Linking s-domain and phasor analysis...

-------------------- As a start: [w/(s^2+w^2)] *(1/(sRC+1) does not give the expression that you have for Vout/Vin (algebra error) in the s domain. Assuming T=RC the time domain response would be a transient term + the steady state sinusoidal with a phase shift. The phasor analysis only gives information as to the latter. For your information

1/(1+Ts)(1+s^2/w^2) transforms to (T*w^2*e^-t/T)/(1+(Tw)^2) +[w*sin(wt-phi)]/(1+(Tw)^2)^0.5 where phi= arctan (Tw)

Compare the result of the sinusoidal value to the phasor result.

Don Kelly snipped-for-privacy@shawcross.ca remove the X to answer

Clickety Doo Da... :-)

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Thanks for the reply Don. Can you point out the algebra error for me?

I used [w/(s^2+w^2)] because this is the Laplace transform of sin(w*t).

From your answer it looks like I should have used something else here?

Both the Laplace and the phasor (Euler) approaches are frequency domain solutions to time domain problems (essentially linear differential equations) The Laplace works in a complex frequency domain where s= sigma

+jw and leads to a complete solution to problems such as the one you propose, giving the transient response, as well as steady state DC and AC responses (including multiple frequencies. The phasor model considers the steady state response at one particular frequency corresponding to a single point on the sigma=0 axis.

Consider a series RLC circuit with 0 initial conditions and a voltage applied at t=0 v(t)=Ri(t) +Ldi(t)/dt +(1/C) integral of i(t) dt and knowing v(t) we can solve for i(t) If we use Laplace this results in V(s)=(R+sL +1/sC)I(s) or I(s)= V(s)/(R+sL+1/SC) and knowing v(t) we can find V(s) and solve for I(s) THEN convert back to the time domain through the inverse transform. We will get a complete solution for all times >t=0.

IF the applied voltage v(t) is a sinusoid, and we are only interested in the steady state response we can simply substitute jw for s. V(jw)=(R+jwL +1/jwC)I(jw)

There is a bit more math background here but we don't need to go into it at present.

The point of both is that it is easier to solve algebraic equations using complex numbers than it is to solve differential equations- particularly when you may have simultaneous equations. Neither phasor not Laplace voltages and currents exist except mathematically but they are convenient. The phasor approach is easier than the Laplace because you are looking for only a specific part of the total solution.

In your particular case, you are applying a sinusoid to a circuit which has a transfer function Vo/Vi =1/(1+sT) for any actual signal (within limits). In particular, for a sinusoid of any particular frequency, Vo/Vi=1/(1+jwT)in steady state.

Thanks for the reply Don. Can you point out the algebra error for me?

I used [w/(s^2+w^2)] because this is the Laplace transform of sin(w*t).

From your answer it looks like I should have used something else here?

What you have done, mathematically correctly, is to expand the denominator so it is expressed as a cubic. This complicates things as you do need the roots of this. you already have one root from sT+1 s=-and the others from s^2+w^2 =(s+jw)(s-jw) You can then write Vo/Vi =w/[(s+jw)(s-jw)(s+1/T)] where T is RC You can then use partial fractions to expand this in three terms K1/(s+jw) +K2/(s-jw) +K3(s+1/T) K2= the complex conjugate of K1 and K3 is real Note that K1=w(s+jw)/(s+jw)(s-jw)(s+1/T) evaluated at s=-jw K3 =w(s^2+w^2) evaluated at s=-1/T Alternatively you can use (As+B)/(s^2+w^2) +C(s+1/T) and expand the numerator and solve for A,B C considering that the terms in s, s^2 are 0 and the other term =w

As for not getting a sinusoid from mathcad- the Laplace will give the transient as well as the Steady state results. The transient in this case is exponentially decaying and the steady state is a phase shifted sinusoid.

I'm sorry to not get back to you earlier but my Oct 10 reply was the last I got out before my mail server went out and it has just come back on.

Dear Professor:

I got the Schaum's Electric Circuits, Basic Electrical Engineering and Lyshevski's Electromechanical Systems,Electric Machines and Applied Mechatronics books per your recommendations. I can understand the first few chapters of the Schaums books but Lyshevski starts something like "derive the differential equation for ....." The preface says it is a textbook for one or two semester graduate classes.

Apparently the rust that has accumulated on my brain over the last 28 years is much more substantial than the original estimate. I went to Barns and Nobel and found this book which is helping to get through the several inches of loose rust that breaks off easily.

Forgotten Algebra. by Barbara Lee Bleau. ISBN 10: 0 7641

2008-5 When I complete the forgotten lessons, hopefully I will begin to remember a little more about differential equations than which pretty girls were in the class.

A Devoted Fan

Peace Dawg

The key block here is ""graduate classes" I am sorry for the reference which looked good to me. Go back and look for texts such as Fitzgerald, Kingley and Umans . Basic machines are covered well- even though the higher level approaches are not. Another that looks good is Electric machinery by Chapman (at least the part presented on line looks not too difficult) or one by Charles Gross. Unfortunately calculus is involved and this is beyond forgotten algebra.

------------ I wrote " You can then write Vo/Vi =w/[(s+jw)(s-jw)(s+1/T)] where T is RC" Correction:

Vo/Vi =1/(Ts+1) where T=RC" Vo =Vo/Vi) =Vo [1/(1+sT)]={w/(s^2+w^2)(1+sT)} so the solution that you get will be for Vo(s) leading to vo(t).

Thanks Don, I'm working my way through this.

Before performing the partial fraction decomposition you wrote s^2+w^2 =3D (s+jw)(s-jw)

I intuitivey think to write (s+w)(s-w) ... Why does the imaginary term end up in this expression?

David

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in

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froms^2+w^2 =3D(s+jw)(s-jw)

Before performing the partial fraction decomposition you wrote s^2+w^2 = (s+jw)(s-jw)

I intuitivey think to write (s+w)(s-w) ... Why does the imaginary term end up in this expression?

David

------------ (s+w)(s-w) = s^2+ws-ws-w^2 =s^2-w^2

(s+jw)(s-jw) =s^2 +jsw -jsw +w^2 = s^2+w^2

------- You could avoid complex numbers by using the alternative approach As+B/(s^2+w^2) but it will be just as much work. For some development : The Laplace transform of e^-at =1/(s+a) so that of e^jwt =1/(s-jw) and that of e^-jwt =1/(s+jw) consider that sin(wt) = (1/2j) [e^jwt -e^-jwt] and work it out to get the Laplace transform for sin(wt) --

Don Kelly snipped-for-privacy@shawcross.ca remove the X to answer

I've been working with phasors ands-domainanalysis for a little bit now. My mathematical skills aren't the best and I'm not seeing how these two analysis techniques are linked together. I analyzed the simple RC circuit here:

Is my math at the end correct in finding the sinusoidal response? There should be ans^3 term in the denominator? How do we manage to find frequency response so much easier using Eulers formulas? I'm not asking how to do it with Euler's, I'm asking to understand intuitively how the two are linked. Is there a certain proof out there that is commonly used to make this "click" ? Thanks.

-------------------- As a start: [w/(s^2+w^2)] *(1/(sRC+1) does not give the expression that you have for Vout/Vin (algebra error) in thesdomain. Thanks for the reply Don. =A0Can you point out the algebra error for me? I used [w/(s^2+w^2)] because this is the Laplace transform of sin(w*t). From your answer it looks like I should have used something else here? What you have done, mathematically correctly, is to expand the denominator so it is expressed as a cubic. This complicates things as you do need the roots of this. you already have one root from sT+1 =A0s=3D-and the others froms^2+w^2 =3D(s+jw)(s-jw) You can then write Vo/V= i =3Dw/[(s+jw)(s-jw)(s+1/T)] where T is RC You can then use partial fractions to expand this in three terms K1/(s+jw) +K2/(s-jw) +K3(s+1/T) K2=3D the complex conjugate of K1 and K3 is real Note that K1=3Dw(s+jw)/(s+jw)(s-jw)(s+1/T) evaluated ats=3D-jw K3 =3Dw(s^2+w^2) evaluated ats=3D-1/T Alternatively you can use (As+B)/(s^2+w^2) +C(s+1/T) and expand the numerator and solve for A,B C considering that the terms ins,s^2 are 0 and the other term =3Dw As for not getting a sinusoid from mathcad- the Laplace will give the transient as well as the Steady state results. The transient in this case is exponentially decaying and the steady state is a phase shifted sinusoid. I'm sorry to not get back to you earlier but my Oct 10 reply was the last I got out before my mail server went out and it has just come back on.

I am a bit reluctant to respond because I think the Ray on Web TV might really be a Roy in disguise.

Your intuition is wrong. (s+w)(s-w) = s^2 - w^2. Rely less upon your intuition and more on mathematical accuracy.

Bill