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So I have an unknown circuit with a plot like this:
enter image description here

From this I have to determine a likely transfer function model. It looks to me like there is a pole at 50 rad/sec and a zero at 1500 rad/sec, with a dc gain of 1.

When I plug those into a transfer function I get: H(s)= (S+1500)/(S+50) or as H(f) = (j2πf+1500)/(j2πf+50)

How do I find the magnitude of this transfer function?

I've been doing |H(f)| = √((2πf)^2+1500^2) / √((2πf)^2+80^2) but I am not getting the same magnitude as my plot.

Please Help!!!!!

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    \$\begingroup\$ I assume 80 is a typo. Are you entering in radians and Hz properly? \$\endgroup\$
    – DKNguyen
    Oct 24, 2021 at 22:54
  • \$\begingroup\$ Curiously same plot found in Russian site , 5fan-ru.translate.goog/files/13/… \$\endgroup\$ Oct 25, 2021 at 0:54
  • \$\begingroup\$ It's j \$2\pi f\$, not \$2\pi f\$. You're not calculating the complex absolute value correctly. \$\endgroup\$ Oct 25, 2021 at 7:26

1 Answer 1

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enter image description here

For one, you are not expressing the magnitude in \$dB\$, \$20log10(H)\$ as in your plot. Try plotting in this form.

$$ |H(w)_{dB}| = 20log10(\sqrt{(1+(w/1500)^2)/(1+(w/50)^2)}) $$

edit. per comment below, you could also write the original tf as

$$H(s=jw) = \frac{1}{30}\frac{(s+1500)}{(s+50)}$$ $$ = \frac{50}{1500}\frac{1500}{50}\frac{(\frac{s}{1500}+1)}{(\frac{s}{50}+1)}$$

which is equivalent to above (and unity gain at DC). You could also just find the magnitude (dB) of the second form and plot it, but it's more convenient to put the pole and zero in the first form and plot it, then doing these derivations.

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  • \$\begingroup\$ Is that the rule were each numerator and denominator each needs a +1? \$\endgroup\$
    – DKNguyen
    Oct 24, 2021 at 23:00
  • \$\begingroup\$ Does that re-write give unity gain at DC? \$\endgroup\$ Oct 25, 2021 at 2:33

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