GATE EE Control Systems Formula Sheet: Key Formulas and Traps

Q: How do you find the steady-state error for a ramp input?

Compute the velocity error constant Kv as the limit of sG(s) as s tends to zero, for a unity feedback loop. The error to a unit ramp is 1/Kv. A type 0 loop has infinite ramp error, type 1 a finite error and type 2 zero error. Check first that the closed loop is stable, because the error constants mean nothing for an unstable loop.

Q: What is the Nyquist stability criterion formula?

The criterion is Z = N + P. Here P is the number of open-loop poles in the right half plane, N is the number of clockwise encirclements of the point minus one by the plot of G(jω)H(jω), and Z is the number of closed-loop right half plane poles. The closed loop is stable only when Z is zero. Anticlockwise encirclements count as negative.

By MD ANISH AHAMADUpdated 4 Oct 202610 min read

GATE EE Control Systems Formula Sheet: Key Formulas and Traps

The key GATE EE control systems formulas are the closed-loop gain $G/(1+GH)$ , Mason's formula, the second order specifications, the error constants, the Routh conditions, the root-locus rules, the margins, $Z = N + P$ , the compensator results and $C(sI-A)^{-1}B + D$ . Each comes with a condition that decides whether it applies. This sheet teaches them in learning order, with a worked example and a trap for each part.

In this guide

Key takeaways
The terms this sheet uses
Transfer functions, feedback and Mason's gain formula
First and second order time response
Steady-state error and the error constants
Routh-Hurwitz: counting unstable roots
Root locus: where the closed-loop poles travel
Bode plots, margins and frequency-domain specifications
Nyquist: counting encirclements
Compensators and P, PI and PID controllers
State space and the state transition matrix
Quick revision

Key takeaways

Every time-response and error formula assumes a stable closed loop.
Settling time depends on the real part of the poles, while overshoot depends on $\zeta$ alone.
Error constants apply to a unity feedback loop; the type is the number of open-loop poles at the origin.
Gain margin is read at the phase crossover and phase margin at the gain crossover.
In $Z = N + P$ , clockwise encirclements of $-1$ count as positive.
The poles of a state-space model are the eigenvalues of $A$ .

The terms this sheet uses

The open-loop transfer function $G(s)H(s)$ is the gain around the broken loop, with $G$ forward and $H$ in the feedback path. The closed-loop transfer function $T(s)$ is output over input with the loop closed. Unity feedback means $H = 1$ .

The type is the number of poles of $G(s)H(s)$ at the origin. A second order system has a natural frequency $\omega_n$ and a damping ratio $\zeta$ ; it is underdamped when $0 < \zeta < 1$ .

The gain crossover frequency $\omega_{gc}$ is where the open-loop magnitude is 0 dB. The phase crossover frequency $\omega_{pc}$ is where the open-loop phase is $-180^\circ$ .

Transfer functions, feedback and Mason's gain formula

A transfer function is output over input in the Laplace domain with zero initial conditions. Negative feedback divides the forward gain by one plus the loop gain.

Formula	Watch out for
$\displaystyle T = \dfrac{G}{1 + GH}$ (negative feedback); $\displaystyle T = \dfrac{G}{1 - GH}$ (positive)	Follow the summing junction sign
$\displaystyle S_G^T = \dfrac{1}{1 + GH}$ ; $\displaystyle S_H^T = \dfrac{-GH}{1 + GH}$	High loop gain makes $T$ sensitive to $H$
Cascade multiplies, parallel adds	Moving a take-off point past a block changes that branch
$\displaystyle T = \dfrac{\sum_k P_k \Delta_k}{\Delta}$	$\Delta_k$ drops every loop touching path $k$
$\Delta = 1 - \sum L_i + \sum L_i L_j - \sum L_i L_j L_k + \dots$	Products of non-touching loops only

Worked example. With $G = 20/(s+4)$ and $H = 1$ , $T = 20/(s+24)$ , the DC gain is $20/24 = 0.833$ and the DC sensitivity $1/(1+5) = 0.167$ .

Worked example (Mason). One path of gain 6 touches two non-touching loops of gains $-2$ and $-3$ : $\Delta = 1 - (-5) + 6 = 12$ , $\Delta_1 = 1$ , so $T = 6/12 = 0.5$ .

The trap is the sign: loop gains carry their own minus signs, so subtracting them adds them back.

First and second order time response

A first order system rises with one time constant $\tau$ . A second order system's speed is set by $\omega_n$ and its overshoot by $\zeta$ .

Formula	Watch out for
$\displaystyle \dfrac{1}{\tau s + 1}$ : step response $1 - e^{-t/\tau}$	63.2 per cent at $t = \tau$
$t_r = 2.2\tau$ (10 to 90 per cent); $t_s = 4\tau$ (2 per cent)	$t_r$ is $\tau \ln 9$ , not $\tau$
$\displaystyle \dfrac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}$ ; poles $-\zeta\omega_n \pm j\omega_n\sqrt{1-\zeta^2}$	Read $\omega_n^2$ from the constant term
$\omega_d = \omega_n\sqrt{1-\zeta^2}$ ; $t_p = \pi/\omega_d$	Uses $\omega_d$ , not $\omega_n$
$M_p = 100\,e^{-\pi\zeta/\sqrt{1-\zeta^2}}$ per cent	Depends on $\zeta$ alone
$t_s = 4/(\zeta\omega_n)$ (2 per cent); $3/(\zeta\omega_n)$ (5 per cent)	Check the stated band
$t_r = (\pi - \cos^{-1}\zeta)/\omega_d$ (0 to 100 per cent)	A different rise-time definition

Worked example. $T = 25/(s^2 + 6s + 25)$ gives $\omega_n = 5$ , $\zeta = 6/10 = 0.6$ , $\omega_d = 5 \times 0.8 = 4$ rad/s, $t_p = \pi/4 = 0.785$ s, $M_p = 100\,e^{-0.75\pi} = 9.48$ per cent and $t_s = 4/3 = 1.33$ s.

Remember: settling time is fixed by how far left the poles sit, and overshoot by the angle they make with the negative real axis. Equal real parts mean equal settling times.

The trap is a reduced model. You may drop a pole far to the left of the dominant pair, but keep the DC gain unchanged.

Steady-state error and the error constants

Steady-state error is what remains once the transient dies out. For unity feedback it follows from the type and three constants.

Formula	Watch out for
$K_p = \lim_{s \to 0} G(s)$ ; step error $\displaystyle \dfrac{1}{1 + K_p}$	Finite only for type 0
$K_v = \lim_{s \to 0} sG(s)$ ; ramp error $\displaystyle \dfrac{1}{K_v}$	Type 1: zero step error, finite ramp error
$K_a = \lim_{s \to 0} s^2G(s)$ ; parabola error $\displaystyle \dfrac{1}{K_a}$	Type 2: zero step and ramp errors
Input of size $A$ : multiply the unit error by $A$	A ramp $5t$ gives $5/K_v$
Sinusoid: output $\lvert T(j\omega) \rvert \sin(\omega t + \angle T(j\omega))$	Use the closed-loop $T$ , not $G$

Worked example. For $G = 40(s+2)/\big(s(s+4)(s+5)\big)$ the closed loop $s^3 + 9s^2 + 60s + 80$ is stable since $9 \times 60 > 80$ , so $K_v = 80/20 = 4$ and the unit ramp error is $0.25$ .

The trap is non-unity feedback. Reduce it to a unity feedback form before taking the limits.

Routh-Hurwitz: counting unstable roots

The Routh array counts right-half-plane roots without solving the polynomial. Each first-column sign change marks one.

Formula	Watch out for
Sign changes in column one $=$ right-half-plane roots	Positive coefficients are necessary, not sufficient
$s^3 + a_2 s^2 + a_1 s + a_0$ stable iff all positive and $a_2 a_1 > a_0$	Build the full array for higher orders
Zero in column one: replace it by $\varepsilon$	Let $\varepsilon \to 0^+$ before counting
Row of zeros: auxiliary polynomial from the row above	Its roots give the oscillation frequency

Worked example. For $G = K/\big((s+1)(s+2)(s+3)\big)$ the characteristic equation is $s^3 + 6s^2 + 11s + 6 + K = 0$ , so stability needs $66 > 6 + K$ and $6 + K > 0$ , giving $-6 < K < 60$ ; at $K = 60$ , $6s^2 + 66 = 0$ gives oscillation at $\sqrt{11} = 3.32$ rad/s.

Trap: a right-half-plane pole or zero in the plant can make closed-loop coefficients change sign with the gain. Always build the closed-loop polynomial, then the array.

Root locus: where the closed-loop poles travel

The root locus traces the closed-loop poles as $K$ rises from zero. Here $n$ counts open-loop poles and $m$ zeros.

Formula	Watch out for
Starts at the poles ( $K = 0$ ), ends at the zeros or at infinity	$n - m$ branches go to infinity
Real axis: left of an odd number of real poles and zeros	Count only those to the right of the point
Asymptote angles $\displaystyle \dfrac{(2q+1)180^\circ}{n - m}$ ; centroid $\displaystyle \dfrac{\sum p_i - \sum z_j}{n - m}$	Use signed pole values
Breakaway where $\displaystyle \dfrac{dK}{ds} = 0$	Keep only roots on the locus
Angle condition $\angle G(s)H(s) = \pm 180^\circ(2q+1)$ ; gain $\displaystyle K = \dfrac{\prod \lvert s - p_i \rvert}{\prod \lvert s - z_j \rvert}$	Factors must have unit leading coefficients
Departure from a complex pole $= 180^\circ - \sum \text{(other pole angles)} + \sum \text{(zero angles)}$	Arrival at a zero swaps the roles

Worked example. For $K/\big(s(s+1)(s+5)\big)$ the centroid is $(0 - 1 - 5)/3 = -2$ , the asymptotes lie at $60^\circ$ , $180^\circ$ and $300^\circ$ , $3s^2 + 12s + 5 = 0$ gives breakaway at $s = -0.473$ , and Routh gives the imaginary-axis crossing at $K = 30$ , $\omega = \sqrt{5} = 2.24$ rad/s.

The trap is the other breakaway root, $s = -3.53$ . It lies between $-1$ and $-5$ , where there is no locus, so reject it.

Bode plots, margins and frequency-domain specifications

A Bode plot draws magnitude in dB and phase against log frequency. Each factor adds a straight-line asymptote.

Formula	Watch out for
Each pole $-20$ dB/decade and $-90^\circ$ ; each zero $+20$ dB/decade and $+90^\circ$	$-45^\circ$ and a 3 dB error at a first-order corner
Quadratic pole pair $-40$ dB/decade, resonant peak if $\zeta < 0.707$	The asymptote hides the peak
Type 1: initial slope meets 0 dB at $\omega = K_v$ ; type 0: level $20\log_{10} K_p$	Extend the initial line past any corner
$e^{-sT}$ : magnitude 1, phase $-\omega T$ rad	Convert to degrees before adding
$\displaystyle \text{GM} = \dfrac{1}{\lvert GH(j\omega_{pc}) \rvert}$ ; in dB $-20\log_{10}\lvert GH(j\omega_{pc}) \rvert$	Read at the phase crossover
$\text{PM} = 180^\circ + \angle GH(j\omega_{gc})$	Read at the gain crossover
$\displaystyle M_r = \dfrac{1}{2\zeta\sqrt{1-\zeta^2}}$ at $\omega_r = \omega_n\sqrt{1 - 2\zeta^2}$	Valid only for $\zeta < 0.707$
$\omega_b = \omega_n\sqrt{1 - 2\zeta^2 + \sqrt{4\zeta^4 - 4\zeta^2 + 2}}$	Closed-loop bandwidth

Worked example (gain margin). For $10/\big(s(s+1)(s+5)\big)$ the phase is $-180^\circ$ at $\omega_{pc} = \sqrt{5}$ , where $\lvert G \rvert = 10/30$ , so $\text{GM} = 3$ , or $9.54$ dB, matching the Routh limit $K = 30$ .

Worked example (phase margin). For $2/\big(s(s+1)\big)$ , $\omega^4 + \omega^2 - 4 = 0$ gives $\omega_{gc} = 1.25$ rad/s, so $\text{PM} = 180^\circ - 90^\circ - \tan^{-1}1.25 = 38.7^\circ$ .

Worked example (specifications). With $\zeta = 0.5$ and $\omega_n = 10$ rad/s, $M_r = 1/(2 \times 0.5 \times 0.866) = 1.155$ at $\omega_r = 7.07$ rad/s, and $\omega_b = 10\sqrt{1.618} = 12.72$ rad/s.

These answers are decimals, so read the virtual calculator guide before the exam.

Nyquist: counting encirclements

The Nyquist criterion counts closed-loop right-half-plane poles from the open-loop frequency response, even when the open loop is unstable.

Formula	Watch out for
$Z = N + P$	$N$ counts clockwise encirclements of $-1 + j0$
Stable iff $Z = 0$ , so anticlockwise encirclements must equal $P$	Anticlockwise counts as negative $N$
Strictly proper $GH$ : the infinite arc maps to the origin	A proper $GH$ maps it to $\lim_{s \to \infty} GH$
Pole of order $q$ at the origin: an arc of $q \times 180^\circ$ at infinite radius, clockwise	Depends on the side of the indentation

Worked example. For $G = K/(s-1)$ , $P = 1$ , and for $K > 1$ the plot circles $-1$ once anticlockwise, so $N = -1$ and $Z = -1 + 1 = 0$ : stable, matching the closed loop $s - 1 + K$ .

In one line: Nyquist is bookkeeping, so write down $P$ first, count $N$ with its sign, and only then decide stability.

Compensators and P, PI and PID controllers

A lead compensator adds phase near crossover. A lag compensator raises low-frequency gain relative to crossover.

Formula	Watch out for
Lead $\displaystyle \dfrac{1 + aTs}{1 + Ts}$ , $a > 1$ : $\displaystyle \sin\phi_m = \dfrac{a-1}{a+1}$ at $\displaystyle \omega_m = \dfrac{1}{T\sqrt{a}}$	$\omega_m$ is the geometric mean of the corners
Lead gain at $\omega_m$ : $10\log_{10} a$ dB	Half the high-frequency gain $20\log_{10} a$
Lag $\displaystyle \dfrac{1 + Ts}{1 + \beta Ts}$ , $\beta > 1$ : attenuation $20\log_{10}\beta$ dB	Cuts error, slows the response
PI $K_p + K_i/s$ : raises the type by one	Removes step error, adds phase lag
PD $K_p + K_d s$ : phase lead $\tan^{-1}(K_d\omega/K_p)$	Amplifies noise
PID $K_p + K_i/s + K_d s$ : two zeros and a pole at the origin	Lead and lag together form lead-lag

Worked example. With $a = 4$ and $T = 0.05$ s, $\sin\phi_m = 3/5$ gives $\phi_m = 36.9^\circ$ at $\omega_m = 1/(0.05 \times 2) = 10$ rad/s, where the gain is $10\log_{10}4 = 6.02$ dB.

The trap is the P controller alone. A higher $K_p$ shrinks a type 0 step error but never removes it.

State space and the state transition matrix

The poles and the transfer function of a state-space model both come from $sI - A$ .

Formula	Watch out for
$\dot{x} = Ax + Bu$ , $y = Cx + Du$ ; $G(s) = C(sI - A)^{-1}B + D$	Keep the $D$ term
Poles $=$ roots of $\det(sI - A) = 0$	Eigenvalues of $A$
$\Phi(t) = e^{At} = \mathcal{L}^{-1}\big[(sI - A)^{-1}\big]$	Check $\Phi(0) = I$
$\Phi(t)^{-1} = \Phi(-t)$ ; $\Phi(t_1 + t_2) = \Phi(t_1)\Phi(t_2)$	A product, not a sum
$x(t) = \Phi(t)x(0) + \int_0^t \Phi(t - \tau)Bu(\tau)\,d\tau$	Zero-input plus zero-state
$z = Tx$ : $A_z = TAT^{-1}$ , $B_z = TB$ , $C_z = CT^{-1}$	Eigenvalues and $G(s)$ unchanged
Controllable iff $\text{rank}\,[B \;\, AB \;\, \cdots \;\, A^{n-1}B] = n$ ; observable iff $\text{rank}\,[C;\, CA;\, \cdots;\, CA^{n-1}] = n$	Full rank is the test

Worked example. Below, $\det(sI - A) = s^2 + 5s + 6$ , so the poles are $-2$ and $-3$ and $G(s) = 1/(s^2 + 5s + 6)$ ; partial fractions of $(sI-A)^{-1}$ give $\Phi(t)$ .

A = \begin{bmatrix} 0 & 1 \\ -6 & -5 \end{bmatrix},\; B = \begin{bmatrix} 0 \\ 1 \end{bmatrix},\; C = \begin{bmatrix} 1 & 0 \end{bmatrix},\qquad \Phi(t) = \begin{bmatrix} 3e^{-2t} - 2e^{-3t} & e^{-2t} - e^{-3t} \\ -6e^{-2t} + 6e^{-3t} & -2e^{-2t} + 3e^{-3t} \end{bmatrix}

At $t = 0$ this is the identity. The trap is the direction of the transformation: $x = Tz$ gives $T^{-1}AT$ , while $z = Tx$ gives $TAT^{-1}$ .

The last-minute sheet in the GATE EE 2027 book covers every syllabus topic this way, each result with its conditions. The book has 643 pages, 914 questions with worked solutions and 10 full mock tests, built on all 5 GATE EE papers from 2022 to 2026, counted question by question. For MCQ, MSQ and NAT marking, see the exam pattern and marking scheme, common to every paper, and the timetable in GATE 2027 exam dates.

More formula sheets: all of GATE EE · Electrical Machines · Power Electronics · Power Systems

Quick revision

Negative feedback gives $G/(1+GH)$ ; Mason subtracts the loop sum and adds non-touching products.
Second order: $t_p = \pi/\omega_d$ , overshoot from $\zeta$ alone, $t_s = 4/(\zeta\omega_n)$ at 2 per cent.
Check stability, then use $1/(1+K_p)$ , $1/K_v$ and $1/K_a$ .
A cubic is stable when all coefficients are positive and $a_2 a_1 > a_0$ .
Root locus: centroid $(\sum p - \sum z)/(n-m)$ , breakaway from $dK/ds = 0$ .
Gain margin at the phase crossover, phase margin at the gain crossover.
Nyquist: $Z = N + P$ , clockwise $N$ positive, stable when $Z = 0$ .
Lead: $\sin\phi_m = (a-1)/(a+1)$ ; state-space poles from $\det(sI - A)$ .

Frequently asked questions

What is the formula for peak overshoot in a second order system?

For the standard underdamped second order system the peak overshoot is $100\,e^{-\pi\zeta/\sqrt{1-\zeta^2}}$ per cent. It depends on the damping ratio alone, not on the natural frequency. A damping ratio of 0.6 gives about 9.5 per cent, and 0.5 gives about 16.3 per cent. Use it only for a pure second order system with no zero, or when one pair of poles clearly dominates.

How do you find the steady-state error for a ramp input?

Compute the velocity error constant $K_v$ as the limit of $sG(s)$ as $s$ tends to zero, for a unity feedback loop. The error to a unit ramp is $1/K_v$ . A type 0 loop has infinite ramp error, type 1 a finite error and type 2 zero error. Check first that the closed loop is stable, because the error constants mean nothing for an unstable loop.

What is the Nyquist stability criterion formula?

The criterion is $Z = N + P$ . Here $P$ is the number of open-loop poles in the right half plane, $N$ is the number of clockwise encirclements of the point minus one by the plot of $G(j\omega)H(j\omega)$ , and $Z$ is the number of closed-loop right half plane poles. The closed loop is stable only when $Z$ is zero. Anticlockwise encirclements count as negative.

What is the difference between gain margin and phase margin?

Gain margin is read at the phase crossover frequency, where the phase is minus 180 degrees, and equals one over the open-loop magnitude there. Phase margin is read at the gain crossover frequency, where the magnitude is one, and equals 180 degrees plus the phase there. For a minimum phase system the closed loop is stable when both margins are positive.

How do I revise control systems formulas before GATE EE?

Read a compact sheet daily in the last two weeks and attach one trap to each formula. Work one small example for each sub-topic by hand, such as a Routh array or a margin calculation, so the steps stay quick. Keep full precision on the virtual calculator and round only the final numerical answer.

Sources

Dates, fees and the syllabus are set by the GATE 2027 organising institute and can change. Always confirm at gate2027.iitm.ac.in.