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Frequency response

Bode and polar plots, gain and phase margins

Frequency response describes what a stable linear system does to a sinusoid at each frequency: it scales the sinusoid by $∣ G (j ω) ∣$ and shifts it by $∠ G (j ω)$ . From the open-loop $G (j ω) H (j ω)$ , together with the number of open-loop poles in the right half plane, you can judge whether the closed loop is stable and how close it is to instability, using gain and phase margins, Bode asymptotes and polar plots. The core skills are: sketch or read a Bode plot, find where a polar plot starts, ends and crosses the axes, work out GM, PM, the largest allowable gain or delay, and link $ζ$ to resonant peak, bandwidth and phase margin.

27 previous-year questions · 26 solved · 6 practice questions

The sinusoidal transfer function

Drive a stable LTI system $G (s)$ with $u (t) = A sin ω t$ . Once the transients die out, the output is a sinusoid of the same frequency. It is scaled by $∣ G (j ω) ∣$ and shifted by $∠ G (j ω)$ . The function $G (j ω)$ , which you get by putting $s = j ω$ , is the sinusoidal transfer function, and its behaviour as $ω$ runs from $0$ to $\infty$ is the frequency response. A negative angle is a phase lag.

Quick check: take $G (s) = \frac{10}{s + 5}$ with input $2 sin 5 t$ . Then $G (j 5) = \frac{10}{5 + j 5}$ , so $∣ G ∣ = 2$ and $∠ G = - 4 5^{\circ}$ . The steady-state output is $22 sin (5 t - 4 5^{\circ})$ .

SINUSOIDAL STEADY-STATE RESPONSE

u (t) = A sin ω t \Rightarrow y_{ss} (t) = A ∣ G (j ω) ∣ sin (ω t + ∠ G (j ω))

Valid only for a stable $G (s)$ , after the transients decay. The frequency does not change, only the amplitude and phase.

Stability matters only for this physical reading. An unstable $G (s)$ has no steady-state sinusoid, but you can still compute and plot $G (j ω)$ , and that is exactly what the Nyquist criterion does with the open-loop function. Every margin in this topic is read from the open-loop $G (j ω) H (j ω)$ of a negative-feedback loop, never from the closed-loop $T (j ω)$ .

TIME-CONSTANT FORM: MAGNITUDES ADD IN DB, PHASES ADD

G (s) ∣ G (j ω) ∣_{dB} ∠ G (j ω) = \frac{K \prod _{i} ( 1 + s T _{z i} )}{s ^{N} \prod _{k} ( 1 + s T _{p k} )} = 20 lo g_{10} K + i \sum 20 lo g_{10} 1 + ω^{2} T_{z i}^{2} - 20 N lo g_{10} ω - k \sum 20 lo g_{10} 1 + ω^{2} T_{p k}^{2} = i \sum tan^{- 1} (ω T_{z i}) - 9 0^{\circ} N - k \sum tan^{- 1} (ω T_{p k})

For $K > 0$ and real poles and zeros in the left half plane ( $T_{z i}, T_{p k} > 0$ ). A quadratic factor adds its own magnitude and phase terms (see the Bode table below). $N$ is the system type, the number of poles at the origin. This additive structure is why Bode plots are built factor by factor.

Frequency-domain specifications and their link to ζ

These specifications are read from the closed-loop response $T (j ω) = C (j ω) / R (j ω)$ :

The resonant peak $M_{r}$ is the maximum value of $∣ T (j ω) ∣$ .
The resonant frequency $ω_{r}$ is the frequency where that maximum occurs.
The bandwidth $ω_{B}$ is the frequency where $∣ T (j ω) ∣$ falls to $1/ 2$ of its zero-frequency value $∣ T (0) ∣$ , i.e. 3 dB below it.
The cut-off rate is the slope of the magnitude beyond $ω_{B}$ .

Bandwidth measures speed. For the standard second-order system with $ζ$ fixed, $ω_{B}$ is proportional to $ω_{n}$ while rise time and settling time are proportional to $1/ ω_{n}$ , so doubling the bandwidth halves them. $M_{r}$ measures relative stability, as overshoot does in the time domain.

Take the standard second-order system $T (s) = \frac{ω _{n}^{2}}{s ^{2} + 2 ζ ω _{n} s + ω _{n}^{2}}$ . It is unity feedback around $G (s) = \frac{ω _{n}^{2}}{s ( s + 2 ζ ω _{n} )}$ , so $∣ T (0) ∣ = 1$ . Every specification follows from $ζ$ and $ω_{n}$ . Throughout, write $u = ω / ω_{n}$ .

STANDARD SECOND-ORDER MAGNITUDE AND PHASE

∣ T (j ω) ∣ ∠ T (j ω) = \frac{1}{( 1 - u ^{2} ) ^{2} + ( 2 ζ u ) ^{2}} = - tan^{- 1} (\frac{2 ζ u}{1 - u ^{2}}) (u < 1), - 18 0^{\circ} + tan^{- 1} (\frac{2 ζ u}{u ^{2} - 1}) (u > 1)

$u = ω / ω_{n}$ , $ζ > 0$ . The phase runs continuously from $0^{\circ}$ to $- 18 0^{\circ}$ , so pick the branch by whether $u$ is below or above 1.

VALUE AT THE NATURAL FREQUENCY

u = 1 : ∣ T (j ω_{n}) ∣ = \frac{1}{2 ζ}, ∠ T (j ω_{n}) = - 9 0^{\circ}

Holds for every $ζ > 0$ . In dB the value is $- 20 lo g_{10} (2 ζ)$ , which is 0 dB exactly when $ζ = 0.5$ .

RESONANT FREQUENCY

ω_{r} = ω_{n} 1 - 2 ζ^{2}

Only for $0 < ζ < 1/ 2 \approx 0.707$ . In that range $ω_{r} < ω_{d} < ω_{n}$ , where $ω_{d} = ω_{n} 1 - ζ^{2}$ .

RESONANT PEAK

M_{r} = \frac{1}{2 ζ 1 - ζ ^{2}}

Only for $0 < ζ < 0.707$ . For $ζ \geq 0.707$ there is no peak, $∣ T ∣$ falls monotonically, and $M_{r} = 1$ at $ω = 0$ . $M_{r}$ depends on $ζ$ alone.

BANDWIDTH OF THE STANDARD SECOND-ORDER SYSTEM

ω_{B} = ω_{n} (1 - 2 ζ^{2}) + 4 ζ^{4} - 4 ζ^{2} + 2

The frequency where $∣ T ∣ = 1/ 2$ (−3 dB relative to $∣ T (0) ∣ = 1$ ). Valid for every $ζ > 0$ . $ω_{B} = ω_{n}$ exactly at $ζ = 1/ 2$ and $ω_{B} = 0.644 ω_{n}$ at $ζ = 1$ .

BANDWIDTH OF A FIRST-ORDER SYSTEM

T (s) = \frac{1}{1 + s τ} \Rightarrow ω_{B} = \frac{1}{τ}

The bandwidth equals the corner frequency. A smaller time constant gives a wider bandwidth and a faster response.

PHASE MARGIN OF THE STANDARD SECOND-ORDER LOOP

ω_{g c} = ω_{n} 1 + 4 ζ^{4} - 2 ζ^{2}, PM = tan^{- 1} \frac{2 ζ}{1 + 4 ζ ^{4} - 2 ζ ^{2}}

For open loop $G (s) = ω_{n}^{2} / [s (s + 2 ζ ω_{n})]$ with unity negative feedback. GM is infinite because the phase never reaches $- 18 0^{\circ}$ . Rough approximation: PM in degrees $\approx 100 ζ$ for $0 < ζ \leq 0.6$ , off by at most about 3.5°. Use the exact formula when the options are close.

$ζ$	$M_{r}$ (ratio / dB)	$ω_{r} / ω_{n}$	$ω_{B} / ω_{n}$	PM	$M_{p}$ (%)
0.2	2.55 / 8.14 dB	0.959	1.51	22.6°	52.7
0.3	1.75 / 4.85 dB	0.906	1.45	33.3°	37.2
0.4	1.36 / 2.70 dB	0.825	1.37	43.1°	25.4
0.5	1.15 / 1.25 dB	0.707	1.27	51.8°	16.3
0.6	1.04 / 0.35 dB	0.529	1.15	59.2°	9.5
0.707	1 / 0 dB (peak just vanishes)	0 (peak at $ω = 0$ )	1.00	65.5°	4.3
1.0	no peak ( $M_{r} = 1$ at $ω = 0$ )	—	0.644	76.3°	0

EXAM TIP

Do not mix the two families of quantities. $M_{r}$ , $ω_{r}$ and $ω_{B}$ belong to the closed-loop $T (j ω)$ . $ω_{g c}$ , $ω_{p c}$ , GM and PM belong to the open-loop $G (j ω) H (j ω)$ . For the standard second-order system, $M_{r}$ , $M_{p}$ and PM depend on $ζ$ only, while $ω_{r}$ , $ω_{B}$ and $ω_{g c}$ all scale with $ω_{n}$ .

EXAM TIP

A resonant peak needs $ζ < 0.707$ , but time-domain overshoot needs only $ζ < 1$ . A system with $ζ = 0.8$ overshoots by about 1.5% and still has no resonant peak.

Bode plots: building them factor by factor

A Bode plot is a pair of graphs against $lo g_{10} ω$ : magnitude in dB and phase in degrees. On log scales, products become sums, so you plot each factor on its own and add them. Always rewrite the transfer function in time-constant form first, so that every factor looks like $(1 + s T)$ and the constant in front is the time-constant (Bode) gain $K = lim_{s \to 0} s^{N} G (s)$ . For a type-0 system this is the DC gain.

DECIBELS, DECADES AND SLOPES

M_{dB} = 20 lo g_{10} ∣ G (j ω) ∣, M (ω_{2}) = M (ω_{1}) + S lo g_{10} \frac{ω _{2}}{ω _{1}}

$S$ is the slope in dB/decade, and the relation holds along one straight segment. A decade is a factor of 10 and an octave a factor of 2, so 20 dB/decade $\approx$ 6.02 dB/octave.

Factor	Magnitude asymptote	Phase	Note
$K > 0$	flat at $20 lo g_{10} K$	$0^{\circ}$	$K < 0$ adds $\pm 18 0^{\circ}$
$1/ (j ω)^{N}$ (poles at origin)	straight line, $- 20 N$ dB/dec, through 0 dB at $ω = 1$	$- 9 0^{\circ} N$ (constant)	sets the low-frequency slope
$(j ω)^{N}$ (zeros at origin)	$+ 20 N$ dB/dec through 0 dB at $ω = 1$	$+ 9 0^{\circ} N$	low-frequency slope is positive
$1/ (1 + j ω T)$ , simple pole	0 dB up to $ω_{c} = 1/ T$ , then $- 20$ dB/dec	$0^{\circ} \to - 9 0^{\circ}$ , $- 4 5^{\circ}$ at $ω_{c}$	exact value $- 3.01$ dB at $ω_{c}$
$(1 + j ω T)$ , LHP zero	0 dB up to $1/ T$ , then $+ 20$ dB/dec	$0^{\circ} \to + 9 0^{\circ}$ , $+ 4 5^{\circ}$ at $ω_{c}$	exact value $+ 3.01$ dB at $ω_{c}$
$1/ [1 + 2 ζ (j ω / ω_{n}) + (j ω / ω_{n})^{2}]$	0 dB up to $ω_{n}$ , then $- 40$ dB/dec	$0^{\circ} \to - 18 0^{\circ}$ , $- 9 0^{\circ}$ at $ω_{n}$	exact value $- 20 lo g_{10} 2 ζ$ at $ω_{n}$ ; peak if $ζ < 0.707$
$1 + 2 ζ (j ω / ω_{n}) + (j ω / ω_{n})^{2}$	0 dB up to $ω_{n}$ , then $+ 40$ dB/dec	$0^{\circ} \to + 18 0^{\circ}$ , $+ 9 0^{\circ}$ at $ω_{n}$	exact value $+ 20 lo g_{10} 2 ζ$ at $ω_{n}$ ; dip if $ζ < 0.707$
$e^{- j ω T}$ , time delay	0 dB at every frequency	$- ω T$ rad, grows without bound	changes phase only

SIMPLE-POLE EXACT VALUES NEAR THE CORNER Ω_C = 1/T

ω = ω_{c} ω = ω_{c} /2, 2 ω_{c} ω = 0.1 ω_{c}, 10 ω_{c} : - 3.01 dB, - 4 5^{\circ} : error - 0.97 dB; phase - 26. 6^{\circ}, - 63. 4^{\circ} : error - 0.04 dB; phase - 5. 7^{\circ}, - 84. 3^{\circ}

The errors are measured from the straight-line asymptote. The usual straight-line phase approximation (0° at $0.1 ω_{c}$ , −45°/decade, −90° at $10 ω_{c}$ ) is never off by more than 5.7°. For a zero, flip every sign.

LOW-FREQUENCY ASYMPTOTE AND ITS 0 DB CROSSING

∣ G ∣_{dB, LF} = 20 lo g_{10} K - 20 N lo g_{10} ω, ω_{0 dB} = K^{1/ N} (N \geq 1)

$K$ must be the time-constant-form gain. The line (extended if necessary) reads $20 lo g_{10} K$ at $ω = 1$ . Type 0: the plot starts flat at $20 lo g_{10} K$ .

STATIC ERROR CONSTANTS FROM THE LOW-FREQUENCY ASYMPTOTE

Type 0: K_{p} = K = 1 0^{M_{LF} /20}, Type 1: K_{v} = K = ω_{0 dB}, Type 2: K_{a} = K = ω_{0 dB}^{2}

For unity feedback, where $K_{p} = lim_{s \to 0} G (s)$ , $K_{v} = lim_{s \to 0} s G (s)$ and $K_{a} = lim_{s \to 0} s^{2} G (s)$ . $K$ is the time-constant gain, $M_{LF}$ is the flat low-frequency level in dB (type 0), and $ω_{0 dB}$ is where the low-frequency line, extended if needed, crosses 0 dB.

Write $G (j ω) H (j ω)$ in time-constant form $K \prod (1 + j ω T_{z}) / [(j ω)^{N} \prod (1 + j ω T_{p})]$ and list the corner frequencies $1/ T$ in ascending order.
Draw the low-frequency asymptote with slope $- 20 N$ dB/decade through $20 lo g_{10} K$ at $ω = 1$ . If $ω = 1$ lies beyond the first corner, extend the line to $ω = 1$ to locate it.
Move right. At each corner, change the slope by $- 20$ for a simple pole, $+ 20$ for a simple zero, $- 40$ for a quadratic pole pair and $+ 40$ for a quadratic zero pair. A factor repeated $r$ times multiplies the change by $r$ .
Check that the final slope is $- 20 (n - m)$ dB/decade, where $n$ is the number of poles and $m$ the number of zeros.
If exact values are needed, correct the asymptote at each corner: $- 3$ dB for a simple pole, $+ 3$ dB for a simple zero ( $\mp 3 r$ dB for a factor repeated $r$ times), $- 20 lo g_{10} (2 ζ)$ for a quadratic pole pair and $+ 20 lo g_{10} (2 ζ)$ for a quadratic zero pair. Neighbouring corners add their own small errors, so this is exact only when corners are well separated.
Phase: add the factor phases at a few chosen frequencies. For a minimum-phase function with $K > 0$ it starts at $- 9 0^{\circ} N$ and ends at $- 9 0^{\circ} (n - m)$ .

EXAM TIP

The $K^{1/ N}$ rule gives where the low-frequency line (extended) crosses 0 dB. If a corner lies below that frequency, the actual plot crosses 0 dB somewhere else, so walk along the segments instead.

EXAM TIP

A classic slip is using the pole-zero gain instead of the time-constant gain. For $G (s) = \frac{100 ( s + 2 )}{s ( s + 10 )}$ the time-constant gain is $K = 100 \times 2/10 = 20$ , not 100. The low-frequency line therefore reads $26$ dB at $ω = 1$ and, extended, crosses 0 dB at $ω = 20$ rad/s. The actual asymptotic plot does not cross there: it levels off at 20 dB between the zero corner at 2 rad/s and the pole corner at 10 rad/s, then falls at −20 dB/decade and crosses 0 dB at 100 rad/s.

Reading a transfer function; minimum phase, non-minimum phase and time delay

Reading a Bode plot means running the construction backwards. Slopes identify the factor types, the frequencies where the slope changes are the corners, and the low-frequency line gives the type and $K$ . A magnitude plot fixes the transfer function only if you are told it is minimum phase, because an RHP zero has exactly the same magnitude as its LHP mirror image.

Initial slope $- 20 N$ dB/decade gives type $N$ . A slope of $+ 20$ means a zero at the origin.
Find $K$ from any point on the low-frequency line: $20 lo g_{10} K = M (ω) + 20 N lo g_{10} ω$ . For type 0, $K$ is the flat starting level.
A slope change of $+ 20$ at $ω_{c}$ is a zero $(1 + s / ω_{c})$ . A change of $- 20$ is a pole $(1 + s / ω_{c})^{- 1}$ . A change of $\pm 40$ is a double factor, or a quadratic if the plot shows a peak or dip; then $ζ$ follows from the height $- 20 lo g_{10} 2 ζ$ at $ω_{n}$ .
If the options are in pole-zero form, convert: $K_{pole-zero} = K \times \prod p_{k} / \prod z_{i}$ , where the $p_{k}$ and $z_{i}$ are the corner frequencies (magnitudes of the poles and zeros).

WORKED EXAMPLE

The asymptotic Bode magnitude plot of a minimum-phase open-loop transfer function $G (s)$ (unity negative feedback) has slope −40 dB/decade for $ω < 2$ rad/s, −20 dB/decade for $2 < ω < 10$ rad/s and −40 dB/decade for $ω > 10$ rad/s. The asymptote reads 20 dB at $ω = 2$ rad/s. Find $G (s)$ , the gain crossover frequency, the phase margin and the gain margin, and state whether the closed loop is stable.

Solution

Initial slope −40 dB/decade, so the system is type 2. The +20 change at 2 is a zero $(1 + s /2)$ and the −20 change at 10 is a pole $(1 + s /10)$ . So $G (s) = \frac{K ( 1 + s /2 )}{s ^{2} ( 1 + s /10 )}$ .
On the low-frequency line $∣ G ∣ \approx K / ω^{2}$ . At $ω = 2$ this is $K /4 = 10$ (20 dB), so $K = 40$ .
Pole-zero form: $G (s) = \frac{40 \cdot \frac{s + 2}{2}}{s ^{2} \cdot \frac{s + 10}{10}} = \frac{200 ( s + 2 )}{s ^{2} ( s + 10 )}$ .
Asymptotic crossover: for $2 < ω < 10$ , $∣ G ∣ \approx 20/ ω$ , which is 2 (6.02 dB) at $ω = 10$ . For $ω > 10$ , $∣ G ∣ \approx 200/ ω^{2} = 1$ at $ω = 200 = 14.1$ rad/s.
Exact crossover: set $200 ω^{2} + 4 = ω^{2} ω^{2} + 100$ and put $x = ω^{2}$ . This gives $x^{3} + 100 x^{2} - 40000 x - 160000 = 0$ , so $x \approx 158.6$ and $ω_{g c} \approx 12.6$ rad/s.
Phase: $∠ G = - 18 0^{\circ} + tan^{- 1} (ω /2) - tan^{- 1} (ω /10)$ . At 12.6 rad/s this is $- 18 0^{\circ} + 80.9 8^{\circ} - 51.5 5^{\circ} = - 150. 6^{\circ}$ , so $PM = 29. 4^{\circ}$ . The asymptotic 14.1 rad/s would give 27.2°, a fair estimate.
For every $ω > 0$ , $ω /2 > ω /10$ , so $∠ G > - 18 0^{\circ}$ at all finite frequencies. There is no phase crossover, so GM $= \infty$ .
The open loop has no RHP poles and PM > 0, so the closed loop is stable. Check with Routh: $s^{3} + 10 s^{2} + 200 s + 400$ has $10 \times 200 > 400$ .

Answer: $G (s) = \frac{200 ( s + 2 )}{s ^{2} ( s + 10 )}$ . $ω_{g c} \approx 12.6$ rad/s (asymptotic estimate 14.1 rad/s), PM $\approx 29. 4^{\circ}$ , GM $= \infty$ . The closed loop is stable.

A minimum-phase transfer function has no poles or zeros in the right half s-plane and no time delay. For it, the magnitude plot fixes the phase uniquely, and the high-frequency slope $- 20 (n - m)$ dB/decade always comes with a final phase of $- 9 0^{\circ} (n - m)$ (for $K > 0$ ). A non-minimum-phase function has at least one RHP zero, or a delay. It has the same magnitude as its minimum-phase counterpart but more phase lag. Some texts also call a function with RHP poles non-minimum phase. For Bode reading, what matters is the RHP zero or the delay.

Property	Minimum phase	Non-minimum phase
Pole/zero location	all poles and zeros in the LHP, no delay	at least one RHP zero, or $e^{- s T}$
Magnitude vs phase	phase fixed by the magnitude plot	same magnitude as the mirror-image LHP function, extra lag
High-frequency phase	$- 9 0^{\circ} (n - m)$ , consistent with the slope $- 20 (n - m)$ dB/dec	more negative than $- 9 0^{\circ} (n - m)$ ( $- 18 0^{\circ}$ extra per RHP zero); with a delay it goes to $- \infty$
Typical factor	$(1 + s T)$ : phase $+ tan^{- 1} ω T$	$(1 - s T)$ : phase $- tan^{- 1} ω T$

RHP ZERO

(1 - j ω T) : ∣ \cdot ∣ = 1 + ω^{2} T^{2}, ∠ = - tan^{- 1} (ω T)

Same magnitude as an LHP zero, but the phase is a lag. $T > 0$ .

FIRST-ORDER ALL-PASS

\frac{1 - s T}{1 + s T} : ∣ \cdot ∣ = 1, ∠ = - 2 tan^{- 1} (ω T)

$T > 0$ . Magnitude is 0 dB everywhere and the phase goes from $0^{\circ}$ to $- 18 0^{\circ}$ , even though $n - m = 0$ .

TIME DELAY

e^{- j ω T} : ∣ \cdot ∣ = 1, ∠ = - ω T rad = - \frac{18 0 ^{\circ}}{π} ω T

The magnitude plot is unchanged; the phase lag grows linearly with $ω$ , which curves downward on a log-frequency axis. Remember 1 rad = 57.3°.

FIRST-ORDER PADÉ APPROXIMATION OF A DELAY

e^{- s T} \approx \frac{1 - s T /2}{1 + s T /2}

An all-pass, non-minimum-phase rational approximation. It is accurate at low frequencies ( $ω T$ small).

INTEGRATOR WITH DELAY

G H = \frac{K e ^{- s T}}{s} : ω_{p c} = \frac{π}{2 T}, GM = \frac{π}{2 K T}, stable ⟺ 0 < K < \frac{π}{2 T}

Negative feedback, $K > 0$ , $T > 0$ . The phase crossover comes from $- 9 0^{\circ} - ω T \cdot \frac{18 0 ^{\circ}}{π} = - 18 0^{\circ}$ . Here $∣ G H ∣ = K / ω$ crosses 1 only once, at $ω_{g c} = K$ , and the phase falls monotonically, so the margins decide stability. Equivalently PM $= 9 0^{\circ} - K T \cdot \frac{18 0 ^{\circ}}{π}$ must be positive.

EXAM TIP

Test for minimum phase on a given plot (with $K > 0$ ): if the magnitude settles to $- 20 k$ dB/decade but the high-frequency phase ends below $- 9 0^{\circ} k$ , or keeps falling without limit, there is an RHP zero or a delay. This test cannot detect an RHP pole, because at high frequency an RHP pole contributes $- 9 0^{\circ}$ , the same as an LHP pole.

EXAM TIP

A delay leaves the magnitude curve alone, so $ω_{g c}$ does not move, and it cuts PM by exactly $ω_{g c} T$ radians. It also pulls $ω_{p c}$ lower, so when $∣ G H ∣$ falls with frequency (the usual case) GM falls too. With a delay, the phase always reaches $- 18 0^{\circ}$ eventually, so GM is finite even for a first-order plant.

Polar plots

A polar plot is the locus of $G (j ω) H (j ω)$ in the complex plane as $ω$ goes from $0$ to $\infty$ . The distance from the origin is the magnitude, and the angle is measured from the positive real axis, counter-clockwise positive. When the phase lag grows with $ω$ (the usual case), the locus therefore moves clockwise as $ω$ increases. The Nyquist plot adds the mirror image for negative $ω$ (and the detour arcs). To sketch a polar plot, find three things: the start ( $ω \to 0$ ), the end ( $ω \to \infty$ ) and the axis crossings.

START AND END POINTS

ω \to 0 : ω \to \infty : G \to K ∠ 0^{\circ} (N = 0), G \to \infty∠ (- 9 0^{\circ} N) (N \geq 1) G \to 0∠ (- 9 0^{\circ} (n - m)) (n > m)

For a minimum-phase function with $K > 0$ in time-constant form. The plot reaches the origin tangent to the axis at angle $- 9 0^{\circ} (n - m)$ . If $n = m$ , it ends at the finite point equal to the ratio of the leading coefficients.

AXIS CROSSINGS

Im G (j ω) = 0 \Rightarrow real-axis crossing; Re G (j ω) = 0 \Rightarrow imaginary-axis crossing

A crossing of the negative real axis happens at the phase crossover frequency $ω_{p c}$ . If the plot crosses at $- a$ , then GM $= 1/ a$ .

LOW-FREQUENCY ASYMPTOTE OF A TYPE-1 POLAR PLOT

G (s) = \frac{K \prod ( 1 + s T _{z} )}{s \prod ( 1 + s T _{p} )} : ω \to 0 lim Re G (j ω) = - K (\sum T_{p} - \sum T_{z})

The plot starts at infinity along −90°, running beside this vertical line. For $K / [s (1 + s T)]$ the asymptote is $Re = - K T$ .

$G (s)$ (all $T > 0$ , $K > 0$ )	Start ( $ω \to 0$ )	End ( $ω \to \infty$ )	Shape and crossings
$\frac{K}{1 + s T}$	$K ∠ 0^{\circ}$	$0∠ - 9 0^{\circ}$	semicircle in the 4th quadrant, centre $K /2$ , diameter $K$ ; $(K / 2) ∠ - 4 5^{\circ}$ at $ω = 1/ T$
$\frac{K}{( 1 + s T _{1} ) ( 1 + s T _{2} )}$	$K ∠ 0^{\circ}$	$0∠ - 18 0^{\circ}$	crosses the negative imaginary axis at $ω = 1/ T_{1} T_{2}$ , value $- j K T_{1} T_{2} / (T_{1} + T_{2})$ ; never crosses the negative real axis
$\frac{K}{( 1 + s T _{1} ) ( 1 + s T _{2} ) ( 1 + s T _{3} )}$	$K ∠ 0^{\circ}$	$0∠ - 27 0^{\circ}$	crosses the negative real axis at $ω = (T_{1} + T_{2} + T_{3}) / (T_{1} T_{2} T_{3})$ , so GM is finite; with $T_{1} = T_{2} = T_{3}$ the crossing is at $- K /8$ , so the closed loop is stable only for $K < 8$
$\frac{K}{s}$	$\infty∠ - 9 0^{\circ}$	$0∠ - 9 0^{\circ}$	the whole negative imaginary axis
$\frac{K}{s ( 1 + s T )}$	$\infty∠ - 9 0^{\circ}$	$0∠ - 18 0^{\circ}$	vertical asymptote $Re = - K T$ ; GM $= \infty$
$\frac{K}{s ( 1 + s T _{1} ) ( 1 + s T _{2} )}$	$\infty∠ - 9 0^{\circ}$	$0∠ - 27 0^{\circ}$	crosses the negative real axis at $ω = 1/ T_{1} T_{2}$ , at $- K T_{1} T_{2} / (T_{1} + T_{2})$
$\frac{K}{s ^{2} ( 1 + s T )}$	$\infty∠ - 18 0^{\circ}$	$0∠ - 27 0^{\circ}$	lies entirely in the 2nd quadrant; closed loop unstable for every $K > 0$
$\frac{K e ^{- s T}}{1 + s T _{1}}$	$K ∠ 0^{\circ}$	spirals into the origin	crosses the negative real axis infinitely many times

CONSTANT-M CIRCLES (UNITY FEEDBACK)

\frac{G}{1 + G} = M (M \neq = 1) : centre (- \frac{M ^{2}}{M ^{2} - 1}, 0), radius \frac{M}{M ^{2} - 1}; M = 1 : the line Re G = - \frac{1}{2}

Unity negative feedback only. These are the loci in the $G (j ω)$ plane where the closed-loop magnitude $∣ T (j ω) ∣$ equals $M$ . For $M > 1$ the circles lie left of $Re G = - 1/2$ , each encloses $- 1 + j 0$ , and they shrink onto $- 1 + j 0$ as $M \to \infty$ . The resonant peak $M_{r}$ is the $M$ of the circle the polar plot is tangent to (the highest $M$ it reaches), and $ω_{r}$ is the frequency at the point of tangency.

EXAM TIP

Multiplying by $1/ (j ω)$ rotates every point of the plot by $- 9 0^{\circ}$ and scales it by $1/ ω$ . Each extra simple LHP pole adds $- 9 0^{\circ}$ to the final approach angle. This is how you get the start and end of any plot in the table without computing anything.

EXAM TIP

To find where a polar plot crosses the negative real axis, rationalise $G (j ω)$ , set the imaginary part to zero and solve for $ω$ , then substitute back. For two simple poles plus an integrator, you can skip the algebra: the crossing is at $ω = 1/ T_{1} T_{2}$ .

Gain margin and phase margin

Convention: negative feedback, open-loop $G (s) H (s)$ , $K > 0$ , and phase written as a lag (a negative angle such as $- 15 0^{\circ}$ ). The gain crossover frequency $ω_{g c}$ is where $∣ G H ∣ = 1$ (0 dB). The phase crossover frequency $ω_{p c}$ is where $∠ G H = - 18 0^{\circ}$ . Gain margin is the factor by which the loop gain can be multiplied before the closed loop reaches the edge of instability. Phase margin is the extra phase lag at $ω_{g c}$ that would bring it to the same edge. Both measure the distance of the plot from the critical point $- 1 + j 0$ .

CROSSOVER FREQUENCIES

∣ G (j ω_{g c}) H (j ω_{g c}) ∣ = 1 (0 dB), ∠ G (j ω_{p c}) H (j ω_{p c}) = - 18 0^{\circ}

Both are open-loop quantities, found on the Bode plot or on the polar plot (unit circle and negative real axis).

GAIN MARGIN

GM = \frac{1}{∣ G ( j ω _{p c} ) H ( j ω _{p c} ) ∣}, GM_{dB} = - 20 lo g_{10} ∣ G (j ω_{p c}) H (j ω_{p c}) ∣

If the polar plot crosses the negative real axis at $- a$ , then GM $= 1/ a$ . A positive dB value means stable, in the standard case described below.

PHASE MARGIN

PM = 18 0^{\circ} + ∠ G (j ω_{g c}) H (j ω_{g c})

Use the phase as a negative angle: $- 15 0^{\circ}$ gives PM $= 3 0^{\circ}$ . On the polar plot, PM is the angle from the negative real axis to the point where the plot meets the unit circle.

LARGEST GAIN AND LARGEST DELAY

K_{marginal} = K \cdot GM, T_{m a x} = \frac{PM \times π /180}{ω _{g c}} (PM in degrees)

$K$ is the gain at which GM was computed. $T_{m a x}$ is the delay that uses up the whole phase margin; it assumes a single gain crossover, which a delay does not move.

TYPE-1, THIRD-ORDER LOOP

G H = \frac{K}{s ( 1 + s T _{1} ) ( 1 + s T _{2} )} : ω_{p c} = \frac{1}{T _{1} T _{2}}, ∣ G H (j ω_{p c}) ∣ = \frac{K T _{1} T _{2}}{T _{1} + T _{2}}, K_{marginal} = \frac{T _{1} + T _{2}}{T _{1} T _{2}}

$T_{1}, T_{2} > 0$ . Comes from $tan^{- 1} a + tan^{- 1} b = 9 0^{\circ} ⟺ ab = 1$ for $a, b > 0$ . It agrees with the Routh result for $T_{1} T_{2} s^{3} + (T_{1} + T_{2}) s^{2} + s + K = 0$ . In pole-zero form $G H = K / [s (s + a) (s + b)]$ with $a, b > 0$ : $ω_{p c} = ab$ and $K_{marginal} = ab (a + b)$ .

The margins alone decide stability when the open loop has no RHP poles, and $∣ G H ∣$ and $∠ G H$ each cross 1 and $- 18 0^{\circ}$ only once (the usual minimum-phase case).
Reading off a Bode plot: GM (dB) is how far the magnitude curve lies below 0 dB at $ω_{p c}$ , and PM is how far the phase curve lies above $- 18 0^{\circ}$ at $ω_{g c}$ .
Stable: GM > 1 (GM $_{dB}$ > 0) and PM > 0, which is the same as $ω_{g c} < ω_{p c}$ .
Marginally stable: GM = 1 (0 dB) and PM = 0, so $ω_{g c} = ω_{p c}$ . The closed loop oscillates steadily at that frequency.
Unstable: GM < 1 (negative dB) and PM < 0, so $ω_{g c} > ω_{p c}$ .
If the phase stays above $- 18 0^{\circ}$ at every finite frequency (for example any first-order loop, $K / [(1 + s T_{1}) (1 + s T_{2})]$ or $K / [s (1 + s T)]$ with all $T > 0$ ), there is no phase crossover and GM $= \infty$ . A double integrator $K / s^{2}$ is different: its phase is exactly $- 18 0^{\circ}$ at every frequency, PM $= 0$ , and the closed loop is marginally stable. If $∣ G H ∣ < 1$ at every frequency, there is no gain crossover and PM is taken as $\infty$ .
For open-loop-unstable or conditionally stable loops, the margins can mislead, so apply the Nyquist criterion instead: $Z = N + P$ . Here $N$ counts clockwise encirclements of $- 1 + j 0$ by the plot of $G (s) H (s)$ as $s$ traverses the Nyquist contour clockwise, $P$ is the number of open-loop RHP poles, and the closed loop is stable only if $Z = 0$ .

NYQUIST CRITERION (CLOCKWISE CONVENTION)

Z = N + P, stable ⟺ Z = 0

$Z$ counts closed-loop RHP poles, $P$ open-loop RHP poles and $N$ clockwise encirclements of $- 1 + j 0$ , with the Nyquist contour traversed clockwise. Texts that count counter-clockwise encirclements write $Z = P - N$ .

WORKED EXAMPLE

For the loop $G (s) H (s) = \frac{4}{s ( 1 + 0.5 s ) ( 1 + 0.1 s )}$ with negative feedback, find the gain margin (in dB), the phase margin, the gain at which the closed loop becomes marginally stable, and the largest pure time delay that can be inserted before instability.

Solution

Phase: $∠ G H = - 9 0^{\circ} - tan^{- 1} (0.5 ω) - tan^{- 1} (0.1 ω)$ . Setting it to $- 18 0^{\circ}$ needs $tan^{- 1} (0.5 ω) + tan^{- 1} (0.1 ω) = 9 0^{\circ}$ , so $(0.5 ω) (0.1 ω) = 1$ and $ω_{p c} = 20 = 4.47$ rad/s.
$∣ G H (j ω_{p c}) ∣ = \frac{4}{4.47 \times 1 + 5 \times 1 + 0.2} = \frac{4}{4.47 \times 2.449 \times 1.095} = \frac{4}{12} = \frac{1}{3}$ .
GM $= 3$ , i.e. $20 lo g_{10} 3 = 9.54$ dB. The gain can rise to $4 \times 3 = 12$ . Check: $(T_{1} + T_{2}) / (T_{1} T_{2}) = 0.6/0.05 = 12$ . Routh on $0.05 s^{3} + 0.6 s^{2} + s + K$ also needs $0.6 > 0.05 K$ , i.e. $K < 12$ .
Gain crossover: solve $4 = ω 1 + 0.25 ω^{2} 1 + 0.01 ω^{2}$ by trial. At $ω = 2.45$ the right side is $2.45 \times 1.581 \times 1.030 \approx 3.99$ , so $ω_{g c} \approx 2.45$ rad/s (more precisely 2.454 rad/s).
$∠ G H (j 2.45) = - 9 0^{\circ} - 50. 8^{\circ} - 13. 8^{\circ} = - 154. 6^{\circ}$ , so PM $\approx 25. 4^{\circ}$ .
PM $= 25. 4^{\circ} = 0.443$ rad, so $T_{m a x} = 0.443/2.45 \approx 0.18$ s.
$ω_{g c} = 2.45 < ω_{p c} = 4.47$ , GM > 0 dB and PM > 0, consistent with a stable closed loop.

Answer: GM = 3 (9.54 dB), PM ≈ 25.4°, marginal gain K = 12, largest delay ≈ 0.18 s.

EXAM TIP

GM in dB is $- 20 lo g_{10} ∣ G H (j ω_{p c}) ∣$ . If the polar plot cuts the negative real axis at $- 0.25$ , GM $= 4$ = +12.04 dB, not −12 dB.

EXAM TIP

Raising $K$ lifts the magnitude curve only. The phase curve and $ω_{p c}$ stay put, $ω_{g c}$ moves right, PM falls, and GM (dB) falls by $20 lo g_{10}$ of the factor. A delay acts on the other curve: magnitude and $ω_{g c}$ stay put, the phase curve drops, $ω_{p c}$ moves left, and PM falls by $ω_{g c} T$ radians.

EXAM TIP

If a phase comes out as, say, $+ 21 0^{\circ}$ (for example from a calculator's $tan^{- 1}$ branch), convert it to the equivalent lag $- 15 0^{\circ}$ before using PM $= 18 0^{\circ} + ∠ G H$ . Track the quadrant from the factor phases, not from a single $tan^{- 1}$ of the ratio.

PREVIOUS-YEAR QUESTIONS

27 questions from the EC, EE and IN papers

Try each question first, then open the answer. One shows the official answer only. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE ECE 2025 · GATE ECE 2024 · GATE ECE 2023 · GATE ECE 2020 · GATE ECE 2018 · GATE EE 2026 · GATE EE 2024 · GATE EE 2023 · GATE EE 2022 · GATE EE 2020 · GATE IN 2026 · GATE IN 2024 · GATE IN 2023 · GATE IN 2022 · GATE IN 2020 · GATE IN 2018

GATE 2026 EE Q.29

GATE 2026 · EEQ.29 · 1 mark · MCQ

The asymptotic Bode magnitude plot of a system is shown.

[Figure]

Which one of the following options best represents the transfer function of the system?

GATE 2026 EE question 29: Asymptotic Bode magnitude plot. — Figure from the GATE 2026 EE paper. It is part of the question.

(A)
$G (s) = \frac{1 + \frac{s}{ω _{0}}}{\frac{s}{ω _{0}}}$
(B)
$G (s) = \frac{\frac{s}{ω _{0}}}{1 + \frac{s}{ω _{0}}}$
(C)
$G (s) = \frac{1 + \frac{s}{ω _{0}}}{1 - \frac{s}{ω _{0}}}$
(D)
$G (s) = \frac{1 - \frac{s}{ω _{0}}}{1 + \frac{s}{ω _{0}}}$

Answer and solution

GATE 2026 EE Q.29: answer and solution

GATE 2026 IN Q.60

GATE 2026 · INQ.60 · 2 marks · NAT

The system $G (s) = \frac{K}{( s + 1 ) ^{4}}$ has a gain margin of 20 dB. The value of $K$ is ______. (rounded off to two decimal places)

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GATE 2026 IN Q.60: answer and solution

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GATE 2025 EC Q.16

GATE 2025 · ECQ.16 · 1 mark · MCQ

The Nyquist plot of a system is given in the figure below. Let $ω_{P}$ , $ω_{Q}$ , $ω_{R}$ , and $ω_{S}$ be the positive frequencies at the points $P$ , $Q$ , $R$ , and $S$ , respectively.

Which one of the following statements is TRUE?

GATE 2025 EC question 16: Nyquist plot in the complex plane. — Figure from the GATE 2025 EC paper. It is part of the question.

(A)
$ω_{S}$ is the gain crossover frequency and $ω_{P}$ is the phase crossover frequency
(B)
$ω_{Q}$ is the gain crossover frequency and $ω_{R}$ is the phase crossover frequency
(C)
$ω_{Q}$ is the gain crossover frequency and $ω_{S}$ is the phase crossover frequency
(D)
$ω_{S}$ is the gain crossover frequency and $ω_{Q}$ is the phase crossover frequency

Answer and solution

GATE 2025 EC Q.16: answer and solution

GATE 2024 EC Q.12

GATE 2024 · ECQ.12 · 1 mark · MCQ

In the context of Bode magnitude plots, 40 dB/decade is the same as _______.

(A)
12 dB/octave
(B)
6 dB/octave
(C)
20 dB/octave
(D)
10 dB/octave

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GATE 2024 EC Q.12: answer and solution

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GATE 2024 EE Q.58

GATE 2024 · EEQ.58 · 2 marks · NAT

Consider the stable closed-loop system shown in the figure. The asymptotic Bode magnitude plot of $G (s)$ has a constant slope of $- 20$ dB/decade at least till $100$ rad/sec with the gain crossover frequency being $10$ rad/sec. The asymptotic Bode phase plot remains constant at $- 9 0^{\circ}$ at least till $ω = 10$ rad/sec. The steady-state error of the closed-loop system for a unit ramp input is ________ (rounded off to 2 decimal places).

GATE 2024 EE question 58: Unity negative feedback loop. — Figure from the GATE 2024 EE paper. It is part of the question.

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GATE 2024 EE Q.58: answer and solution

GATE 2024 EE Q.59

GATE 2024 · EEQ.59 · 2 marks · NAT

Consider the stable closed-loop system shown in the figure. The magnitude and phase values of the frequency response of $G (s)$ are given in the table. The value of the gain $K_{I} (> 0)$ for a $5 0^{\circ}$ phase margin is ________ (rounded off to 2 decimal places).

| $ω$ in rad/sec | Magnitude in dB | Phase in degrees |
|---|---|---|
| 0.5 | $- 7$ | $- 40$ |
| 1.0 | $- 10$ | $- 80$ |
| 2.0 | $- 18$ | $- 130$ |
| 10.0 | $- 40$ | $- 200$ |

GATE 2024 EE question 59: Unity negative feedback loop. — Figure from the GATE 2024 EE paper. It is part of the question.

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GATE 2024 EE Q.59: answer and solution

GATE 2024 IN Q.18

GATE 2024 · INQ.18 · 1 mark · MCQ

Q.18 The sinusoidal transfer function corresponding to the polar plot shown in the figure, for $T > 0$ , is

GATE 2024 IN question 18: Polar plot on Re–Im axes (horizontal axis labelled Re, vertical axis labelled Im, both drawn with arrows at both ends). — Figure from the GATE 2024 IN paper. It is part of the question.

(A)
$1 - j ω T$
(B)
$\frac{1 - j ω T}{1 + j ω T}$
(C)
$1 + j ω T$
(D)
$\frac{1}{1 + j ω T}$

Answer and solution

GATE 2024 IN Q.18: answer and solution

GATE 2023 EC Q.25

GATE 2023 · ECQ.25 · 1 mark · MCQ

The open loop transfer function of a unity negative feedback system is

G (s) = \frac{k}{s ( 1 + s T _{1} ) ( 1 + s T _{2} )}

where $k$ , $T_{1}$ and $T_{2}$ are positive constants. The phase cross-over frequency, in rad/s, is

(A)
$\frac{1}{T _{1} T _{2}}$
(B)
$\frac{1}{T _{1} T _{2}}$
(C)
$\frac{1}{T _{1} T _{2}}$
(D)
$\frac{1}{T _{2} T _{1}}$

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GATE 2023 EC Q.25: answer and solution

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GATE 2023 EC Q.57

GATE 2023 · ECQ.57 · 2 marks · NAT

The asymptotic magnitude Bode plot of a minimum phase system is shown in the figure. The transfer function of the system is $(s) = \frac{k ( s + z ) ^{a}}{s ^{b} ( s + p ) ^{c}}$ , where $k, z, p, a, b$ and $c$ are positive constants. The value of $(a + b + c)$ is ________ (rounded off to the nearest integer).

GATE 2023 EC question 57: Asymptotic Bode magnitude plot. — Figure from the GATE 2023 EC paper. It is part of the question.

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GATE 2023 EC Q.57: answer and solution

GATE 2023 EE Q.36

GATE 2023 · EEQ.36 · 2 marks · MCQ

The magnitude and phase plots of an LTI system are shown in the figure. The transfer function of the system is

GATE 2023 EE question 36: Two plots stacked vertically. — Figure from the GATE 2023 EE paper. It is part of the question.

(A)
$2.51 e^{- 0.032 s}$
(B)
$\frac{e ^{- 2.514 s}}{s + 1}$
(C)
$1.04 e^{- 2.514 s}$
(D)
$2.51 e^{- 1.047 s}$

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GATE 2023 EE Q.36: answer and solution

GATE 2023 IN Q.21

GATE 2023 · INQ.21 · 1 mark · MCQ

A system has the transfer-function

\frac{Y ( s )}{X ( s )} = \frac{s - π}{s + π}

Let $u (t)$ be the unit-step function. The input $x (t)$ that results in a steady-state output $y (t) = sin π t$ is ______.

(A)
$x (t) = sin (π t) u (t)$
(B)
$x (t) = sin (π t + \frac{π}{2}) u (t)$
(C)
$x (t) = sin (π t - \frac{π}{2}) u (t)$
(D)
$x (t) = cos (π t + \frac{π}{4}) u (t)$

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GATE 2023 IN Q.21: answer and solution

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GATE 2023 IN Q.43

GATE 2023 · INQ.43 · 2 marks · MCQ

The magnitude and phase plots shown in the figure match with the transfer-function ________.

GATE 2023 IN question 43: A plot titled 'Bode Diagram' with two stacked panels that share a LINEAR frequency axis labelled 'Frequency (rad/s)'. — Figure from the GATE 2023 IN paper. It is part of the question.

(A)
$\frac{10000}{s ^{2} + 2 s + 10000}$
(B)
$\frac{10000}{s ^{2} + 2 s + 10000} e^{- 0.05 s}$
(C)
$\frac{10000}{s ^{2} + 2 s + 10000} e^{- 0.5 \times 1 0^{- 12} s}$
(D)
$\frac{100}{s ^{2} + 2 s + 100}$

Answer and solution

GATE 2023 IN Q.43: answer and solution

GATE 2023 IN Q.50

GATE 2023 · INQ.50 · 2 marks · NAT

The phase margin of the transfer function $G (s) = \frac{2 ( 1 - s )}{( 1 + s ) ^{2}}$ is ________ degrees. (rounded off to the nearest integer).

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GATE 2023 IN Q.50: answer and solution

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GATE 2022 EE Q.11

GATE 2022 · EEQ.11 · 1 mark · MCQ

The transfer function of a real system, $H (s)$ , is given as:

H (s) = \frac{A s + B}{s ^{2} + C s + D},

where $A, B, C$ and $D$ are positive constants. This system cannot operate as

(A)
low pass filter.
(B)
high pass filter.
(C)
band pass filter.
(D)
an integrator.

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GATE 2022 EE Q.17

GATE 2022 · EEQ.17 · 1 mark · MCQ

The Bode magnitude plot of a first order stable system is constant with frequency. The asymptotic value of the high frequency phase, for the system, is $- 18 0^{\circ}$ . This system has

GATE 2022 EE question 17: Sketch of a Bode plot. — Figure from the GATE 2022 EE paper. It is part of the question.

(A)
one LHP pole and one RHP zero at the same frequency.
(B)
one LHP pole and one LHP zero at the same frequency.
(C)
two LHP poles and one RHP zero.
(D)
two RHP poles and one LHP zero.

Answer and solution

GATE 2022 EE Q.17: answer and solution

GATE 2022 EE Q.37

GATE 2022 · EEQ.37 · 2 marks · MCQ

An LTI system is shown in the figure where

G (s) = \frac{100}{s ^{2} + 0.1 s + 10} .

The steady state output of the system, to the input $r (t)$ , is given as $y (t) = a + b sin (10 t + θ)$ . The values of ' $a$ ' and ' $b$ ' will be

GATE 2022 EE question 37: A single block labelled G(s). — Figure from the GATE 2022 EE paper. It is part of the question.

(A)
$a = 1, b = 10$
(B)
$a = 10, b = 1$
(C)
$a = 1, b = 100$
(D)
$a = 100, b = 1$

Official answer

Official answer: ASource: Official answer key, GATE 2022 EE (IIT Kharagpur)

A NOTE ON THIS QUESTION

As printed, none of the four options is correct. The paper gives $G (s) = \frac{100}{s ^{2} + 0.1 s + 10}$ and $r (t) = 1 + 0.1 sin (10 t)$ . The DC term gives $a = G (0) = \frac{100}{10} = 10$ . At $ω = 10$ : $G (j 10) = \frac{100}{( 10 - 100 ) + j 1} = \frac{100}{- 90 + j 1}$ , so $∣ G (j 10) ∣ = \frac{100}{8101} \approx 1.111$ and $b = 0.1 \times 1.111 \approx 0.111$ . No option has $a = 10$ with $b \approx 0.111$ . The official key is (A), $a = 1$ and $b = 10$ , which is what the same working gives if the constant term of the denominator is $100$ instead of $10$ : then $G (0) = 1$ and $G (j 10) = \frac{100}{j 1}$ , so $∣ G (j 10) ∣ = 100$ and $b = 0.1 \times 100 = 10$ . The printed question appears to contain a misprint. We show the official answer, and we do not publish a worked solution that the printed numbers do not lead to.

GATE 2022 EE Q.37: official answer

GATE 2022 IN Q.43

GATE 2022 · INQ.43 · 2 marks · MCQ

The Nyquist plot of a stable open-loop system $G (j ω)$ is plotted in the frequency range $0 \leq ω < \infty$ as shown. It is found to intersect a unit circle with center at the origin at the point $P = - 0.77 - 0.64 j$ . The points $Q$ and $R$ lie on $G (j ω)$ and assume values $Q = 14.40 + 0.00 j$ and $R = - 0.21 + 0.00 j$ . The phase margin (PM) and the gain margin (GM) of the system are ________

GATE 2022 IN question 43: Nyquist plot on axes Re(G(jω)) (horizontal, arrow to the right) and Im(G(jω)) (vertical, arrow upward). — Figure from the GATE 2022 IN paper. It is part of the question.

(A)
$PM = 39. 7^{\circ}$ and $GM = 4.76$
(B)
$PM = 39. 7^{\circ}$ and $GM = 0.07$
(C)
$PM = - 39. 7^{\circ}$ and $GM = 4.76$
(D)
$PM = - 39. 7^{\circ}$ and $GM = 0.07$

Answer and solution

GATE 2022 IN Q.43: answer and solution

GATE 2022 IN Q.53

GATE 2022 · INQ.53 · 2 marks · NAT

In a unity-gain feedback control system, the plant
$P (s) = \frac{0.001}{s ( 2 s + 1 ) ( 0.01 s + 1 )}$
is controlled by a lag compensator
$C (s) = \frac{s + 10}{s + 0.1}$
The slope (in dB/decade) of the asymptotic Bode magnitude plot of the loop gain at $ω = 3$ rad/s is ________ (in integer)

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GATE 2020 EC Q.49

GATE 2020 · ECQ.49 · 2 marks · NAT

A system with transfer function $G (s) = \frac{1}{( s + 1 ) ( s + a )}$ , $a > 0$ is subjected to an input $5 cos 3 t$ . The steady state output of the system is $\frac{1}{10} cos (3 t - 1.892)$ . The value of $a$ is ______.

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GATE 2020 EE Q.41

GATE 2020 · EEQ.41 · 2 marks · MCQ

A stable real linear time-invariant system with single pole at $p$ , has a transfer function $H (s) = \frac{s ^{2} + 100}{s - p}$ with a dc gain of 5. The smallest positive frequency, in rad/s, at unity gain is closest to:

(A)
8.84
(B)
11.08
(C)
78.13
(D)
122.87

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GATE 2020 IN Q.14

GATE 2020 · INQ.14 · 1 mark · MCQ

The closed loop transfer function of a control system is given by $\frac{C ( s )}{R ( s )} = \frac{1}{s + 1}$ . For the input $r (t) = sin t$ , the steady state response $c (t)$ is ____

(A)
$1$
(B)
$\frac{1}{2} cos t$
(C)
$\frac{1}{2} sin (t + \frac{π}{4})$
(D)
$\frac{1}{2} sin (t - \frac{π}{4})$

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GATE 2020 IN Q.55

GATE 2020 · INQ.55 · 2 marks · NAT

The loop transfer function of a negative feedback system is $G (s) H (s) = \frac{2 ( s + 1 )}{s ^{2}}$ . The phase margin of the system (in degrees, rounded off to one decimal place) is ____

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GATE 2018 EC Q.41

GATE 2018 · ECQ.41 · 2 marks · NAT

For a unity feedback control system with the forward path transfer function

G (s) = \frac{K}{s ( s + 2 )}

The peak resonant magnitude $M_{r}$ of the closed-loop frequency response is 2. The corresponding value of the gain $K$ (correct to two decimal places) is ________.

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GATE 2018 EC Q.42

GATE 2018 · ECQ.42 · 2 marks · NAT

The figure below shows the Bode magnitude and phase plots of a stable transfer function $G (s) = \frac{n _{0}}{s ^{3} + d _{2} s ^{2} + d _{1} s + d _{0}}$ .

Consider the negative unity feedback configuration with gain $k$ in the feedforward path. The closed loop is stable for $k < k_{0}$ . The maximum value of $k_{0}$ is ______.

GATE 2018 EC question 42: Left: Bode plots sharing a common frequency axis. — Figure from the GATE 2018 EC paper. It is part of the question.

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GATE 2018 EC Q.42: answer and solution

GATE 2018 IN Q.12

GATE 2018 · INQ.12 · 1 mark · NAT

An input $p (t) = sin (t)$ is applied to the system $G (s) = \frac{s - 1}{s + 1}$ . The corresponding steady state output is $y (t) = sin (t + φ)$ , where the phase $φ$ (in degrees), when restricted to $0^{\circ} \leq φ \leq 36 0^{\circ}$ , is ______.

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GATE 2018 IN Q.13

GATE 2018 · INQ.13 · 1 mark · MCQ

The approximate phase response of $\frac{10 0 ^{2} e ^{- 0.01 s}}{s ^{2} + 0.2 s + 10 0 ^{2}}$ is

GATE 2018 IN question 13: All four options are phase plots. — Figure from the GATE 2018 IN paper. It is part of the question.

(A)
Figure: phase falls in a straight line from 0 to a little below the -50 label just before 100 rad/s, drops vertically to the -250 label, then keeps falling with the same slope to the bottom of the plot at 200 rad/s.
(B)
Figure: phase is flat at 0 up to 100 rad/s, steps down to -180 degrees and stays flat there out to 200 rad/s.
(C)
Figure: phase rises linearly from 0 to about +80 degrees at 100 rad/s, drops vertically to about -120 degrees, then rises linearly to about -20 degrees at 200 rad/s.
(D)
Figure: phase is flat at 0 up to 100 rad/s, steps up to +180 degrees and stays flat there out to 200 rad/s.

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GATE 2018 IN Q.13: answer and solution

GATE 2018 IN Q.14

GATE 2018 · INQ.14 · 1 mark · NAT

Consider the transfer function $G (s) = \frac{2}{( s + 1 ) ( s + 2 )}$ . The phase margin of $G (s)$ in degrees is ______.

Answer and solution

GATE 2018 IN Q.14: answer and solution

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GATE 2018 IN Q.14: answer and solution

PRACTICE

6 questions to test yourself

PRACTICE 12 marks · MCQ

The asymptotic Bode magnitude plot of a minimum-phase transfer function $G (s)$ has slope −20 dB/decade for $ω < 1$ rad/s, 0 dB/decade for $1 < ω < 5$ rad/s and −20 dB/decade for $ω > 5$ rad/s. The flat segment lies at 20 dB. Which option gives $G (s)$ and the gain crossover frequency read from the asymptotic plot?

(A)
$G (s) = \frac{50 ( s + 1 )}{s ( s + 5 )}$ , $ω_{g c} \approx 50$ rad/s
(B)
$G (s) = \frac{10 ( s + 1 )}{s ( s + 5 )}$ , $ω_{g c} \approx 10$ rad/s
(C)
$G (s) = \frac{50 ( s + 1 )}{s ( s + 5 )}$ , $ω_{g c} \approx 10$ rad/s
(D)
$G (s) = \frac{50 ( s + 5 )}{s ( s + 1 )}$ , $ω_{g c} \approx 50$ rad/s

Answer and solution

Answer: (A)

Initial slope −20 dB/decade means type 1. The +20 change at 1 rad/s is a zero $(1 + s)$ and the −20 change at 5 rad/s is a pole $(1 + s /5)$ .
The low-frequency line $K / ω$ meets the flat 20 dB segment at $ω = 1$ , so $K = 10$ .
$G (s) = \frac{10 ( 1 + s )}{s ( 1 + s /5 )} = \frac{50 ( s + 1 )}{s ( s + 5 )}$ .
For $ω > 5$ , $∣ G ∣ \approx \frac{10 ω}{ω \cdot ω /5} = \frac{50}{ω}$ , which equals 1 at $ω = 50$ rad/s. Equivalently, the line falls 20 dB in one decade from 20 dB at 5 rad/s, reaching 0 dB at 50 rad/s. (The exact crossover is 49.8 rad/s.)

PRACTICE 22 marks · NAT

A negative-feedback loop has $G (s) H (s) = \frac{12}{s ( s + 2 ) ( s + 4 )}$ . Find the gain margin in dB, correct to two decimal places.

Answer and solution

Answer: $12.04$ (accepted range $\pm 0.1$ )

Phase: $∠ G H = - 9 0^{\circ} - tan^{- 1} (ω /2) - tan^{- 1} (ω /4)$ . For $- 18 0^{\circ}$ you need $(ω /2) (ω /4) = 1$ , so $ω_{p c} = 8 = 2.83$ rad/s.
$∣ G H (j ω_{p c}) ∣ = \frac{12}{8 8 + 4 8 + 16} = \frac{12}{2304} = \frac{12}{48} = 0.25$ .
GM $= 1/0.25 = 4$ , so GM $_{dB} = 20 lo g_{10} 4 = 12.04$ dB.
Check with Routh: $s^{3} + 6 s^{2} + 8 s + K = 0$ is stable for $K < 48$ , and $48/12 = 4$ .

PRACTICE 31 mark · NAT

A unity-feedback system has closed-loop transfer function $T (s) = \frac{25}{s ^{2} + 4 s + 25}$ . Find the resonant peak $M_{r}$ of $∣ T (j ω) ∣$ as a ratio (not in dB), correct to two decimal places.

Answer and solution

Answer: $1.36$ (accepted range $\pm 0.01$ )

Compare with $ω_{n}^{2} / (s^{2} + 2 ζ ω_{n} s + ω_{n}^{2})$ : $ω_{n} = 5$ rad/s and $2 ζ ω_{n} = 4$ , so $ζ = 0.4$ .
$ζ = 0.4 < 0.707$ , so a resonant peak exists.
$M_{r} = \frac{1}{2 ζ 1 - ζ ^{2}} = \frac{1}{0.8 0.84} = \frac{1}{0.7332} = 1.364$ .
It occurs at $ω_{r} = 5 1 - 0.32 = 4.12$ rad/s.

PRACTICE 41 mark · MCQ

Consider $G_{1} (s) = \frac{1 + s}{1 + 10 s}$ and $G_{2} (s) = \frac{1 - s}{1 + 10 s}$ . Which one of the following statements is correct?

(A)
$∣ G_{1} (j ω) ∣ = ∣ G_{2} (j ω) ∣$ for all $ω$ , and $∠ G_{2} (j ω) \to - 18 0^{\circ}$ as $ω \to \infty$
(B)
$∠ G_{1} (j ω) = ∠ G_{2} (j ω)$ for all $ω$
(C)
$∠ G_{1} (j ω) \to - 9 0^{\circ}$ as $ω \to \infty$
(D)
$∣ G_{2} (j ω) ∣ > ∣ G_{1} (j ω) ∣$ at high frequencies

Answer and solution

Answer: (A)

$∣1 - j ω ∣ = ∣1 + j ω ∣ = 1 + ω^{2}$ , so the magnitudes are identical at every $ω$ . This rules out D.
$∠ G_{1} = tan^{- 1} ω - tan^{- 1} 10 ω$ , which goes to $9 0^{\circ} - 9 0^{\circ} = 0^{\circ}$ as $ω \to \infty$ . This rules out C.
$∠ G_{2} = - tan^{- 1} ω - tan^{- 1} 10 ω$ , which goes to $- 18 0^{\circ}$ . The phases differ, which rules out B. $G_{2}$ is non-minimum phase because of its RHP zero at $s = 1$ .
So A is correct.

PRACTICE 52 marks · MSQ

A unity negative-feedback system has open-loop transfer function $G (s) = \frac{5}{s ( 1 + 0.2 s )}$ . Which of the following statements is/are correct?

(A)
The polar plot of $G (j ω)$ starts at infinite magnitude along $- 9 0^{\circ}$ and reaches the origin tangent to the negative real axis (angle $- 18 0^{\circ}$ ).
(B)
The gain margin is infinite.
(C)
As $ω \to 0$ , the polar plot approaches the vertical asymptote $Re {G (j ω)} = - 1$ .
(D)
Inserting a pure time delay of 0.1 s in the loop makes the closed loop unstable.

Answer and solution

Answer: (A), (B), (C)

A: the system is type 1, so it starts at $\infty∠ - 9 0^{\circ}$ . $n - m = 2$ , so it ends at $0∠ - 18 0^{\circ}$ . Correct.
B: $∠ G = - 9 0^{\circ} - tan^{- 1} (0.2 ω)$ stays above $- 18 0^{\circ}$ at every finite $ω$ , so there is no phase crossover and GM $= \infty$ . Correct.
C: the type-1 asymptote is $Re = - K T = - 5 \times 0.2 = - 1$ . Correct.
D: $∣ G ∣ = 1$ gives $ω^{2} (1 + 0.04 ω^{2}) = 25$ , i.e. $0.04 x^{2} + x - 25 = 0$ with $x = ω^{2}$ . So $x = 15.45$ and $ω_{g c} = 3.93$ rad/s.
$∠ G (j 3.93) = - 9 0^{\circ} - 38. 2^{\circ} = - 128. 2^{\circ}$ , so PM $= 51. 8^{\circ} = 0.904$ rad and $T_{m a x} = 0.904/3.93 = 0.23$ s.
A 0.1 s delay removes only $0.393$ rad $= 22. 5^{\circ}$ , leaving PM $\approx 29. 3^{\circ} > 0$ . The magnitude still crosses 1 only once and the phase still falls monotonically, so the system stays stable and D is wrong.

PRACTICE 61 mark · NAT

A unity negative-feedback loop has open-loop transfer function $G (s) = \frac{K e ^{- 0.1 s}}{s}$ with $K > 0$ . Find the largest value of $K$ for which the closed loop is stable, correct to two decimal places.

Answer and solution

Answer: $15.71$ (accepted range $\pm 0.05$ )

The magnitude is $∣ G (j ω) ∣ = K / ω$ , unaffected by the delay. It falls monotonically and crosses 1 once, at $ω_{g c} = K$ .
The phase is $∠ G = - 9 0^{\circ} - 0.1 ω \cdot \frac{18 0 ^{\circ}}{π}$ , which falls monotonically. It reaches $- 18 0^{\circ}$ when $0.1 ω = π /2$ , so $ω_{p c} = \frac{π}{0.2} = 15.71$ rad/s.
$∣ G (j ω_{p c}) ∣ = K /15.71$ , so GM $= 15.71/ K$ . The open loop has no RHP poles, so the closed loop is stable while GM > 1, i.e. $K < 15.71$ .
Cross-check with PM: $PM = 9 0^{\circ} - 0.1 K \cdot \frac{18 0 ^{\circ}}{π}$ , which is positive only for $K < \frac{π}{0.2} = 15.71$ .

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