Circuit Analysis Example

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En provenance du cours de University of Rochester

Fundamentals of Audio and Music Engineering: Part 1 Musical Sound & Electronics

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University of Rochester

Fundamentals of Audio and Music Engineering: Part 1 Musical Sound & Electronics

299 notes

In this course students learn the basic concepts of acoustics and electronics and how they can applied to understand musical sound and make music with electronic instruments. Topics include: sound waves, musical sound, basic electronics, and applications of these basic principles in amplifiers and speaker design.

À partir de la leçon

Week 3 - Some Necessary Mathematics MB & AC Circuit Analysis

Transistors, vacuum tubes, opamps, amplification, power gain, single-stage amplifiers. Converting sound to electrical signals –microphones and guitar pickups, converting electrical signals to sound – loudspeakers

Circuit Analysis Example 8:43

Impedance Analysis of RL Circuit 6:57

Filters as Frequency Dependant Voltage Dividers 7:20

Rencontrer les enseignants

Robert Clark
Provost and Senior Vice President for Research and Professor
Mechanical Engineering
Mark Bocko
Distinguished Professor and Chair
Electrical and Computer Engineering

Now we're in a position to use the machinery that we've developed earlier,

to do an AC analysis of a simple circuit. Now, the one that we're going to look at

is just an RC circuit like we looked at before.

Now the difference here is we're not going to assume that there is a switch,

but we're just going to assume that this circuit has a source of alternating

current, voltage, AC voltage driving this circuit.

Now again, keep in mind, this is any old frequency omega.

It doesn't matter what frequency it is, it's just a general frequency omega.

And so, if I can solve this circuit for the, a general frequency, I know the

solution of this circuit for all frequencies.

So, what we're looking for is the current as a function of frequency.

Now, after we know the current, we can then find the voltage across this

capacitor as a function of frequency. Now, we're going to use our phasor

notation and write Kirchhoff's voltage law here, just like we did before.

And we'll start at the ground and then go up through the AC source of voltage.

So, you pick up a voltage V, and then when we pass through the resistor, we

drop a voltage equal to I times R. And then when we go through the

capacitor, there's a voltage drop, which is the impedance of the capacitor times

I. So, we're treating the impedance here

just like resistance. And so instead of this equation being,

having derivatives in it like it did before, actually the derivative was over

here. instead of having derivatives, these are

just algebraic equations now. Now that is the great simplification in

AC circuit analysis, is that these differential equations where we had the a

function, like the current and derivatives of the current, and voltage

and derivative of the voltage in one equation.

Instead of having those differential equations in the frequency domain here.

We have simple algebraic equation, and we can just solve these using the rules of

algebra, but we have to keep the track of the complex numbers.

So, I can solve this for I. So, I'll put the I terms on one side and

the voltage terms on the other side, and then divide through by the factors,

multiplying I and I get I is V over this factor.

And then the voltage across the capacitor is the impedance of the capacitor times

the current. Now, this current here, this is a current

going through both elements. It's the same current going through R and

C. If I want the voltage across the C, I

just multiply this current times 1 over j omega, C.

So I plug this in here and I get this expression, and now I'm going to take and

just multiply this j omega C through the denominator.

And so it cancels out this term and I get a 1, and then I get j omega C times R.

So here's the voltage across the capacitor, is equal to this factor times

the voltage driving the circuit. Now, what I want to do is compute the

magnitude of VC. So, we introduced back when we were

talking about complex numbers, we introduced the magnitude of a complex

number. So, I don't care about the phase, I just

want to know the magnitude of the voltage.

And so, the magnitude of VC over V is just the magnitude of this factor here.

And I just rewrote it as one plus j omega CR.

So, I can write this magnitude of 1 over 1 plus j omega CR is, the factor times

its complex conjugate. And I get the complex conjugate just by

changing the sign in front of the j, then I take the square root of that entire

expression. And so, if I multiply these two, then I

get 1 plus omega squared, C squared, R squared.

And the cross terms, here's plus j omega CR, and this is minus J omega CR, they

cancel. So here is the expression for the

voltage, magnitude of the voltage across the capacitor divided by the voltage

driving the circuit. So if I plot that, VC over V.

And now we're plotting this as a function of frequency.

So remember it's very, it's very important to make a distinction between

this and when we did the transient analysis of the circuit, we plotted the

response of the circuit as a function of time.

Now, we're plotting the response of the circuit as a function of frequency.

Now, I'm taking and there's I'm expressing the frequency as the frequency

divided by 1 over RC. Now RC, R times C is the so called time

constant of the circuit, it has units of time.

So if I multiply a frequency times a time, this is dimension less.

so what I'm really doing is I'm taking my frequency and I'm normalizing it by 1

over that time, that characteristic time of the circuit.

Now, this point when omega is 1 over RC, then that makes omega times RC equal to

1. And at that point, this factor, 1 over 1,

becomes 1 over 1 plus 1. So that's 1 over the square root of 2 and

that's 0.707, and that's this point right here.

So this is the frequency when omega equals 1 over RC.

The response of this circuit is down to 1 over the square root of two, times the

response at DC. So this is an example of a low pass

filter. What happens here, and it's much easier

to see here in the frequency domain than it was when we were talking about this in

the time domain. The low frequencies pass through with

very little attenuation. The voltage across the capacitor is

almost equal to the voltage driving this entire circuit.

As I, the frequency goes up, the size of the voltage appearing here across the

capacitor gets smaller. And when omega is 1 over RC, it's 1 over

the square root of 2 times this driving voltage.

when omega is much bigger than 1 over RC, the response goes down considerably.

So high frequencies do not pass through this circuit very efficiently.

Low frequencies are passed through this circuit very efficiently.

And there is a very solid connection between this and the transient analysis

we were talking about. The, at low frequencies, there's plenty

of time for this capacitor to charge up and this voltage to become quite large.

At high frequencies, there isn't enough time for the capacitor to charge up and

for this voltage to become very significant.

And so, high frequencies are attenuated by the circuit, low frequencies pass

through this circuit.

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