Question 1

How is it that at 2:44 you set I = e^jwt and this also happens at 4:33.
But how then, at 6:47 can U = e^jwt ? 
Doesn't that imply that U = I ? How can current and voltage be the same?

Or is the e^jwt simply arbitrarily substituted for I in the case of the resistor and the inductor is substituted for U in the case of the capacitor?

Accepted Answer

Sorry for the confusion. The value e^jwt can be assigned to anything, just like any other number, like 3 or -9. If I say, "Let v = 3," there is an understood unit of volts based on the context of the equation. It would be better if I said "Let v = 3 V," but I left that part out in the video. Later on, I might starting talking about current, and I could say, "Let i = 3". Again, I neglected to state the units, which is not the greatest teaching technique, but it is a good bet I'm assuming the units are amperes. In these two examples with "3", where I didn't state the units, you probably wouldn't think that 3 volts = 3 ampere. The same thing is happening in the video when I write I = e^jwt and then U (voltage) = e^jwt. The e^jwt is just a numerical function (that happens to be a function of time). Stated by itself, e^jwt does not have units, just like "3" doesn't have units. e^jwt becomes a value with units from the context of the equation where its used.

Let me know if that helped answer your question.

Let me know if that helped answer your question.

Question 2

We substitute e^(jwt) for the current instead of the cosine function, to figure out impedance, and I get that but I don't understand how we derive general equations for the impedance using only part of our AC signal. Since the AC signal is made up of two exponential functions, shouldn't a general rule for the impedance also involve the superimposed functions rather that just one?

Accepted Answer

Good question. This answer evaded me for many years. It's easy to forget the logic behind some of the powerful cleverness that happens right at the moment you are learning a new technique the first time.

Cosine can be broken into two complex exponentials. The exponentials are conjugates. That's important.

The first time you do a problem you draw it up for superposition. You create two versions of the circuit and drive each one with one of the two complex exponentials.  At the end you add the two answers together (superimpose) to get the response to cosine. Right at that moment you notice, "Hey! The two circuit outputs were nearly the same, they were just complex conjugates of each other." Then your next thought is, "Wait a minute. The next time I do a problem like this, I will just do one complex exponential and find the output, since I know the circuit response to other one comes out as the conjugate of the first answer. Saves me half the work!".

From then on, we just talk about one exponential input drive signal. All the extra analysis of the conjugate exponential is not done and not mentioned.

Question 3

ohms are a type of measurement for impedance right?

Accepted Answer

Yes. The unit of impedance is the ohm.  Impedance is a generalized notion of resistance that applies to resistors, capacitors, and inductors.

Question 4

So in this video, the situation is both v and I are sinusoid? When there is no current source and V is sinusoid, does that mean I must be sinusoid too? I am very confused.

Accepted Answer

Let's do a simple example. Suppose you connect a resistor to a voltage source. Let the voltage source be defined to produce a sinusoid, v = A sin wt. What is the current in the resistor? Using Ohm's Law, i = v/R gives us i = (A sin wt) / R. So when the voltage is a sinusoid, the current is a sinusoid, too.

In general, if you define either v or i to be sinusoidal, the other will follow along and be sinusoidal, too.

In a resistor, the sinusoidal i and v are lined up with each other in time. The interesting thing about inductors and capacitors is i and v *don't* line up in time, they get a 90 degree offset time delay from each other.

Any time we talk about Impedance, there is an underlying assumption that we've restricted the driving signal to a sinusoid. When we make this restriction, all of a sudden the i-v equations for inductors and capacitors start to look just like Ohm's Law for resistors. This is fantastic because it gives us a way to analyze RLC circuits using the same methods we learned about for just-Resistor circuits.

Question 5

The current and voltages for passive devices are assumed to be e^jwt but for a complete sinusoid, we also need e^-jwt. Both exponents of combine to form sinusoidal, but just taking one exponents would make it something else, won't it ?

Accepted Answer

In this video (https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/modal/v/ee-ac-analysis-superposition) we showed how the circuit response to e^-jwt always comes out as the complex conjugate of whatever e^+jwt causes.

That means if I know what e^+jwt does then I know what e^-jwt does. AND, if I know what both of those do then I know what happens if I apply sine or cosine to the same circuit.

We focus on the response to e^+jwt for 2 reasons. 
1. We know how to solve forced differential equations when the forcing in put is an exponential.
2. We know finding the response to e^+jwt instantly leads to the answer for sine and cosine inputs, which is what we care about. 
3. (Bonus feature) We know how to construct many other types of signal shape (square, triangle, any periodic function, most non-periodic functions) as a sum of sines and cosines. (This is what Fourier series and Fourier transforms and Laplace transforms are all about.)

When you work with e^+jwt you pretty much learn everything about an analog electronic system.

Question 6

The part between 2:28 and 3:45 seems a circular reasoning to me.

At 2:28 `v =  i*R`. Then after some steps we get that, indeed, `R = v/i`.

(I mean we would get that no matter how `i` is defined, it does not matter that we used `e^jwt` in this case, it would work for any other kind of `i` as well.)

Accepted Answer

Yes. The discussion of the resistor is circular. It's just two forms of Ohm's Law, and it is valid for any form of i. This is intended to emphasize the perspective of Ohm's Law as the ratio of v to i. It helps set up what comes next for L and C.

When we consider L and C, AND we assume the current or voltage is a complex exponential, the concept of Impedance emerges. It applies to L, C, _and_ R. The R case is what you already know (Ohm's Law) so it seems super simple.

Question 7

Quick question, throughout the video you use jwt, and so far I know that t stands for time, w stands for frequency, but I never caught on as to what j stood for. Could you elaborate on that please? :)

Accepted Answer

In electrical engineering, we use the letter j instead of i as the imaginary unit, sqrt(-1). In our field, i is used as the variable for current (from the French intensité du courant électrique, coined by Ampère).

Question 8

Is AC resistance measured in reactance or impedance?

Accepted Answer

Hi Chloe. Impedance (Z) is defined as the ratio of Voltage to Current (V/I). In the most general sense impedance has a complex value. Z = real part + j imaginary part. For a resistor, it has a real part and no imaginary part. The real part is called Resistance and it has units of Ohms.

A capacitor or inductor have imaginary impedance (no real part, just an imaginary part. The imaginary part is called "Reactance", and L and C are called "reactive elements". Reactance is the ratio of V/I, so it has the units of Ohms, just like resistance.

The impedance of an inductor is Z = jwL. The reactance of an inductor is the imaginary part, wL. To get the reactance you multiply frequency (w) times Henrys to get Ohms of reactance.

I never found it very useful to talk about a capacitor or inductor in terms of its reactance in Ohms. This value is different at every different frequency. It was much more helpful to model a capacitor as 1/jwC and an inductor as jwL. You end up making plots with w as the independent horizontal axis (frequency response plots).

Question 9

For inductor and resistor, current was assmed to be e^jwt. And for Capacitor, voltage was assumed to be e^jwt.
What are intuitions behind these assumptions, how were these assumptions made ? And how can the assumed value used interchangeably between voltage and current ?

Accepted Answer

Here is a high-level review of why we do AC analysis this way.
1. Circuits with L and C have a natural response modeled with differential equations.
2. The solutions to these equations are exponential functions. Therefore, we love exponential functions.
3. In the real world a large number of signals resemble sine waves (talking, music) and we want to amplify those signals or put them into filters to modify their frequency content. (Side note: sine waves can be used to construct a lot of other signals that don't look like sines, see for example Sal's video sequence on Fourier Series.)
4. We build amplifiers where we apply external sine-like signals (a forced input) and we want to know what the amplifier (with its R's, L's, and C's modeled as differential equations) does to the signal. The answer will be a natural response plus a forced response.
5. By the miracle of Euler's formulas we know sine and cosine can be represented as exponential functions. Those exponentials happen to have imaginary exponents, which makes them totally weird, but it makes the math totally work because now the forced input has the same general shape as the natural response (both are exponentials) and we can solve the differential equations.

That is why we picked e^jwt for our assumed input. We know how to solve the differential equation, and we know how to reconstruct a sine wave. That's the winning combination at the heart of AC analysis.

As for picking voltage or current interchangeably, it isn't too tricky. For a resistor that obeys Ohm's Law, the shape of the voltage and current are the same, they are just scaled versions of each other, so I could start with either v or i. For L and C, when signals are exponentials, I know I get another exponential when I take a derivative. I like to work with the derivative form of the i-v equation rather than the integral form, so I chose to assume the variable inside the derivative. For L that would be current (v = L di/dt), and for I that would be voltage (i = C dv/dt).

Electrical engineering

Unit 2: Lesson 5

Impedance

Want to join the conversation?

Video transcript