# Finding phase angle of transfer function for hi and lo pass filters

Discussion in 'Electronics Homework Help' started by Jodles, Nov 19, 2010.

1. ### Jodles

13
0
Nov 19, 2010
Hi!

I'm trying to derive the argument of H(jw) for hi and lo pass filters.

So far I've gotten the following:
for low pass:
From the voltage dividor I've gotten: H(jw) = 1/ (1 + j(w/w_c)) (w_c = ang. cutoff freq.)
Then I found the angle by converting it to polar form and I get:
arg H(jw) = arctan (w / w_c)

But to my knowledge, it should be H(jw) = - arctan(w/w_c). Where is this minus coming from?

Same thing with hi pass. Following the same procedure as above, starting with H(jw) = 1 / (1-j( w_c / w ) ), I get:
arg H(jw) = arctan (w_c / w)

I also know I "should" be getting H(jw) 90 - arctan (w_c / w).

I can see this by recognizing how the filters work and how the output voltage lags the input voltage. But I don't see how this happens mathematically?

Any help would be much appreciated!

2. ### barathbushan

223
0
Sep 26, 2009
step 1 ) factorise H(jw) = 1/ (1 + j(w/w_c))

i mean multiply both numerator and denominator by 1 - j(w/w_c) , and then
separate real and imaginary terms

step 2) angle = tan (imaginary/real)

Last edited: Nov 19, 2010
3. ### Jodles

13
0
Nov 19, 2010
Thanks! I tried that earlier but ended up with a huge equation. Will try again!

4. ### Jodles

13
0
Nov 19, 2010
Ok, I've now gotten the right answer for the lo pass. But for the hi pass filter, I get:
1 / (1-j w_c/w) = (1+jw_c/w) / (1+w_c^2 /w^2), then separating these and taking the arctan of b/a, I get w_c /w as before. Where do I find the 90 - arctan w_c/w?