Welcome back to our continuing series on linear circuits work.
Today, we talk about AC circuit analysis. Today, we're going to be identifying how
past techniques that we applied to resistive circuits can be applied to
circuits that are sinusoidal, and how we can use impedances and phasors, to be
able to do analyses, for those types of systems.
In the previous lesson we talked about impedance and its relationship to
phasors. And then we looked at how impedance
changes with frequency in reactive circuits like capacitors and in
inductors. So now we're going to apply that to doing
analyses to sinusoidal systems and that will lead us next time to talk about
transfer functions. The objectives for today's lesson are to
apply techniques from DC analysis to sinusoidal systems, find equivalent
impedances for devices that are placed in series or parallel, use superposition for
analysis, particularly for systems that have multiple frequencies.
And then be able to analyze a system using all of these techniques.
So first of all, we want to identify that impedance is a linear property.
So if you remember back when we talked about linearity, we use, as an example,
Ohm's law, where we have V is equal to I times R.
We said that this is a linear property, because it's basically multiplying by
constant value. But now, we're doing the exact same
thing. In phasors, we have the property V is
equal to I times Z. Now V and I are phasors, so they're
complex valued things representative of. They assign certain functions.
And Z is the impedance, or the relationship between the two.
And because of this setup, this is also linear linear properties, just the same
as Ohm's law, but now with complex numbers.
And so consequently, some of the same implications will occur.
So if I wanted to do an analysis of impedances in series, her I have voltage
V and I believe a phaser I. And I have these three impedances.
Using this relationship of impedance to voltage, if I wanted to find what I
happens to be, I know that I. It's going to be equal to V over Z.
And if I look at each of the individual voltages, here, here and here.
They'd be one, this would be two and this would be three.
The voltage here is going to be the current, I, times Z1, V2 is the current I
times Z2, and V3 is the current I times Z3.
They're all in series so all of these I's are the same so adding them all together
because we know that using Kircoff's voltage that V is equal V1 + V2 + V3.
So adding all these up and factoring out the I we see that.
V is equal to i times z1 plus z2 plus z3. So we can basically come to find what z
happens to be right looking at this format.
We see that our equivalent z is just going to be equal to the sum of each of
the individual z's, and here I'll just use the summation notation for gravity.
You might recognize this as being the exact same analysis we used for resistors
except now we have complex numbers in our calculation and we get the same result
The equivalent of impedance devices in series is just the sum of the impedances.
And the natural next step is to look at it when we have devices in parallel.
And doing the exact same technique that we used before for resistors, we, again,
basically come up with the same results. We'll see that this Zeq is going to be
equal to the sum of the inverses of each of the individual impedances, and then
you invert the whole thing. So again, for brevity, I put it in
summation notation. So it's exactly what you would expect
with resistors. Kirchhoff's laws also apply in phasors,
and before well, all of the input currents and the output currents kind of
had to match up. Now we have the additional property that
they have to match up in the real and the imaginary parts, because the sum has to
be zero. And you can add up all the real parts
which you want, but you're never going to get an imaginary.
So we see that, that gives us one extra little bit of help, while we're doing
analysis with phasor systems, is we can look at the real parts and the imaginary
parts, and they all have to match up. So just as a review, Kirchhoff''s current
law, the input currents have to equal the output currents, even the phasor, and for
Kirchhoff's voltage law, which we already used.
The, if I do a sum around the circle, we see that Vs has to be equal to V1 plus V2
plus V3. Source transformations also work, but now
we're going to be having sources that are phasors, and, instead of having a
Thevenin resistance, we have a Thevenin impedance.
But the same basic relationship applies, that V Thevenin is equal to I Norton
times Z Thevenin. Superposition also works.
When we initially presented the superposition, you might have thought it
was kind of a lot of work to do analysis for something because you'd have to break
it up into each individual piece and then add up all the results from the
individual analysis together. Sometimes it's useful if, by breaking it
up into individual places and zeroing out sources, the system becomes trivial, but
that's not always the case. However, one time where superposition is
extraordinarily useful is if we have systems with different frequencies
because its linear superposition still holds.
But because of the frequencies are different, you can't immediately relate
one system to the other. Because that, we know that as frequency
changes, impedances change in reactive components.
So let's look at this example. Again, I zero out sources the same way
that was presented before. So these current sources are going to go
to open circuits. So I will do them one at a time if I
first get rid of i2 by zeroing it out and leaving i1 intact.
We get i1, which is now a phasor, and going down to this, we see square root of
two, with an angle of zero degrees. And that would come over, here we have a
capacitor, a resistor, and the impedance of the capacitor is going to be equal to
1 over j omega c. Here, our omega is 10 to the 6th.
So this is equal to 1 over j times 10 to the 6th.
The capacitance is 1 nanofarad, which is 10 to the negative 9th, put all that
together, you get minus j 10 to the third ohms.
And then here the impedance of our resistor is simply the resistance.
So this is also equal to minus j times one kilo ohm.
Now, what I'm trying to solve is the voltage v here.
So here I label it as v1, because it's the voltage generated by our first
source, and v1 is going to be equal to i times z.
So i is squared of 2.0, z is going to be the sum of these 2, which it turns out to
be the the square root of 2, arc minus pi fourths, so multiply this, taking two
values together. That gives us 2 with the phase angle of
minus pi fourths. In other words the voltage from the first
element, it's going to be equal to 2 cosine 10 to the 6th t minus pi 4th
volts. Now that we finish that part, we are
going to now 0 out source 1. Probably should indicate that something
was removed here. So when we remove source one we'll have
something like this. Again we have our capacitance, our
resistance. Now this, again, is, 1 over j omega c.
This is r, which is just one kilo ohm, and the second source.
Now here, notice that this is a sine function.
So, this phaser is actually going to be three with a phase angle of minus, oops,
minus pi halves. Now coming back to this one over j omega
c, well this has a different frequency, so this time omega is equal to 1 over the
square root of 3 times 10 to the 6th. So I take 1 over j times 1 over the
square root of 3, times 10 to the 6th times c, again, 10 to the negative 9th.
This equals minus j square root of 3 times 10 to the 3rd ohms, or minus j
square root of 3 kilo ohms. So now that we can clearly see that
capacitance is different for the system than it was in the system.
Now remember the phases always have this implied frequency, and we can't ignore
it. It's very important that we don't.
So if I put all of this together, I discover that again we're finding this v
here. So I'm calling, going to call it v2 is
equal to i from 2 times this impedance. So that gives us 3, phase angle of minus
pi halves, times. And here we if we look at this we see
that, adding 1k and minus square root of 3k in this format that, the amplitude
here amplitude here is 2. And this is minus pi thirds, the angle.
So we're going to be multiplying this by 2, an angle of minus pi 3rds.
Oh my mistake. This should be kilovolts, not volts,
because remember there's these, these k's here.
This should be 2k. This should be 2k.
So, multiplying this together we get 6k, with an angle of minus 5 pi 6ths.
And if we put this back into the other format, that means that v2 is going to be
equal to 6 cosine of 1 over the square root of 3 times 10 to the 6th t minus 5
pi 6ths kilovolts. Then my final v is going to be equal to
v1 Plus v2. It turns out that in this particular
example, we are not able to directly solve it in one step, we have to use
superposition because the two sources are operating at different frequencies.
So the consequently the capacitance leads to different impedances for each
different piece. So, superposition is the only technique
to use. But, you can see that it's not
particularly difficult to do the solution.
To review valid impedance techniques, we can use Kirchhoff's Laws, Superposition,
Node-voltage analysis, Mesh-current analysis, Thevenin and Norton Equivalent
Circuits and Source Transformations. And the only real difference from what we
were doing with resistors is that now we are using complex-valued impedances.
One more example that we will do, is this RLC circuit, and we're going to want, to
find the value of vc. So to do this, we will, again put it into
a phasor form. So, the input will be a V, and that is
going to be equal to 1 with a base angle of zero.
We're going to be operating at 780 hertz. So omega's two pi times 780.
The impedance for our resistance is 20 kilo-ohms.
The impedance for our inductance is equal to j times omega times l, and for the
impedance with capacitance, it's 1 over j times omega times c.
We can use a voltage divider here to find the value of VC.
This is just going to be Zc over Zr + Zl + Zc.
All times v, and obviously this is after the switch is closed.
this is the same circuit that we analyzed when we were doing the differential
equations for RLC circuits. But in this case we are, going to be
looking at it and analyzing it with a sinusoidal input, as opposed to just a
constant value that switches on. Now since v is 1 angle 0 this can
basically be somewhat ignored. it's important because of the units but
that just basically means that vc is going to be equal to zc over zr plus zl
plus zc. Plugging in our values for both various
things we get 1 over J omega C divided by R plus J I make it L plus 1 over J we'll
make it C. Now, plugging in the values of the
capacitance, the inductance, and the omegas here, and doing a little bit of
calculation with complex numbers. We'll discover that this turns out to be
about 0.51, minus j 0.5 zero, which is pretty close to 1 over the square root of
2 with a phase of minus phi fourths. Putting this back into our time domain
Would be c of t. This is going to be 1 over the square
root of 2, times the cosine of 2 pi times 780 times t, minus pi 4ths volts.
And so you an see that this is quite a bit simpler than doing the calculation
using differential equations. It is entirely possible to go through and
do a derivation to get the same results using differential equations.
But using phasors gives us an excellent tool at being able to apply complex
number number to, to give a solution to see how the system will respond to a
particular sinusoidal input. In summary, we showed how DC analysis
techniques can be directly applied to sinusoidal systems by using phasors and
impedances. Then we use superposition to analyze the
system that had multiple frequencies, and then solve the example system using these
techniques. In the next lesson, there will be a demo
showing sinusoidal response, and then a discussion of transfer functions.
Transfer functions are essentially how systems react across different
frequencies for various inputs. Until next time.