link is still poles and zeros. You should also be aware that the 1/sqrt(2) ratio (i.e. -3dB) at the corner frequency, is the case when you have a change in slope e.g. from flat to -20dB/decade. (and some other assumptions too, but it is a common simple case). If you have something like a real double pole, going from flat to -40dB/decade change in slope, you will have -6dB at the breakpoint \$\endgroup\$ – Pete W Feb 11 at 17:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.