# Phase shift of a sinusoidal signal

Discussion in 'The Projects Forum' started by Enforcer83, Mar 18, 2011.

1. ### Enforcer83 Thread Starter Member

Oct 29, 2010
25
0
Alright everyone, here is my situation. I am playing around with a design I found on true sine wave inverters, found here: http://www.wpi.edu/Pubs/E-project/Available/E-project-042507-092653/unrestricted/MQP_D_1_2.pdf, to see how I can make a 3 phase inverter. At the time I found the above paper I was looking into inverters and PWM systems trying to learn more about them and ultimately SMPSs. I am looking at books to purchase in the relatively near future on SMPS and PWM systems to gain a deeper understanding of them.

This "project" is mostly just theory for now so I wanted to make sure I was understanding everything properly. I altered the design slightly to use a wein bridge oscillator versus a bubba oscillator. So with that basis, here is the meat.

Using an op-amp phase shift circuit I found here, http://www.tedpavlic.com/teaching/osu/ece209/lab1_intro/lab1_intro_phase_shifter.pdf, I began puting information regarding phase angle (e.g. 60, 120, 180, 240, and 300) into excel and using the equation $\frac{tan(\frac{x \cdot \pi}{180})}{2 \cdot \omega \cdot y}$ (where y either = c in farads or r in ohms depending what you wish to solve for and x = phase shift in degrees). I noticed something intriguing. for a phase shift of 120° and 240°, the values I solve for are the same only one is negative. If I am understanding this right, this means that after I phase shift the signal by 120° I should use a unity gain inverting amplifier to get my 240° phase shift. I assume for sinusoidal signals and square waves, an inversion is equivalent to a phase shift of 180° (inversion not always being equivalent with certain signals).

Is my thinking correct with this? It has been some time since I worked with op-amps and signals so I want to make sure.

2. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
The numbers don't add up. 120 + 180 = 300, not 240. If you make a shift of 60, then the 180 shift works.

3. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
You can only get -180° maximum per stage with the circuit you referenced. As shown, you will get phase lag (negative). If you swap the R and C, you will get lead (positive). Your best bet is probably to make a -120° shifter and a +120° shifter from the reference input. You can tune them by making a trim pot part of the resistor.

4. ### Enforcer83 Thread Starter Member

Oct 29, 2010
25
0
Ah, I see. I should have realized that when I saw the limits on $\phi$ $( \angle H(j\omega) )$.

I had seen a circuit like that earlier where the R and C were switched and at the time didn't realize its significance. I did, however, know it affected the signal differently.

I had intended on using a trimpot in series with a fixed value resistor to provide greater resolution per turn for just that purpose.

Thank you all for the quick response. I appreciate the time to help me clarify those hazy spots in my memory.