About H(jw) question

What happens if you insert w=0 into the equation? Can you determine the remaining value?

 

You have asked for the transfer function - thus, I mean H(jw), of course.

 

The magnitude plot shows for small w values (down to w=0) a level of -10dB. Can you calculate the corresponding absolute value?
Then, you must compare this result with both transfer function for the case w=0. Which one is valid?

 

Don`t you see it in one of your transfer functions?

 

I repeat: Don`t you see it in one of your transfer functions?

 

At the bottom of your first post I see TWO functions.
Are you able to insert w=0 into BOTH functions - and decide, which one results in 0.318?

 

Sorry - I resign. Are you kidding me?

 

Sometimes teachers forget how strange it all seems to beginners.
The mechanics of reading frequency response charts ( we called them Bode plots in my day)
becomes so natural
that your lecturer may have completely forgot to mention it,
or his mention may have been so brief you missed it....

Some of the filter's attenuation is not frequency dependent. Its "Gain", "Attenuation", "Transfer Function" , call it what you like ,
has two parts, a constant part and a frequency dependent part
so
it is of the form K X f(frequency)
and your graph shows you at its left edge that the constant term K is -10db not 0db
and 10^-0.5 = .316, not 1
(if it started at zero db then K would be 1)

That's how transfer functions work.

I don't know how he got .318 instead of .316. You should ask him.

 

as explained above, it's K X f(frequency) ,, 0.316 X 1/[1+(jw/612)]

 

First find the "DC gain" (for w = 0) for this transfer function 1/[1+(jw/612)]
Next do the same for this one 0.318/[1+(jw/612)], then you need to compare the results. And you will see that only one of the results will fits the "graphical representation".

 

I think if you do that division and expand the polynomial you'll find in it terms that are not multiplied by jw

whenever you see that , there's DC gain

as you become accustomed to these exercises you'll come to recognize the "shape" of commonly occuring polynomials and your mind will leap immediately to an equivalent circuit.

That one's a "first order lag"



R2/R1 is the "K" above.

 

Hi Jim - perhaps you are right in some cases, however, after 25 years experience in teaching electronics, I know about the problems some beginners might have.
But considering that
* both possible transfer functions were given at the bottom of the first post
* and reading again my several answers (in particular my post#6) and the OP`s answer (mentioning the magic value of 0.318)
you certainly will come to the conclusion that I am not "guilty".

In post#12 I have asked him to compare both given functions - he simply has ignored this hint.

I simply have assumed that sombody who presents a magnitude plot together with two possible transfer functions (same denominator, only the constant factor is different) should be able to decide which function belongs to the graph.
Or - do you really think I have overstrained the questioner?