Transfer functions and block diagram reduction

Why it matters

A real mechatronic system — a DC-motor drive with a tachometer loop inside a position loop, a hydraulic actuator with pressure feedback — is a web of interacting sub-systems. Transfer functions let you describe each piece as one algebraic block, and block diagram reduction collapses the whole web into a single closed-loop transfer function whose poles tell you speed, damping and stability. Almost every later topic (time response, steady-state error, root locus, Bode) starts from the closed-loop transfer function you get here.

Key ideas

Transfer function. For a linear time-invariant (LTI) system, the transfer function is the ratio of the Laplace transform of the output to the Laplace transform of the input, with all initial conditions zero:

• It is a property of the system alone, not of the input applied.
• It is usually a ratio of polynomials in s, G(s) = N(s)/D(s). It is not "output polynomial over input polynomial": N(s) and D(s) both come from the system's differential equation.
• Roots of N(s) are zeros; roots of D(s) are poles. D(s) = 0 is the characteristic equation. Pole locations fix the natural modes (e^(pt)); all poles in the open left half of the s-plane means the system is BIBO stable.
• The order of the system is the degree of D(s). A system is proper if deg N ≤ deg D (all physical systems are).
• G(s) is also the Laplace transform of the impulse response g(t).
• Units: G(s) carries output units per input unit, e.g. a motor speed transfer function is in (rad/s)/V.

A transfer function cannot represent non-zero initial conditions, nonlinearities (saturation, dead zone, friction) or time-varying parameters. For those, use the state-space model or linearise about an operating point first.

Block diagram. Signals flow along arrows; a block multiplies its input signal by its transfer function; a summing point adds or subtracts signals (signs must be marked); a take-off (pickoff) point copies a signal without loading it. Block diagrams assume blocks do not load each other — if connecting two RC stages changes the first stage's behaviour, you must derive the combined transfer function, not just multiply.

Reduction rules.

1. Blocks in cascade (series) multiply.
2. Blocks in parallel (same input, outputs summed) add, with the signs at the summing point.
3. A negative feedback loop with forward path G and feedback path H reduces to G/(1 + GH); positive feedback gives G/(1 − GH).
4. Moving a summing point ahead of (upstream of) a block G: put 1/G in the moved branch. Moving it past (downstream of) G: put G in the moved branch.
5. Moving a take-off point past a block G: put 1/G in the branch. Moving it ahead of G: put G in the branch.
6. Adjacent summing points can be interchanged; adjacent take-off points can be interchanged. A summing point and a take-off point cannot simply be swapped.

Strategy. Reduce the innermost loop first; when loops overlap, move a summing or take-off point so that loops become nested; then reduce from inside out. For several inputs (reference R and disturbance D), set all inputs but one to zero, find each output, and add — superposition holds because the system is linear. For heavily interlaced loops, Mason's gain formula on the signal flow graph (next topic) is quicker.

Loop gain and sensitivity. The product G(s)H(s) is the open-loop (loop) transfer function. Where |GH| ≫ 1, the closed-loop transfer function approaches 1/H, which is why feedback makes the result insensitive to changes in G.

Formulas

G(s) = Y(s) / U(s) (zero initial conditions)

• Y(s): Laplace transform of output; U(s): Laplace transform of input; G(s): units of output per unit input.

G_series = G₁(s) · G₂(s)

• Applies to non-loading cascaded blocks.

G_parallel = G₁(s) ± G₂(s)

• Sign as marked at the summing point.

T(s) = C(s)/R(s) = G(s) / (1 + G(s)·H(s)) (negative feedback) T(s) = G(s) / (1 − G(s)·H(s)) (positive feedback)

• G: forward-path transfer function; H: feedback-path transfer function; G·H: loop transfer function.

1 + G(s)·H(s) = 0 — characteristic equation of the closed loop.

E(s) = R(s) / (1 + G(s)·H(s)) — actuating error for unity-type comparison, used later for steady-state error.

T(s) = G(s) / (1 + G(s)) — unity feedback (H = 1).

Worked examples

Example 1 (standard). Two blocks G₁(s) = 2/(s + 1) and G₂(s) = 3/(s + 2) are in cascade in the forward path of a negative feedback loop with H(s) = 1/(s + 3). Find the closed-loop transfer function, its poles and its DC gain.

1. Series: G = G₁·G₂ = 6 / ((s + 1)(s + 2)).
2. Feedback formula: T = G / (1 + G·H).
3. Multiply numerator and denominator by (s + 1)(s + 2)(s + 3): T = 6(s + 3) / ((s + 1)(s + 2)(s + 3) + 6).
4. Expand: (s + 1)(s + 2) = s² + 3s + 2; times (s + 3) gives s³ + 6s² + 11s + 6; add 6.
5. T(s) = 6(s + 3) / (s³ + 6s² + 11s + 12).
6. Poles: s³ + 6s² + 11s + 12 = (s + 4)(s² + 2s + 3), so the poles are s = −4 and s = −1 ± j1.414. All are in the left half-plane, so the loop is stable. The zero at s = −3 is the pole of H.
7. DC gain: T(0) = 6 × 3 / 12 = 1.5 (output units per input unit).

Example 2 (GATE level, minor loop). A position servo has an amplifier gain K = 10 in the forward path, followed by the motor-and-load G_p(s) = 1/(s(s + 1)) (output in rad, input in V). A tachometer feeds back H_t(s) = 0.5s from the output to a summing point placed after the amplifier (negative sign), and the outer loop has unity negative feedback. Find the closed-loop transfer function, ωn, ζ and the peak overshoot, and compare with the system without the tachometer.

1. Inner loop: G_i = G_p / (1 + G_p·H_t) = [1/(s(s + 1))] / [1 + 0.5s/(s(s + 1))].
2. Multiply through by s(s + 1): G_i = 1 / (s² + s + 0.5s) = 1 / (s² + 1.5s).
3. Forward path of outer loop: G = 10 / (s² + 1.5s).
4. Unity feedback: T = G / (1 + G) = 10 / (s² + 1.5s + 10).
5. Compare with ωn² / (s² + 2ζωn·s + ωn²): ωn = √10 = 3.16 rad/s; 2ζωn = 1.5, so ζ = 1.5 / (2 × 3.162) = 0.237.
6. Peak overshoot Mp = exp(−πζ/√(1 − ζ²)) = exp(−π × 0.237 / 0.972) = 46.4 %.
7. Without the tachometer: T = 10/(s² + s + 10), ζ = 1/(2 × 3.162) = 0.158, Mp = 60.5 %. The rate feedback raised the damping without changing ωn — the standard reason for a tacho minor loop.

Common mistakes

• Writing G/(1 + GH) for positive feedback, or mixing the sign at the summing point with the sign in the formula. Read the sign on the summing junction first.
• Multiplying blocks that load each other (two passive RC sections connected directly). Their combined transfer function is not the product of the separate ones.
• Moving a take-off or summing point and forgetting the compensating 1/G or G block, or putting it on the wrong branch.
• Leaving compound fractions unsimplified, then reading poles off the wrong polynomial. Always clear fractions to get one numerator over one denominator.
• Assuming the closed-loop zeros are the open-loop zeros: zeros of T(s) are the zeros of G plus the poles of H.
• Including initial conditions in a transfer function, or calling the denominator "the input".

For GATE ME

Expect: reducing a two- or three-loop block diagram to C/R; finding C/D for a disturbance input with the reference set to zero; minor-loop (tachometer) problems asking for ζ, ωn or overshoot after reduction; identifying the characteristic equation; and conceptual MCQs on moving summing and take-off points. Practise clearing fractions quickly and checking your answer with the DC gain (set s = 0) and with the order of the denominator.

Quick check

1. What must be true of initial conditions when you define a transfer function?
2. A take-off point is moved from before a block G to after it. What block must be inserted in the take-off branch?
3. Find the closed-loop transfer function for G = 4/(s + 2) with unity negative feedback.
4. A loop has G = 5 and H = 0.2 with positive feedback. What happens to the closed-loop gain?
5. Why can two RC low-pass sections connected directly not be modelled as the product of their individual transfer functions?

Answers: 1. They are all zero. 2. 1/G. 3. 4/(s + 6). 4. 1 − GH = 0, so the gain becomes unbounded (the loop is on the edge of instability). 5. The second section loads the first, changing its output; the combined circuit must be analysed together.