Transfer function help

t_n_k

Joined Mar 6, 2009
5,455
It looks wrong. Why not show your working in detail - rather than just giving an answer. It's easier then to pin point where the mistake has arisen.
 

t_n_k

Joined Mar 6, 2009
5,455
I just wrote the equation

(Vout-Vin)/1000 + Vout/(-1/(s*.0000001))+Vout/250 = 0

and solved for Vout/Vin

Your approach is correct so the error is in your re-arranging to solve for Vout/Vin.

The correct answer in a simpler form could be

Vout/Vin=1/(5(1+0.0002jω))

or perhaps

Vout/Vin=0.2/(1+0.0002jω)

Your solution does not reduce to either of these.

It comes close - where you would have for instance

Vout/Vin=1/(5(1+j0.0002/ω)) - the ω being misplaced as a denominator term in the overall denominator - if you get my drift.
 

hgmjr

Joined Jan 28, 2005
9,027
I just wrote the equation

(Vout-Vin)/1000 + Vout/(-1/(s*.0000001))+Vout/250 = 0

and solved for Vout/Vin
Your starting equation looks reasonable for the s-domain transfer function with one exception. The term Vout/(-1/(s*.0000001)) needs a revisit. There are two errors that jump out at me. One is the sign of 1 should not be negative. I will leave the second error for you to discover.

HINT: Double check the decimal number you have written.

hgmjr
 

Thread Starter

jstrike21

Joined Sep 24, 2009
104
Thanks for the help from both of you! On 7b i need a little help. My book doesn't go over fourier series so I'm not really sure how to find the coefficients from what I read on the internet i think you have to put the equation into an integral of some sort but im not sure what the formula is.
 

t_n_k

Joined Mar 6, 2009
5,455
Thanks for the help from both of you! On 7b i need a little help. My book doesn't go over fourier series so I'm not really sure how to find the coefficients from what I read on the internet i think you have to put the equation into an integral of some sort but im not sure what the formula is.
Both functions are the simple addition of two sinusoids of different frequency. The two are almost identical, save for the phase shift of 45° in the second term of the second function V2(t).

While the two will be different when viewed as time based waveforms, their individual component magnitudes at the two frequencies will be the same when plotted in the frequency domain.
 
Top