# v(t) = V*cos(wt+phi) = Re{[V*e^(j*phi)]*e^(jwt)}

Discussion in 'Homework Help' started by screen1988, Jul 14, 2013.

1. ### screen1988 Thread Starter Member

Mar 7, 2013
310
4
Hi,
v(t) = V*cos(wt+phi) = Re{[V*e^(j*phi)]*e^(jwt)}
Then what step I have to do next?

Here is the method that WBahn mentioned:http://forum.allaboutcircuits.com/showpost.php?p=471718&postcount=8
Let's assume that I have a linear circuit (resistors, inductors, and capacitors along with some simple voltage and current sources) and all of my sources are putting out steady sinusoidal voltages/currents and have been on for a long time (long enough for all of the elements to settle into their steady-state response). Let's focus on one such voltage signal, v(t), which has the form:

v(t) = V*cos(wt+phi)

While it may not be obvious why we would do this, we could express this signal as:

v(t) = V*cos(wt+phi) = Re{[V*e^(j*phi)]*e^(jwt)}

The factor V*e^(j*phi) is what you are used to working with as the "phasor" for this voltage when working with complex impedances. While very hand-wavy, this expression represents the transformation between the time domain and the complex frequency domain and back. In the complex frequency domain, everything has the frequency exponential factor, e^(jwt), multiplying it, so we simply divide it out and work with the phasors to get an answer and then multiply by the frequency exponential factor again and take the real part to get our final time-domain answer. Again, the steps involved in going from the time domain equation to the phasor representation is so simple that we usually do it by inspection.

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2. ### WBahn Moderator

Mar 31, 2012
23,109
6,956
The first thing you need to do is establish some foundational stuff.

Let's first take each of our three main components, a resistor, a capacitor, and an inductor, and determine the relationship between the voltage across them and the current through them under the constraint that the voltage is a steady state sinusoid.

So, if you have

v(t) = V*cos(wt+phi)

applied across a resistor, what is the current?

What about if it is applied across a capacitor?

What about if it is applies across an inductor?

Now write each of those in the form:

Re{[I*e^(j*phi)]*e^(jwt)}

taking care to write 'I' in terms of 'V' and the component parameter (i.e., R, C, or L}

Do that much and let's get it settled and then we can move on.

3. ### screen1988 Thread Starter Member

Mar 7, 2013
310
4
iR = V/R *cos(wt+phi)
iC = CdV(t)/dt = - wCV sin(wt+phi) = wCV cos(wt+phi +π/2)
iL = 1/L ∫V(t)dt = V/(wL) sin(wt+phi) = V/(wL) cos(wt+phi -π/2)

iR = Re{[V/R*e^(j*phi)]*e^(jwt)}
iC = Re{[wCV*e^(j*(phi + π/2)]*e^(jwt)}
iL = Re{[V/(wL)*e^(j*(phi - π/2)]*e^(jwt)}

But I don't understand your implication. In the circuit the voltage v(t) = V*cos(wt+phi) is applied to a system including both R and C not to individual element.

4. ### WBahn Moderator

Mar 31, 2012
23,109
6,956
Patience. Remember what I said -- first we have to step back and build up the foundation.

Now, take each of the three components and write the impedance, which is the complex-frequency equivalent of the resistance, by taking the ratio of the voltage to the current.

5. ### screen1988 Thread Starter Member

Mar 7, 2013
310
4
v(t) = V*cos(wt+phi)= Re{[V*e^(j*phi)]*e^(jwt)}
iR = Re{[V/R*e^(j*phi)]*e^(jwt)}

=> R = V(t)/iR = Re{[V*e^(j*phi)]*e^(jwt)}/ Re{[V/R*e^(j*phi)]*e^(jwt)}

iC = Re{[wCV*e^(j*(phi + π/2)]*e^(jwt)}
=> Zc = V(t)/iC = Re{[V*e^(j*phi)]*e^(jwt)}/ Re{[wCV*e^(j*(phi + π/2)]*e^(jwt)}

iL = Re{[V/(wL)*e^(j*(phi - π/2)]*e^(jwt)}
=>ZL= V(t)/iL = Re{[V*e^(j*phi)]*e^(jwt)}/ Re{[V/(wL)*e^(j*(phi - π/2)]*e^(jwt)}

I am confused because Re (a + jb) / Re (c + jd) ≠ Re ((a + jb)/(c + jd)) and all expressions above cannot be simplified.

6. ### screen1988 Thread Starter Member

Mar 7, 2013
310
4
On second thought it seems that you meant this:

R = V*e^(j*phi)]*e^(jwt)/ [V/R*e^(j*phi)]*e^(jwt)] = R
Zc = [V*e^(j*phi)]*e^(jwt)}/ {[wCV*e^(j*(phi + π/2)]*e^(jwt)} = 1/(wC) * e^j(-π/2)
ZL= [V*e^(j*phi)]*e^(jwt)}/ {[V/(wL)*e^(j*(phi - π/2)]*e^(jwt)} = wL* e^j(π/2)

Last edited: Jul 14, 2013
7. ### screen1988 Thread Starter Member

Mar 7, 2013
310
4
I am still not sure how this method work. The interested signal is the real part of the complex signal. In this case we calculate on complex signal and the take the real part of interested result. How this work?

8. ### screen1988 Thread Starter Member

Mar 7, 2013
310
4
Can I solve problem by continuing the the method in post #5?

9. ### screen1988 Thread Starter Member

Mar 7, 2013
310
4
No one help???
I got stuck for a week and struggling to understand it.

10. ### WBahn Moderator

Mar 31, 2012
23,109
6,956
Yes.

Yes, but you need to carry it one step further:

e^(-j∏/2) = -j = 1/j
Z_C = -j/(ωC) = 1/(jωC)

Same here

e^(j∏/2) = j
Z_L = jωL

Now, what this enables you to do is to replace all of the capacitors and inductors with their impedances and then solve the circuit the same why you would solve it if you were working with nothing but resistors. In general, the voltages and currents you end up with will be complex values. But then you just take those complex values, multiply them by e^(jwt), and take the real part to get your final answer.

The thing to keep in mind is that this approach only works for linear systems in steady state operation.

11. ### screen1988 Thread Starter Member

Mar 7, 2013
310
4
Thanks,
There is a question that I don't not quite understand. All interested signals initial are real; then we transform it into complex-frequency equivalent and do all calculation in this. Finally we take the real part of the complex-frequency signal( after multiplying them by e^(jwt)) to get interested result. How can we know that this method work?
I don't know why but I think it is related to my question above. If possible, please help me explain it.

12. ### WBahn Moderator

Mar 31, 2012
23,109
6,956
The underlying method requires a background in both differential equations and complex variables to really understand the validity. But what you are doing is taking the differential equations that describes the circuit and applying the Fourier transform to them. What that does is turn differential equations into algebraic ones. You then solve the algebraic equations and transform the results back. The transform relies on being able to represent a signal in the time domain as the superposition of signals in the frequency domain, and since superposition only works for linear systems, the Fourier transform inherents that same limitation.