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)]?
20log(Vout/Vin)=-10The 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?
from the graph (a) 0 to w0 ≅-10dBAt 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?
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?
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".
as explained above, it's K X f(frequency) ,, 0.316 X 1/[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?
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)
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.
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.
I dont see where you're guilty of that.and sorry for the misunderstanding I have caused.