Bode Plot Method For a Transfer Function

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1. May 22, 2016

jendrix

Hi ,

I was hoping you could look at this link and let me know if I have interpreted the method correctly

http://i.imgur.com/5axTiBN.jpg

As I understand it the transfer function has had s replaced by jw but also the top and bottom are divided by 10 -
This is so that the brackets on the bottom fit a first order lag? Though it looks like they forgot to divide the top by 10.

When it gets put into dB form 20log10 is used for the first term though?

Is that correct and the 100 in the numerator is a typo?

Thanks

2. May 27, 2016

Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. May 28, 2016

Staff: Mentor

Yes, they factorized the denominator from (jω + 10) to 10(jω/10 + 1) but accidently lost that first 10.

The reason for standardising on the form (jω/N + 1) is so that for ω<<N this term contributes a value of approx unity to the T.F. so can be "ignored" until ω→N.

I don't like their use of 10‧log |(jω)2 + 1|
I think the j should not be included like that; we don't want to square the j.

The full expression for the amplitude of each term does correctly involve 20‧log
but because each frequency-dependent expression is of the form,
e.g., 20‧log √((1/10)2 + 12)
the rules of mathematics allow them to combine the 20 and the √ and write it as 10 without the √.

you should satisfy yourself of the validity of this.

A little confusing!