Help Analyzing Phase Shifting Ckt

[likephysics]

Homework Statement

This is not a homework problem. But I still need answers/help.

I need help analyzing this phase shifting ckt.
Basically, its a sine wave fed to one end of a Resistor and an inverted sine wave (180 deg) fed to one end of a capacitor.
The other 2 ends of capacitor and resistor are tied together.
As you vary the resistance of the resistor, the phase of the resulting output at the intersection of R & C shifts with respect to the input.
I don't understand how this ckt works. I have no clue how to start analyzing this.
I thought the voltage at R and the capacitor voltage (which lags by 90 deg) add together.
But after simulating I found that's not the case.

In the attached ckt, the 2 opamps feed the sine wave to R and C.

Homework Equations

The Attempt at a Solution

 

[vk6kro]

If you connect the sinewave to the left side of the open circuited resistor at the bottom LHS, (just a typo), the opamp circuits do no more than generate a sinewave across the RC network.

The voltages across the components do add, but they add vectorially.

In the following diagram, the voltage from the opamps is shown as V total and the voltages across the resistor and capacitor are shown as sides of a triangle.

[PLAIN]http://dl.dropbox.com/u/4222062/R%20C%20vector.PNG

Start with the bottom triangle.
The voltage applied to the series RC network is V total, and the current is in phase with the voltage across the resistor. So, the current leads the applied voltage by angle X. (Rotation is anticlockwise).

If you now rotate the applied voltage (v total) by 90 degrees anticlockwise, then the other vectors will also rotate by 90 degrees, but keep the same relationship to each other.

As you might expect from your knowledge of right angled triangles,
V total ²= V resistor ² + V capacitor ²

The frequency affects the reactance of the capacitor and this affects the total current and the phase relationship between the voltages across the capacitor and resistor and the applied voltage.

 

[likephysics]

vk6kro,

Thanks for the reply.
I read your explanation and tried to apply it to a particular case. But things are not adding up.

Let's say, the frequency is 1KHz sine wave, 1Vp-p amplitude.
Xc is 160 ohms.
So I made R also 160 ohms.

Now, I expect both Vr and Vc to be equal in magnitude.

Vr will have a phase angle of 0 degrees.
Vc will be -180-90 = -270 degrees.

So the resulting Vout will be sqrt (Vr^2+Vc^2)
Phase will be -315 deg.

When I simulated, I get a phase lag of 90 degrees.
See attachments.

Also, thinking from vectors point, what combination of 2 vectors will make the resultant vector 90 degrees.

 

[vk6kro]

If V total is 1 volt and the resistor and capacitive reactance are equal then the triangle will have angles of 45 deg, 45 deg and 90 deg.

So, work out the two voltages across the capacitor and the resistor.
1 ² = Vr ² + Vc ²

so Vr = 0.707 volts and Vc = 0.707 volts.

The current leads the supply voltage by 45 degrees.

 

[likephysics]

By V total do you mean the output at the intersection of R & C?

What exactly is Vr. Is it the voltage across R?
I put a differential probe across R and C.
Vr and Vc were 1.4V.
Vt = sqrt(1.4^2+1.4^2) = 1.96
But output voltage is 1V in the simulation.

I don't see any voltage that's 0.707

 

[vk6kro]

The triangle diagram refers only to the R-C combination.

If you measure the voltage between the two opamp outputs, then the voltages across the R and C should be 0.707 (1 / SQRT(2)) times that voltage.

Ignoring the opamps for a moment, do you understand how the series R-C combination divides the voltage?

Phase shifts like this are usually described as minus 90 to plus 90 degrees rather than 0 to 360 degrees, because phase shifts of more than 90 degrees are not possible with simple networks of two components.

So, in this case, the voltage across the resistor leads the total voltage by 45 degrees and the voltage across the capacitor lags the total voltage by 45 degrees.

 

[likephysics]

I measured the voltage across the op amp outputs. It's 2Vp-p
Voltage across R & C is also 2Vp-p

The output shifts from 0 to 180 with constant amplitude (which is same as input).

I think I understand RC votlage divider. The voltage is divided according to R and Xc

 

[vk6kro]

I simulated this by using two voltage sources producing 1000 Hz sinewaves at 100 volts peak but 180 degrees out of phase with each other.

The result suggests that the output would be a sinewave 90 degrees out of phase with either input.

[PLAIN]http://dl.dropbox.com/u/4222062/RC%20phases.PNG

 

[likephysics]

So V1 is 1000V, and V2 is 1000V, but -180 phase lag.
V1=Vr and V2=Vc, Correct?
So how do they added up vectorially.
Both voltages are 1000V, but out of phase by 180 degrees. Don't they cancel out?

 

[vk6kro]

So V1 is 1000V, and V2 is 1000V, but -180 phase lag.
V1=Vr and V2=Vc, Correct?
So how do they added up vectorially.
Both voltages are 1000V, but out of phase by 180 degrees. Don't they cancel out?

Both are 100 Volts peak, but at a frequency of 1000 Hz.

No, they don't cancel out because they are like the two leads coming from a transformer winding.

They are 180 degrees out of phase (like the outputs from your opamps) but they are not joined together except through the 160 ohms and the 1 uF capacitor.

Vectorially, I think they are like this:

[PLAIN]http://dl.dropbox.com/u/4222062/vector.PNG

The output is Vout which is at right angles to the two inputs and equal in amplitude to both of them.
The two inputs add together like this because they are out of phase with each other relative to ground which would be the centre point on the hypotenuse.

 

[uart]

Where you have a node like that, driven from multiple voltage sources (Vi) each through its respective admittance (Yi), you can find the node voltage as a linear combination of the driving voltages. That is,

$$V = \frac{\sum Y_i \, V_i}{\sum Y_i}$$.

In this case,

$$V_o = \frac{(1/R) V_1 + jwc (-V_1)}{1/R + jwc}$$

$$V_o = \frac{1 - jw \tau}{1 + jw \tau} V_i$$

$$H(s) = \frac{1-\tau s}{1 + \tau s}$$

This is the classical TF of a phase shift network. It is easy to show that,

$$|H(jw)| = 1$$

and

$$\arg(H(jw)) = -2 \, \tan^{-1}(w \tau)$$

So the output amplitude is independent of frequency and the output phase varies continuously from 0 degrees to -180 degrees as $\omega$ varies from DC to infinity.

 

[likephysics]

Thanks, uart.
How did you get the magnitude and phase.

I tried to solve by rationalizing the denominator and solving. But did not get 1.

 

[uart]

Thanks, uart.
How did you get the magnitude and phase.

I tried to solve by rationalizing the denominator and solving. But did not get 1.

Yeah you could rationalize the denominator but there's no need to. That magnitude is simply the magnitude of the numerator divided by the magnitude of the denominator. Similarly the phase is the angle of the numerator subtract the angle of the denominator.

 

[likephysics]

Yeah you could rationalize the denominator but there's no need to. That magnitude is simply the magnitude of the numerator divided by the magnitude of the denominator. Similarly the phase is the angle of the numerator subtract the angle of the denominator.

When I tried to solve the magnitude, I got

$$\sqrt{1-\tau^{2}\omega^{2}-2j\omega\tau}$$ / $$\sqrt{1-\tau^{2}\omega^{2}+2j\omega\tau}$$

 

[uart]

Yeah I think I know where you're going wrong. It looks like you've tried to rationalize the denominator and then expand the numerator and have messed up the maths somewhere.

You're just making it difficult for yourself to expand terms (or rationalize the denom) here.

- The magnitude of a product is equal to the product of the magnitudes.
- The magnitude of a quotient is equal to the quotient of the magnitudes.
- The magnitude of a square is equal to the square of the magnitudes etc.

So don't make it harder than it need be.

 

[uart]

It looks like you've rationalized the denom like this :

$$\frac{1 - (w \tau)^2 - j 2 w \tau} {1 + (wt)^2}$$

Then tried to find the magnitude of numerator like this,

$$\sqrt{ ((1 - (w \tau)^2)^2 + 4 (w \tau)^2 }$$

$$= \sqrt{ ((1 + (w \tau)^2)^2 }$$

$$= 1 + (w \tau)^2$$

Well obviously that's doing it properly and it gives the expected result. You must have slipped up somewhere by trying to do it the "hard way".

 

[likephysics]

uart, I forgot to take the square after rationalizing.
Now I am stuck at finding the angle.
Its arctan (Im part/real part)
So in this case (after rationalizing) -

arctan (-2wt/(1-(wt)²)

where am I screwing up?

 

[uart]

ARH You keep doing it the hard way. Forget about rationalizing the denominator just go with the original transfer function.

$$\frac{1 - jw \tau}{1 + jw \tau}$$

That magnitude is simply the magnitude of the numerator divided by the magnitude of the denominator. Similarly the phase is the angle of the numerator subtract the angle of the denominator.
If you want to work with the rationalized denominator version it will work, but it's more difficult to simplify and you'll need to recognize where you can use the double angle tan formula.

As you correctly wrote,

$$\phi = \tan^{-1}\left(\frac{-2 w \tau}{ 1 - (w \tau)^2}\right)$$

To simplify that you need to use the double angle tan formula as follows.

Let

$$\tan(a) = w \tau$$

Then you can write,

$$\tan(\phi) = \frac{- 2 \tan(a)}{1 - \tan^2(a)} = \tan(-2a)$$

So it follows that $\phi = -2a$

 

[likephysics]

Thanks again.
How do you do it without rationalizing the denominator?
The magnitude will be
Sqrt(1-(wt)²-2jwt)/sqrt(1-(wt)²+2jwt)

Angle will be

arctan (-jwt)-arctan(jwt) ?

 

[uart]

Thanks again.
How do you do it without rationalizing the denominator?
The magnitude will be
Sqrt(1-(wt)²-2jwt)/sqrt(1-(wt)²+2jwt)

No it's much simpler than that!

$$a + jb = \sqrt{a^2 + b^2}$$

So the magnitude is just,

$$\sqrt{1 + (-w \tau)^2} / \sqrt{1 + (w \tau)^2}$$

Angle will be arctan (-jwt)-arctan(jwt) ?

That's correct.

 

[likephysics]

Where you have a node like that, driven from multiple voltage sources (Vi) each through its respective admittance (Yi), you can find the node voltage as a linear combination of the driving voltages. That is,

$$V = \frac{\sum Y_i \, V_i}{\sum Y_i}$$.

uart, how do you derive this?

 

[uart]

uart, how do you derive this?

It's actually just KCL applied to a single node.

Rearrange the following, that's all it is.

$$\sum Y_i \, (V - V_i) = 0$$Where "V" is the voltage of the node in question, and the sum of "V_i" runs over all the connecting nodes.

 

[likephysics]

I tried to use KCL and derive the equation.
But again, got stuck.

So there are 2 voltage sources V1, V2 across RC. One, 180 degrees out of phase(V2) with the other.
Let the voltage between R & C be V, which is the required voltage.

So I1 is flowing from one source and I2 from the other.
KCL says

I1+I2=0

I1 = (V1-V)/R

I2 = -(V2-V)/Xc

I tried to plug this to the I1+I2 equation, but got nowhere. I am not sure if my I2 equation is correct.
How do you account for the 180 deg phase shift in V2.

Also, why can't we solve this by finding total current I and then the voltage across capacitor C?

 

[uart]

I tried to use KCL and derive the equation.
But again, got stuck.

So there are 2 voltage sources V1, V2 across RC. One, 180 degrees out of phase(V2) with the other.
Let the voltage between R & C be V, which is the required voltage.

So I1 is flowing from one source and I2 from the other.
KCL says

I1+I2=0

I1 = (V1-V)/R

I2 = -(V2-V)/Xc

I tried to plug this to the I1+I2 equation, but got nowhere. I am not sure if my I2 equation is correct.

That should work, but you've got the sign wrong on I2. Since we are summing currents to zero then we must have either both currents into the node or both currents out.

Use,

I1 = (V1-V)/R : directed into the node

I2 = (V2-V)/Xc : directed into the node

How do you account for the 180 deg phase shift in V2.

It inverts the signal That's the same thing as a 180 degree phase shift for some signals, but fundamentally it just negates the signal.

V2 = -V1

Also, why can't we solve this by finding total current I and then the voltage across capacitor C?

You can. I've checked that method and it produces exactly the same solution as before. Remember however that you don't just want to solve for the voltage across the capacitor. The output voltage is V2 plus the voltage across the cap.

Solve for "I" then use,

Vc = Xc I
Vo = V2 + Vc
Vo = Vc - V1

This also gives the correct answer.

 

[uart]

Edit the above (just realized that you where using reactance instead of impedance in your equation). You have to use impedance as phase is definitely important when summing currents at a node.

Use,

I1 = (V1-V)/R : directed into the node

I2 = (V2-V)/Zc : directed into the node

 

[likephysics]

Edit the above (just realized that you where using reactance instead of impedance in your equation). You have to use impedance as phase is definitely important when summing currents at a node.

Use,

I1 = (V1-V)/R : directed into the node

I2 = (V2-V)/Zc : directed into the node

Isn't Zc same as Xc for a capacitor?

I tried to solve for V using I1+I2=0

(V1-V)/R + (V2-V)/Xc = 0

V2=-V1

(V1-V)/R - (V1+V)/Xc = 0

V1/R-V1/Xc = V/R +V/Xc

V1(1/R-1/Xc) = V(1/R+1/Xc)

V1(Xc-R) = V (R+Xc)

V = V1(Xc-R)/(R+Xc)

ah! got it.

I don't think this would work for R & L.
What if V2 is 90 deg out of phase with V1.
V2= -jV1?

 

[uart]

Isn't Zc same as Xc for a capacitor?

No, technically the reactance "X" doesn't include the "j".

Xc = - 1 / (w C)

Zc = - j / (w C)

I tried to solve for V using I1+I2=0

(V1-V)/R + (V2-V)/Xc = 0

V2=-V1

(V1-V)/R - (V1+V)/Xc = 0

V1/R-V1/Xc = V/R +V/Xc

V1(1/R-1/Xc) = V(1/R+1/Xc)

V1(Xc-R) = V (R+Xc)

V = V1(Xc-R)/(R+Xc)

ah! got it.

Yep just make sure you include the "j" in the Xc and that's it. :)