Modelling a Control System: Feedback, Transfer Functions and Response

For a linear time-invariant system the transfer function is the ratio of the Laplace transform of the output to that of the input with all initial conditions zero, and that last clause is not decoration: it is the reason the transfer function describes the forced response and says nothing directly about where the system started. Its denominator set to zero is the characteristic equation, whose roots are the poles, and the poles alone decide the shape of the natural response — a real pole at −a contributes e^(−at), a complex pair at −σ ± jω contributes e^(−σt) times a sinusoid at ω, and a pole in the right half plane contributes a growing exponential. Zeros do not change which exponentials appear; they change how much of each appears, and a zero in the right half plane produces the characteristic initial undershoot of a non-minimum-phase system.

Almost every examinable result in this section is stated for the standard second-order form T(s) = ω_n²/(s² + 2ζω_n s + ω_n²), so learn to read ω_n and ζ off a given denominator before doing anything else. Take T(s) = 25/(s² + 6s + 25). Matching the constant term gives ω_n² = 25, so ω_n = 5 rad/s; matching the s term gives 2ζω_n = 6, so ζ = 6/10 = 0.6. The poles are then at −ζω_n ± jω_n√(1 − ζ²) = −3 ± j4, and the damped natural frequency ω_d = 4 rad/s is the imaginary part — a useful check, since 3² + 4² = 5² means the poles lie on a circle of radius ω_n at an angle whose cosine is ζ. That geometric reading is worth keeping: distance from the origin is ω_n, and the cosine of the angle from the negative real axis is ζ, which is how a root-locus plot in the next chapter is read for damping at a glance.

Three rules do most of the work. Blocks in cascade multiply. Blocks in parallel into a summing junction add. And a negative feedback loop of forward gain G and feedback gain H collapses to G/(1 + GH), with the sign in the denominator opposite to the sign at the summing junction — positive feedback gives G/(1 − GH), which is where a sign slip costs a whole question. The order to apply them in is the part worth practising: always reduce the innermost loop first, because an inner loop that is still open blocks every outer simplification, and because reducing outwards means you never have to move a block past anything.

Work a two-loop example all the way to a standard form. The forward path is an amplifier-integrator G_1 = 10/s followed by a plant G_2 = 1/(s + 2). Around G_2 alone there is a minor negative feedback loop with H_1 = 0.5 — a tachometer, say. Around the whole thing there is unity negative feedback. Reduce the inner loop first: G_2/(1 + G_2H_1) = [1/(s + 2)]/[1 + 0.5/(s + 2)] = 1/(s + 2.5), which is the same plant with its pole pushed left, from −2 to −2.5. That is what minor-loop feedback does physically, and it is worth noticing before the algebra buries it. The forward path is now 10/[s(s + 2.5)], and closing the unity loop gives T(s) = 10/[s(s + 2.5) + 10] = 10/(s² + 2.5s + 10).

Read the standard form off it at once. ω_n = √10 = 3.162 rad/s and 2ζω_n = 2.5, so ζ = 2.5/6.325 = 0.395 — underdamped, and badly enough that the step response overshoots by exp(−ζπ/√(1 − ζ²)) = 25.9 %. That number is the whole reason the minor loop was there: raise H_1 and the inner pole moves further left, ζ rises and the overshoot falls, at the cost of a slower loop. Two reduction rules for the awkward diagrams are worth stating exactly, because their asymmetry is a favourite trap. To move a block G backwards past a summing junction, the other branch entering that junction must be divided by G. To move a take-off point backwards past a block G, the branch leaving it must be multiplied by G. Those two go opposite ways, and a diagram that has been rearranged in the wrong direction usually still looks plausible.

A signal flow graph is the same information as a block diagram with the blocks turned into labelled arrows and the summing junctions into nodes, and its point is that Mason's gain formula then reads the answer off without any rearrangement. The formula is T = Σ_k (P_k Δ_k) / Δ, where P_k is the gain of the kth forward path, Δ = 1 − Σ(loop gains) + Σ(products of pairs of non-touching loops) − Σ(products of triples of non-touching loops) + …, and Δ_k is that same Δ computed with every loop that touches path k deleted. Two definitions carry all the risk: a loop gain includes the sign at its summing junction, so a negative feedback loop has a negative loop gain, and non-touching means sharing no node at all, not merely looking separate on the page.

Apply it to the very system reduced above, as a check on both methods. There is one forward path, P_1 = G_1G_2. There are two loops: the minor loop L_1 = −G_2H_1 and the outer loop L_2 = −G_1G_2. They share the node between G_1 and G_2, so they touch, and there is no non-touching product to add. Hence Δ = 1 + G_2H_1 + G_1G_2, Δ_1 = 1, and T = G_1G_2/(1 + G_2H_1 + G_1G_2). Substitute G_1 = 10/s, G_2 = 1/(s + 2) and H_1 = 0.5: the numerator is 10/[s(s + 2)], the denominator is [s(s + 2) + 0.5s + 10]/[s(s + 2)], and the common factor cancels to leave T = 10/(s² + 2.5s + 10) — identical to the block-diagram result, which is the point of doing it twice.

Mason earns its keep when loops do not touch, because that is exactly the case block-diagram reduction handles clumsily. Consider a chain G_1 → G_2 → G_3 → G_4 from input to output with a local feedback −H_1 around G_2 and another −H_2 around G_4. The loops are L_1 = −G_2H_1 and L_2 = −G_4H_2, and they share no node, so the pair product enters with a plus sign: Δ = 1 + G_2H_1 + G_4H_2 + G_2H_1G_4H_2 = (1 + G_2H_1)(1 + G_4H_2). The single forward path touches both loops, so Δ_1 = 1 and T = G_1G_2G_3G_4/[(1 + G_2H_1)(1 + G_4H_2)] — which is just the product of two independently closed loops, exactly as intuition demands. Put numbers on it: with G_1 = 2, G_2 = 5, G_3 = 3, G_4 = 4, H_1 = 0.2 and H_2 = 0.5, the denominators are 1 + 1 = 2 and 1 + 2 = 3, so T = 120/6 = 20.

For a first-order system 1/(1 + sτ) there is only one number to know: the step response reaches 63.2 % of its final value in one time constant, 86.5 % in two, 95 % in three and 98.2 % in four, so a settling time to two per cent is 4τ and there is never any overshoot. For the standard second-order system everything follows from ζ and ω_n, and the six results below are the ones questions ask for. Compute them all for the system met earlier, T(s) = 25/(s² + 6s + 25), with ζ = 0.6, ω_n = 5 rad/s and ω_d = ω_n√(1 − ζ²) = 4 rad/s.

Two structural readings of that table matter more than the arithmetic. Overshoot depends only on ζ — ω_n is nowhere in the formula — so scaling a system in time changes every duration and leaves the overshoot untouched: 9.48 % for ζ = 0.6 whether the system settles in a second or a microsecond. And settling time depends only on the product ζω_n, which is the real part of the pole pair, so a specification on settling time is a vertical line in the s-plane and a specification on overshoot is a pair of rays from the origin. That is exactly how a design region is drawn on a root locus, and it is the reason the two chapters of this section join where they do. The commonly quoted damping values follow the same logic: ζ = 0.707 gives about 4.3 % overshoot and is the usual compromise, while ζ = 0.5 gives about 16 %.

Where transient analysis asks how the response gets there, steady-state analysis asks where it ends up, and for a unity-feedback loop the answer needs only one limit. The error is E(s) = R(s)/[1 + G(s)], and the final value theorem gives e_ss = lim_(s→0) sR(s)/[1 + G(s)]. Evaluating that for the three standard inputs produces the three error constants: K_p = lim G(s) for a step, K_v = lim sG(s) for a ramp, and K_a = lim s²G(s) for a parabola. The system type is simply the number of poles G(s) has at the origin, and it decides which constants are finite: type 0 has a finite K_p, type 1 an infinite K_p and a finite K_v, type 2 an infinite K_v and a finite K_a.

Two worked cases fix the pattern. Take the type-1 loop G(s) = 20/[s(s + 2)(s + 5)] with unity feedback. Then K_p = lim G(s) = ∞, so the step error is 1/(1 + K_p) = 0 — a type-1 system tracks a step exactly, eventually. Next K_v = lim sG(s) = 20/(2 × 5) = 2, so a unit-ramp input leaves a constant error of 1/K_v = 0.5: the output runs parallel to the ramp, half a unit behind, for ever. And K_a = lim s²G(s) = 0, so a parabolic input leaves an error that grows without bound. Now remove the pole at the origin to get the type-0 loop G(s) = 20/[(s + 2)(s + 5)]. Now K_p = 20/10 = 2, so the step error is 1/(1 + 2) = 0.333 — a third of the command is never delivered — and the ramp error is infinite.

The frequency response of an LTI system is T(jω), obtained by substituting s = jω, and it is the steady-state gain and phase seen by a sinusoid at each frequency. Three numbers describe a closed-loop response and all three are computable from ζ and ω_n. The resonant peak M_r = 1/[2ζ√(1 − ζ²)] is the maximum of the magnitude, occurring at the resonant frequency ω_r = ω_n√(1 − 2ζ²). The bandwidth ω_B is where the magnitude falls to 1/√2 of its low-frequency value, that is by 3 dB, and for the standard second-order system it is ω_B = ω_n√[1 − 2ζ² + √(4ζ⁴ − 4ζ² + 2)]. For the running example with ζ = 0.6 and ω_n = 5: ω_r = 5√0.28 = 2.646 rad/s, M_r = 1/(2 × 0.6 × 0.8) = 1.042, which is 0.35 dB, and ω_B = 5√1.3185 = 5.74 rad/s.

Three consequences of those formulas are examinable in their own right. First, ω_r is real only when 1 − 2ζ² > 0, that is when ζ < 1/√2 = 0.707, so a system damped at 0.707 or more has no resonant peak at all and its magnitude falls monotonically — which is the frequency-domain statement of why ζ = 0.707 is the standard compromise. Second, ω_r is always below ω_d, and both are below ω_n, so the three frequencies must never be used interchangeably: here they are 2.646, 4 and 5 rad/s. Third, bandwidth and speed measure the same thing from two sides — ω_B is roughly proportional to ω_n for fixed damping, so a faster rise time and a wider bandwidth are the same requirement, and a demand for both a fast response and a narrow noise bandwidth is a demand to be traded, not satisfied.