Block diagrams & SFG formulas

Modelling, block diagrams and signal flow graphs. 21 formulas for GATE EC, EE, IN Control Systems, each with when to use it. Each is explained in the topic notes.

All 21 Block diagrams & SFG formulas

CLOSED-LOOP TRANSFER FUNCTION (NEGATIVE FEEDBACK)

T (s) = \frac{C ( s )}{R ( s )} = \frac{G ( s )}{1 + G ( s ) H ( s )}

Here the feedback signal is subtracted at the comparator. The characteristic equation is $1 + G (s) H (s) = 0$ . After clearing fractions, its roots are the closed-loop poles.

CLOSED-LOOP TRANSFER FUNCTION (POSITIVE FEEDBACK)

T (s) = \frac{G ( s )}{1 - G ( s ) H ( s )}

Here the feedback signal is added at the comparator. The characteristic equation is $1 - G (s) H (s) = 0$ .

ACTUATING-SIGNAL TRANSFER FUNCTION

\frac{E ( s )}{R ( s )} = \frac{1}{1 + G ( s ) H ( s )}

This uses $E = R - H C$ with negative feedback. For unity feedback ( $H = 1$ ), $E$ is the true error $R - C$ . With non-unity $H$ , the actuating signal is not the reference-minus-output error.

DEFINITION OF THE TRANSFER FUNCTION

G (s) = \frac{Y ( s )}{U ( s )}_{zero initial conditions} = L {g (t)}

Applies to LTI systems only. $g (t)$ is the unit-impulse response.

TRANSFER FUNCTION FROM A LINEAR DIFFERENTIAL EQUATION

a_{n} y^{(n)} + \dots + a_{1} \overset{y}{˙} + a_{0} y = b_{m} u^{(m)} + \dots + b_{1} \overset{u}{˙} + b_{0} u G (s) = \frac{Y ( s )}{U ( s )} = \frac{b _{m} s ^{m} + \dots + b _{1} s + b _{0}}{a _{n} s ^{n} + \dots + a _{1} s + a _{0}}

With all initial conditions zero, each $d^{k} / d t^{k}$ becomes $s^{k}$ . The denominator polynomial set to zero is the characteristic equation.

POLE-ZERO (ROOT-LOCUS) FORM

G (s) = K \frac{( s - z _{1} ) ( s - z _{2} ) \dots ( s - z _{m} )}{( s - p _{1} ) ( s - p _{2} ) \dots ( s - p _{n} )}

The zeros $z_{i}$ are the values of $s$ that make $G (s) = 0$ . The poles $p_{j}$ are the roots of the denominator, where $G (s) \to \infty$ . $n$ is the order of the system. $G$ is proper if $m \leq n$ and strictly proper if $m < n$ ; physical systems are proper.

TIME-CONSTANT (BODE) FORM

G (s) = \frac{K _{d c} ( 1 + s T _{1} ) ( 1 + s T _{2} ) \dots}{s ^{N} ( 1 + s τ _{1} ) ( 1 + s τ _{2} ) \dots}

$N$ is the number of poles at the origin. For the loop $G (s) H (s)$ , $N$ is the type. With $N = 0$ , $K_{d c} = G (0)$ is the DC gain, which is the steady-state output per unit step if the system is stable.

MASS-SPRING-DAMPER

\frac{X ( s )}{F ( s )} = \frac{1}{M s ^{2} + B s + K}

Derived from $M \frac{d ^{2} x}{d t ^{2}} + B \frac{d x}{d t} + K x = f$ . The natural frequency is $ω_{n} = K / M$ , the damping ratio is $ζ = \frac{B}{2 K M}$ , and the DC gain (static compliance) is $1/ K$ .

SERIES RLC, OUTPUT ACROSS C

\frac{V _{C} ( s )}{V _{i} ( s )} = \frac{1}{L C s ^{2} + R C s + 1}

Series $R$ , $L$ and $C$ driven by $v_{i}$ . Here $ω_{n} = 1/ L C$ and $ζ = \frac{R}{2} C / L$ .

INVERTING OP-AMP STAGE

\frac{V _{o} ( s )}{V _{i} ( s )} = - \frac{Z _{f} ( s )}{Z _{i} ( s )}

This assumes an ideal op-amp (infinite open-loop gain, zero input current) with its non-inverting input grounded. $Z_{i}$ is in series with the input and $Z_{f}$ connects the output to the inverting input. With $Z_{i} = R_{1}$ and $Z_{f} = 1/ (s C)$ the stage is an integrator, $- 1/ (s R_{1} C)$ . A second inverting stage restores the sign. An ideal op-amp has zero output impedance, so cascaded stages do not load each other.

ARMATURE-CONTROLLED DC MOTOR

\frac{Θ ( s )}{V _{a} ( s )} = \frac{K _{t}}{s [ ( R _{a} + s L _{a} ) ( J s + B ) + K _{t} K _{b} ]}

Motor torque is $T_{m} = K_{t} i_{a}$ and back emf is $e_{b} = K_{b} d θ / d t$ , with $J$ and $B$ referred to the motor shaft. The back emf is a built-in negative velocity feedback. In SI units, $K_{t} = K_{b}$ numerically. For speed output, drop the factor $s$ : $Ω/ V_{a} = K_{t} / [(R_{a} + s L_{a}) (J s + B) + K_{t} K_{b}]$ . With a load torque $T_{L}$ opposing the motor, superposition gives $Ω (s) = \frac{K _{t} V _{a} ( s ) - ( R _{a} + s L _{a} ) T _{L} ( s )}{( R _{a} + s L _{a} ) ( J s + B ) + K _{t} K _{b}}$ .

DC MOTOR SPEED WITH NEGLIGIBLE ARMATURE INDUCTANCE

\frac{Ω ( s )}{V _{a} ( s )} = \frac{K _{m}}{τ _{m} s + 1}, K_{m} = \frac{K _{t}}{R _{a} B + K _{t} K _{b}}, τ_{m} = \frac{R _{a} J}{R _{a} B + K _{t} K _{b}}

Valid when $L_{a} \approx 0$ . $K_{m}$ is the steady-state speed per volt. The inertia $J$ affects only the time constant, not the final speed.

GEAR TRAIN AND REFLECTED LOAD

\frac{θ _{2}}{θ _{1}} = \frac{N _{1}}{N _{2}}, J_{e q} = J_{1} + (\frac{N _{1}}{N _{2}})^{2} J_{2}, B_{e q} = B_{1} + (\frac{N _{1}}{N _{2}})^{2} B_{2}

Gear 1 (motor side) has $N_{1}$ teeth and gear 2 (load side) has $N_{2}$ . To refer a load-side $J_{2}$ , $B_{2}$ or stiffness to the motor side, multiply by $(N_{1} / N_{2})^{2}$ . For ideal lossless gears the torque scales as $T_{2} = (N_{2} / N_{1}) T_{1}$ , so a load torque $T_{L}$ referred to the motor side becomes $(N_{1} / N_{2}) T_{L}$ .

OUTPUT WITH REFERENCE, DISTURBANCE AND SENSOR NOISE

C (s) = \frac{G _{1} G _{2}}{1 + G _{1} G _{2} H} [R (s) - N (s)] + \frac{G _{2}}{1 + G _{1} G _{2} H} D (s)

Negative feedback. The noise term has the same transfer function as the reference, with a minus sign, so no amount of loop gain can separate the two.

SENSITIVITY (DEFINITION)

S_{G}^{T} = \frac{\partial T / T}{\partial G / G} = \frac{G}{T} \frac{\partial T}{\partial G}

The fractional change in $T$ per fractional change in the parameter $G$ , equal to $\partial ln T / \partial ln G$ . It is valid for small changes. For a large change, compute $T$ before and after instead.

CLOSED-LOOP SENSITIVITY TO G AND TO H

S_{G}^{T} = \frac{1}{1 + G H}, S_{H}^{T} = \frac{- G H}{1 + G H}

For negative feedback, $T = G / (1 + G H)$ . An open loop ( $T = G$ ) has $S_{G}^{T} = 1$ . When $∣ G H ∣ ≫ 1$ , $S_{G}^{T} \to 0$ but $S_{H}^{T} \to - 1$ , so a feedback-path error reaches the output almost one-for-one. For the frequency-dependent sensitivity, evaluate at $s = j ω$ .

DISTURBANCE TRANSFER FUNCTION

\frac{C ( s )}{D ( s )} = \frac{G _{2}}{1 + G _{1} G _{2} H} \approx \frac{1}{G _{1} H} (∣ G_{1} G_{2} H ∣ ≫ 1)

For a disturbance entering after $G_{1}$ . Compared with open loop, where $C / D = G_{2}$ , feedback reduces the disturbance's effect by the factor $∣1 + G_{1} G_{2} H ∣$ . In the high-gain limit, only the gain upstream of the disturbance point (and $H$ ) sets the residual effect $1/ (G_{1} H)$ ; raising $G_{2}$ does not reduce it. For a disturbance added directly at the output, $C / D = 1/ (1 + G H)$ .

GAIN-BANDWIDTH TRADE (FIRST-ORDER PLANT)

T (s) = \frac{K}{1 + K H} \cdot \frac{1}{1 + s \frac{τ}{1 + K H}}

Plant $K / (1 + s τ)$ with a constant negative-feedback gain $H$ (use $H = 1$ for unity feedback). The DC gain falls by the factor $(1 + K H)$ . The bandwidth rises from $1/ τ$ to $(1 + K H) / τ$ rad/s, and the time constant shrinks by the same factor. The gain-bandwidth product stays $K / τ$ .

EQUIVALENT UNITY-FEEDBACK FORWARD PATH

G_{e q} (s) = \frac{G ( s )}{1 + G ( s ) H ( s ) - G ( s )}

A unity negative-feedback loop around $G_{e q}$ has the same closed-loop transfer function $G / (1 + G H)$ . Use it when a steady-state-error formula assumes unity feedback, so that $E = R - C$ is the true error.

MASON'S GAIN FORMULA

T = \frac{\sum _{k} P _{k} Δ _{k}}{Δ}

The gain from a source node to any other node. $P_{k}$ is the gain of the $k$ -th forward path. $Δ$ is the graph determinant. $Δ_{k}$ is $Δ$ recomputed with every loop that touches the $k$ -th forward path removed.

GRAPH DETERMINANT

Δ = 1 - i \sum L_{i} + i, j \sum L_{i} L_{j} - i, j, k \sum L_{i} L_{j} L_{k} + \dots

The second sum runs over every pair of mutually non-touching loops, the third over every triple, and so on. Loop gains carry their own signs, so a negative-feedback loop ( $L = - G H$ ) adds $+ G H$ to $Δ$ . $Δ = 0$ is the characteristic equation, and $Δ$ is the same for every input-output pair in a given graph.

Where this was asked: 2018 (1), 2020 (2), 2022 (2), 2023 (2), 2024 (1), 2025 (1), 2026 (1). See the 10 previous-year questions on Block diagrams & SFG. This shows what was asked, not what will be asked.

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Block diagrams & SFG: the notes and 10 solved previous-year questions