The transfer function of a linear time-invariant system is defined as the ratio of the Laplace transforms of the output and the input:

with all initial conditions zero

with the initial conditions of the output only

The main effect of negative feedback on the sensitivity of the closed-loop gain to changes in the forward-path gain G is that it:

reduces it by the factor 1 + GH

increases it by the factor 1 + GH

The poles of a stable second-order system are complex with negative real parts. Its unit-step response is:

an underdamped oscillation settling to a final value

a monotonic rise with no overshoot

a sustained oscillation of constant amplitude

an exponentially growing oscillation

Which of the following statements about a standard underdamped second-order system are true?

The percentage overshoot depends only on the damping ratio

Increasing the damping ratio at fixed natural frequency reduces the overshoot

The 2% settling time is approximately 4/(ζω_n)

The percentage overshoot depends only on the damping ratio

Increasing the damping ratio at fixed natural frequency reduces the overshoot

The 2% settling time is approximately 4/(ζω_n)

Doubling the natural frequency at fixed damping ratio doubles the overshoot

A unity-feedback loop has G(s) = K/[s(s + 4)]. If K is increased from 4 to 16, then:

the ramp error falls by a factor of 4 and the damping ratio falls from 1 to 0.5

the ramp error is unchanged and the damping ratio falls from 1 to 0.5

the ramp error falls by a factor of 4 and the damping ratio rises

both the ramp error and the damping ratio are unchanged

The type of a control system is determined by:

the number of poles of the open-loop transfer function at the origin

the total number of poles of the closed-loop transfer function

the number of zeros of the open-loop transfer function

the damping ratio of the dominant poles

Control Systems I: Modelling, Transfer Functions, Feedback, Block Diagrams, Signal Flow Graphs, and Transient and Steady-State Response

Section B1.3 of the Robotics and Automation paper is the control theory that closes Part B1, and it is one section in two chapters. This first chapter takes the first half of the paragraph: mathematical modelling and representation of systems, the feedback principle, the transfer function, block diagrams and signal flow graphs, and the transient and steady-state analysis of linear time-invariant systems. The second chapter takes stability by the Routh–Hurwitz and Nyquist criteria, Bode plots, root loci, the lag, lead and lead–lag compensators, and the P, PI and PID controllers. The two are separated because the first asks what a loop does and the second asks whether, and how, it can be made to do it well.

1. Mathematical modelling and the transfer function

A model is the differential equation that relates output to input, and the transfer function is what it becomes under the Laplace transform with zero initial conditions: G(s) = Y(s)/U(s). A mass m on a spring k and damper c, driven by a force F, obeys m ẍ + c ẋ + kx = F, so X(s)/F(s) = 1/(ms² + cs + k): natural frequency ω_n = √(k/m) and damping ratio ζ = c/(2√(km)). An RLC series circuit with output across the capacitor gives 1/(LCs² + RCs + 1), the same form with the electrical analogues, and an armature-controlled DC motor gives a second-order lag from voltage to speed. The roots of the denominator are the poles, the roots of the numerator the zeros, and the denominator set to zero is the characteristic equation; its degree is the order of the system.

Analogies between mechanical and electrical systems (force–voltage)
Mechanical	Electrical
Force F	Voltage v
Mass m	Inductance L
Damper c	Resistance R
Spring stiffness k	Reciprocal capacitance 1/C
Velocity ẋ	Current i

A transfer function describes only the linear, time-invariant part with zero initial conditions, and it hides internal states that do not reach the output. A system is called type N by the number N of poles at the origin in the open-loop transfer function, which becomes central in the steady-state analysis below.

2. The feedback principle

A feedback loop measures the output, compares it with the reference, and drives the plant with the difference. With forward path G and feedback path H, negative feedback gives the closed-loop transfer function T = G/(1 + GH), and 1 + GH = 0 is the closed-loop characteristic equation. The sensitivity of T to a change in G is S = 1/(1 + GH): a large loop gain GH makes the closed-loop gain depend on H, a stable, accurate component, rather than on G, a motor or amplifier that drifts. With G = 50 and H = 0.1 the closed-loop gain is 50/(1 + 5) = 8.33, and it is close to 1/H = 10 for larger G.

Benefits. Lower sensitivity to plant-parameter changes, rejection of disturbances, better steady-state accuracy, and a wider bandwidth for a faster response.
Costs. Lower overall gain by the factor 1 + GH, the risk that the loop oscillates or goes unstable, more components and measurement noise entering the loop.
Positive feedback, 1 − GH in the denominator, raises gain and drives the loop towards instability; it is used deliberately in oscillators and Schmitt triggers.

3. Block diagrams and signal flow graphs

A block diagram reduces by three rules. Blocks in series multiply, G₁G₂; blocks in parallel, summed at a junction, add, G₁ ± G₂; and a feedback loop collapses to G/(1 ± GH) (minus for negative feedback, so the sign of the loop is the opposite of the sign in the denominator). To move a summing point or pick-off point past a block, keep every path’s gain unchanged: moving a pick-off point forward past G means the branch it feeds must be multiplied by G, and moving it back past G divides by G. For a unity negative feedback loop around G(s) = 10/(s + 2), T = 10/(s + 2 + 10) = 10/(s + 12).

A signal flow graph draws the same equations as nodes joined by directed branches with gains, and Mason’s gain formula reads the overall gain without reduction: T = Σ P_k Δ_k / Δ. P_k is the gain of the k-th forward path; Δ = 1 − (sum of all individual loop gains) + (sum of products of gains of every pair of non-touching loops) − (products of triples of non-touching loops) + …; and Δ_k is Δ with every loop that touches the k-th forward path removed. Worked: two forward paths, P₁ = 6, which touches both loops, and P₂ = 2, which touches neither; two non-touching loops of gain −1 each. Δ = 1 − (−1 − 1) + (−1)(−1) = 4; Δ₁ = 1 (both loops touch P₁); Δ₂ = Δ = 4. So T = (6 × 1 + 2 × 4)/4 = 3.5.

⚠️ Δ_k removes touching loops only

Two loops touch when they share a node, and a loop touches a forward path when it shares a node with it. Δ₁ for a path touching every loop is 1, but for a path touching none it is the full Δ. Forgetting the second case, or including a loop’s gain in Δ_k when it touches the path, are the two commonest slips in Mason’s formula.

4. Transient response of first- and second-order systems

A first-order system G = 1/(τs + 1) has the step response 1 − e^−t/τ: 63.2% at t = τ, and the time constant τ is 1/(the pole’s distance from the origin). The 2% settling time is about 4τ and the 5% settling time about 3τ; the 10–90% rise time is 2.2τ. A pole further left is a faster mode.

The standard second-order system is ω_n²/(s² + 2ζω_n s + ω_n²), with poles at −ζω_n ± jω_n√(1 − ζ²). It is undamped for ζ = 0, underdamped for 0 < ζ < 1, critically damped at ζ = 1 and overdamped for ζ > 1. For an underdamped step response, with the damped frequency ω_d = ω_n√(1 − ζ²):

Second-order step-response measures (underdamped)
Measure	Formula
Peak time	t_p = π/ω_d
Percentage overshoot	%M_p = 100 exp(−πζ/√(1 − ζ²))
Settling time (2%)	t_s ≈ 4/(ζω_n)
Rise time	t_r ≈ (π − cos⁻¹ζ)/ω_d, about 1.8/ω_n for ζ ≈ 0.5
Oscillation frequency	ω_d

The overshoot depends on ζ alone: ζ = 0.5 gives exp(−π × 0.5/0.866) = 0.163, that is 16.3%, whatever ω_n. Raising ω_n at fixed ζ makes the response faster with the same shape. For ω_n = 10 and ζ = 0.4 (the system 100/(s² + 8s + 100)), ω_d = 10√0.84 = 9.165 rad/s and t_p = π/9.165 = 0.343 s; for ζ = 0.5 and ω_n = 10, t_s = 4/5 = 0.8 s. Moving the poles vertically changes ω_d, radially changes ω_n, and along a line from the origin holds ζ = cos θ fixed.

5. Steady-state error and system type

For a stable unity-feedback loop with open-loop transfer function G(s), the steady-state error to a standard input follows from the error constants: position K_p = lim_s→0 G(s), velocity K_v = lim_s→0 sG(s) and acceleration K_a = lim_s→0 s²G(s). The type is the number of integrators (poles at s = 0) in G.

Steady-state error of a unity-feedback loop
System type	Unit step	Unit ramp	Unit parabola
Type 0	1/(1 + K_p)	∞	∞
Type 1	0	1/K_v	∞
Type 2	0	0	1/K_a

Worked. G(s) = 10/[s(s + 2)] is type 1 with K_v = lim sG = 10/2 = 5, so the error to a unit ramp is 1/5 = 0.2 and to a step, zero. G(s) = 20/[(s + 1)(s + 4)] is type 0 with K_p = 20/4 = 5, so the error to a unit step is 1/(1 + 5) = 0.167, and to a ramp, unbounded. A ramp of slope 3 into the first system gives 3 × 0.2 = 0.6; the errors scale with the input’s size.

The trade-off is the heart of loop design. For G = K/[s(s + a)] the closed-loop characteristic equation is s² + as + K, so ω_n = √K and ζ = a/(2√K): raising K cuts the ramp error (K_v = K/a) but lowers ζ and raises the overshoot. An integrator improves accuracy and costs stability, which is what the compensators of the next chapter re-balance. The error formulas hold only for a stable closed loop, so stability is checked first.

⚠️ Check stability before the error constants

For G = 5/[s(s − 1)] the limit K_v = lim sG = −5 exists, but the closed loop s² − s + 5 is unstable, and the error formula means nothing. The limit theorem needs every closed-loop pole in the left half-plane; a finite K_p or K_v from an unstable loop is a trap answer.

Key takeaways

G(s) = Y/U with zero initial conditions; poles are roots of the characteristic equation; a mass–spring–damper has ω_n = √(k/m), ζ = c/(2√(km)).
Negative feedback: T = G/(1 + GH) and S = 1/(1 + GH); it trades gain for insensitivity and accuracy, at the risk of instability.
Mason: T = ΣP_kΔ_k/Δ, with Δ_k the Δ that leaves out every loop touching path k.
First order: t_s ≈ 4τ. Second order: %M_p = 100 exp(−πζ/√(1 − ζ²)) depends on ζ only; t_p = π/ω_d; t_s ≈ 4/(ζω_n).
Type = number of integrators; K_p, K_v, K_a give errors 1/(1 + K_p), 1/K_v, 1/K_a for step, ramp, parabola, valid only for a stable loop.

Practice questions (15)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The transfer function of a linear time-invariant system is defined as the ratio of the Laplace transforms of the output and the input:
1. with all initial conditions zero
2. with the initial conditions of the output only
3. for a unit step input only
4. for a sinusoidal input only
Show answer
Answer: A — with all initial conditions zero
The definition is for the zero-state response, so initial conditions are set to zero; the result is then a property of the system alone and holds for any input. It is not restricted to a step or a sinusoid.
A mass of 2 kg is attached to a spring of stiffness 50 N/m and a viscous damper of 8 N·s/m. The damping ratio of the system is ____.
Numerical answer — type the value.
Show answer
Answer: 0.4
ζ = c/(2√(km)) = 8/(2√(50 × 2)) = 8/(2 × 10) = 0.4, with ω_n = √(50/2) = 5 rad/s. Using c/(2m) = 2 (the decay rate σ = ζω_n) is a different quantity, the real part of the pole.
A forward path with gain 50 has a negative feedback path of gain 0.1. The overall closed-loop gain, correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 8.33
T = G/(1 + GH) = 50/(1 + 50 × 0.1) = 50/6 = 8.33. The approximation 1/H = 10 is off by 20% because the loop gain GH = 5 is not very large.
The main effect of negative feedback on the sensitivity of the closed-loop gain to changes in the forward-path gain G is that it:
1. reduces it by the factor 1 + GH
2. increases it by the factor 1 + GH
3. leaves it unchanged
4. makes it exactly zero
Show answer
Answer: A — reduces it by the factor 1 + GH
S = (dT/T)/(dG/G) = 1/(1 + GH), so a loop gain of 100 makes the closed-loop gain about a hundred times less sensitive. It is never exactly zero for a finite loop gain.
In a signal flow graph the forward paths have gains 6 and 2. Two non-touching loops each have gain −1; both loops touch the first forward path, and neither touches the second. The overall transfer function by Mason’s formula is ____.
Numerical answer — type the value.
Show answer
Answer: 3.5
Δ = 1 − (−1 − 1) + (−1)(−1) = 1 + 2 + 1 = 4. Δ₁ = 1 because both loops touch path 1; Δ₂ = Δ = 4 because none touches path 2. T = (6 × 1 + 2 × 4)/4 = 14/4 = 3.5. Using Δ₂ = 1 gives 2 and forgets that path 2 is untouched by any loop.
A unity negative feedback loop has the forward-path transfer function G(s) = 10/(s + 2). The closed-loop transfer function is:
1. 10/(s + 12)
2. 10/(s − 8)
3. 10/(s + 2)
4. 1/(s + 12)
Show answer
Answer: A — 10/(s + 12)
T = G/(1 + G) = [10/(s + 2)]/[1 + 10/(s + 2)] = 10/(s + 2 + 10) = 10/(s + 12). The pole moves from −2 to −12, so the loop is faster. 10/(s − 8) comes from a positive-feedback sign error.
A first-order system has the transfer function 1/(0.5s + 1). The time its step response takes to settle within 2% of the final value, using t_s = 4τ, is ____ s.
Numerical answer — type the value.
Show answer
Answer: 2
τ = 0.5 s, so t_s = 4 × 0.5 = 2 s; the response is at 1 − e⁻⁴ = 98.2% by then. The pole is at −2, and 4/2 = 2 gives the same value directly.
A standard second-order system with damping ratio 0.5 is given a unit step input. The percentage peak overshoot, correct to one decimal place, is ____ %.
Numerical answer — type the value.
Show answer
Answer: 16.3
%M_p = 100 exp(−πζ/√(1 − ζ²)) = 100 exp(−π × 0.5/0.866) = 100 exp(−1.814) = 16.3%. It depends on ζ only. exp(−πζ) without the square root gives 20.8%.
The closed-loop transfer function of a system is 100/(s² + 8s + 100). The peak time of its unit-step response, correct to two decimal places, is ____ s.
Numerical answer — type the value.
Show answer
Answer: 0.34
ω_n = 10 and 2ζω_n = 8, so ζ = 0.4. ω_d = 10√(1 − 0.16) = 9.165 rad/s, and t_p = π/ω_d = 0.3428 ≈ 0.34 s. Using ω_n, t = π/10 = 0.31 s, forgets that the oscillation is at the damped frequency.
The poles of a stable second-order system are complex with negative real parts. Its unit-step response is:
1. an underdamped oscillation settling to a final value
2. a monotonic rise with no overshoot
3. a sustained oscillation of constant amplitude
4. an exponentially growing oscillation
Show answer
Answer: A — an underdamped oscillation settling to a final value
Complex conjugate poles give an oscillation at ω_d, and the negative real part −ζω_n multiplies it by a decaying exponential, so it settles. A sustained oscillation needs poles on the imaginary axis and a growing one needs the right half-plane; real poles give the monotonic rise.
A unity-feedback system has the open-loop transfer function G(s) = 10/[s(s + 2)]. The steady-state error to a unit ramp input is ____.
Numerical answer — type the value.
Show answer
Answer: 0.2
The system is type 1 with K_v = lim_s→0 sG(s) = 10/2 = 5, so e_ss = 1/K_v = 0.2. The closed loop, s² + 2s + 10, is stable, so the result is valid. 1/10 confuses K_v with the numerator gain.
A unity-feedback system has G(s) = 20/[(s + 1)(s + 4)]. The steady-state error to a unit step input, correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.17
The system is type 0, K_p = G(0) = 20/4 = 5, and e_ss = 1/(1 + K_p) = 1/6 = 0.167 ≈ 0.17. The closed loop s² + 5s + 24 is stable. Taking 1/K_p = 0.2 forgets the 1 in 1 + K_p.
Which of the following statements about a standard underdamped second-order system are true?
1. The percentage overshoot depends only on the damping ratio
2. Increasing the damping ratio at fixed natural frequency reduces the overshoot
3. The 2% settling time is approximately 4/(ζω_n)
4. Doubling the natural frequency at fixed damping ratio doubles the overshoot
Show answer
Answer: A — The percentage overshoot depends only on the damping ratio; B — Increasing the damping ratio at fixed natural frequency reduces the overshoot; C — The 2% settling time is approximately 4/(ζω_n)
(A) %M_p = 100 exp(−πζ/√(1 − ζ²)). (B) The exponent grows in magnitude with ζ, so the overshoot falls. (C) The envelope e^{−ζω_n t} reaches 2% at t = 4/(ζω_n). (D) False: doubling ω_n at fixed ζ halves the peak and settling times with the same overshoot.
A unity-feedback loop has G(s) = K/[s(s + 4)]. If K is increased from 4 to 16, then:
1. the ramp error falls by a factor of 4 and the damping ratio falls from 1 to 0.5
2. the ramp error is unchanged and the damping ratio falls from 1 to 0.5
3. the ramp error falls by a factor of 4 and the damping ratio rises
4. both the ramp error and the damping ratio are unchanged
Show answer
Answer: A — the ramp error falls by a factor of 4 and the damping ratio falls from 1 to 0.5
K_v = K/4 goes from 1 to 4, so the ramp error 1/K_v goes from 1 to 1/4. The characteristic equation s² + 4s + K has ω_n = √K and ζ = 4/(2√K) = 2/√K: 1 at K = 4 and 0.5 at K = 16. More gain buys accuracy at the price of a livelier, less damped response.
The type of a control system is determined by:
1. the number of poles of the open-loop transfer function at the origin
2. the total number of poles of the closed-loop transfer function
3. the number of zeros of the open-loop transfer function
4. the damping ratio of the dominant poles
Show answer
Answer: A — the number of poles of the open-loop transfer function at the origin
Type is the number of integrators in the loop, that is of open-loop poles at s = 0; the order is the total number of poles. The type sets which inputs (step, ramp, parabola) are followed with zero error.