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schematic

simulate this circuit – Schematic created using CircuitLab

Hi,

In the follwoing two images, I am having a problem seeing the confusion the author is talking about and more importantly how he uses the phasor diagram in figure 1.61 to clear this confusion up? He describes the confusion the reader is likely to see in the second image.

But first, Note the last line in the first image, "Lets take an example, namely the fact that.."

He is talking about an RC filter and the 3db attenuation you get at f = 1/(2 * pi * R * C).

In the second image, he states that if you change the capacitor to a resistor equal to R, you get 6db attenuation. R / (R + R) = 50%. I see that, simple voltage divider.

He says that if you go back to the origional RC circuit at frequency f = 1/(2 * pi * R * C), the capicitors impediance equals R and a confused person, like me, would expect to see 6db attenuation. He descibes figure 1.61 and the reactive parts of C as an explanation to why one sees 3db attenuation instead of 6db attenuation. Note, he says that at f = 1/(2 * pi * R * C) the impediance of the capacitor equals R.

Questions: 1) How does figure 1.61 show this clearly? I don't see its explanation.

2) In the text of the second image, the author says the input voltage (applied across the series RC pair) is proportional to the hypotnuse. Why is this so, the input is from en external source Vin (wall outlet)?

3) He then says that the output voltage (across R only) is proportional to the length of the R leg of the triangle. Again, I'm not seeing how the phasor diagram explains all this.

Thanks for the great help. I've never been this far in this book as yet and think the other chapters are home free. :)

enter image description here

enter image description here

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  • \$\begingroup\$ These are not so much things to derive as they are axioms for everything else. \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 5 '17 at 23:22
  • \$\begingroup\$ TL/DR : Pythagoras. \$\endgroup\$ – Brian Drummond Jan 6 '17 at 10:03
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This is because db does account for phase, db is a magnitude.

for this circuit in phasor notation:

V=(-jwC)/(R-jwC)

at the corner frequency:

V=(-jR)/(R-jR)

||V||=(R)/(R*sqrt(2))=-3db

To relate it to the resistor divider circuit, if you were to separate the equation to its real and imaginary components you would get this:

V=(1/2)*(1-j)

Separate out the Real (in phase power):

Re{V}=(1/2)

||Re{V}||=1/2=-6db

So for this circuit at the -3db corner frequency you would see 1/sqrt(2) the voltage (-3db) with a 45deg phase shift on an oscilloscope. The 45deg phase shift reduces the in phase voltage by another 1/sqrt(2). So the REAL in phase voltage magnitude at the output is half (-6db).

Ive noticed Phasor noticed is often explained poorly and used incorrectly.

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  • \$\begingroup\$ Can you explain how V = V=(-jwC)/(R-jwC)? Is that V=I*R? When the text (in the second picture) says that the input voltage is proportional to the hypotenuse, does that imply that the circuit is fed with a current source and whatever voltage is delivered is a function of the resistance to ground (the complete series circuit)? \$\endgroup\$ – Jeffrey Edward Messikian Jan 6 '17 at 14:12
  • \$\begingroup\$ Yes, I made a mistake on that equation. It should be:V=(-j/wC)/(R-j/wC) with corner frequency at: w=1/RC. The result will be the same. \$\endgroup\$ – Tony Jan 7 '17 at 2:12
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That figure is confusing. I'm pretty sure they are just trying to show this circuit is seperated into real and imaginary components.

Phasor notation is basically taking the fourier transform of the circuit. Everything is written in terms of a completely continuous sinusoidal input, no DC component whatsoever. Once you do this reactive components (capacitors/inductors) can be treated like having resistors with a reactive impedence. So for capacitor: i=C(dv/dt) time domain ==> I=jwCV fourier transform

I/V=Z(impedence)=-j/wC

When you convert to phasor notation vin,vout or what ever voltage no longer means the same thing.

vout != (Vout)phase_domain.

*IF the input voltage is a continuous sinusoid (no DC) then the magnitude of the time domain voltage equals the magnitude of the phasor domian voltage:

||v1||==||{V1}||

So Example 1.23 is not always true if Vout is a time domain voltage and is never true if Vin is not completely sinusoidal.

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  • \$\begingroup\$ In the second picture, what is meant, or what is the significance of of saying that the input voltage is proportional to the hypotenuse? The length of the hypotneuse should be the total impedance of the circuit. Does he mean to say that if you add more impediance, you can increase the voltage. This could only happen if it was driven by a current source. Why is output proportional to the R leg of the triangle? I am assuming that the Vout is read between R and C and C connects to ground. \$\endgroup\$ – Jeffrey Edward Messikian Jan 7 '17 at 4:31
  • \$\begingroup\$ Ok, the length of the hypotenous is equal the magnitude of the input sinusoid. Then on a phase plot they put each series component as a vector. The length of each compenent vector is equal to magnitude of the sinosoidal voltage drop across the compenent. So that is a typo when they said Vout is proportional to the length of R... it is proportional to length wC. At the corner frequency they are the same though. \$\endgroup\$ – Tony Jan 7 '17 at 15:08
  • \$\begingroup\$ Is Vout inversely proportional to (R - (j/wC))? Like a voltage divider. How does the magnitude of the input signal affect the hypotneuse? Everything in Fugure 1.61 (A) seems to indicate resistance or reactance. I owuld think that if you increased the magnitude of the input signal, all you would do is create more current, given impediance stayed the same. \$\endgroup\$ – Jeffrey Edward Messikian Jan 7 '17 at 18:44
  • \$\begingroup\$ Also, I drew the circuit at the top. I drew this after reading the section. Would it have been better if I switched the R and the C? If I did that and assumed the circuit was supplied with a current source, it would all make sense for me. It should not matter at the corner frequency right? \$\endgroup\$ – Jeffrey Edward Messikian Jan 8 '17 at 1:55
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Impedance is the complex sum of Resistance and Reactance.

This is because the relationship between voltage and current gets phase shifted in time the more reactance you add to a circuit, such as your basic filter.

Pure capacitors shift the voltage with respect to current by -90 degrees or in other words 1/4 of a cycle delayed or 'lagged'. Thats where the -j comes into it and now we have to use the maths of complex numbers to add real and imaginary components.

Pure inductors shift the voltage with respect to current by +90 degrees or 1/4 of a cycle ahead or 'leading'.

The 3dB/6dB thing is a red herring and only comes out of the maths when comparing magnitudes. Actually the -3dB point is called the 'critical' frequency where the reactance in ohms is equal to the resistance in ohms.

A reminder of basic complex impedance: $$ Z = R + jX = |Z|\angle\theta $$ $$ X = j\omega L - \frac{j}{\omega C} $$ $$ \omega = 2\pi f $$ $$ |Z| = \sqrt{R^2+X^2} $$ $$ \theta = tan^{-1}\frac{X}{R} $$

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I could not fit this in a comment..

Switching R and C will not make difference. changing the supply to current source can be done when using phasors. You will get the same answer if the current magnitude is scaled such that the power to circuit is the same and you will have to invert the phase angle when converting back to voltage.

It is unnecessary to do this. All you need to do it solve this problem is put '-j/(wC)' as the impedance for C and solve it like a normal voltage divider. That Z plot is confusing and not useful. No body solves like that unless they doing it as an exercise.

Vout=(-j/wC)/(R-j/wC)*Vin...

At corner frequecny:

Vout=(-jR)/(R-jR)*Vin=

(1-j)/(2)*Vin=

(1/sqrt(2))e^(-jpi/4)*Vin.

If vin is a cosine with no phase angle and magnitude V then:

Vcos(wt) =Re{Ve^(jwt)}

^^^^^^ this is basis of phasor notation. You make the input to the system totally sinusoidal then use a simplified eulers formula to write the sinusoids as phasors.

So now:

Vout=[(1/sqrt(2))e^(-jpi/4)][Re{(V)*e^(jwt)}]

=Re{(V/sqrt(2))*e^(j(wt-pi/4))}

Convert back to time domian by using cos(wt+a)=Re{e^(jwt+a)}

vout=(V/sqrt(2))*cos(wt-pi/4) time domain

Pretty much everybodys does not write the 'Re' portion.

So this shows that vout has magnitude of V/sqrt(2) and is 45deg out of phase.

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