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)]
Sorry - I resign. Are you kidding me?
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.
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.