Prior-Knowledge Diagnostic

Why B is correct

For an ideal viscous damper, the force is proportional to the relative velocity across the damper: F_d = −c(v_mass − v_wall), where c > 0. The minus sign is the important physical statement: the damper resists relative motion.

Here the wall is fixed, so v_wall = 0 and F_d = −c ẋ. If the mass moves to the right, the damping force points left; if the mass moves to the left, the damping force points right. A passive damper removes mechanical energy because F_d ẋ = −c ẋ² ≤ 0.

Why A is correct

Hooke's law is F_s = −kΔx, where k > 0 is the spring stiffness and Δx is deformation relative to the chosen reference. For an unpreloaded spring that reference is its natural length; after a model is shifted about a loaded equilibrium, Δx may instead denote displacement from that equilibrium. The magnitude k|Δx| grows linearly with deformation.

The minus sign makes the force restoring: stretching the spring produces a force toward the undeformed position, and compressing it produces a force in the opposite direction. The spring can store energy, V = ½k(Δx)², and return it later; unlike an ideal damper, it does not dissipate that energy.

Why C is correct

The free response is the motion generated by the system's own stored energy when the external input is set to zero. In state-space language, ẋ = Ax with x(0) ≠ 0; in a mechanical system, the stored energy may come from an initial displacement or velocity.

Its time dependence is determined by the system's natural modes, which are connected to eigenvalues or poles. By contrast, the forced response is caused by a nonzero input. A complete response can contain both pieces when initial conditions and an external input are both present.

Why D is correct

Let S{u} denote the output produced by input u under the same initial-condition convention. A linear system satisfies S{a u₁ + b u₂} = aS{u₁} + bS{u₂}. This combines additivity and homogeneity.

Choice D states additivity: S{u₁ + u₂} = S{u₁} + S{u₂}. It is therefore the best answer among the choices. The stronger complete test of linearity also includes scaling, S{a u} = aS{u}.

Why A is correct

The unilateral Laplace transform changes differentiation into algebraic multiplication while keeping initial conditions explicit. For example, L{ẋ} = sX(s) − x(0⁻), and L{ẍ} = s²X(s) − sx(0⁻) − ẋ(0⁻). A differential equation can therefore be rearranged as an algebraic equation in s.

This makes it easier to solve for an input-output relation, combine subsystems, identify poles and zeros, and recover the time response with an inverse transform. The transform is a representation tool: it simplifies analysis, but it does not replace modeling or guarantee good behavior.

Why B is correct

For a continuous-time LTI model, the transfer function is G(s) = Y(s)/U(s) evaluated with all initial conditions set to zero. This definition separates the input-driven, or zero-state, response from the free response caused by stored initial energy.

If initial conditions are nonzero, the transformed differential equation contains additional terms such as x(0) and ẋ(0). Those terms must be included when computing the total response, but they are not placed inside G(s).

Why C is correct

A transfer function specifies how one chosen input affects one chosen output in an LTI model when initial conditions are zero. In the transform domain, Y(s) = G(s)U(s). The same physical device can have different transfer functions for different input-output selections.

Because it is an external input-output description, a transfer function need not reveal every internal state or mode. A state-space model retains internal coordinates explicitly and can represent multiple inputs and outputs more naturally.

Why A is correct

If the first block produces V(s) = G₁(s)U(s) and the second produces Y(s) = G₂(s)V(s), substitution gives Y(s) = G₂(s)G₁(s)U(s). Therefore G_total(s) = G₂(s)G₁(s).

For scalar SISO transfer functions, multiplication commutes, so G₂G₁ = G₁G₂ and choice A is correct. This simple multiplication assumes the connection has no additional summing junction, loading effect, or feedback path.

Why D is correct

For G(s) = N(s)/D(s), poles are values p for which the reduced denominator satisfies D(p) = 0. These values generate natural terms such as e^{pt} in the impulse or transient response when the corresponding residues are nonzero.

Pole location therefore carries dynamic information: negative real parts produce decaying modes, positive real parts produce growing modes, and imaginary parts produce oscillation. Numerator roots are zeros, which shape how inputs are transmitted and can suppress particular responses.

Why B is correct

A continuous-time pole p = σ + jω contributes a mode proportional to e^{pt} = e^{σt}e^{jωt}. Its real part σ sets the exponential envelope. When σ < 0, the envelope decays; when σ > 0, it grows; when σ = 0, it does not decay.

Asymptotic stability requires every natural mode to vanish as t → ∞. Therefore every relevant pole must lie strictly in the open left-half plane, Re(p) < 0.

Why C is correct

Write the pole pair as p = σ ± jω with σ < 0. Combining the conjugate modal terms gives a real response of the form e^{σt}[C₁ cos(ωt) + C₂ sin(ωt)].

The sine and cosine create oscillation at angular frequency |ω|, while e^{σt} is a decaying envelope because σ is negative. The result is an underdamped-looking oscillation whose amplitude shrinks over time.

Why A is correct

A stable real pole p = −a contributes e^{−at}. Smaller a means slower decay and a larger time constant 1/a. For complex poles, the same comparison is made using their real parts.

Therefore the pole nearest the imaginary axis—while still in the left-half plane—usually remains visible longest and dominates the late transient. Poles farther left decay more quickly.

Why D is correct

A response is often interpreted as y(t) = y_transient(t) + y_steady-state(t). The transient contains natural-mode terms created by initial conditions and by the adjustment from the initial state to the input-driven behavior.

For a stable system, those modal terms decay, so y_transient(t) → 0 as t → ∞. What remains is the steady-state response associated with the continuing input—for example, a constant under a step input or a sinusoid under sinusoidal forcing.

Why B is correct

Complex exponentials are eigenfunctions of LTI systems. If u(t) is represented by e^{jωt}, the steady-state output is G(jω)e^{jωt}. Taking the real part gives a sinusoid at the same angular frequency ω.

The complex number G(jω) changes two things: |G(jω)| scales the amplitude, and ∠G(jω) shifts the phase. The system's transient response may contain other frequencies, but those terms decay in a stable system before steady state is reached.

Why C is correct

The state x(t₀) is the information needed at the present time so that, together with the future input u(t) for t ≥ t₀ and the model, the future state and output can be determined. The past trajectory does not need to be retained once a complete state is known.

For a mass-spring-damper, position and velocity form a natural state because they specify stored configuration and motion. An output may measure only position, while the input may be an applied force. State, input, and output therefore play different roles.