# integrator phase shift

Discussion in 'Homework Help' started by pietj1212, Oct 10, 2015.

1. ### pietj1212 Thread Starter New Member

Sep 10, 2015
10
2
i know something about phasor diagrams but ive never read anything about a serie parallel duo. Can i just imagine C on the +j axes and R2 on the -x axes and r1 on the +x axes and then use simple math?

2. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,968
616
Um... sort of.

You can convert capacitor into its complex form: $\frac{1}{jwC}=\frac{-j}{wC}=-\frac{1}{wC}j$
So now capacitor is sort of "resistor" that depends on frequency w.
Since you now have two resistors in parallel, you can find their equivalent resistance, it will be complex quantity, and you can map this complex quantity, Real part goes on x axis, imaginary part goes on the y axis.

3. ### pietj1212 Thread Starter New Member

Sep 10, 2015
10
2
i do understand that but the complex impedance will be (((1/sC)*R2)/((1/sC)+R2)) im not sure how to draw that.

4. ### pietj1212 Thread Starter New Member

Sep 10, 2015
10
2
I guess i got the answer maybe somebody cares to check.
I choose R1 to be 1k
I rewrite the formula above to R2/(jwR2C+1)
i calculate the complex vanlues of z2 which turns out to be 1j+ 1.73
R2/(jwR2C+1) = 1j+ 1.73
i rewrite to j/R2 + 1,73/R2 = jwR2C + 1
i rewrite to jwc + 1/R2 = j/R2^2 + 1/73/R2^2
1/R2 must be equal to 1.73/R2^2 so r2 = 1.73
plug R2 in formula above c = 5e^-6