Symbulator 9

the best portable symbolic simulator of linear circuits

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Symbulator 9 9

Lesson 7

Alternating current analysis

Learn how to run an alternating current analysis (via phasor analysis). Learn how to describe capacitors, inductors, impedances and admittances, and sources for AC analysis.

Last updated 2023-07-08

In this lesson you will learn how to use the AC analysis to solve AC circuits in Symbulator. You will also learn how to describe the elements you already know so they suit AC analysis, and how to find equivalents in AC.

7.1AC analysis in Symbulator

The main thing to know is that in AC mode Symbulator understands phasors — complex numbers. It uses them in the input, in the analysis and in the output.

The AC analysis

Set Analysis to AC — alternating current. Unlike DC and TR, it needs one more input: the frequency in radians per second, typed into the ω — angular frequency box beside the menu.

AC mode in the other tools

The equivalence tools work in AC the same way: choose Find equivalent as usual, set Analysis to AC — alternating current, and give the ω box a frequency where the circuit needs one.

The Mini-Tools card

One more thing before the first AC problem, because AC is where it earns its place. Under the results sits the Mini-Tools card: four small calculators that work on values you type rather than on the circuit, and are ready before and after any solve. Choose one in its Tool menu, fill in the boxes it shows, and press Run:

  • aa — amplitude and angle turns a complex value into polar form. It takes one value: an answer's name, such as ir1, or any expression, such as v1 - v2. This lesson uses it throughout.
  • pf — power factor takes one value: a complex power, such as se, read as given, or the name of an element, such as e, read on the power a source delivers or an impedance consumes. It answers with the power factor and whether it leads or lags, and for a name says which power it read. Lesson 8 introduces it.
  • gain — voltage, current and power gain takes the voltage and current at a two-port's input and at its output. Lesson 13 introduces it.
  • pz — poles and zeros takes one value: a transfer function, such as v2/vg after an FD run with the source left as a symbol, or any ratio of polynomials in s. It answers with the roots of the denominator, the poles, and the roots of the numerator, the zeros, after cancelling any factor the two share.

They are mini-tools, not analyses: none of them solves the circuit, and none needs the circuit to have been solved, though an answer's name only means something once there are answers to read.

Describing elements for AC

When describing a circuit for AC analysis, element values have to be written the way this analysis expects. Valueless elements, like the short circuit and the op amp, are unchanged. Here's how to describe the ones with values:

  • Current sources j and voltage sources e accept complex values. You can declare their value in rectangular form, e.g. 10-3𝐢, or in angular form, e.g. (100∠120°).
  • The element r no longer represents just a resistor: it now represents an impedance. If you are wondering why we didn't use the letter z, wait until we get to two-ports. In AC analysis, r accepts complex numbers as its value, such as 10-3𝐢. We use r to describe resistors (real values) and conductances (real values, inverted), but also impedances (complex values) and admittances (the inverse of their complex value).
  • The elements c and l are used to describe capacitors and inductors only when their values are given in farads and henries, respectively. For AC analysis their description no longer requires the fifth field for the initial condition, since alternating current analysis focuses on the steady state after transient effects are long gone. So we only use the name, the two nodes, and the value.

The values of all of these elements can use SI prefixes.

What answers do you get

You get the same answers you did in DC, except that now they are phasors. The one difference is that now, besides the real power consumed, you also get the complex power consumed. We will discuss this in Lesson 8.

Rectangular or polar

A phasor can be written two ways, and circuit problems use both: in rectangular form as 3 + 4j, or in polar form as an amplitude and an angle, 5∠53.13°. They are the same number. Which one you want depends on the question — rectangular adds and subtracts easily, polar multiplies and divides easily, and textbook answers are usually quoted in polar.

Symbulator answers in rectangular form unless you ask otherwise. In Settings, under Display, tick Show AC answers as polar phasors and every answer is shown as an amplitude and an angle. The circuit is re-solved as soon as you tick it.

Three things it leaves alone:

  • Average power stays a plain number. It is real, so an angle would be meaningless. Complex power is converted: its polar form is the apparent power and the power-factor angle. See Lesson 8.
  • Symbolic answers are left as they are. There is no angle to take of \(v_{in} r_b/(r_a + r_b)\).
  • It applies to AC only. DC answers are real, and TR and FD answers are functions of t or s. The checkbox greys out there.

An answer that happens to be real still gets its angle, 0° or 180°, as the aa mini-tool does.

The aa mini-tool is still the better choice for converting one value, or an expression that is not an answer by itself, such as a difference between two node voltages. Lesson 9 uses it that way throughout.

7.2Solved numerical examples

With values in F and H

AS7's Example 9.9

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Find v(t) and i(t).

AS7's Example 9.9

Solution

Since this is the first AC problem we will solve, I will explain every step, including the manual ones, in full detail.

Step 1: sinusoids to phasors. We have to express the value of the source in terms Symbulator can understand, so we convert it from a sinusoid to a phasor: 10∠0°. We can pass this as (10∠0°) or, since the angle is zero, simply as 10.

Step 2: note the frequency. The circuit has a capacitor with a value given in farads. Because of this, Symbulator will need the frequency, given by the problem as 4 rad/s, to convert that value into an impedance in ohms.

Step 3: describe the circuit.

Circuit Description
e,1,0,10
r1,1,2,5
c,2,0,.1

Choose AC — alternating current, and put 4 in the ω — angular frequency box that appears beside it.

Notice that, in our circuit description, the order we give to the nodes of the resistor and the capacitor is chosen to be convenient for the answers we will read.

Step 4: read the answers. Once Symbulator is done, you should take a look at what it found:

  • the usual voltages in the nodes, voltage drops in the elements, and currents through the elements, under the names that should be familiar by now
  • the average power consumed in the source and the resistor, in ape and apr1. None is given for the capacitor, since capacitors and inductors do not consume real power.
  • the complex power consumed in all elements, in sc, se and sr1
  • the equivalent impedance of the rest of the circuit as seen by the source, in ze

The answers this problem wants are ir1 and vc. It is likely that Symbulator will give you the answers in rectangular form, with a real part and an imaginary part.

If we want to see them as an amplitude and angle, open the Mini-Tools card, leave the mini-tool set to aa — amplitude and angle, and give it the answer's name:

Value
ir1

It reads 1.789∠26.57° A.

To see every answer in polar form, use the setting described in Rectangular or polar, above.

This is correct. You manually convert it to a sinusoid, by putting it back in the same terms the input was given in:

\[i(t) = 1.789 \cos(4t + 26.57°)\ \mathrm{A}\]

Asking for the capacitor voltage the same way gives 4.472∠-63.43°, that is,

\[v(t) = 4.472 \cos(4t - 63.43°)\ \mathrm{V}\]

which is correct.

Future solved problems will not include this level of detail in the solution, only the circuit description and the settings we choose.

AS7's Practice Problem 9.9

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Determine v(t) and i(t).

AS7's Practice Problem 9.9

Solution

Circuit Description
e,1,0,(20∠30°)
r1,1,2,4
l,2,0,.2

The source is given in polar form and typed as written: magnitude, the angle sign, then degrees.

Then AC, with 10 for ω. Mini-Tools with aa reads ir1 as 4.472∠3.43° A and vl as 8.944∠93.43° V. Both are correct.

AS7's Example 9.10

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Find the input impedance of the circuit. Assume that the circuit operates at ω = 50 rad/s.

AS7's Example 9.10

Solution

Since this is a passive circuit, to reduce it we find its equivalent impedance.

Circuit Description
ca,1,2,2'm
r1,2,3,3
cb,3,0,10'm
l1,2,4,.2
r2,4,0,8

Set Type of analysis to Find equivalent and Type of equivalent to Resistance / impedance, with nodes 1 and 0, in AC at ω 50. The answer is called zeq.

equivalent impedance
\[Z_{eq} = 3.22 - 11.07\,\text{j}\ \Omega\]

This is correct.

AS7's Problem 9.35

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Find the steady-state current i in the circuit when vs(t) = 50 cos 200t V.

AS7's Problem 9.35

Solution

The capacitor is in farads and the inductor in henries, so Symbulator needs the frequency, and the source is entered as its phasor: 50 cos 200t is 50 at an angular frequency of 200.

Circuit Description
e,1,0,50
r,1,2,10
c,2,3,5'm
l,3,0,20'm

AC, with ω — angular frequency set to 200.

Read it with Mini-Tools set to aa: aa(ir) gives 4.789∠-16.70° A. Correct.

With values in Ω only

When the values of all the capacitors and inductors in the circuit are given in imaginary ohms, they must be entered in the circuit description as impedances, r. In these cases, Symbulator will not need a frequency.

AS7's Problem 9.37

Determine the admittance Y for the circuit.

AS7's Problem 9.37

Solution

Since the values are in ohms, these are all impedances. This is a passive circuit, so we could find its equivalent. However, the structure of this circuit is so simple that we can reduce it by hand. The problem asks for the equivalent admittance, so our answer will be the inverse of the equivalent impedance:

Evaluate
1/pr(4,8j,-10j)
\[0.25 - 0.025\,\text{j}\ \mathrm{S}\]

This is correct.

AS7's Problem 9.39

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For the circuit shown, find the equivalent impedance, and use that to find the current I. Let omega = 10 rad/s.

AS7's Problem 9.39

Solution

The frequency the question gives is superfluous: every value is already in ohms. Nothing here needs a full simulation either; the equivalent impedance is one expression, using the parallel shorthand:

Evaluate
4+20j+pr(16,-14j+25j)
\[9.135 + 27.47\,\text{j}\ \Omega\]

That is 9.135 + j27.47 Ω. The current is the source voltage divided by it:

Put the whole thing in Mini-Tools with aa: aa(12/(4+20j+pr(16,-14j+25j))) gives 0.4145∠-71.60° A. Correct.

AS7's Problem 9.73

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Determine the equivalent impedance for the circuit.

AS7's Problem 9.73

Solution

Eight impedances and no source: a job for Find equivalent rather than a solve. The names carry the two nodes each element bridges, for convenience.

Circuit Description
r10,1,0,6j
r20,2,0,8j
r30,3,0,8j
r40,4,0,12j
r12,1,2,2
r23,2,3,-6j
r34,3,4,4
r14,1,4,-4j

Set Type of analysis to Find equivalent and Type of equivalent to Resistance / impedance, with nodes 1 and 0, in AC. The answer is called zeq.

equivalent impedance
\[Z_{eq} = 0.3794 + 1.46\,\text{j}\ \Omega\]

That is 0.3794 + j1.46 Ω, which is correct.

With dependent sources

A dependent source is described exactly like an independent one — the value is just an expression naming another element's answer instead of a number.

The name is the one the answers use: the current through cx is icx, the voltage across rx is vrx.

AS7's Example 10.1

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Find ix in the circuit.

AS7's Example 10.1

Solution

The current source is controlled by the current through the capacitor cx, which is what the last field of j1 says.

Circuit Description
e,1,0,20
r1,1,2,10
cx,2,0,.1
l1,2,3,1
j1,0,3,2*icx
l2,3,0,.5

AC, with ω — angular frequency set to 4.

aa(icx) gives 7.589∠108.4° A. Correct.

AS7's Practice Problem 10.1

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Find v1 and v2 in the circuit.

AS7's Practice Problem 10.1

Solution

This one is controlled by a voltage rather than a current — three times the voltage across rx.

Circuit Description
j,0,1,10
rx,1,0,2
c,1,2,.2
l,2,0,2
r,2,3,4
e,3,0,3*vrx

AC, with ω — angular frequency set to 2.

aa(v1) gives 11.33∠60.02° V and aa(v2) gives 33.02∠57.13° V.

Both are correct.

AS7's Example 10.13

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Obtain vo and io in the circuit.

AS7's Example 10.13

Solution

The book wants its answers in terms of cosine rather than sine, so the source is taken as 8 cos(1000t − 40°) — as a phasor, (8∠-40°).

Note the decimal points on the values. They make the arithmetic approximate, which here is what you want: an exact solve of this circuit carries unresolved square roots through every step for no benefit.

Circuit Description
e,1,0,(8∠-40°)
r1,1,2,4'k
co,2,0,2'µ
l1,2,3,50'm
j1,0,3,.5*ico
ro,3,0,2'k

AC, with ω — angular frequency set to 1000.

aa(vro) gives 1.550∠-95.18° V and aa(ico) gives 0.003264∠-3.743° A. Both correct.

AS7's Example 10.14

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Find V1 and V2 in the circuit.

AS7's Example 10.14

Solution

Every value here is already in ohms, so no frequency matters. Again the decimal points keep the arithmetic approximate.

Circuit Description
j1,0,1,3
r1,1,0,1
rx,1,0,-1j
r3,1,2,-2j
j2,1,2,.2*vrx
r4,1,2,2+2j
r5,2,0,-1j
r6,2,3,2+2j
e,3,0,(18∠30°)

AC. The frequency is asked for but never used, so anything will do.

aa(v1) gives 2.708∠-56.73° V and aa(v2) gives 6.914∠-80.70° V. Both correct.

7.3Solved numerical-from-symbolic examples

AS7's Problem 9.89

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Calculate the value of C so that the net impedance is purely resistive at 2 kHz.

AS7's Problem 9.89

Solution

The answer given by the book is 25 µF. But I think this is one of those unfortunately common instances where there is a mistake in the book. I believe they meant to ask "at 2k rad/s", because that is the frequency at which the answer they give is right. Let me tell you how I solved this problem and I will let you be the judge.

Circuit Description
c,1,0,c
r1,1,2,10
l,2,0,5'm

Find equivalent, Resistance / impedance, nodes 1 and 0, AC. The problem gives 2000 Hz and the ω box wants radians per second, so type 2*pi*2e3 straight into it — it takes an expression, not only a number.

Once the simulation finishes, solve for the value of c that makes the imaginary part of the equivalent impedance zero:

Now open the Solve card and ask for the capacitance that leaves no imaginary part:

Equation(s) to solve in terms of the results
im(zeq) = 0
Unknown(s) to solve for
c

Tick real solutions only before running it. Without it the answer comes back carrying an im(c) term, because nothing has told Symbulator that a capacitance is a real number.

The answer is c = 0.000001235, or 1.235 µF. This is not the answer the book gives.

I first solved this over the complex numbers, which also works, but my friend Qifan Wang — who verified my answers — pointed out, correctly, that it is not necessary: focusing on the imaginary part removes every reference to the complex operator, so we are solving an equation in real terms only.

Let's now repeat the process, using what I suspect is the frequency they meant: 2000 rad/s. I also want to show you something cool. Now that we know the answer is in the range of µF, we can declare the value of the capacitor as c'µ, so that the value of c we get will be in that scale:

Circuit Description
c,1,0,c'µ
r1,1,2,10
l,2,0,5'm

at 2000 rad/s, as it was meantOpen in app ↗Open in split view

Find equivalent, Resistance / impedance, nodes 1 and 0, in AC at ω 2000. That gives zeq = \(\frac{500 i}{- c + 25 + 25 i}\), and the condition goes to the Solve card:

Equation(s) to solve in terms of the results
im(zeq) = 0

with c as the unknown and real solutions only ticked — the same equation as before, now against the new zeq — which answers c = 25.

The answer is c = 25, that is to say 25 µF. This is the right answer in the book. So they asked the wrong question — they asked it using the wrong frequency unit.

7.4Solved symbolic examples

AS7's Problem 10.69

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Find Vo/Vs.

AS7's Problem 10.69

Solution

Circuit Description
e,1,0,vs
c,1,2,c
r1,2,o,r
o,0,2,o

The ω box takes a name as readily as a number: type omega into it and the answers come back as functions of it.

Evaluate gives vo/vs as −c·ω·r·𝐢, which is correct.