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MAE 4182Review & practiceCourse calendar

Electromechanical Control System Design

Core Concepts Practice

The complete practice sheet, plus three short additional questions.
Try each question first, then read its answer and explanation.

40 original questions24 multiple choice16 two-part fill-ins+3 additional questions

Foundations · Laplace transform · Local linearization · Block diagrams

Start with question 01 ↓New: questions 41–43 ↓Review notes · 20-page PDF ↗Download the original practice PDF

Multiple choice

Questions 01–24 · Choose one answer per question.

Question 01Foundations

A motor receives a voltage command and a sensor measures its shaft angle. Which quantity is the applied input?

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Answer · B

The voltage command.

Why this works

Begin at the boundary of the system being modeled. Here, a voltage command enters the motor, and a sensor reports the resulting shaft angle. The input is the voltage; the output is the measured angle.

A desired angle is a different signal: it tells a controller what we want. The controller may use that reference to choose a voltage, but the reference and the voltage are not the same quantity.

Check the other choices

A. The angle is the observed output, not the quantity applied to the motor.

C. The sampling interval describes when measurements are recorded; it is not the motor command.

D. The desired angle is a reference. The stated input to the motor itself is voltage.

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Question 02Foundations

To predict future motion using a given dynamic model, which information is generally needed?

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Answer · C

The present state and future inputs.

Why this works

Imagine two identical motors receiving zero voltage. One is stopped and the other is spinning. Their current commands match, but their subsequent motion need not match because their present states differ.

For a known deterministic model, the state contains the present information needed to continue the prediction. Future inputs are also needed: applying a torque next will generally produce a different motion from applying no torque.

Check the other choices

A. The same current input can act on a stopped system or an already moving one.

B. A desired output tells us the goal, not the current motion or the applied inputs.

D. Sensor count does not tell us the present state or the future commands.

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Question 03Foundations

Which input-output relation satisfies superposition?

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Answer · A

y = −3u

Why this works

Superposition means that scaling an input scales its output, and adding inputs adds their outputs. For y = −3u, an input αu1 + βu2 produces −3αu1 − 3βu2, exactly the corresponding combination of the two outputs.

A negative sign does not make a rule nonlinear. Squaring and sine fail the general scaling test; an added constant fails the zero-input test. For example, u + 4 gives 4 even when u = 0.

Check the other choices

B. Doubling the input quadruples its square, rather than doubling the output.

C. The offset gives output 4 at zero input, which a linear map cannot do.

D. Sine does not generally preserve addition or scaling of its argument.

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Question 04Foundations

Why is y = 3u + 1 not a linear input-output map?

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Answer · D

Zero input produces a nonzero output.

Why this works

Test the simplest input first: setting u = 0 gives y = 1. A linear map must send zero input to zero output. This offset therefore proves that the stated map is not linear.

You can also check scaling. At u = 1, the output is 4; at u = 2, it is 7, not 8. The graph is straight, but the added constant prevents the output from doubling with the input.

Check the other choices

A. Its graph is a straight line, but that alone does not establish superposition.

B. The multiplier 3 is compatible with linearity; the added 1 is the problem.

C. A one-input model may be linear or nonlinear; input count is a separate classification.

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Question 05Foundations

A model has fixed parameters and no explicit absolute-time dependence. Its state is moving. Which statement is correct?

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Answer · B

The model can still be time-invariant.

Why this works

A swinging pendulum changes angle and angular velocity continuously. If its length, gravity, and governing rule stay the same, starting the same experiment later does not change the rule. A moving state is fully compatible with a time-invariant model.

Time variation concerns an explicit change in the governing law, such as a prescribed coefficient a(t). Linearity asks a different question: whether scaling and addition work. A model can be nonlinear and time-invariant at the same time.

Check the other choices

A. Motion changes the state, not necessarily the rule governing the state.

C. An input acts on the system; the state describes its present condition. They need not be the same.

D. A nonlinear rule, such as one containing sin θ, can still have no explicit time dependence.

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Question 06Foundations

One aircraft has four independently commanded motors and several measured attitude outputs. The chosen model is:

Choose an answer to question 6
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Answer · A

MIMO.

Why this works

Count signals, not objects. Four independent motor commands are four input channels. Several measured attitude quantities are several output channels. That makes the chosen model multiple-input, multiple-output: MIMO.

The same aircraft could also be studied through a simpler single-input, single-output channel, if we deliberately selected only one input and one output. The classification belongs to the chosen model, not permanently to the vehicle.

Check the other choices

B. SISO counts input and output channels, not the number of physical vehicles.

C. Motion does not prevent classification by the number of inputs and outputs.

D. The number of input and output channels does not establish linearity.

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Question 07Foundations

A pendulum is recorded by a digital logger. Which statement is correct?

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Answer · D

The physical motion is continuous-time; the stored samples are discrete-time.

Why this works

Think of filming the pendulum one frame at a time. The images are separate observations, but the pendulum does not freeze between frames. It keeps moving whether or not the logger is taking a measurement.

We can describe the physical angle as θ(t) and regularly sampled data as θ[k] = θ(kTs). Here Ts is the sampling interval. The sequence represents selected observations of the continuous motion.

Check the other choices

A. The pendulum also moves between the times at which the logger records it.

B. Recording samples changes the data representation, not the physical pendulum's motion.

C. A digital sample record contains values at selected instants, not a value at every time.

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Question 08Foundations

In feedback control, what is the reference?

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Answer · C

The desired value or behavior.

Why this works

Suppose you ask a motor to reach a particular angle. That requested angle is the reference. The sensor reports the actual angle, which may be different while the motor is moving toward the target.

In a direct comparison, the error is e = r − y: desired angle minus measured angle. A controller uses this difference to choose an input. Keep the target, the observation, and their difference as three separate signals.

Check the other choices

A. The measured output is what happens; it need not equal what was requested.

B. The error is formed by comparing the reference with the returned measurement.

D. The state describes the system's present condition, not necessarily its target.

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Question 09Laplace transform

Using a Laplace transform can help us:

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Answer · A

Rewrite a linear differential equation as algebra while keeping initial-condition terms.

Why this works

For the linear differential equations with constant coefficients used in this course, a derivative becomes an expression such as sY(s) − y(0+). This lets us rearrange and solve an algebraic equation in s instead of directly solving the time-domain differential equation.

The physical system has not changed. Its starting condition remains in the calculation. After solving for the output transform, an inverse Laplace transform recovers the time-domain response. Linearizing a nonlinear model is a separate step.

Check the other choices

B. Laplace transformation is not a linearization method; nonlinear physical terms do not disappear.

C. The initial state still affects the response and appears in the derivative-transform terms.

D. The transform is defined by an integral, not by renaming the variable t as s.

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Question 10Laplace transform

What does an inverse Laplace transform recover from F(s)?

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Answer · B

The time-domain signal.

Why this works

The forward transform takes a time signal f(t) to its transformed representation F(s). The inverse goes back: it tells us which time signal corresponds to the expression we have found in s.

For example, the table pairs 1 divided by s + 2 with e−2t for t ≥ 0. Reading that row backward recovers a signal that decreases over time; it does not create a new plant or choose a new input.

Check the other choices

A. Finding an equilibrium input is a separate modeling calculation, not the definition of the inverse transform.

C. Sensor placement is a physical design choice, not something an inverse transform supplies.

D. Changing representations does not replace the physical system.

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Question 11Laplace transform

Which description matches a unit step signal?

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Answer · C

It switches on to 1 and stays there.

Why this works

The standard unit step us(t) is zero before t = 0 and one from t = 0 onward. It models a command that is suddenly switched on and then held constant.

A motor-speed command can be a step even though the motor's actual speed takes time to change. The step describes the requested input signal; the motor dynamics determine the resulting output.

Check the other choices

A. That describes an ideal unit impulse, not a step that remains on.

B. A steadily increasing straight-line signal is a ramp.

D. A sinusoid oscillates; a unit step stays at its new level.

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Question 12Laplace transform

When a rectangular pulse becomes very narrow to model a unit impulse, what must be preserved?

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Answer · D

An area of 1.

Why this works

For a rectangular pulse, area equals height times width. A unit-impulse approximation preserves area 1. Halving the width therefore requires doubling the height: a shorter pulse is not a weaker pulse if its area is unchanged.

For a force pulse, the area measures the delivered impulse, or change in momentum. This is why a very brief hit can still have a finite effect. The ideal impulse is a limiting model with unit area, not an ordinary signal with a finite peak of one.

Check the other choices

A. Keeping height 1 while shrinking the width would make the area approach zero.

B. The width must shrink in the impulse approximation, not remain fixed.

C. A delay locates the pulse in time; it does not determine its unit strength.

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Question 13Laplace transform

For an ordinary signal, why should the term f(0+) not automatically be removed from sF(s) − f(0+)?

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Answer · B

It records the actual initial condition.

Why this works

For the ordinary derivative rule used here, ℒ{ḟ} = sF(s) − f(0+). The notation 0+ means immediately after the starting time. The subtraction carries that actual starting value into the transformed equation.

If the starting value is 2, the result is sF(s) − 2, not simply sF(s). Dropping the term would silently replace the stated initial condition with zero and can produce a different predicted response.

Check the other choices

A. The starting value and a possible eventual value are different pieces of information.

C. An initial-condition term describes the starting state and does not by itself make the governing equation nonlinear.

D. A time delay appears as an exponential factor in s, not as this initial-value subtraction.

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Question 14Laplace transform

Sine and cosine transforms can have the same denominator. What still needs to be checked when using a transform table?

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Answer · A

The numerator and any scale factor.

Why this works

For the same frequency ω, the sine row has ω divided by s2 + ω2, while the cosine row has s divided by s2 + ω2. The numerator distinguishes the two signals even though the denominators agree.

Match the scale factor too. For example, 1 divided by s2 + 4 corresponds to ½ sin(2t), because the unscaled sine row has numerator 2. Looking only at the denominator would miss the factor of one half.

Check the other choices

B. A shared quadratic denominator is not enough to distinguish sine from cosine or set the amplitude.

C. The transform pair depends on the signal's mathematical form, not its chosen label.

D. Their denominators may match, but their numerators are different.

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Question 15Laplace transform

Which transform-domain expression represents delaying a causal signal by 2 seconds?

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Answer · D

e−2sF(s)

Why this works

Delaying a causal waveform by 2 seconds gives f(t − 2)us(t − 2). It is zero before 2 seconds, then follows the original waveform shifted to the right. Its transform is e−2sF(s).

Keep a time delay separate from an exponential change in amplitude. F(s + 2) corresponds to multiplying the original time signal by e−2t. That changes its envelope rather than postponing the waveform.

Check the other choices

A. Shifting s by 2 corresponds to exponential weighting by e⁻²ᵗ, not a delay.

B. This doubles the signal amplitude without moving its starting time.

C. This halves the signal amplitude without delaying it.

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Question 16Modeling & linearization

Two pendulums have the same angle but different angular velocities. Under the same future torque, their future motion:

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Answer · C

Can differ because their states differ.

Why this works

Picture two pendulums at the same angle: one is moving left and the other is moving right. Even before a new torque has much effect, they move to different positions because their angular velocities differ.

The pendulum state includes both θ and θ̇. Giving the same future input does not erase a difference in the starting state. A snapshot of angle alone therefore cannot determine the future motion.

Check the other choices

A. Matching angles alone does not match the complete mechanical states.

B. A suitable model can predict the motion when both initial angles and angular velocities are supplied.

D. Neither a shared angle nor a shared torque forces the acceleration to stay zero.

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Question 17Modeling & linearization

Which condition defines an equilibrium under a constant input?

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Answer · A

All state rates are zero.

Why this works

An equilibrium is a state that remains constant under its corresponding constant input. In the model ẋ = f(x, u), this means f(x*, u*) = 0: every component of the state derivative must be zero.

For a pendulum, both angular velocity and angular acceleration must be zero. Its angle need not be zero, and an applied torque may be needed to hold that angle. Check the whole state, not just whether one acceleration vanishes.

Check the other choices

B. A constant nonzero position can be an equilibrium; it is the rates that must vanish.

C. A nonzero constant input may be needed to balance other forces or torques.

D. A nonzero velocity can still change position even when acceleration is zero.

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Question 18Modeling & linearization

Can a constant nonzero torque hold a pendulum at a nonzero equilibrium angle?

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Answer · B

Yes, when the torque balances gravity and angular velocity is zero.

Why this works

Gravity tends to rotate a pendulum away from some nonzero angles. An applied torque can oppose and exactly balance that gravitational torque. Then the net torque is zero, so the angular acceleration is zero.

The pendulum must also have zero angular velocity. Otherwise its angle keeps changing despite the instant of balanced torque. A nonzero equilibrium angle and nonzero applied torque are therefore possible, but both state rates must vanish.

Check the other choices

A. The applied torque may be nonzero while the net torque is zero.

C. The applied torque must have the appropriate magnitude and sign to balance gravity at the chosen angle.

D. Equilibrium means a constant state, not necessarily a zero state.

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Question 19Modeling & linearization

A pendulum passes through the bottom with nonzero angular velocity. Its angular acceleration is zero at that instant. It is:

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Answer · D

Not at equilibrium because its angle is still changing.

Why this works

The pendulum state is angle and angular velocity. Its first state rate is θ̇. Because that rate is nonzero, the angle changes immediately after the pendulum passes through the bottom, so the state is not an equilibrium.

Here, the zero acceleration refers to angular acceleration: angular velocity is not changing at that instant, but it is not zero. The pendulum is still moving through the bottom. Equilibrium requires every state rate to vanish, not just the angular acceleration.

Check the other choices

A. Zero angular acceleration alone is insufficient when the angle is still changing.

B. Being at the bottom is a position, not a guarantee that the pendulum will stay there.

C. An instant of zero angular acceleration does not remove the system's motion or dynamics.

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Question 20Modeling & linearization

A local linear approximation is most appropriate for:

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Answer · C

Sufficiently small changes near the selected operating point.

Why this works

A curved force-versus-displacement graph can look almost straight over a small neighborhood. Linearization keeps the local slope at the chosen operating point. It predicts nearby changes without claiming that the entire curve is straight.

For example, a car suspension is already compressed by the car's weight. Small additional movements can be described using the local stiffness there. Large compression may move into a region with a different slope, where that same approximation becomes inaccurate.

Check the other choices

A. A local approximation need not remain accurate far from its operating point.

B. We can approximate small deviations around a nonzero operating point too.

D. Linearization changes the mathematical approximation, not the hardware.

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Question 21Modeling & linearization

For θ = π + δθ, what does δθ = 0 mean?

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Answer · A

θ = π.

Why this works

Substitute directly into the definition: θ = π + 0 = π. The deviation measures a change from the chosen operating angle. It does not use zero absolute angle as its reference in this example.

Think of setting a ruler's zero at the upright position. Reading zero on that ruler means no displacement from upright. It does not mean that the pendulum has moved to the hanging position, and it does not specify the future input.

Check the other choices

B. Zero deviation is not zero absolute angle when the operating angle is π.

C. The value of an angle deviation does not specify all future input commands.

D. A zero deviation at one instant does not make a local approximation globally exact.

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Question 22Modeling & linearization

If D = 0 in an angle-output model, which statement is correct?

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Answer · B

The input may affect the output through the state dynamics.

Why this works

The output equation is δy = Cδx + Dδu. Setting D = 0 removes only Dδu, the direct input contribution. The state contribution Cδx remains.

A torque can change a pendulum's angular acceleration, then its angular velocity and angle. The input therefore affects the angle through the evolving state, even though the output equation has no separate direct torque term. B describes the input's effect on the state rate; D describes direct feedthrough to the output.

Check the other choices

A. The input can still change the state, which then changes the measured angle.

C. B and D describe different paths; D = 0 does not require B = 0.

D. With D = 0, the output equation still contains the state term Cδx.

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Question 23Block diagrams

What connection is shown below?

Signal connections for question 23U enters G1. The output of G1 enters G2. The output of G2 is Y.UG₁G₂Y
Original signal connections from the practice sheet.
Choose an answer to question 23
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Answer · D

Series.

Why this works

Follow the arrows from left to right. The input passes through the first block, and that block's output becomes the input to the second. There is one forward path through both blocks, with no branch or returning signal.

If the first block gives V = G1U and the second gives Y = G2V, substituting yields Y = G2G1U. This is why series blocks multiply.

Check the other choices

A. Parallel blocks would receive a shared input on separate paths and have their outputs added.

B. A negative-feedback loop would return a downstream signal to a subtracting summing point.

C. A positive-feedback loop would return a downstream signal to an adding summing point.

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Question 24Block diagrams

In the diagram below, what signal comes back through the feedback path to the summing point?

Signal connections for question 24R enters the plus side of a summing point. E enters G, whose output is Y. A takeoff from Y returns through H to the minus side of the summing point.RΣ+−EGYH
Original signal connections from the practice sheet.
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Answer · C

HY

Why this works

Start at the output Y and follow the return arrow. It passes through the block H before reaching the summing point. By the block rule, the signal leaving that block is HY.

The minus sign belongs to the comparison at the summing point: E = R − HY. The signal arriving through the return path is HY; the summing point subtracts it from the reference. Reading these two steps separately avoids confusing the returned signal with the error.

Check the other choices

A. A block multiplies its input by H; it does not divide by H.

B. The feedback path starts from output Y and passes through H, not directly from R through G.

D. R is the reference entering the other side of the summing point.

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Fill in the blank

Questions 25–40 · Complete both parts. Use the formulas provided.

Question 25Foundations

Enter SISO or MIMO for each chosen model.

Compare your answers with the explanation below. Equivalent expressions are fine.

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Answer (a)

SISO

There is one input channel and one output channel: single-input, single-output (SISO).

Answer (b)

MIMO

Four independent commands are multiple inputs, and several measured attitudes are multiple outputs: multiple-input, multiple-output (MIMO).

Why this works

SISO and MIMO describe the input and output signals chosen for a model. Count the independent commands and the measured outputs.

Do not count physical machines instead. One aircraft can have multiple input and output channels, while one chosen voltage-to-angle channel can be modeled as SISO.

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Question 26Laplace transform

Use the supplied transform pairs.

Given
us(t) ↔ 1 divided by s, e−at ↔ 1 divided by s + a

Compare your answers with the explanation below. Equivalent expressions are fine.

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Answer (a)

2 divided by s

L{us(t)} = 1 divided by s. Multiplying the time signal by 2 multiplies its transform by 2, so 2 × 1 divided by s = 2 divided by s.

Answer (b)

1 divided by s + 5

Match e−5t to e−at: here a = 5. Substituting into 1 divided by s + a gives 1 divided by s + 5.

Why this works

A transform pair connects a time-domain signal on the left to its Laplace transform on the right. Match the signal's form before changing any numbers.

The first blank uses a constant scale factor. The second uses the exponential parameter. These are separate operations; neither requires solving a differential equation.

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Question 27Laplace transform

For t ≥ 0, use the supplied pairs in reverse.

Given
us(t) ↔ 1 divided by s, e−at ↔ 1 divided by s + a

Compare your answers with the explanation below. Equivalent expressions are fine.

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Answer (a)

us(t), or 1 for t ≥ 0

The pair us(t) ↔ 1 divided by s says directly that the inverse is the unit step. On the stated time interval, its value is 1.

Answer (b)

e−2t

Compare s + 2 with s + a, giving a = 2. The matching time signal is therefore e−2t for t ≥ 0.

Why this works

An inverse Laplace transform recovers the time-domain signal. A table works in either direction: find the supplied transform on the right, then read its partner on the left.

Both expressions already match complete table entries. No partial-fraction decomposition or additional algebra is needed, and the stated causal-signal convention remains in force.

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Question 28Laplace transform

There is no impulse at the starting time. Apply the supplied derivative rule.

Given
L{ḟ} = sF(s) − f(0+)

Compare your answers with the explanation below. Equivalent expressions are fine.

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Answer (a)

sF(s)

Insert the given initial value: sF(s) − f(0+) = sF(s) − 0 = sF(s).

Answer (b)

sF(s) − 3

This is a separate initial-value case. Substitute 3 in the same rule: sF(s) − f(0+) = sF(s) − 3.

Why this works

Transforming a derivative introduces the actual starting value of the signal. The transform changes the mathematical representation; it does not reset the system's initial condition.

The initial term disappears only in the first case because its given value is zero. The no-impulse condition lets us use the supplied ordinary derivative rule at the starting time.

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Question 29Laplace transform

Write L{sin(2t)} = a divided by s2 + b. Use the sine pair below.

Given
L{sin(ωt)} = ω divided by s2 + ω2

Compare your answers with the explanation below. Equivalent expressions are fine.

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Answer (a)

2

Match sin(2t) with sin(ωt), so ω = 2. The numerator is ω, hence a = 2.

Answer (b)

4

The constant term in the denominator is ω2, not ω. Thus b = 22 = 4.

Why this works

The number multiplying time inside the sine is the angular frequency. Here that number is 2.

Use that same value in both parts of the supplied pair: L{sin(2t)} = 2 divided by s2 + 4. Matching only the denominator would miss the numerator scale factor.

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Question 30Modeling & linearization

At an operating point, a spring force is 10 N and its local stiffness is 2 N/cm. Use the additional compression of 0.1 cm.

Given
ΔF ≈ klocalΔx, F ≈ F(x*) + ΔF

Compare your answers with the explanation below. Equivalent expressions are fine.

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Answer (a)

0.2 N

Use the local slope with the change in compression: ΔF ≈ (2 N/cm)(0.1 cm) = 0.2 N. The centimeter units cancel.

Answer (b)

10.2 N

Add the predicted change to the existing force: F ≈ F(x*) + ΔF = 10 N + 0.2 N = 10.2 N.

Why this works

The spring already carries 10 N at its operating point. Local stiffness predicts how much that force changes when the compression changes slightly.

The first blank asks only for the extra force; the second includes the baseline. This is a local prediction near the operating point, not a claim that the same stiffness describes every possible compression.

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Question 31Modeling & linearization

The operating position is x* = 5 m. Use δx = x − x*.

Compare your answers with the explanation below. Equivalent expressions are fine.

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Answer (a)

0.2 m

Subtract the operating value from the actual position: δx = 5.2 m − 5 m = 0.2 m.

Answer (b)

4.9 m

Rearrange the definition to x = x* + δx. Then x = 5 m + (−0.1 m) = 4.9 m.

Why this works

A deviation measures the difference from the selected operating value. A positive deviation is above that value; a negative deviation is below it.

Keep the absolute position and the deviation separate. Subtract the baseline to find a deviation, or add the deviation to the baseline to recover the position.

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Question 32Block diagrams

Treat these as two separate signal operations.

Compare your answers with the explanation below. Equivalent expressions are fine.

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Answer (a)

4

A branch point copies the signal, so each branch carries 4. It does not divide 4 between the two outgoing paths.

Answer (b)

7

Follow the signs at the summing point: 9 + (−2) = 9 − 2 = 7.

Why this works

A branch and a summing point perform different jobs. A branch makes the same signal available to more than one path; a summing point combines incoming signals with their indicated signs.

The two blanks describe independent operations. Do not feed the answer from the branch question into the summing-point question.

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Question 33Block diagrams

Two scalar gains are G1 = 3 and G2 = 4. Use the supplied rules.

Given
Gseries = G2G1, Gparallel = G1 + G2

Compare your answers with the explanation below. Equivalent expressions are fine.

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Answer (a)

12

The first block multiplies the input by 3 and the next multiplies that result by 4. Thus Gseries = 4 × 3 = 12.

Answer (b)

7

Both blocks receive the same input. Adding their outputs gives 3U + 4U = (3 + 4)U, so Gparallel = 7.

Why this works

In series, a signal passes through the blocks one after another. In parallel, the same input reaches both blocks and their outputs are then combined.

The given rules represent these different connections. Use multiplication for this series path and addition for the stated parallel sum; the same gain values do not imply the same overall gain.

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Question 34Modeling & linearization

The pendulum states are x1 = θ and x2 = θ̇. Use the word bank: angular velocity | angular acceleration.

Compare your answers with the explanation below. Equivalent expressions are fine.

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Answer (a)

angular velocity

Differentiate the first state: ẋ1 = θ̇ = x2. The rate of change of angle is angular velocity.

Answer (b)

angular acceleration

Differentiate the second state: ẋ2 = θ̈. The rate of change of angular velocity is angular acceleration.

Why this works

A dot denotes a time derivative. Read each state definition first, then ask what the rate of that physical quantity represents.

The two state equations describe successive rates: angle changes at angular velocity, and angular velocity changes at angular acceleration. A state and its rate are not the same quantity.

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Question 35Modeling & linearization

Complete the equilibrium statements.

Compare your answers with the explanation below. Equivalent expressions are fine.

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Answer (a)

0

The state equation is ẋ = f(x, u). At equilibrium every state rate is zero, so evaluating at (x*, u*) gives f(x*, u*) = 0.

Answer (b)

constant

Starting at the exact equilibrium and keeping the matching input leaves every state derivative zero. The state therefore remains at its equilibrium value.

Why this works

An equilibrium is a constant state under a corresponding constant input. The equilibrium test applies to every component of the state rate.

The state values themselves do not have to be zero. Also, this statement concerns the exact equilibrium; it does not say what happens after a disturbance away from it.

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Question 36Modeling & linearization

In δẋ = Aδx + Bδu, use the word bank: state | input | output.

Compare your answers with the explanation below. Equivalent expressions are fine.

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Answer (a)

state

A multiplies δx, the state deviation. The term Aδx describes how the current state deviation contributes to the state rate.

Answer (b)

input

B multiplies δu, the input deviation. The term Bδu describes how a change in the applied input contributes to the state rate.

Why this works

Read each matrix together with the variable beside it. The equation tells us how the local state changes, with one contribution from the current state and another from the input.

The delta symbols indicate deviations from the selected operating point. This is a state-rate equation; a separate output equation describes what is measured.

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Question 37Modeling & linearization

Let δx1 be angle deviation and δx2 be angular-velocity deviation. With D = 0, δy = Cδx.

Compare your answers with the explanation below. Equivalent expressions are fine.

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Answer (a)

δx1 (angle deviation)

Multiply the row by the state components: δy = 1 · δx1 + 0 · δx2 = δx1. This selects the angle deviation.

Answer (b)

δx2 (angular-velocity deviation)

Here δy = 0 · δx1 + 1 · δx2 = δx2. This selects the angular-velocity deviation.

Why this works

The output matrix tells us which state quantities are measured. A coefficient of 1 keeps a component, while a coefficient of 0 removes its contribution to this output.

These are two separate choices of measurement. Because the variables are deviations, the selected output is a deviation too. Either the state symbol or its stated physical meaning is an acceptable answer.

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Question 38Modeling & linearization

Complete the first-order linearization statements.

Compare your answers with the explanation below. Equivalent expressions are fine.

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Answer (a)

2δx

2δx is proportional to the first power of the deviation. The term (δx)2 is second order, so it is omitted in a first-order linearization.

Answer (b)

small

Close to the operating point, higher-order deviation terms can be neglected locally. For large deviations, those omitted terms may matter, so the same approximation is not generally reliable.

Why this works

First-order linearization keeps the terms proportional to small deviations and drops higher-order products and powers. The order refers to the approximation, not the number of states.

Keeping fewer terms gives a simpler model near the operating point. It does not make the original nonlinear system globally linear or guarantee an accurate prediction far from that point.

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Question 39Block diagrams

Let R be the reference and B the returned measurement signal. Here B is a signal, not a matrix. Enter + or -.

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Answer (a)

-

Negative feedback subtracts the returned measurement from the reference: E = R − B. The required sign is minus.

Answer (b)

+

Positive feedback adds the returned measurement to the reference: E = R + B. The required sign is plus.

Why this works

The feedback sign tells us how the returned signal enters the summing point. Write that signal relationship before considering any block-reduction formula.

In this question B names a returned signal. It is not the input matrix from a state-space model, so no matrix multiplication or linearization is needed.

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Question 40Block diagrams

Choose an operation from: multiply | add | divide.

Compare your answers with the explanation below. Equivalent expressions are fine.

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Answer (a)

multiply

If V = G1U and Y = G2V, substitute the first relation into the second: Y = G2G1U. The equivalent gain is the product.

Answer (b)

add

The same input goes through both branches. Adding the outputs gives Y = G1U + G2U = (G1 + G2)U, so add the gains.

Why this works

Choose the operation from the way the blocks are connected. Series applies one gain after another; parallel applies the gains to the same input on separate paths.

The parallel statement explicitly says both outputs are added. A different sign at the summing point would change that combination, so always read the signs as well as the block values.

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Additional practice

Questions 41–43 · Three short variations on the same course concepts. These are web-only additions, not part of the original PDF.

Question 41Block diagrams

With zero initial conditions, we replace a connected group of blocks by one equivalent block. What must remain unchanged?

Choose an answer to question 41
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Answer · B

The overall output produced by each external input.

Why this works

Think of comparing the two diagrams from outside. Apply the same input to each, with zero initial conditions. If the replacement is equivalent, both predict the same output, not just for one test input but for every input allowed by the model.

For example, two scalar gains 2 and 3 in series produce Y = 3(2U) = 6U. One block with gain 6 gives the same result. The intermediate signal 2U no longer needs to appear in the reduced drawing.

Check the other choices

A. An equivalent block can hide internal signals. It must preserve the relationship at the chosen external input and output.

C. Reduction deliberately changes the number of blocks; equivalence concerns their overall behavior.

D. A different drawing can describe exactly the same external input-output relationship.

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Question 42Block diagrams

Which statement correctly describes this diagram?

Signal connections for question 42U splits at a takeoff point into two paths. The upper path passes through gain 5 into the plus input of a summing point. The lower path passes through gain 2 into the minus input. The sum is Y. No path returns from Y to an earlier point.U52Σ+−Y
Follow each arrow and read the signs at the summing point.
Choose an answer to question 42
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Answer · C

The blocks are in parallel and the lower output is subtracted, giving gain 3.

Why this works

Follow the input U into both branches. The upper block produces 5U, and the lower block produces 2U. Read the signs where the paths meet: Y = 5U − 2U = 3U. The equivalent gain is 3.

The minus sign means subtraction at that summing point. It does not, by itself, mean feedback. Feedback requires a signal to return from a later point in the system to an earlier comparison point; this diagram has no such return path.

Check the other choices

A. The output of one block does not enter the other. Both blocks receive the same input, so their gains are not multiplied.

B. The paths are parallel, but the lower output enters with a minus sign, not a plus sign.

D. A feedback loop needs a returning signal. Here both paths travel from the input toward the output; neither returns from downstream.

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Question 43Modeling & linearization

For ẋ = f(x, u) = −x2 + u, the equilibrium is x* = 2, u* = 4. Which procedure correctly finds A?

Given
Use A = ∂f divided by ∂x|*. Differentiate with respect to x, holding u fixed. Reminder: ∂(x2) divided by ∂x = 2x.
Choose an answer to question 43
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Answer · B

First find ∂f/∂x = −2x; then substitute x = 2, giving A = −4.

Why this works

First differentiate the model while keeping x as a variable: ∂f/∂x = −2x. Only then evaluate at x* = 2: A = −2(2) = −4.

The fact that f(2, 4) = 0 tells us the state does not move at the exact equilibrium. It does not say the rate stays zero when the state changes slightly. The coefficient A describes that nearby change, which is why we must keep the variable until after differentiation.

Check the other choices

A. This gives the slope of the constant 0, not the slope of f near the equilibrium. Substituting first removes the x-dependence we need.

C. The derivative with respect to u gives B, the input coefficient. A uses the derivative with respect to x.

D. The equilibrium value tells us where to evaluate the derivative. It is not itself the derivative.

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Questions 01–40 follow the original practice PDF in order; question 19 specifies angular acceleration to clarify the original wording. Their explanations expand the supplied practice answer key. Questions 41–43 are additional web-only practice. This is a study resource, not a separate announcement of exam scope or exam rules.

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