Transfer Function and Block Diagram Reduction
Closed-Loop Transfer Function
T(s) = \frac{G(s)}{1 + G(s)H(s)}
| Symbol | Description | Unit |
|---|---|---|
| T(s) | Closed-loop transfer function | dimensionless |
| G(s) | Forward path transfer function | dimensionless |
| H(s) | Feedback transfer function | dimensionless |
Worked example
Find T(s) for G(s) = 10/(s+2), H(s) = 1 (unity feedback).
Given: G(s) = 10/(s+2), H(s) = 1
- T(s) = G(s) / (1 + G(s)·H(s))
- = [10/(s+2)] / [1 + 10/(s+2)]
- = [10/(s+2)] / [(s+2+10)/(s+2)]
- = 10 / (s+12)
Answer: T(s) = 10/(s+12)
Characteristic Equation
1 + G(s)H(s) = 0
Worked example
Write characteristic equation for G(s) = K/(s(s+4)), H(s) = 1.
Given: G(s)H(s) = K/(s(s+4))
- 1 + K/[s(s+4)] = 0
- s(s+4) + K = 0
- s² + 4s + K = 0
Answer: Characteristic equation: s² + 4s + K = 0
Signal Flow Graph — Mason's Gain Formula
T = \frac{\sum_k P_k \Delta_k}{\Delta}
| Symbol | Description | Unit |
|---|---|---|
| P_k | k-th forward path gain | dimensionless |
| \Delta | Graph determinant | dimensionless |
| \Delta_k | Cofactor of k-th path | dimensionless |
Worked example
SFG with one forward path P1 = G1·G2·G3, one loop L1 = G2·H1. Find T.
Given: P1 = G1G2G3, L1 = G2H1
- Δ = 1 - L1 = 1 - G2H1
- Δ1 = 1 (loop L1 touches forward path)
- T = P1·Δ1 / Δ = G1G2G3 / (1 - G2H1)
Answer: T = G1G2G3 / (1 - G2H1)
Time-Domain Response Specifications
Second-Order System Standard Form
T(s) = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}
| Symbol | Description | Unit |
|---|---|---|
| \omega_n | Natural frequency | rad/s |
| \zeta | Damping ratio | dimensionless |
Worked example
T(s) = 25/(s²+6s+25). Find ωn and ζ.
Given: T(s) = 25/(s²+6s+25)
- ωn² = 25 → ωn = 5 rad/s
- 2ζωn = 6 → ζ = 6/(2×5) = 0.6
Answer: ωn = 5 rad/s, ζ = 0.6 (underdamped)
Peak Overshoot
M_p = e^{-\pi\zeta / \sqrt{1-\zeta^2}} \times 100\%
| Symbol | Description | Unit |
|---|---|---|
| M_p | Percent peak overshoot | % |
Worked example
Find peak overshoot for ζ = 0.6.
Given: ζ = 0.6
- √(1-ζ²) = √(1-0.36) = √0.64 = 0.8
- Exponent = -π × 0.6 / 0.8 = -1.8850 / 0.8 = -2.356
- M_p = e^{-2.356} × 100 = 0.0948 × 100
Answer: M_p ≈ 9.48%
Settling Time (2% criterion)
t_s \approx \frac{4}{\zeta \omega_n}
| Symbol | Description | Unit |
|---|---|---|
| t_s | Settling time | s |
Worked example
Find settling time for ζ = 0.6, ωn = 5 rad/s.
Given: ζ = 0.6, ωn = 5 rad/s
- t_s = 4 / (ζ × ωn)
- = 4 / (0.6 × 5)
- = 4 / 3
- = 1.33 s
Answer: t_s ≈ 1.33 s
Rise Time (0% to 100%)
t_r = \frac{\pi - \phi}{\omega_d}, \quad \phi = \cos^{-1}\zeta, \quad \omega_d = \omega_n\sqrt{1-\zeta^2}
| Symbol | Description | Unit |
|---|---|---|
| t_r | Rise time | s |
| \omega_d | Damped natural frequency | rad/s |
| \phi | Phase angle | rad |
Worked example
Find rise time for ζ = 0.6, ωn = 5 rad/s.
Given: ζ = 0.6, ωn = 5
- ωd = 5 × √(1-0.36) = 5 × 0.8 = 4 rad/s
- φ = cos⁻¹(0.6) = 0.9273 rad
- t_r = (π - 0.9273) / 4
- = 2.2143 / 4 = 0.5536 s
Answer: t_r ≈ 0.55 s
Stability — Routh-Hurwitz Criterion
Routh Array Construction
\text{All roots in LHP} \Leftrightarrow \text{all elements of first column} > 0
Worked example
Check stability of s³ + 6s² + 11s + 6 = 0.
Given: Coefficients: a3=1, a2=6, a1=11, a0=6
- Row s³: 1 11
- Row s²: 6 6
- Row s¹: b1 = (6×11 - 1×6)/6 = (66-6)/6 = 10
- Row s⁰: c1 = (10×6 - 6×0)/10 = 6
- First column: 1, 6, 10, 6 — all positive
Answer: System is stable (all first-column elements positive)
Routh Marginal Stability (Row of Zeros)
\text{If row of zeros appears, form auxiliary polynomial from row above}
Worked example
s⁴+s³+3s²+3s+2=0. Identify if purely imaginary roots exist.
Given: Row s²: 3·1 - 1·3 / 1 ... (check for zero row)
- Row s⁴: 1 3 2
- Row s³: 1 3
- Row s²: (3×1 - 1×3)/1 = 0 → auxiliary from s³ row
- Auxiliary: s³ + 3s = 0 → s(s²+3) = 0 → s = ±j√3
- Replace s² row using d/ds(s³+3s) = 3s²+3 → row: 3 3
Answer: System has purely imaginary roots at s = ±j√3 (marginally stable)
Frequency Response — Bode Plots
Gain Margin
GM = -20\log|G(j\omega_{pc})H(j\omega_{pc})| \text{ dB}
| Symbol | Description | Unit |
|---|---|---|
| GM | Gain margin | dB |
| \omega_{pc} | Phase crossover frequency (∠GH = -180°) | rad/s |
Worked example
At ω_pc = 2 rad/s, |G(j2)H(j2)| = 0.5. Find gain margin.
Given: |GH| at ω_pc = 0.5
- GM = -20 log(0.5)
- = -20 × (-0.301)
- = 6.02 dB
Answer: GM = +6 dB (stable)
Phase Margin
PM = 180° + \angle G(j\omega_{gc})H(j\omega_{gc})
| Symbol | Description | Unit |
|---|---|---|
| PM | Phase margin | degrees |
| \omega_{gc} | Gain crossover frequency (|GH| = 1) | rad/s |
Worked example
At ω_gc, ∠GH = -135°. Find phase margin.
Given: ∠GH at ω_gc = -135°
- PM = 180° + (-135°)
- PM = 45°
Answer: PM = 45° (well-damped, stable)
Bode Magnitude of First-Order Factor
|1 + j\omega/\omega_c|_{dB} \approx 0 \text{ dB for } \omega \ll \omega_c;\quad +20\log(\omega/\omega_c) \text{ for } \omega \gg \omega_c
| Symbol | Description | Unit |
|---|---|---|
| \omega_c | Corner frequency = 1/τ | rad/s |
Worked example
G(s) = 1/(s+10). Find Bode magnitude at ω = 1, 10, 100 rad/s.
Given: G(jω) = 1/(10+jω), corner at ωc = 10 rad/s
- At ω = 1: |G| = 1/√(100+1) ≈ 1/10.05 → -20.04 dB ≈ -20 dB
- At ω = 10 (corner): |G| = 1/√200 = 0.0707 → -23 dB
- At ω = 100: |G| = 1/√(100²+10²) ≈ 1/100 → -40 dB (slope -20 dB/dec)
Answer: Bode plot: flat at -20 dB below corner; rolls off at -20 dB/decade above ωc = 10
Root Locus
Angle Condition (Root Locus)
\angle G(s)H(s) = (2k+1) \times 180°, \quad k = 0, \pm1, \pm2, \ldots
Worked example
Check if s = -1+j1 is on root locus of G(s) = K/[s(s+2)].
Given: s = -1+j, G(s) = K/[s(s+2)]
- ∠(s) = angle of (-1+j) from origin = 135°
- ∠(s+2) = angle of (1+j) from -2 = 45°
- ∠G(s) = 0 - 135° - 45° = -180°
- -180° = (2×0+1)×(-180°) ✓
Answer: s = -1+j is on the root locus
Centroid of Asymptotes
\sigma_a = \frac{\sum \text{poles} - \sum \text{zeros}}{n - m}
| Symbol | Description | Unit |
|---|---|---|
| \sigma_a | Centroid on real axis | dimensionless |
| n | Number of poles | dimensionless |
| m | Number of zeros | dimensionless |
Worked example
G(s) = K/[s(s+2)(s+4)]. Find centroid of asymptotes.
Given: Poles: 0, -2, -4; no zeros
- Σ poles = 0 + (-2) + (-4) = -6
- Σ zeros = 0
- n - m = 3 - 0 = 3
- σ_a = (-6 - 0) / 3 = -2
Answer: Centroid σ_a = -2
Angle of Asymptotes
\theta_a = \frac{(2k+1) \times 180°}{n - m}
Worked example
Find asymptote angles for G(s) = K/[s(s+2)(s+4)].
Given: n = 3, m = 0
- k=0: θ = 1×180°/3 = 60°
- k=1: θ = 3×180°/3 = 180°
- k=2: θ = 5×180°/3 = 300° (= -60°)
Answer: Asymptote angles: 60°, 180°, 300°
Steady-State Error
Steady-State Error (Unity Feedback)
e_{ss} = \frac{1}{1 + K_p} \text{ (step)};\quad \frac{1}{K_v} \text{ (ramp)};\quad \frac{1}{K_a} \text{ (parabolic)}
| Symbol | Description | Unit |
|---|---|---|
| K_p | Position error constant = lim_{s→0} G(s) | dimensionless |
| K_v | Velocity error constant = lim_{s→0} sG(s) | 1/s |
| K_a | Acceleration constant = lim_{s→0} s²G(s) | 1/s² |
Worked example
G(s) = 10/[s(s+2)]. Find steady-state error to unit ramp.
Given: G(s) = 10/[s(s+2)]
- K_v = lim_{s→0} s·G(s) = lim_{s→0} s·10/[s(s+2)]
- = lim_{s→0} 10/(s+2)
- = 10/2 = 5
- e_ss = 1/K_v = 1/5 = 0.2
Answer: e_ss = 0.2 for unit ramp input
System Type Number
\text{Type} = \text{number of pure integrators (poles at } s=0\text{) in } G(s)H(s)
Worked example
Identify type of G(s) = 5(s+3)/[s²(s+10)].
Given: G(s) = 5(s+3)/[s²(s+10)]
- Count poles at s = 0: s² contributes 2 poles at origin
- Type = 2
Answer: Type 2 system — zero steady-state error to ramp, finite error to parabolic
State-Space Representation
State-Space Equations
\dot{\mathbf{x}} = A\mathbf{x} + B\mathbf{u}, \quad \mathbf{y} = C\mathbf{x} + D\mathbf{u}
| Symbol | Description | Unit |
|---|---|---|
| A | System matrix (n×n) | dimensionless |
| B | Input matrix (n×1) | dimensionless |
| C | Output matrix (1×n) | dimensionless |
| D | Feed-forward matrix | dimensionless |
Worked example
Write state-space for G(s) = 1/(s²+3s+2).
Given: Transfer function: 1/(s²+3s+2)
- Choose states x1 = y, x2 = dy/dt
- ẋ1 = x2
- ẋ2 = -2x1 - 3x2 + u
- A = [[0,1],[-2,-3]], B = [[0],[1]], C = [1,0], D = 0
Answer: A = [0 1; -2 -3], B = [0; 1], C = [1 0], D = 0
Characteristic Equation from State Matrix
\det(sI - A) = 0
Worked example
Find eigenvalues of A = [[0,1],[-2,-3]].
Given: A = [0 1; -2 -3]
- det(sI - A) = det([[s,-1],[2,s+3]])
- = s(s+3) - (-1)(2) = s² + 3s + 2
- s² + 3s + 2 = 0 → (s+1)(s+2) = 0
Answer: Eigenvalues: s = -1 and s = -2 (system is stable)
Quick reference
| Formula | Expression |
|---|---|
| Closed-Loop TF | T(s) = G(s)/(1+G(s)H(s)) |
| Characteristic Equation | 1 + G(s)H(s) = 0 |
| Peak Overshoot | Mp = exp(-πζ/√(1-ζ²)) × 100% |
| Settling Time (2%) | t_s = 4/(ζωn) |
| Rise Time | t_r = (π - φ)/ωd |
| Gain Margin | GM = -20 log|GH(jω_pc)| dB |
| Phase Margin | PM = 180° + ∠GH(jω_gc) |
| Velocity Error Constant | Kv = lim_{s→0} sG(s) |
| Centroid of Asymptotes | σ_a = (Σpoles - Σzeros)/(n-m) |
| Asymptote Angles | θ_a = (2k+1)×180°/(n-m) |
| State Equation | ẋ = Ax + Bu |
| Eigenvalues | det(sI - A) = 0 |
| ωn from TF | ωn² = constant term in denominator |
| Damping Ratio from TF | ζ = (middle coeff)/(2ωn) |
| Mason's Gain | T = Σ(Pk Δk)/Δ |
Exam tips
- GATE examiners test Routh-Hurwitz by asking how many roots are in the RHP — count sign changes in the first column, not just whether all elements are positive.
- For Bode plot gain margin, find the phase crossover frequency first (where phase = -180°), then read the magnitude — do not confuse ω_gc with ω_pc.
- Phase margin of 45°–60° corresponds roughly to ζ = 0.45–0.6 — examiners often ask you to infer damping from frequency-domain specs.
- In root locus, the breakaway point between two poles on the real axis satisfies dK/ds = 0 — always solve this before sketching the locus.
- Steady-state error requires identifying system type first — Type 0 has infinite error to ramp, Type 1 has zero error to step, finite error to ramp.
- For state-space controllability, verify rank of [B AB A²B ... A^{n-1}B] = n; for observability check rank of [C; CA; CA² ...; CA^{n-1}] = n.