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## Main Question or Discussion Point

**Determine the transfer function of the system H(jw)**

why the answer is not 1/[1+(jw/612)] but equal to 0.318/[1+(jw/612)]?

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why the answer is not 1/[1+(jw/612)] but equal to 0.318/[1+(jw/612)]?

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

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

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Then, you must compare this result with both transfer function for the case w=0. Which one is valid?

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20log(Vout/Vin)=-10

Then, you must compare this result with both transfer function for the case w=0. Which one is valid?

丨Vout/Vin丨=10^(-0.5)=0.316? then what is the meaning of this number

for the equation of 1/[1+jw/612] 丨Vout/Vin丨=0.999998 they are not the same,then how can i solve this problem?

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

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

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Are you able to insert w=0 into BOTH functions - and decide, which one results in 0.318?

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from the graph (a) 0 to wTWOfunctions.

Are you able to insert w=0 into BOTH functions - and decide, which one results in 0.318?

from the graph (b) phase differnet ≅-40 degree so the equation

so V

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Sorry - I resign. Are you kidding me?

- #15

jim hardy

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The

becomes so natural

that your lecturer may have completely forgot to mention it,

or his mention may have been so brief you missed it....

丨Vout/Vin丨=10^(-0.5)=0.316? then what is the meaning of this number

for the equation of 1/[1+jw/612] 丨Vout/Vin丨=0.999998 they are not the same,then how can i solve this problem?

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

and your graph shows you at its left edge that the constant term K is -10db not 0db

and 10

(

That's how transfer functions work.

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

- #16

jim hardy

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as explained above, it'swhy the answer is not 1/[1+(jw/612)] but equal to 0.318/[1+(jw/612)]?

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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".

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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".

as explained above, it'sK Xf(frequency) ,,0.316 X1/[1+(jw/612)]

thank for your,since my lecture notes just have the formula 1/[1+(jw/RC)],so i do not know that it have Dc gain K, but after search at google and read your comment, i got the point now, thank so much.Sorry - I resign. Are you kidding me?

- #19

jim hardy

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I think if you do that division and expand the polynomial you'll find in it terms that are not multiplied by jw1/[1+(jw/RC)

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.

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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.Sometimes teachers forget how strange it all seems to beginners.

.

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?

Last edited:

- #21

jim hardy

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Sorry LVW, my remark was aimed at his lecturer not at your comments.however, after 25 years experience in teaching electronics, I know about the problems some beginners might have.

That he didn't seem to pick up on yours is why i belabored the mechanistic starting point. I well remember being so abjectly confused that i needed a "Reset" button , and that's what i tried to give him.

If it came across as criticism - that was the opposite of my intent,

indeed i noticed your .posts and used them for a springboard

i should have opened (as i often do) with "LVW is on the right track here, " because you were.

I think he came around

Did i go too far in the direction of spoon-feeding ?

Same question re my post #4 here:

https://www.physicsforums.com/threads/opamp-transfer-function-problem.869001/

old jim

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It was such a simple problem - and I really could not belief that somebody could have difficulties to decide which of the given functions correspondes to the proper DC gain. Finally, I was not sure if it was a kind of trolling.

LvW

- #23

jim hardy

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I dont see where you're guilty of that.and sorry for the misunderstanding I have caused.

Your science is consistently very good

and we all struggle for words to express our thoughts

there's two translations in any communication -

one our mental image to our words

and one from those words to another mental image in mind of the listener...

Then listener tries to communicate his summation back to us .........

remember the early days of language translation by computer ?

They handed the English sentence

"The spirit is strong but the flesh is weak"

to a computer for translation into Russian then back to English

and it came back

"The wine's great but the meat is just awful."

With Highest Regards

old jim

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to a computer for translation into Russian then back to English

and it came back

"The wine's great but the meat is just awful."

Perhaps I should try to repeat this procedure?

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