Transient analysis

Bo2's Example 5.1

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Solution. Here we have two intervals. In the first, t < 0s, we assume things have been steady for a long time and any transient has passed, so we run a DC simulation to find the initial conditions for the second.

e,1,0,V
r1,1,2,r1
c,2,0,c
r,2,0,r

Run it in DC and read vc, the voltage in the capacitor c. It is the expression shown below, which is correct.

For the second interval, when 0s ≤ t, we do a transient simulation. We give the capacitor as initial condition its voltage from the previous interval.

c,2,0,c,r*v/(r+r1)
r,2,0,r

TROpen in app ↗Open in split view

Run it in TR and read vc again. It is the expression below, which is correct.

Bo2's Drill Exercise 5.1

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Solution. Again, we have two intervals: the first for DC, the second for TR.

For the first interval, when t < 0s, the current source is 2A. Simulate in DC.

j,0,1,2
r,1,0,3
c,1,0,1/12

The answers you want are in Results:

So the voltage in the capacitor is 6V. We give this value the name vc₀ in Define, to use it as the initial condition of the capacitor for the second interval.

For the second interval, when 0s ≤ t, we do a transient simulation. The current source is 0A, and the capacitor has an initial condition: voltage vc₀.

j,0,1,0
r,1,0,3
c,1,0,1/12,vc0

vc0 = 6

TROpen in app ↗Open in split view

Set Analysis to TR — transient / time domain.

The answers you want are vc and ic, in Results.

Run it, and the two answers are the expressions below. They are right.

Bo2's p224 5.2

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Solution. Again, we have two intervals: the first for DC, the second for TR.

For the interval when t < 0s, the voltage source is 12 V. Simulate in DC.

e,1,0,12
r3,1,2,3
r6,2,0,6
r4,2,3,4
c,3,0,1/12

The answers you want are in Results:

For the interval when 0s ≤ t, the voltage source is 0V. Simulate in TR, giving the capacitor as initial condition its voltage from the previous interval.

e,1,0,0
r3,1,2,3
r6,2,0,6
r4,2,3,4
c,3,0,1/12,8

TROpen in app ↗Open in split view

Set Analysis to TR — transient / time domain.

The answers you want are in Results:

Bo2's Drill Exercise 5.2

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Solution. For t < 0s, the 6A source is part of the circuit. Simulate in DC.

j,0,1,6
r10,1,0,3
r12,1,2,3
r20,2,0,3
c,2,0,1/10

The answers you want are in Results:

For 0s ≤ t, the source is no longer relevant and can be excluded from the circuit. Simulate in TR, giving the capacitor its proper initial condition.

r10,1,0,3
r12,1,2,3
r20,2,0,3
c,2,0,1/10,6

TROpen in app ↗Open in split view

Set Analysis to TR — transient / time domain.

The answers you want are in Results:

Bo2's Figure 5.10a

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Look at the simple resistor-inductor (RL) circuit shown, where at time t=0 the inductor current is i_L(0). Determine v_L(t), i_L(t) and v_R(t) for t≥0.

Since we are already given the initial condition, we only run the transient simulation for t≥0. In our circuit description below, the initial condition is il0.

l,1,0,l,il0
r,1,0,r

Set Analysis to TR — transient / time domain.

The answers you want are vl, il and vr, in Results.

We get the right answers:

Bo2's Example 5.3

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Determine i_L(t) for all t.

Solution. For t < 0s, simulate in DC.

e,1,0,v
r1,1,2,r1
l,2,0,l
r2,2,0,r

The answer you want is il, in Results.

For t ≥ 0s, simulate in TR, giving the inductor its initial condition.

l,2,0,l,v/r1
r2,2,0,r

TROpen in app ↗Open in split view

Set Analysis to TR — transient / time domain.

The answer you want is in Results:

Bo2's Drill Exercise 5.3

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Find i_L(t), v_L(t) and i(t) for all t.

Solution. For t < 0s, simulate in DC.

e,1,0,8
r4,1,2,4
l,2,0,1
r12,2,0,12

The answers you want are in Results:

For t ≥ 0s, simulate in TR, giving the inductor its initial condition of 2A.

e,1,0,0
r4,1,2,4
l,2,0,1,2
r12,2,0,12

TROpen in app ↗Open in split view

Set Analysis to TR — transient / time domain.

The answers you want are in Results:

Bo2's p230 (Dependent source)

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Find i_L(t) for t≥0, given that i_L(0) = 5A.

We only run the transient simulation for t≥0, with initial condition 5A.

r,v,0,r
j,v,0,2*ir
l,v,0,l,5

Set Analysis to TR — transient / time domain.

The answer you want is in Results:

Bo2's Drill Exercise 5.4 (Dependent source)

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Find i_L(t) and v_L(t) for all t.

Solution. For t < 0s, simulate in DC. This gives us the initial condition.

ei,1,0,12
r6,1,2,6
l,2,0,16
r8,2,3,8
ed,3,0,3*vr8

The answers you want are in Results:

For t ≥ 0s, simulate in TR, giving the inductor its initial condition.

l,2,0,16,2
r8,2,3,8
ed,3,0,3*vr8

Dependent source, TROpen in app ↗Open in split view

Set Analysis to TR — transient / time domain.

The answers you want are in Results:

Bo2's Example 5.5 (Op Amp)

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Determine v_C(t), i_C(t) and v_o(t) for all t.

Solution. For t < 0s, simulate in DC. This gives us the initial condition.

e,3,0,4
r2,3,1,2
r5,1,o,5
c,1,o,1/20
o,0,1,o

The answers you want are in Results:

For t ≥ 0s, simulate in TR, giving the capacitor its initial condition.

r2,0,1,2
r5,1,o,5
c,1,o,1/20,10
o,0,1,o

Op Amp, TROpen in app ↗Open in split view

Set Analysis to TR — transient / time domain.

The answers you want are in Results:

Bo2's Drill Exercise 5.5 (Op Amp)

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Determine v_C(t), i_C(t) and v_o(t) for all t.

Solution. For t < 0s, simulate in DC. This gives us the initial condition.

e,2,0,4
o,2,1,o
c,o,1,1/20
r5,o,1,5
r2,1,0,2

The answers you want are in Results:

For t ≥ 0s, simulate in TR, giving the capacitor its initial condition.

o,0,1,o
c,o,1,1/20,10
r5,o,1,5
r2,1,0,2

Op Amp, TROpen in app ↗Open in split view

Set Analysis to TR — transient / time domain.

The answers you want are in Results:

Bo2's Example 5.6

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In the circuit shown, there are two switches: one that opens at time t=0 and one that closes at time t=1 second. Determine v_C(t) and i_C(t) for all t.

As we mentioned before, every time a switch moves marks the end of one interval and the beginning of another interval for Symbulator.

For the first interval, which corresponds to t<0, we simulate in DC.

e,2,0,10
r1,2,1,1
c,1,0,1/4
r4,1,0,4

The answers you want are in Results:

The voltage in the capacitor at the end of this interval will serve as the initial condition of the capacitor for the next interval. The second interval runs between 0 and 1 second, i.e. 0 < t ≤ 1 second. We simulate in TR.

c,1,0,1/4,8
r4,1,0,4

TR, 0 < t <= 1 sOpen in app ↗Open in split view

Set Analysis to TR — transient / time domain.

The answers you want are in Results:

The capacitor's voltage at the end of this second interval is the initial condition for the third. We can use its exact value, 8*exp(-1), but the textbook prefers using its approximate value, i.e. 2.943.

The third interval corresponds to t > 1 second. We simulate in TR. To make things easier for Symbulator, we replace the resistors by their equivalent.

c,1,0,1/4,2.943
re,1,0,[4,6]

TR, t > 1 sOpen in app ↗Open in split view

The expressions we get are equivalent to:

These are the right answers, as can be seen by checking the book's answers.

Bo2's Drill Exercise 5.6

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Let’s see another example where a TR simulation is done for an interval that starts at a time other than t=0. Determine i_L(t) and v_L(t) for all t.

For the first interval, which corresponds to t<0, we simulate in DC.

e,1,0,9
r9,1,0,9
r3,1,2,3
l,2,0,6

The answers you want are in Results:

The 3A current in the inductor is its initial condition for the second interval, which goes from 0 to 2 seconds, i.e. 0 < t ≤ 2 seconds. We simulate in TR.

r,1,0,9+3
l,1,0,6,3

TR, 0 < t <= 2 sOpen in app ↗Open in split view

Set Analysis to TR — transient / time domain.

The answers you want are in Results:

The inductor's current at the end of this second interval is the initial condition for the third. Find its approximate value thus:

Put il in Evaluate with t = 2 in its Conditions box. It reads

.055

The third interval corresponds to t > 2 second. We simulate in TR. To make things easier for Symbulator, we replace the resistors by their equivalent.

r,1,0,[9+3,4]
l,1,0,6,.055

TR, t > 2 sOpen in app ↗Open in split view

Set Analysis to TR — transient / time domain.

The answers you want are in Results:

These are the kind of expressions your book or professor are looking for.

Bo2's Figure 5.19

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In the following circuit, assume that all the initial conditions are zero. The source is a step function u(t) with value V volts. Find i_R, v_R, i_C and v_C.

As discussed before, a step source with symbolic value requires that we use the u(t) nomenclature to describe it. Since the value is V volts, we describe the source as V*u(t).

e,1,0,V*u(t)
r,1,2,r
c,2,0,c,0

Set Analysis to TR — transient / time domain.

The answers you want are vc, vr and ic, in Results.

There is no need to look for i_R, since i_R=i_C. We get the following expressions:

Bo2's Figure 5.24

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In the following circuit, assume that all the initial conditions are zero. The source is a step function u(t) with value V volts. Find i_R, v_R, i_L and v_L.

e,1,0,V*u(t)
l,1,2,l,0
r,2,0,r

Set Analysis to TR — transient / time domain.

The answers you want are vl, vr and il, in Results.

There is no need to look for i_R, since in a series circuit it will be identical to i_L.

Bo2's Drill Exercise 5.7

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For the circuit, with a step source of 12 volts, find i_L, v_L, i_L and v.

As discussed before, a step source with numerical value can be described without the u(t) nomenclature. Thus, we describe the source as 12.

e,1,0,12
r4,1,2,4
l,2,3,2,0
r2,3,0,2

Set Analysis to TR — transient / time domain.

The answers you want are in Results:

Bo2's Figure 5.26

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In the following circuit, assume that all the initial conditions are zero. The source is a step function u(t) with value I amperes. Find i_C, i_R, and v.

As discussed before, a step source with symbolic value requires that we use the u(t) nomenclature to describe it. Since the value is I amperes, we describe the source as I*u(t). Thus:

j,0,1,i*u(t)
c,1,0,c,0
r,1,0,r

Set Analysis to TR — transient / time domain.

The answers you want are in Results:

Bo2's Example 5.7 (Op Amp)

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Suppose that v_s(t)=u(t). Find v_C, i_C, and v_o.

This source is a step source with a value of 1 volt. Since its value is numerical, the u(t) nomenclature is not needed. We describe the source plainly as 1.

e,1,0,u(t)
o,1,2,o
c,2,o,1/8,0
r2,2,o,2
r1,2,0,1

Set Analysis to TR — transient / time domain.

The answers you want are in Results:

Bo2's Drill Exercise 5.8 (Op Amp)

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Suppose that v_s(t)=u(t). Find v_C, i_C, and v_o.

Again, following the same logic as above, we describe the source plainly as 1.

e,1,0,u(t)
r1,1,2,1
r2,2,o,2
c,2,o,1/8,0
o,0,2,o

Set Analysis to TR — transient / time domain.

The answers you want are in Results:

Bo2's p249 F5.28

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For the circuit left, the source is v_s(t) = V u(t) - V u(t-t₀). This is shown in the plot to the right. Given this, find the voltage drop in the capacitor for all t.

Because the source has two steps, this problem has to be analyzed in two intervals. For the first interval, the initial conditions are zero and the source value is V*u(t). Our circuit description for the first interval is given below.

e,1,0,V*u(t)
r,1,2,r
c,2,0,c,0

Set Analysis to TR — transient / time domain.

The answer you want is vc, in Results.

The expression for the capacitor's voltage drop in the first interval is:

The second interval starts at t₀. In it, the initial condition of the capacitor is given by the value of the expression above when t=t₀. Here is how we find it:

Put vc in the Evaluate card and write t = to in its Conditions box, to evaluate this expression under this condition. The answer comes back \(V - V e^{- \frac{to}{c r}}\).

The circuit description for the second interval uses that expression as the capacitor's initial condition. The source is now 0 volts — another way of saying it becomes a short — so it need not appear in the description.

r,0,2,r
c,2,0,c,v-e^(-to/(c*r))*v

TR, second intervalOpen in app ↗Open in split view

Set Analysis to TR — transient / time domain.

The answer you want is vc, in Results.

This is the expression for the capacitor's voltage drop in the second interval.

Bo2's Drill Exercise 5.9

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Find the current through the inductor if the source is v_s(t) = V u(t) – V u(t-t_o).

Because the source has two steps, this problem has to be analyzed in two intervals. For the first interval, the initial conditions are zero and the source value is V*u(t). Our circuit description for the first interval is given below.

e,1,0,V*u(t)
l,1,2,l,0
r,2,0,r

Set Analysis to TR — transient / time domain.

The answer you want is il, in Results.

The expression for the inductor's current in the first interval is:

The second interval starts at t₀. In it, the initial condition of the inductor is given by the value of the expression above when t=t₀. Here is how we find it:

Put il in the Evaluate card and write t = to in its Conditions box. The answer comes back \(\frac{V}{r} - \frac{V e^{- \frac{r to}{l}}}{r}\), which is the same expression gathered differently.

The circuit description for the second interval uses that expression as the inductor's initial condition. The source becomes a short, so it need not appear in the description.

l,0,2,l,v/r-e^(-to*r/l)*v/r
r,2,0,r

TR, second intervalOpen in app ↗Open in split view

Set Analysis to TR — transient / time domain.

The answer you want is il, in Results.

To put this in terms of the same t as the first interval, we replace t with t-t_o:

This is the expression for the inductor's current in the second interval.

Bo2's Figure 5.36

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For the circuit shown in (a), find the voltage drop in the capacitor, given that the voltage source is as shown in (b).

This problem requires two intervals. In the first interval, a DC analysis of the circuit with a source of 2 V gives us the voltage in the capacitor.

e,1,0,2
r3,1,2,3
r5,2,3,5
c,3,0,1
j,0,2,2*ir3

The answer you want is vc, in Results. It reads

2

This is the initial condition for the capacitor in the second interval, which we analyze using TR and a source value of -4 V.

e,1,0,-4*u(t)
r3,1,2,3
r5,2,3,5
c,3,0,1,2
j,0,2,2*ir3

TR, second intervalOpen in app ↗Open in split view

Set Analysis to TR — transient / time domain.

The answer you want is in Results:

This is the expression for the voltage drop in the capacitor after t=0.

Bo2's Drill Exercise 5.11

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For the circuit of Bo2's Example 5.7, change the value of the capacitor to ¼ F. Find v_C(t) for the case that the source is v_S(t)=1 V for t<0 and 3V for t≥0.

We can tell Symbulator that we are only interested in one answer, vc.

Optional in Symbulator 9, which solves quickly enough: if you want to, tick Do you want to limit the results to save time? in Settings and list vc in the field it reveals.

Then we run the simulation for the first interval, just as we did before, but with the new capacitor and source values.

e,1,0,1
o,1,2,o
c,2,o,1/4
r2,2,o,2
r1,2,0,1

The answer you want is vc, in Results. It reads

-2

This is the initial condition for the next interval. Again, we can limit the results to the one answer, vc.

Optional in Symbulator 9, which solves quickly enough: if you want to, tick Do you want to limit the results to save time? in Settings and list vc in the field it reveals.

Then we run the simulation for the second interval, just as we did before, but with the new capacitor and source values.

e,1,0,3*u(t)
o,1,2,o
c,2,o,1/4,-2
r2,2,o,2
r1,2,0,1

TROpen in app ↗Open in split view

Set Analysis to TR — transient / time domain.

The answer you want is in Results:

Bo2's Figure 5.38 (Impulse)

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Find the voltage drop and current in the capacitor, given a source of I δ(t) A, where δ(t) is the impulse function and I is a constant.

As we mentioned before, impulse sources must be described using the δ(t) nomenclature. Typing delta(t) spells it without the Greek.

j,0,1,i*δ(t)
c,1,0,c,0
r,1,0,r

Once it has run, the answers we want are:

They are already in Results: vc is \(\frac{i e^{- \frac{t}{c r}}}{c}\) and ic is \(i \delta\left(t\right) - \frac{i e^{- \frac{t}{c r}}}{c r}\).

The impulse survives in the capacitor's current, which is right — the charge arrives all at once, and the delta is that instant.

Bo2's Drill Exercise 5.13 (Impulse)

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For the circuit of Bo2's Example 5.7, find v_C(t), i_C(t) and v_o(t) due to an impulse voltage source of v_S(t)= δ(t) (that is to say, a 1V impulse in t=0.)

Nothing new here. Again we use the δ(t) nomenclature for the source.

e,1,0,δ(t)
o,1,2,o
c,2,o,1/8,0
r2,2,o,2
r1,2,0,1

Set Analysis to TR — transient / time domain.

The answers you want are in Results:

Bo2's Figure 5.39 (Ramp)

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Find i_L(t) and v_L(t) for a circuit that has in parallel a resistor R, an inductor L and a current source with a ramp value I r(t). Initial conditions are zero.

A ramp value I r(t) means a value of I t. We describe this source simply as i*t.

j,0,1,i*t
r,1,0,r
l,1,0,l,0

Set Analysis to TR — transient / time domain.

The answers you want are in Results:

Bo2's Drill Exercise 5.14 (Ramp)

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For the circuit of Bo2's Example 5.7, find v_C(t) and i_C(t) due to a ramp input voltage of v_S(t)= r(t).

Nothing new here. The description below should be clear to you by now.

e,1,0,t
o,1,2,o
c,2,o,1/8,0
r2,2,o,2
r1,2,0,1

Set Analysis to TR — transient / time domain.

The answers you want are in Results:

Bo2's Example 5.12 (Exponential)

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Assume initial conditions zero. Find i_L(t) and v_L(t), given that i_S(t) = 2 e ^-4t A.

Describe the source as given to you by the problem, using an exponential.

j,0,1,2e^(-4t)
r,0,1,6
l,1,0,2,0

Set Analysis to TR — transient / time domain.

The answers you want are in Results:

Bo2's p265 (Exponential)

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Repeat Bo2's Example 5.12 if the current source is now i_S(t) = 2 e ^-3t A.

All we do is change one digit in the circuit description: replace 4 with 3.

j,0,1,2e^(-3t)
r,0,1,6
l,1,0,2,0

Set Analysis to TR — transient / time domain.

The answers you want are in Results:

Bo2's Example 5.13 (Exponential)

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Find the voltage drop in the capacitor and the current in the 3 Ω resistor, if the source value is v_S(t)=18e^-t/2.

e,1,0,18e^(-t/2)
r3,1,2,3
r5,2,3,5
c,3,0,1,0
j,0,2,2*ir3

Set Analysis to TR — transient / time domain.

The answers you want are in Results:

Bo2's Drill Exercise 5.16 (Exponential)

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For the circuit of Bo2's Example 5.7, find v_C(t) and i_C(t) for the case when the source's value is v_S(t)= 2 e ^-4t. As before, assume initial conditions are zero.

e,1,0,2e^(-4t)
o,1,2,o
c,2,o,1/8,0
r2,2,o,2
r1,2,0,1

Set Analysis to TR — transient / time domain.

The answers you want are in Results:

Bo2's Example 6.1

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Find the current in the inductor i(t) and the voltage drop in the capacitor v(t).

Since we have switches changing in time t=0, this is a two interval problem. The first interval is for t<0. We can analyze the first interval using DC.

e,3,0,9
r2,3,1,5
r1,1,0,5
l,1,2,1/2
c,2,0,1/8
r3,2,0,2

The answers you want are in Results:

These are the initial conditions for the second interval. The second is for t≥0. We can analyze the second interval using TR. We only care about vc and il.

r1,1,0,5
l,1,2,1/2,1
c,2,0,1/8,2

TR, second orderOpen in app ↗Open in split view

Bo2's Drill Exercise 6.1

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For the circuit shown, find v(t) and i(t) for t≥0.

Nothing new here. This is a two interval problem, as others seen before. The first interval, for t<0, is analyzed in DC to find the initial conditions.

j,0,1,1
r3,1,0,3
r6,1,0,6
l,1,2,1/4
c,2,0,1/3

The answers you want are in Results:

These conditions are used in the TR analysis of the second interval, for t≥0.

The answers are vc = \(3 e^{- 2 t} - e^{- 6 t}\) and il = \(- 2 e^{- 2 t} + 2 e^{- 6 t}\), both in Results.

r3,1,0,3
r6,1,0,6
l,1,2,1/4,0
c,2,0,1/3,2

TROpen in app ↗Open in split view

Bo2's Example 6.2

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Find i(t) and v(t) for the circuit shown.

I decided to simulate this using only fractions, not decimals.

Write the value as the fraction it is, 7/2 rather than 3.5: a fraction stays exact, a decimal point makes the arithmetic approximate.

7/2

Because of the switches, this is a two interval problem. The first interval is to be analyzed using DC. Notice that for t<0 the circuit is such that we basically have two separate circuits. We simulate each one of them separately:

e,0,1,7/2
r1,1,2,4
c,2,0,2
r2,2,0,1/6

The answer you want is vc, in Results.

-7/50

r3,3,0,1/6
l,3,0,1/50
r4,3,4,3
e2,4,0,3

DC, the right halfOpen in app ↗Open in split view

The answer you want is il, in Results.

1

The second interval, for t≥0, is analyzed using TR. I could have described the circuit again from scratch, but out of laziness I copied the descriptions from the DC simulations. To avoid renaming nodes, I simulated the right switch as a short between nodes 2 and 3.

vc comes back \(\frac{\left(- \sin{\left(4 t \right)} - 7 \cos{\left(4 t \right)}\right) e^{- 3 t}}{50}\) and il \(\left(- \sin{\left(4 t \right)} + \cos{\left(4 t \right)}\right) e^{- 3 t}\).

c,1,0,2,-7/50
r,1,0,[1/6,1/6]
l,1,0,1/50,1

TR, both halves joinedOpen in app ↗Open in split view

These are the right answers.

Bo2's Example 6.3

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Find i(t) and v(t) for the circuit shown.

Since the sources change value, this is a two interval problem. For t<0, in DC:

e,1,0,5
r3,1,2,3
l,2,3,1
c,3,0,1
r1,3,0,1
j,0,3,1

The answers you want are in Results:

Using these as initial conditions, now we use TR for t≥0:

r3,0,2,3
l,2,3,1,1
c,3,0,1,2
r1,3,0,1

TROpen in app ↗Open in split view

Bo2's Drill Exercise 6.4

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My copy of Bo2 is second-hand, and this problem had scribbles on it. The previous owner evidently struggled with it and concluded that the inductor had to be ½ H, not the 1 H the schematic says. Symbulator can show he is right and the book wrong. First, solve the circuit as printed, with a 1 H inductor.

e,2,0,2
r,1,2,1
l,1,0,1
c,1,0,1/8

The answers you want are in Results:

r,1,0,1
l,1,0,1,2
c,1,0,1/8,0

TR, the book's 1 HOpen in app ↗Open in split view

The answer we get doesn't match the answer provided by the book. Now let's repeat the TR simulation using the ½ H value for the inductor:

r,1,0,1
l,1,0,1/2,2
c,1,0,1/8,0

TR, 1/2 H insteadOpen in app ↗Open in split view

Now the answer matches the book's. An expert user can reach for FD analysis and the t2s tool, which we have not discussed yet, to find the inductor's value from the book's answer. This is how I found the ½ H myself, working backwards:

t2s(-16*t*e^(-4*t))

which gives \(- \frac{16}{\left(s + 4\right)^{2}}\).

-16/(s+4)^2

2*(s+8)/(s+4)^2

r,1,0,1
l,1,0,l,2
c,1,0,1/8,0

FD, solving for the inductorOpen in app ↗Open in split view

Set Analysis to FD — complex frequency domain.

The answer you want is vc, in Results.

-16*l/(l*s^2+8*l*s+8)

That is what the Solve card is for — a system that is not a circuit:

-16/(s+4)^2 = -16*l/(l*s^2+8*l*s+8)

with l as the unknown. It answers l = 1/2 H.

l=1/2

FD analysis and the t2s tool are used for frequency domain analysis, and will be discussed separately in a subsequent part of Symbulator's documentation.

Bo2's Drill Exercise 6.5

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In the circuit, suppose that v_S(t)= 12 – 12 u(t) V. Find i₂(t) for t≥0.

e,3,0,12
r6,3,1,6
l1,1,0,6
r3,1,2,3
l2,2,0,4

The answers you want are in Results:

r6,0,1,6
l1,1,0,6,2
r3,1,2,3
l2,2,0,4,0

TROpen in app ↗Open in split view

Bo2's Drill Exercise 6.6 (Op Amp)

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In the circuit, suppose that v_S(t)= 2 – 2 u(t) V. Find v_o(t) for t≥0.

Simulate the first interval in DC to find the initial conditions of the capacitors.

e,3,0,2
r1,3,1,1
r2,1,2,2
r3,1,o,2
ca,1,0,1
cb,2,o,1/4
o,0,2,o

The answers you want are in Results:

Then simulate the second interval in TR, to find the voltage in node o.

r1,0,1,1
r2,1,2,2
r3,1,o,2
ca,1,0,1,0
cb,2,o,1/4,4
o,0,2,o

Op Amp, TROpen in app ↗Open in split view

Bo2's Example 6.5 (Plot)

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In the circuit, use v_S=2/5V, R=12Ω, L=2H and C=1/50F. Find v(t) and i(t), and plot them for time 0<t<1.5 seconds.

e,1,0,2/5*u(t)
r,1,2,12
l,2,3,2,0
c,3,0,1/50,0

vc is the first of those and il the second.

These are the right answers. To plot them, open Plotting Tools, enter vc as the variable, 0 as the start and 1.5 as the end. Compare to the graph from the book. Repeat the plot for il, and compare that one to the book's graph too.

Bo2's Drill Exercise 6.7

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For the circuit in Bo2's Example 6.5, use v_S=3V, R=5Ω, L=1/2H and C=1/8F. Find v(t) and i(t).

e,1,0,3*u(t)
r,1,2,5
l,2,3,1/2,0
c,3,0,1/8,0

Bo2's Drill Exercise 6.8 (Ramp)

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For the circuit in Bo2's Drill Exercise 6.7, find the voltage drop in the capacitor v(t) if the source has a value v_S(t) = 3 r(t).

e,1,0,3t
r,1,2,5
l,2,3,1/2,0
c,3,0,1/8,0

Bo2's Example 6.6 (Plot)

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For the circuit, plot the voltage drop in the inductor for 0 < t < 5ms.

First, find the initial conditions for t<0 by running a DC simulation.

e,1,0,12
r,1,2,3
l,2,3,1
c,3,0,1'µ
s1,3,0

The answers you want are in Results:

Then, find the vl for t≥0 by running a TR simulation.

e,1,0,12*u(t)
r,1,2,3
l,2,3,1,4
c,3,0,1'µ,0

Plot, TROpen in app ↗Open in split view

Finally, plot vl in Plotting Tools from 0 to 0.005 s. In the resulting plot you will see something outstanding: around t = 1.57 ms, the voltage drop across the inductor is 3991 volts!

Bo2's Drill Exercise 6.9

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For the circuit, R=1Ω, L=2H, C=1/2F and i_S(t)=u(t). Find i(t) and v(t).

j,0,1,u(t)
r,1,0,1
l,1,0,2,0
c,1,0,1/2,0

Bo2's Figure 6.23 (Op Amp)

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For the circuit, find v_o(t) if v_S(t) = u(t).

e,3,0,u(t)
r1,3,1,1
r2,1,2,1
ca,2,0,1/5,0
cb,1,o,1,0
o,2,o,o

Bo2's Drill Exercise 6.10 (Op Amp)

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In the previous circuit, change the capacitor's value from 1/5 F to 25/16 F. Find v_o(t) if v_S(t) = 3 u(t).

e,3,0,3*u(t)
r1,3,1,1
r2,1,2,1
ca,2,0,25/16,0
cb,1,o,1,0
o,2,o,o

-4e^-2t/5+e^-8t/5+3

Bo2's Drill Exercise 6.11 (p307)

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For the circuit, suppose that R₁ = R₂ = 1Ω, L = 1H, C = 1F and v_S(t)= 2e^-2t u(t). Find i(t) if all initial conditions are zero.

e,3,0,2e^(-2t)
r2,3,2,1
r1,2,1,1
c,2,0,1,0
l,1,0,1,0

-(e^t cos(t)-e^t sin(t)-1) e^-2t

A more complex problem

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The circuit comes from the circuits analysis class of Professor Eliane Boulet de Cabrera, from Universidad Tecnologica de Panama. Find v₁ and v.

Because of the switches, this is a two interval problem. The first is run in DC.

e,3,0,18
r8,3,1,8
ca,1,0,1/6
r12,1,0,12
r18,1,2,18
r6,2,0,6
cb,2,0,1/3

The answers you want are in Results:

These serve as initial conditions for the second interval, which is run in TR. Since we only want v1 and v2, it pays off to let Symbulator know this:

ca,1,0,1/6,9
r12,1,0,12
r18,1,2,18
r6,2,0,6
cb,2,0,1/3,9/4
j,0,2,10e^(-t)*sin(2t+30*pi/180)

TROpen in app ↗Open in split view

Set Analysis to TR — transient / time domain.

Symbulator 9 answers this one in about a second; v1 and v2 are in Results.

These are the correct answers, which explains why Professor Boulet – who made the problem up - is a living legend among circuit students at UTP.

Bo2's Drill Exercise 4.5 (Expert)

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For this circuit, we know that the capacitor’s initial condition is zero, that v_s is an unknown step source (i.e. of the form A u(t), where A is a constant value in volts) and that the voltage drop in the capacitor for t>0 is found to be 1-e^(-t/2). Find the voltage drop in the resistor, the current through the capacitor and the value of the source.

This problem is a match made in heaven for the expert tool, because we have one unknown value in the circuit (i.e. the value of the step source) and we have one known answer (i.e. the voltage drop in the capacitor.) So, the game plan here is to run this circuit through Symbulator’s Expert Mode, add one new equation and one new unknown, and then solve. First, let’s generate the new equation.

The new equation is simply 1-e^(-t/2) = vc: Expert Mode reads it in the time domain, like the answers on screen.

If you would rather convert by hand, t2s is still there and an equation already written in s is left alone rather than transformed twice:

t2s(1-e^(-t/2))

which gives \(\frac{1}{s \left(2 s + 1\right)}\) — the same statement, one domain over.

Now let’s run the Expert simulation of the circuit. Let’s define the value of the source as a*u(t), since we know it’s a step source; the variable a will serve as the unknown value, for which we will solve in Expert Mode.

e,1,0,a*u(t)
r,1,2,2
c,2,0,1,0

Choose TR, then open Expert Mode. Add equation(s) takes one per line:

1-e^(-t/2) = vc

Add unknown(s) takes the name:

a

Let Symbulator solve the equations now. Once it is done, we want to do a quick sanity check. If the system solved correctly, we should see that Symbulator has as voltage drop in the capacitor the expression we already know from the problem statement:

vc in Results reads \(1 - e^{- \frac{t}{2}}\), which it does.

This is correct, and the answers we want are on screen with it: ic is \(\frac{e^{- \frac{t}{2}}}{2}\), and the voltage drop vr is \(e^{- \frac{t}{2}}\).

These are the correct answers. The value of the step source is 1 volt.

Bo2's Drill Exercise 4.13 (Expert)

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This problem is solved just like the one we already saw. First, find the new equation:

No transform this time either — write what you know:

vr = e^(-t)

Then run the Expert simulation. I decided to use vs as the step value of the source:

e,1,0,vs*u(t)
r,1,2,1
c,2,0,1,0

Choose TR, and put vs in Add unknown(s) beside the equation above.

Run it. The sanity check is already on screen: vr reads \(e^{- t}\), which is what we told it.

And the answers with it: ir is \(e^{- t}\) and the source's value vs is 1 V.

Ain’t that a beauty? Yes, ma’am! Atta boy, Symbulator! Atta boy!

Bo2's 164 (Expert)

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This is the same thing as the previous two, so there is no need for commentary.

e,1,0,vs*u(t)
r,1,2,2
l,2,0,1,0

Choose TR, put vs in Add unknown(s), and give Add equation(s) the answer you know — in time, as always here:

1-e^(-2*t) = il

Run it, then put the sanity check into Evaluate:

It does add up — il reads \(1 - e^{- 2 t}\) — and the answers are on screen with it: vl is \(2 e^{- 2 t}\), vr is \(2 - 2 e^{- 2 t}\), and the source's value vs is 2 V.

This is the end of the TR section.