Greetings,

I spent a little time reviewing Bode plots. (it's been a while)

I was playing with the phase response of a plot and it did not seem obvious to me that the phase response should be (minus arctan())

Now, when the transfer function has no poles in the denominator, the book states that the phase response is arctan() (no minus sign).

I've attached a sketch of my thoughts.

Can you explain why the minus sign?

By the way, my book simply states the equation with no reason, a second book simply stated, "note the minus sign"

Thanks
-Sparky

 

It gets worse, wait untl you run into imaginary numbers (advancd stuff), which is the square root of a -1. Basically phase shift can go either direction, inductive and capcitive shifting are opposites, which is plus or minus on the direction of shifting.

 

When you have a transfer function, in order to get the phase of the whole thing, you get the phase of the numerator minus the phase of the denominator... rules of complex numbers..

the arc tan comes from the idea that inorder to get a phase of a complex number, you want the angle between the imaginary and real parts, which are orthogonal.. therefore we have arctan (opposite over adjacent)

now..if you have a transfer function which was say 1/s+2 for example... the numerator is purely REAL and positive.. therefore , has a phase of 0 .. once you get the phase of the bottom part, you will have 0 minus something .. hence the negative...

hope that clarifies

 

Thanks.

I'll dig out my old complex variables book and review those rules.

you're saying it's more than simply starting with the sketch I attached on the first post - correct?

 

OK Sparky, I have attached my effort at explaining the reason why the sign of the phase portion of transfer function of the first-order expression that your question contained bears a negative sign.

hgmjr

 

Thanks

I see how to apply the 'rule' now. (that is - the phase angle in the denominator subtracts from the phase angle in the
numerator). It seems straightforward once the rule is applied (and if you know the rule in the first place.)

I’m embarrassed to say I have completely forgotten that rule about phase angles subtracting when in the denominator.

I do know that when dealing with log’s – they behave this way – (log a/b = log a – log b)

It must have made since in class because I don’t recall having problems with Bode plots.

I guess not knowing the rule doesn’t count – if it does, I’ll try it next time I’m pulled over.

Thanks for all the help.

also - HGMJR - thanks for the detailed pdf!!
-Sparky

 

Just for completion -

I've attached a derivation of sorts showing why phase angles add or subtract depending on whether multiplication or division is involved.

 

Sparky,

Your book is correct. And it does not need to explain the minus sign because the algebra automatically takes care of it. By the way, you do not have to waste time and effort converting to polar form for such a simple expression. Now let's get started.

Starting with the s=jw substitution.

1/(s+so) ==> 1/(so + jw)

Now multiplying both the upper and lower term by the conjugate of the denominator to eliminate any complex terms in the denominator.

(so-jw)/(so^2 +w^2)

So it can be seen that the /_ is Arctan(-w/so) .

Which easily converts to -Arctan(w/so) .

Ratch

 

Thanks Ratch,

I agree (and would not) go through all the effort for the simple expressions.

For me, this topic turned into a curiousity that I wouldn't drop.

When I was led to my complex variables book by Bill and Mazaag and others I found a list of "rules" for various operation of complex numbers. Two of those rules involved phase angles adding or subtracting (the two I've attached the derivation for).

The book stated that it will leave the deivation as an exercise.

I spent a little time playing with it and when I got it, I wanted to share it.

I do agree as you have shown that the algebra takes care of the sign. I did not see that when I first started trying to knock off the rust on my Bode techniques.

Thanks
-Sparky