Answers and explanations

The present state gives the starting angle and angular velocity. The future input torque tells us how the pendulum will be driven. With this information and the model, we can predict its motion.

The voltage is applied to the motor, so it is the input. The measured shaft angle is the output. A desired angle would be a reference: what we want the output to become.

Multiplication by a fixed constant obeys superposition: scaling or adding inputs scales or adds their outputs in the same way. The square and sine do not obey this rule. The added constant in y = 2u + 1 also breaks linearity: even a zero input gives a nonzero output.

Time-invariant means that the governing rule does not change with time. The angle and angular velocity can still change: a changing state is not the same as a changing model. Time invariance does not imply linearity.

For the linear constant-coefficient equations studied in class, the Laplace transform changes derivatives into algebraic expressions in s. Initial-condition terms remain unless they are zero. This makes the equation easier to solve without removing the system's dynamics.

The factor e^−3s makes the causal waveform start 3 seconds later, without changing its shape or amplitude. Before that delay has passed, the delayed signal is zero.

A unit impulse represents a unit total area, not a unit height. For a rectangle, area = height × width. If the width is halved, the height must double to keep that area equal to 1.

The pendulum state includes both angle and angular velocity. At the same angle, one pendulum could be moving left while the other moves right. Matching the angle alone does not give the same starting state.

At equilibrium, every state derivative must be zero. Here the angular velocity is nonzero, so the angle is changing even though the angular acceleration is zero at that instant. Passing through an equilibrium angle is not the same as being in an equilibrium state.

Linearization approximates the nonlinear model near the selected operating point. For small motions about the hanging equilibrium, sin θ is close to θ. For large swings or complete rotations, that approximation is no longer reliable.

First-order linearization keeps terms proportional to the small deviation. The term (δx)² is second order and is dropped. As δx becomes small, its square becomes smaller relative to the first-order term.

In δy = Cδx + Dδu, setting D = 0 removes the direct input term from the output equation. The torque can still affect the state through B in the state equation. As the state changes, the measured angle can change too.

The output Y passes through the feedback block H, so the returned signal is HY. It enters the minus side of the summing point, while R enters the plus side. The error is therefore the reference minus the returned measurement.

An equivalent block must produce the same external output for the same input under the stated zero initial conditions. Its internal drawing may have fewer blocks and arrows. The connection's behavior, not its appearance, is what must be preserved.

Both blocks receive the same input U, so they are in parallel. The summing point adds the upper output and subtracts the lower output: Y = G₁U − G₂U. A minus sign alone does not make a feedback loop; no output is returned upstream here.

SISO means single input, single output: one voltage channel and one measured-angle channel. The digital record contains values at separate sample times, so it is discrete-time. The physical pendulum can still move continuously between those samples.

For (a), the factor 4 stays outside the transform: multiply the unit-step result 1/s by 4. For (b), compare e^−3t with the supplied causal exponential e^−at and set a = 3. The denominator becomes s + 3, not s − 3.

Substitute the stated initial value into the supplied rule. In (a), subtracting zero leaves sY(s); in (b), the initial-condition term is −2. A derivative becomes just sY(s) only when the relevant initial value is zero.

The symbol δx is the state deviation, and δu is the input deviation. Read each matrix together with the quantity beside it: Aδx describes the state contribution, and Bδu describes the input contribution to the state rate.

The model is ẋ = f(x, u), so f(x*, u*) = 0 means that every state rate is zero at the equilibrium. If the state starts exactly there and the input stays at u*, the state remains constant. For a vector state, 0 means the zero vector; “unchanging” means the same as “constant.”

The deviation measures how far the height is from the 1 m operating point. In (a), 1.1 − 1 = 0.1 m. In (b), zero deviation means the drone is at the operating height of 1 m, not at an absolute height of zero.

A branch point copies a signal; it does not divide its value between the branches. Each branch therefore carries 6. The separate summing point combines its signed inputs: 8 − 5 = 3.

In series, the second gain acts on the first block's output, so the gains multiply: 3 × 2 = 6. In additive parallel paths, both gains receive the same input and their outputs are added, so the gains add: 2 + 3 = 5.

The row [1 0] gives δy = 1·δx₁ + 0·δx₂ = δx₁. The row [0 1] gives δy = 0·δx₁ + 1·δx₂ = δx₂. Either the state symbol or its corresponding physical quantity identifies the same answer.

Negative feedback subtracts the returned measurement from the reference: E = R − B. Positive feedback adds the returned measurement: E = R + B. Read the sign at the summing point, rather than guessing from the position of the return arrow.