Q.1 What does the Nyquist plot represent in control systems?
The time response of the system
The frequency response of the open‑loop transfer function
The root locus of the closed‑loop poles
The step response of the closed‑loop system
Explanation - A Nyquist plot is a polar plot of the complex value of the open‑loop transfer function G(jω) as ω varies from 0 to ∞ (and –∞ for the complete contour).
Correct answer is: The frequency response of the open‑loop transfer function
Q.2 According to the Nyquist stability criterion, a closed‑loop system is stable if the Nyquist plot of G(s)H(s) encircles the point (‑1,0) how many times?
Exactly once clockwise
Exactly once counter‑clockwise
Zero times
The number of clockwise encirclements equals the number of right‑half‑plane poles of G(s)H(s)
Explanation - The Nyquist criterion states N = Z – P, where N is the net number of clockwise encirclements of (‑1,0), Z is the number of closed‑loop poles in the right‑half plane, and P is the number of open‑loop poles in the right‑half plane. For stability Z must be zero, so N must equal P.
Correct answer is: The number of clockwise encirclements equals the number of right‑half‑plane poles of G(s)H(s)
Q.3 If an open‑loop transfer function has two poles in the right‑half s‑plane, how many clockwise encirclements of (‑1,0) are required for the closed‑loop system to be stable?
0
1
2
3
Explanation - For stability Z = 0, so the Nyquist plot must have N = P = 2 clockwise encirclements of (‑1,0) to cancel the two right‑half‑plane poles.
Correct answer is: 2
Q.4 Which of the following statements about the Nyquist contour is true?
It always encloses the entire left‑half s‑plane.
It must avoid any poles of G(s)H(s) that lie on the imaginary axis.
It is drawn only for ω > 0.
It is a circle of radius 1 in the s‑plane.
Explanation - When a pole lies on the imaginary axis, the contour is indented to avoid it, ensuring the mapping is well‑defined.
Correct answer is: It must avoid any poles of G(s)H(s) that lie on the imaginary axis.
Q.5 A system has the open‑loop transfer function G(s) = \frac{10}{s(s+2)}. What is the gain margin of the system?
0 dB
2 dB
5 dB
Infinite
Explanation - The phase never reaches –180° because the phase of G(jω) approaches –90° at low frequency and –180° only asymptotically as ω → ∞, but the magnitude is zero at that frequency, giving an infinite gain margin.
Correct answer is: Infinite
Q.6 For the transfer function G(s) = \frac{K}{s(s+1)(s+2)}, the Nyquist plot will cross the negative real axis at which frequency (in rad/s) when K = 1?
ω = 0.5
ω = 1
ω = √2
ω = 2
Explanation - Setting Im{G(jω)} = 0 and solving for ω yields ω = √2 rad/s for K = 1.
Correct answer is: ω = √2
Q.7 Which of the following open‑loop transfer functions will produce a Nyquist plot that encircles (‑1,0) once in the clockwise direction?
G(s) = \frac{1}{s+1}
G(s) = \frac{10}{s(s+2)}
G(s) = \frac{100}{(s+5)^2}
G(s) = \frac{1}{(s-1)(s+1)}
Explanation - The pole at s = 1 lies in the right half‑plane (P = 1). The Nyquist plot of this unstable open‑loop system makes one clockwise encirclement of (‑1,0) to satisfy N = P for stability (Z = 0).
Correct answer is: G(s) = \frac{1}{(s-1)(s+1)}
Q.8 In a Nyquist plot, the point where the plot crosses the real axis corresponds to:
Zero phase angle
Zero gain
Phase of –180°
Frequency where the imaginary part of G(jω) is zero
Explanation - Crossing the real axis means Im{G(jω)} = 0, which occurs at specific frequencies depending on the transfer function.
Correct answer is: Frequency where the imaginary part of G(jω) is zero
Q.9 A system with open‑loop transfer function G(s) = \frac{K}{s(s+2)} has a phase margin of 45°. What is the approximate value of K?
K ≈ 1
K ≈ 2
K ≈ 4
K ≈ 8
Explanation - Phase margin of 45° occurs near the gain crossover frequency where |G(jω)| = 1. Solving |G(jω_g)| = 1 with ω_g ≈ 2 rad/s gives K ≈ 4.
Correct answer is: K ≈ 4
Q.10 Which of the following statements is true regarding the relationship between Nyquist and Bode plots?
Both plots provide the same information, so only one is needed.
Nyquist plots can be directly derived from Bode magnitude plots.
Nyquist plots give phase information, while Bode plots do not.
Both plots represent the frequency response of the same transfer function.
Explanation - Nyquist and Bode plots are two different visualizations of the same complex frequency response G(jω).
Correct answer is: Both plots represent the frequency response of the same transfer function.
Q.11 Consider G(s) = \frac{1}{(s+1)(s+3)}. How many times does its Nyquist plot encircle the point (‑1,0)?
0
1 clockwise
1 counter‑clockwise
2 clockwise
Explanation - All poles are in the left half‑plane (P = 0). The Nyquist plot of a stable minimum‑phase system does not encircle (‑1,0), giving N = 0.
Correct answer is: 0
Q.12 The Nyquist plot of a pure time delay e^{‑jωτ} is:
A circle of radius 1 centered at the origin
A spiral moving inward
A point rotating on the unit circle
A straight line on the negative real axis
Explanation - e^{‑jωτ} has magnitude 1 for all ω; its phase changes linearly with ω, so the plot is a point moving around the unit circle.
Correct answer is: A point rotating on the unit circle
Q.13 If an open‑loop transfer function has a pole at the origin, how does this affect the Nyquist contour?
The contour must be deformed to bypass the pole with a small semicircle.
The Nyquist plot will start at the origin.
The contour does not need to be modified.
The Nyquist plot will be symmetric about the real axis.
Explanation - A pole on the imaginary axis (including the origin) requires an indentation in the Nyquist contour to avoid singularity.
Correct answer is: The contour must be deformed to bypass the pole with a small semicircle.
Q.14 Which of the following open‑loop transfer functions has an infinite phase margin?
G(s) = \frac{10}{s(s+5)}
G(s) = \frac{1}{s+1}
G(s) = \frac{100}{(s+2)^2}
G(s) = \frac{K}{s(s+2)} with K = 0.1
Explanation - A first‑order system never reaches –180° phase; thus the phase margin is infinite.
Correct answer is: G(s) = \frac{1}{s+1}
Q.15 For the system G(s) = \frac{K}{(s+1)(s+2)}, the gain crossover frequency ω_gc is defined as the frequency where:
|G(jω_gc)| = 0 dB
∠G(jω_gc) = –180°
Re{G(jω_gc)} = –1
Im{G(jω_gc)} = 0
Explanation - Gain crossover frequency is where the magnitude of the open‑loop transfer function equals unity (0 dB).
Correct answer is: |G(jω_gc)| = 0 dB
Q.16 If the Nyquist plot of an open‑loop system passes exactly through the point (‑1,0), what can be said about the closed‑loop system?
It is stable.
It is marginally stable.
It is unstable.
Stability cannot be determined from this information.
Explanation - Passing through (‑1,0) indicates a closed‑loop pole on the imaginary axis, leading to marginal stability.
Correct answer is: It is marginally stable.
Q.17 A Nyquist plot that lies entirely in the right half of the complex plane indicates:
The system has positive gain margin.
The system has negative phase margin.
The system is unstable.
The system has zero phase lag.
Explanation - If the plot never approaches (‑1,0), the gain can be increased considerably before reaching the critical point, implying a positive (often infinite) gain margin.
Correct answer is: The system has positive gain margin.
Q.18 Which of the following is NOT a requirement for applying the Nyquist stability criterion?
The open‑loop transfer function must be rational.
All poles must lie in the left half‑plane.
The contour must encircle the entire right half‑plane.
The system must be linear and time‑invariant.
Explanation - The Nyquist criterion can handle open‑loop poles in the right half‑plane; the number of such poles (P) is accounted for in the N = Z – P relation.
Correct answer is: All poles must lie in the left half‑plane.
Q.19 For G(s) = \frac{K}{s(s+4)(s+6)}, what is the minimum value of K that makes the closed‑loop system unstable?
K = 48
K = 96
K = 144
K = 192
Explanation - Using the Nyquist plot, the critical gain where the plot passes through (‑1,0) occurs at K ≈ 144. Larger K pushes the plot past (‑1,0), causing encirclements that violate stability.
Correct answer is: K = 144
Q.20 When constructing a Nyquist plot, why is the frequency range extended to –∞?
To capture the behavior of the system at negative frequencies.
Because the transfer function is defined for negative ω.
To ensure the contour closes in the complex plane.
It is not necessary; only positive frequencies are used.
Explanation - The Nyquist contour includes both the positive‑frequency path (0 → +∞) and the negative‑frequency path (–∞ → 0) to form a closed loop in the G(jω) plane.
Correct answer is: To ensure the contour closes in the complex plane.
Q.21 If a system has a phase margin of 0°, what does this imply about its Nyquist plot?
It passes through (‑1,0).
It encircles (‑1,0) once clockwise.
It stays entirely to the right of (‑1,0).
It never reaches the unit circle.
Explanation - Zero phase margin means the phase reaches –180° at the gain crossover frequency, and the magnitude is unity, placing the point exactly at (‑1,0).
Correct answer is: It passes through (‑1,0).
Q.22 For a second‑order system with transfer function G(s) = \frac{ω_n^2}{s^2+2ζω_n s+ω_n^2}, the Nyquist plot will be a:
Circle passing through (‑1,0).
Ellipse centered at the origin.
Closed curve that never reaches the negative real axis.
A semicircle in the right half of the complex plane.
Explanation - A stable second‑order system with ζ > 0 has a Nyquist plot that stays in the right half of the complex plane and does not cross (‑1,0).
Correct answer is: Closed curve that never reaches the negative real axis.
Q.23 Which of the following best describes the effect of increasing gain K on the Nyquist plot of G(s) = K·F(s)?
The plot expands radially outward from the origin.
The plot rotates clockwise.
The plot translates upward on the imaginary axis.
The shape of the plot changes but its size remains constant.
Explanation - Multiplying the transfer function by a gain scales the magnitude of every point on the plot, stretching it away from (or toward) the origin.
Correct answer is: The plot expands radially outward from the origin.
Q.24 A Nyquist plot of an open‑loop system has exactly one counter‑clockwise encirclement of (‑1,0) while the open‑loop has no right‑half‑plane poles. What is the number of closed‑loop right‑half‑plane poles?
0
1
‑1
2
Explanation - N = Z – P. Here N = –1 (counter‑clockwise is negative), P = 0, so Z = N + P = –1 → one closed‑loop pole in the right half‑plane (unstable).
Correct answer is: 1
Q.25 The term "gain crossover frequency" refers to:
The frequency at which phase = –180°.
The frequency where the magnitude of G(jω) equals 1 (0 dB).
The frequency where the Nyquist plot crosses the real axis.
The frequency where the system’s gain is maximum.
Explanation - Gain crossover frequency ω_gc is defined as the frequency where |G(jω_gc)| = 1.
Correct answer is: The frequency where the magnitude of G(jω) equals 1 (0 dB).
Q.26 In the Nyquist diagram, the point (‑1,0) is often called the:
Critical point
Stability margin
Phase zero
Gain peak
Explanation - The point (‑1,0) is the critical point; its encirclement determines closed‑loop stability.
Correct answer is: Critical point
Q.27 Which of the following modifications to a controller will shift the Nyquist plot to the right (away from the critical point)?
Increasing proportional gain.
Adding a lead compensator.
Adding a lag compensator.
Increasing system delay.
Explanation - A lead compensator adds positive phase at crossover, effectively moving the plot to the right (increasing gain margin).
Correct answer is: Adding a lead compensator.
Q.28 For a system with open‑loop transfer function G(s) = \frac{K}{s(s+3)(s+5)}, what is the smallest K that yields a Nyquist plot intersecting the point (‑1,0)?
K = 30
K = 45
K = 60
K = 75
Explanation - Solving |G(jω)| = 1 and ∠G(jω) = –180° simultaneously gives K ≈ 45.
Correct answer is: K = 45
Q.29 If a Nyquist plot encircles the point (‑1,0) twice clockwise and the open‑loop has one pole in the right half‑plane, how many closed‑loop poles are in the right half‑plane?
1
2
3
0
Explanation - N = 2 (clockwise), P = 1, so Z = N + P = 3 → three closed‑loop poles in the right half‑plane. However, only the net number matters for stability; here Z = 3, which is unstable. (Correction: Since the problem asks for number of right‑half‑plane closed‑loop poles, answer = 3.)
Correct answer is: 1
Q.30 Which of the following is a direct consequence of the Nyquist criterion?
Root locus and Bode plots are equivalent.
Phase margin can be read directly from the Nyquist plot.
Stability can be determined without calculating poles of the closed‑loop system.
Time‑domain response is independent of the Nyquist plot.
Explanation - The Nyquist criterion allows stability assessment by counting encirclements, bypassing explicit pole calculation.
Correct answer is: Stability can be determined without calculating poles of the closed‑loop system.
Q.31 For the transfer function G(s) = \frac{K(s+2)}{s(s+5)}, what type of compensator does the factor (s+2) represent?
Lag compensator
Lead compensator
Proportional compensator
Derivative compensator
Explanation - A zero placed at a higher frequency than the pole (here there is no pole at the same location) provides phase lead, improving stability margins.
Correct answer is: Lead compensator
Q.32 If the Nyquist plot of an open‑loop system lies completely inside a circle of radius 0.5 centered at the origin, what can be said about its gain margin?
Gain margin is less than 6 dB.
Gain margin is exactly 6 dB.
Gain margin is greater than 6 dB.
Gain margin cannot be determined.
Explanation - The distance from the origin to (‑1,0) is 1. If the plot stays within radius 0.5, the gain can be doubled (6 dB) before reaching the critical point, implying a gain margin >6 dB.
Correct answer is: Gain margin is greater than 6 dB.
Q.33 Which of the following best describes the effect of a pure integrator (1/s) on the Nyquist plot of a system?
It shifts the plot upward on the imaginary axis.
It adds a 90° phase lag at all frequencies.
It reduces the magnitude by a factor of ω at each frequency.
It makes the plot symmetric about the real axis.
Explanation - The integrator contributes a magnitude of 1/ω and a constant –90° phase shift.
Correct answer is: It reduces the magnitude by a factor of ω at each frequency.
Q.34 For a system with G(s) = \frac{K}{(s+1)(s+2)(s+3)}, the phase at high frequencies (ω → ∞) approaches:
0°
‑90°
‑180°
‑270°
Explanation - Each first‑order pole contributes –90° at high frequency; three poles give –270° total.
Correct answer is: ‑270°
Q.35 If an open‑loop system has a pole at s = +2, what is the effect on the Nyquist contour?
The contour must be deformed to exclude the right‑half‑plane pole.
The contour is unchanged; poles in the right half‑plane do not affect it.
The contour must include a small semicircle around the pole.
The contour must be reflected about the imaginary axis.
Explanation - The Nyquist contour always encloses the entire right half‑plane; poles there are counted as P in N = Z – P but do not require indentation.
Correct answer is: The contour is unchanged; poles in the right half‑plane do not affect it.
Q.36 A system has a gain margin of 12 dB. By how much can the loop gain be increased before the closed‑loop becomes unstable?
12 dB
6 dB
3 dB
It cannot be increased; the system is already at the limit.
Explanation - Gain margin is the amount (in dB) that the gain can be increased before the Nyquist plot reaches (‑1,0).
Correct answer is: 12 dB
Q.37 In a Nyquist plot, a point that lies exactly on the negative real axis to the left of (‑1,0) indicates:
Positive phase margin
Negative gain margin
Zero phase margin
Infinite gain margin
Explanation - If the plot crosses the negative real axis left of (‑1,0), the magnitude is already greater than 1 when phase is –180°, implying the gain is too high (negative gain margin).
Correct answer is: Negative gain margin
Q.38 Which of the following is true for a minimum‑phase system?
All its zeros are in the right half‑plane.
All its poles and zeros are in the left half‑plane.
It always has an infinite gain margin.
Its Nyquist plot always encircles (‑1,0).
Explanation - A minimum‑phase system has all poles and zeros in the left half‑plane, giving the best possible phase for a given magnitude.
Correct answer is: All its poles and zeros are in the left half‑plane.
Q.39 For the transfer function G(s) = \frac{K}{(s+1)^2}, the Nyquist plot will:
Be a circle passing through the origin.
Be a parabola opening to the right.
Be a curve that starts at infinity on the negative real axis and ends at the origin.
Be a straight line on the imaginary axis.
Explanation - A double pole at –1 causes the magnitude to drop as 1/ω^2 and phase to start at 0° and go to –180°, tracing a curve from far left toward the origin.
Correct answer is: Be a curve that starts at infinity on the negative real axis and ends at the origin.
Q.40 When a Nyquist plot makes a clockwise encirclement of (‑1,0), the direction of encirclement is considered:
Positive
Negative
Zero
Undefined
Explanation - By convention, clockwise encirclements are counted as positive (N > 0).
Correct answer is: Positive
Q.41 If the open‑loop transfer function has a pole at the origin, the system is said to be:
Type‑0
Type‑1
Type‑2
Type‑3
Explanation - The system type is the number of integrators (poles at the origin) in the open loop; a single pole at the origin makes it Type‑1.
Correct answer is: Type‑1
Q.42 A system with G(s) = \frac{K}{s(s+4)} has a phase margin of 0° at K = 8. What is the gain margin at this K?
0 dB
6 dB
Infinite
Cannot be determined
Explanation - Zero phase margin occurs when the Nyquist plot passes through (‑1,0), implying the gain is exactly at the stability limit, so gain margin is 0 dB.
Correct answer is: 0 dB
Q.43 Which of the following statements about the Nyquist plot for a stable, minimum‑phase system is FALSE?
It never encircles (‑1,0).
It lies entirely in the right half of the complex plane.
Its magnitude is always less than 1 for all frequencies.
It starts at (1,0) when ω = 0.
Explanation - A stable system can have magnitudes greater than 1 at some frequencies; only the relationship to (‑1,0) matters for stability.
Correct answer is: Its magnitude is always less than 1 for all frequencies.
Q.44 For a system with G(s) = \frac{K}{s(s+2)(s+4)}, the phase at the gain crossover frequency is approximately:
‑90°
‑135°
‑180°
‑225°
Explanation - At ω_gc, the magnitude is 1. For this third‑order system the phase is roughly –135° (midway between –90° and –180°).
Correct answer is: ‑135°
Q.45 A Nyquist plot that is symmetric about the real axis indicates that the system:
Is linear and time‑invariant.
Has real coefficients in its transfer function.
Is unstable.
Has a pure time delay.
Explanation - If the transfer function has real coefficients, G(jω) and its complex conjugate appear symmetrically, producing a plot symmetric about the real axis.
Correct answer is: Has real coefficients in its transfer function.
Q.46 If a Nyquist plot makes three clockwise encirclements of (‑1,0) and the open‑loop has no right‑half‑plane poles, how many closed‑loop poles are in the right half‑plane?
0
1
2
3
Explanation - N = 3, P = 0 → Z = N + P = 3 closed‑loop poles in the right half‑plane, indicating instability.
Correct answer is: 3
Q.47 Which of the following is a sufficient condition for absolute stability of a nonlinear feedback system using the Nyquist criterion?
The Nyquist plot of the linearized model does not encircle (‑1,0).
The phase margin is greater than 45°.
The gain margin is greater than 10 dB.
The system has no poles on the imaginary axis.
Explanation - If the linearized open‑loop Nyquist plot avoids (‑1,0), the nonlinear system is guaranteed to be absolutely stable (by the circle criterion).
Correct answer is: The Nyquist plot of the linearized model does not encircle (‑1,0).
Q.48 For G(s) = \frac{K}{s(s+1)(s+3)}, the value of K that yields a phase margin of 30° is approximately:
K = 4
K = 8
K = 12
K = 16
Explanation - Using Bode plot approximations, a gain of about 8 gives a phase margin close to 30°.
Correct answer is: K = 8
Q.49 When a system has an infinite gain margin, which of the following must be true?
The Nyquist plot never reaches the unit circle.
The Nyquist plot never approaches (‑1,0) as gain varies.
The phase margin is zero.
The system has no poles.
Explanation - Infinite gain margin means the plot can be scaled arbitrarily without ever reaching (‑1,0).
Correct answer is: The Nyquist plot never approaches (‑1,0) as gain varies.
Q.50 If a Nyquist plot passes through the point (‑1,0) at ω = 5 rad/s, what is the gain crossover frequency?
5 rad/s
Any frequency where the magnitude is 1.
Not defined, because the plot passes through (‑1,0).
0 rad/s
Explanation - When the plot passes through (‑1,0), both magnitude = 1 and phase = –180°, so ω = 5 rad/s is the gain crossover frequency.
Correct answer is: 5 rad/s
Q.51 A system with G(s) = \frac{K(s+2)}{s(s+5)} is compensated with a lag network. What is the primary effect on the Nyquist plot?
It shifts the plot upward.
It moves the plot to the right, increasing gain margin.
It reduces the phase lag at high frequencies.
It compresses the plot toward the origin.
Explanation - A lag compensator reduces high‑frequency gain, pulling the Nyquist plot inward and improving low‑frequency gain without affecting high‑frequency phase much.
Correct answer is: It compresses the plot toward the origin.
Q.52 For G(s) = \frac{K}{(s+1)(s+4)}, the Nyquist plot will intersect the negative real axis at which frequency (in rad/s) when K = 2?
ω = 1
ω = 2
ω = √2
ω = 4
Explanation - Setting Im{G(jω)} = 0 yields ω = √(1·4) = 2 rad/s; with K = 2 the magnitude condition gives the same result.
Correct answer is: ω = √2
Q.53 Which of the following describes the effect of a negative feedback loop on the Nyquist plot of the open‑loop system?
It rotates the plot by 180°.
It scales the plot outward by the feedback gain.
It does not affect the Nyquist plot; only the closed‑loop response changes.
It changes the sign of the real part of the plot.
Explanation - The Nyquist plot is drawn for the open‑loop transfer function G(s)H(s). Adding negative feedback does not alter G(s)H(s) itself; it changes the closed‑loop characteristic.
Correct answer is: It does not affect the Nyquist plot; only the closed‑loop response changes.
Q.54 The number of poles of G(s)H(s) that lie on the imaginary axis is:
Always zero for Nyquist analysis.
Ignored in the Nyquist criterion.
Accounted for by deforming the contour.
Added to the number of encirclements.
Explanation - Poles on the imaginary axis require a small semicircular indentation in the Nyquist contour to avoid singularities.
Correct answer is: Accounted for by deforming the contour.
Q.55 If the Nyquist plot of G(s)H(s) makes a net of –2 (counter‑clockwise) encirclements of (‑1,0), and the open‑loop has P = 1 right‑half‑plane pole, how many closed‑loop poles are in the right half‑plane?
‑1 (impossible)
0
1
3
Explanation - N = –2, P = 1 → Z = N + P = –1 (not possible). The correct interpretation is that the sign convention used makes clockwise positive; thus –2 indicates 2 counter‑clockwise encirclements. Using N = –2, Z = –2 + 1 = –1, which cannot happen, indicating an error in counting. The physically valid answer is that the system is unstable. (Given the options, the only plausible answer is 0, indicating no right‑half‑plane closed‑loop poles if the net encirclement cancels the open‑loop pole.)
Correct answer is: 0
Q.56 A system with transfer function G(s) = \frac{K}{s(s+2)} has a phase margin of 60°. Approximate the gain crossover frequency ω_gc.
ω_gc ≈ 1 rad/s
ω_gc ≈ 2 rad/s
ω_gc ≈ 3 rad/s
ω_gc ≈ 4 rad/s
Explanation - At ω = 2 rad/s the phase of G(jω) ≈ –126°, giving a phase margin of about 54°, close to 60°. The exact value is near 2 rad/s.
Correct answer is: ω_gc ≈ 2 rad/s
Q.57 Which of the following is true about the Nyquist plot of a pure differentiator (s)?
It lies on the positive real axis for all ω.
It is a straight line through the origin at 90°.
It is a circle of radius 1.
It spirals outward as ω increases.
Explanation - s = jω has magnitude ω and phase +90°, so the plot is a ray from the origin at +90°.
Correct answer is: It is a straight line through the origin at 90°.
Q.58 When applying the Nyquist criterion, the contour in the s‑plane must:
Include the entire left half of the s‑plane.
Enclose the entire right half of the s‑plane.
Be a circle of radius 1 centered at the origin.
Exclude all poles of G(s)H(s).
Explanation - The Nyquist contour is chosen to encircle the right half of the s‑plane (including the j‑axis) to capture all poles that could affect stability.
Correct answer is: Enclose the entire right half of the s‑plane.
Q.59 If a Nyquist plot has a single clockwise encirclement of (‑1,0) and the open‑loop has one pole at the origin, the closed‑loop system is:
Stable
Marginally stable
Unstable
Cannot be determined
Explanation - P = 1 (origin pole). N = 1 clockwise. Thus Z = N + P = 2 closed‑loop poles on the right half‑plane? Actually Z = N + P = 2, which indicates instability. However, because a pole at the origin contributes a marginal mode, the system is marginally stable. The most appropriate answer is unstable. (Given the options, the correct classification is unstable.)
Correct answer is: Stable
Q.60 A Nyquist plot that lies entirely within a circle of radius 0.2 centered at the origin indicates:
Gain margin > 14 dB
Phase margin > 90°
Zero gain margin
Infinite phase margin
Explanation - Distance from origin to (‑1,0) is 1. If the plot stays within 0.2, the gain can be increased by a factor of 5 (14 dB) before reaching the critical point.
Correct answer is: Gain margin > 14 dB
Q.61 Which of the following transfer functions will have a Nyquist plot that is symmetric about the real axis and passes through the origin?
G(s) = \frac{1}{s+1}
G(s) = \frac{s}{s^2+1}
G(s) = \frac{1}{s(s+2)}
G(s) = \frac{j}{s}
Explanation - The numerator zero at the origin makes the plot pass through the origin; real coefficients give symmetry.
Correct answer is: G(s) = \frac{s}{s^2+1}
Q.62 A system has a gain crossover frequency of 10 rad/s and a phase margin of 45°. What is the phase of G(j10) in degrees?
‑135°
‑180°
‑225°
‑90°
Explanation - Phase margin = 180° + ∠G(jω_gc). Thus ∠G(j10) = –180° + 45° = –135°.
Correct answer is: ‑135°
Q.63 In the Nyquist plot, the distance from the origin to the point (‑1,0) is:
0
0.5
1
2
Explanation - The point (‑1,0) lies one unit from the origin on the negative real axis.
Correct answer is: 1
Q.64 Which of the following open‑loop transfer functions has a pole on the imaginary axis?
G(s) = \frac{1}{s(s+2)}
G(s) = \frac{1}{(s+1)(s+3)}
G(s) = \frac{1}{s^2+1}
G(s) = \frac{1}{(s-1)(s+1)}
Explanation - s^2 + 1 = 0 gives poles at s = ±j, which lie on the imaginary axis.
Correct answer is: G(s) = \frac{1}{s^2+1}
Q.65 If a Nyquist plot of G(s)H(s) makes two clockwise encirclements of (‑1,0) and the open‑loop has P = 0, what is the minimum number of poles of the closed‑loop system that lie in the right half‑plane?
0
1
2
3
Explanation - N = 2, P = 0 → Z = N + P = 2 closed‑loop poles in the right half‑plane.
Correct answer is: 2
Q.66 When the Nyquist plot of a system passes through the point (‑1,0) at ω = 0, what does this indicate?
The system has an infinite gain margin.
The system has a pole at the origin.
The system is marginally stable at DC.
The system has zero phase margin.
Explanation - Passing through (‑1,0) at ω = 0 means the DC gain is unity with –180° phase, indicating a marginal condition.
Correct answer is: The system is marginally stable at DC.
Q.67 Which of the following statements is true for a system whose Nyquist plot lies entirely in the right half of the complex plane?
It has a negative gain margin.
It is always unstable.
It has an infinite gain margin.
It has zero phase margin.
Explanation - If the plot never approaches (‑1,0), the gain can be increased without limit before instability, giving an infinite gain margin.
Correct answer is: It has an infinite gain margin.
Q.68 A system with transfer function G(s) = \frac{K}{s(s+4)} has a phase margin of 30° when K = 4. What will happen to the phase margin if K is increased to 8?
Phase margin will increase.
Phase margin will decrease.
Phase margin will stay the same.
Phase margin becomes infinite.
Explanation - Increasing gain shifts the Nyquist plot outward, moving the gain crossover to a higher frequency where phase is more negative, reducing phase margin.
Correct answer is: Phase margin will decrease.
Q.69 If a Nyquist plot makes a net clockwise encirclement of (‑1,0) equal to the number of right‑half‑plane poles of G(s)H(s), what can be said about the closed‑loop system?
It is stable.
It is marginally stable.
It is unstable.
Stability cannot be determined.
Explanation - When N = P, Z = N + P = 2P, but for stability we need Z = 0. Actually Nyquist stability requires N = P for Z = 0. Therefore, if N equals P, the closed‑loop system is stable.
Correct answer is: It is stable.
Q.70 The Nyquist plot of a system with an odd number of poles on the right half‑plane will always:
Encircle (‑1,0) clockwise.
Encircle (‑1,0) counter‑clockwise.
Never encircle (‑1,0).
Cross the real axis at (‑1,0).
Explanation - Each right‑half‑plane pole contributes one clockwise encirclement in the Nyquist plot.
Correct answer is: Encircle (‑1,0) clockwise.
Q.71 Which of the following is the correct relationship used in Nyquist analysis?
N = Z + P
N = Z – P
N = P – Z
Z = N – P
Explanation - The Nyquist criterion states N = Z – P, where N is net encirclements of (‑1,0), Z is closed‑loop RHP poles, and P is open‑loop RHP poles.
Correct answer is: N = Z – P
Q.72 If a system has a gain margin of 9 dB, by what factor can the loop gain be multiplied before instability occurs?
≈ 2.8
≈ 3.0
≈ 9.0
≈ 1.5
Explanation - 9 dB corresponds to a gain factor of 10^(9/20) ≈ 2.82.
Correct answer is: ≈ 2.8
Q.73 A Nyquist plot that passes through the point (‑1,0) and then loops back to the right side of the complex plane indicates:
A stable system with infinite gain margin.
A system with zero phase margin and finite gain margin.
A marginally stable system.
An unstable system with multiple encirclements.
Explanation - Crossing (‑1,0) means a closed‑loop pole lies on the imaginary axis, giving marginal stability.
Correct answer is: A marginally stable system.
Q.74 For a system with G(s) = \frac{K}{s(s+2)(s+5)}, the phase at low frequencies (ω → 0) is:
0°
‑90°
‑180°
‑270°
Explanation - At low frequencies the integrator contributes –90°, while the two poles contribute negligible phase, giving a total of –90°.
Correct answer is: ‑90°
Q.75 Which of the following statements about the Nyquist plot of a system with a pure time delay e^{‑jωτ} is FALSE?
Its magnitude is unity for all ω.
Its phase decreases linearly with ω.
It is a circle of radius 1.
It rotates around the origin as ω varies.
Explanation - The plot is a point moving on the unit circle, not a full circle traced out; the set of points is the unit circle, but the trajectory for increasing ω is a single point moving around the circumference.
Correct answer is: It is a circle of radius 1.
Q.76 When applying the Nyquist criterion, why is the contour closed at infinity?
To include all poles of the system.
Because the magnitude of G(jω) becomes zero at infinity.
To satisfy Cauchy's argument principle.
To avoid poles on the imaginary axis.
Explanation - Closing the contour at infinity ensures the contour is a closed loop in the s‑plane, allowing the use of the argument principle to relate encirclements to poles.
Correct answer is: To satisfy Cauchy's argument principle.
Q.77 A system with G(s) = \frac{K}{(s+1)^3} will have how many poles at s = –1?
1
2
3
0
Explanation - The denominator (s+1)^3 indicates a triple pole at s = –1.
Correct answer is: 3
Q.78 If the Nyquist plot of an open‑loop system lies entirely in the right half of the complex plane, what is the phase margin?
Zero degrees
Positive but finite
Infinite
Negative
Explanation - When the plot never reaches the negative real axis, the phase can be increased indefinitely before reaching –180°, implying infinite phase margin.
Correct answer is: Infinite
Q.79 What is the effect of adding a zero to the open‑loop transfer function on the Nyquist plot?
It shifts the plot to the left.
It rotates the plot clockwise.
It pulls the plot away from the origin at high frequencies.
It compresses the plot toward the origin.
Explanation - A zero adds gain at frequencies above its break frequency, stretching the Nyquist plot outward at high ω.
Correct answer is: It pulls the plot away from the origin at high frequencies.
Q.80 For G(s) = \frac{K}{s(s+3)}, the gain crossover frequency ω_gc (where |G(jω)| = 1) satisfies:
ω_gc = \sqrt{3K}
ω_gc = \sqrt{K/3}
ω_gc = \sqrt{K}
ω_gc = \sqrt{3}
Explanation - |G(jω)| = K/(ω\sqrt{ω^2+9}) = 1 → ω^2(ω^2+9) = K^2 → solving yields ω_gc ≈ \sqrt{K/3} for typical K values.
Correct answer is: ω_gc = \sqrt{K/3}
Q.81 If a Nyquist plot makes two counter‑clockwise encirclements of (‑1,0) and the open‑loop has P = 2 right‑half‑plane poles, the closed‑loop system will have:
0 right‑half‑plane poles
2 right‑half‑plane poles
4 right‑half‑plane poles
Cannot be determined
Explanation - N = –2, P = 2 → Z = N + P = 0, indicating no closed‑loop RHP poles (stable).
Correct answer is: 0 right‑half‑plane poles
Q.82 Which of the following open‑loop transfer functions has an infinite phase margin?
G(s) = \frac{1}{s+1}
G(s) = \frac{10}{s(s+5)}
G(s) = \frac{1}{(s+1)^2}
G(s) = \frac{K}{s(s+2)} with K = 1
Explanation - A first‑order lag never reaches –180° phase, giving infinite phase margin.
Correct answer is: G(s) = \frac{1}{s+1}
Q.83 When the Nyquist plot of a system passes through (‑1,0) at a frequency ω_c, the system's gain margin is:
0 dB
Infinite
20·log10(ω_c)
Cannot be determined
Explanation - Crossing (‑1,0) means the gain is exactly at the stability limit; any increase makes it unstable, so gain margin is 0 dB.
Correct answer is: 0 dB
Q.84 A Nyquist plot that never crosses the negative real axis indicates which of the following about the system's stability?
The system is unstable.
The system has a negative phase margin.
The system is stable with infinite gain margin.
The system has zero gain margin.
Explanation - If the plot never reaches the critical point (‑1,0), the gain can be increased indefinitely without causing encirclements, implying infinite gain margin and stability.
Correct answer is: The system is stable with infinite gain margin.
Q.85 For a system with transfer function G(s) = \frac{K}{s(s+4)}, the phase at ω = 2 rad/s is approximately:
‑90°
‑112.5°
‑135°
‑157.5°
Explanation - Phase = –90° (integrator) – arctan(2/4) ≈ –90° – 26.6° ≈ –116.6°, close to –112.5°.
Correct answer is: ‑112.5°
Q.86 If a Nyquist plot of G(s)H(s) has a net clockwise encirclement of (‑1,0) equal to 3, and the open‑loop has P = 1, how many closed‑loop poles are in the right half‑plane?
2
3
4
5
Explanation - N = 3, P = 1 → Z = N + P = 4 closed‑loop RHP poles.
Correct answer is: 4
Q.87 Which of the following is a direct consequence of increasing the phase margin of a control system?
Higher steady‑state error.
Improved transient response (less overshoot).
Decreased gain margin.
Reduced bandwidth.
Explanation - Higher phase margin generally reduces overshoot and improves damping of the closed‑loop response.
Correct answer is: Improved transient response (less overshoot).
Q.88 The Nyquist plot of a system with G(s) = \frac{K}{(s+2)(s+5)} will intersect the real axis at a negative value. This intersection corresponds to:
The gain crossover frequency.
The phase crossover frequency.
A point where Im{G(jω)} = 0.
The point of maximum magnitude.
Explanation - Crossing the real axis means the imaginary part of G(jω) is zero.
Correct answer is: A point where Im{G(jω)} = 0.
Q.89 When a Nyquist plot makes a clockwise encirclement of (‑1,0) and the open‑loop system has no right‑half‑plane poles, the closed‑loop system is:
Stable
Marginally stable
Unstable
Cannot be determined
Explanation - N > 0 with P = 0 yields Z = N > 0, meaning there are closed‑loop poles in the right half‑plane, i.e., instability.
Correct answer is: Unstable
Q.90 A system with transfer function G(s) = \frac{K}{s(s+2)} has a gain margin of 6 dB. What is the maximum value of K that keeps the system stable?
K = 1
K = 2
K = 4
K = 8
Explanation - A gain margin of 6 dB corresponds to a factor of 2. If the critical gain is K_c = 8, the maximum stable gain is K_c / 2 = 4.
Correct answer is: K = 4
Q.91 Which of the following describes the shape of the Nyquist plot for a second‑order under‑damped system with low damping (ζ < 0.2)?
A small circle near the origin.
A large loop that encloses (‑1,0).
A tight spiral approaching the origin.
A figure‑8 shape.
Explanation - Low damping leads to a large resonance peak, causing the Nyquist plot to loop widely and possibly encircle (‑1,0).
Correct answer is: A large loop that encloses (‑1,0).
Q.92 If the Nyquist plot of an open‑loop system stays entirely within a circle of radius 0.1 centered at the origin, what is the minimum gain margin in dB?
20 dB
10 dB
14 dB
6 dB
Explanation - Distance to (‑1,0) is 1. Ratio 1/0.1 = 10 → 20·log10(10) = 20 dB gain margin.
Correct answer is: 20 dB
Q.93 For a system with G(s) = \frac{K}{s(s+3)}, the phase margin is zero when K equals:
3
6
9
12
Explanation - At the gain crossover frequency, the phase is –180° when K ≈ 9, yielding zero phase margin.
Correct answer is: 9
Q.94 Which of the following is true about the Nyquist plot of a system that has a pole at the origin and a zero at the origin?
The plot will start at the origin.
The plot will be undefined.
The plot will be a straight line on the real axis.
The plot will be a circle of radius 1.
Explanation - Both pole and zero at the origin cancel each other, leaving a finite value at ω = 0, causing the plot to start at the origin.
Correct answer is: The plot will start at the origin.
Q.95 If a system has a phase margin of 60° and a gain margin of 10 dB, what can be said about its robustness?
It is not robust.
It has moderate robustness.
It has high robustness.
Robustness cannot be inferred.
Explanation - Both high phase margin (>45°) and gain margin (>6 dB) indicate the system can tolerate considerable variations, signifying high robustness.
Correct answer is: It has high robustness.
Q.96 The Nyquist plot of the transfer function G(s) = \frac{1}{(s+1)(s+2)} will intersect the real axis at:
1
-1
0.5
-0.5
Explanation - Setting Im{G(jω)} = 0 yields ω = √(1·2) = √2, and the real part at this frequency evaluates to -0.5.
Correct answer is: -0.5
Q.97 A Nyquist plot that makes one clockwise encirclement of (‑1,0) and passes through (‑1,0) itself is:
Stable with infinite gain margin.
Unstable with zero gain margin.
Marginally stable with zero gain margin.
Stable with finite gain margin.
Explanation - Passing through (‑1,0) gives zero gain margin, and the single encirclement with P = 0 leads to one closed‑loop pole on the imaginary axis, i.e., marginal stability.
Correct answer is: Marginally stable with zero gain margin.
Q.98 In a Nyquist plot, what does a point at (0,0) represent?
Infinite gain.
Zero gain.
Phase of 0°.
Phase of –90°.
Explanation - The origin corresponds to magnitude zero (|G(jω)| = 0), regardless of phase.
Correct answer is: Zero gain.
Q.99 If the open‑loop transfer function has a pole at s = +2, what is the effect on the Nyquist plot?
The plot will be reflected about the real axis.
The plot will make one clockwise encirclement of (‑1,0).
The plot will be unchanged.
The plot will require a detour around the pole.
Explanation - Poles in the right half‑plane are simply counted as P; they do not require contour deformation.
Correct answer is: The plot will be unchanged.
Q.100 Which of the following is a necessary condition for absolute stability using the Nyquist criterion?
The Nyquist plot must be symmetric about the real axis.
The Nyquist plot must not encircle (‑1,0).
The gain margin must be less than 0 dB.
The phase margin must be exactly 45°.
Explanation - If the plot does not encircle the critical point, the closed‑loop system is guaranteed to be stable (absolute stability).
Correct answer is: The Nyquist plot must not encircle (‑1,0).
Q.101 For G(s) = \frac{K}{s(s+5)}, the phase at high frequency (ω → ∞) approaches:
0°
‑90°
‑180°
‑270°
Explanation - Two poles contribute –180° total at high frequencies.
Correct answer is: ‑180°
Q.102 When the Nyquist plot of a system encircles the point (‑1,0) twice clockwise and the open‑loop has one pole at the origin, the closed‑loop system:
Is stable.
Is marginally stable.
Is unstable.
Cannot be determined.
Explanation - P = 1, N = 2 → Z = N + P = 3 > 0, indicating three closed‑loop RHP poles, hence instability.
Correct answer is: Is unstable.
Q.103 A Nyquist plot that lies entirely in the left half of the complex plane indicates:
Zero gain margin.
Infinite phase margin.
Infinite gain margin.
Negative phase margin.
Explanation - If the plot never reaches the critical point (‑1,0), the gain can be increased without bound, giving infinite gain margin.
Correct answer is: Infinite gain margin.
Q.104 Which of the following statements about the relationship between Nyquist and root‑locus methods is correct?
Both give the same information about closed‑loop poles.
Nyquist can be used for non‑minimum phase systems, root‑locus cannot.
Root‑locus requires knowledge of gain margin, Nyquist does not.
Nyquist provides a graphical test for stability, root‑locus shows pole trajectories as gain varies.
Explanation - Nyquist assesses stability via encirclements; root‑locus plots how closed‑loop poles move with gain.
Correct answer is: Nyquist provides a graphical test for stability, root‑locus shows pole trajectories as gain varies.
Q.105 If a system has a gain margin of 0 dB, what does this imply about its Nyquist plot?
It never approaches (‑1,0).
It passes through (‑1,0).
It encircles (‑1,0) clockwise.
It stays entirely in the right half‑plane.
Explanation - Zero gain margin means the plot is exactly at the critical point; any increase in gain will cause instability.
Correct answer is: It passes through (‑1,0).
Q.106 Which of the following open‑loop transfer functions will have a Nyquist plot that is a perfect circle?
G(s) = \frac{K}{s^2 + 1}
G(s) = \frac{K}{(s+1)(s+2)}
G(s) = \frac{K}{s(s+2)}
G(s) = \frac{K}{s + j}
Explanation - The transfer function 1/(s^2+1) maps the jω axis onto a circle in the G-plane.
Correct answer is: G(s) = \frac{K}{s^2 + 1}
Q.107 When the Nyquist plot of G(s)H(s) makes a net of zero encirclements of (‑1,0) and the open‑loop has P = 2, the closed‑loop system will have:
0 right‑half‑plane poles.
2 right‑half‑plane poles.
4 right‑half‑plane poles.
Cannot be determined.
Explanation - N = 0, P = 2 → Z = N + P = 2 closed‑loop RHP poles.
Correct answer is: 2 right‑half‑plane poles.
Q.108 Which of the following statements about the Nyquist plot of a system with a pole at s = 0 is true?
The plot starts at infinity.
The plot starts at the origin.
The plot is undefined at ω = 0.
The plot has a hole at (‑1,0).
Explanation - A pole at the origin makes G(j0) infinite, so the Nyquist plot is undefined at ω = 0 and requires an indentation.
Correct answer is: The plot is undefined at ω = 0.
Q.109 A system with G(s) = \frac{K}{s(s+4)} has a phase margin of 45° at K = 2. If K is reduced to 1, what happens to the phase margin?
It increases.
It decreases.
It stays the same.
It becomes zero.
Explanation - Reducing gain pulls the Nyquist plot inward, moving the gain crossover to a lower frequency where phase is less negative, increasing phase margin.
Correct answer is: It increases.
Q.110 If a Nyquist plot for G(s)H(s) encloses (‑1,0) three times clockwise and the open‑loop has no RHP poles, the closed‑loop system is:
Stable
Marginally stable
Unstable
Indeterminate
Explanation - N = 3, P = 0 → Z = 3 RHP closed‑loop poles, indicating instability.
Correct answer is: Unstable
Q.111 The point on the Nyquist plot where the magnitude is unity (|G(jω)| = 1) is called:
Phase crossover point
Gain crossover point
Critical point
Zero crossing point
Explanation - The frequency at which the magnitude equals 1 (0 dB) is the gain crossover frequency.
Correct answer is: Gain crossover point
Q.112 For the transfer function G(s) = \frac{K}{(s+1)(s+3)}, the Nyquist plot will cross the real axis at a negative value. This crossing is used to determine:
Gain margin
Phase margin
Stability of the system
Resonant frequency
Explanation - The crossing point helps locate the phase crossover frequency, which is used to compute the phase margin.
Correct answer is: Phase margin
Q.113 If the Nyquist plot of an open‑loop system stays entirely within a circle of radius 0.05 centered at the origin, what is the approximate gain margin in dB?
26 dB
40 dB
46 dB
52 dB
Explanation - Distance to (‑1,0) is 1. Ratio 1/0.05 = 20 → 20·log10(20) ≈ 26 dB. (Correction: Actually 20·log10(20) ≈ 26 dB, not 46 dB. The correct answer should be 26 dB.)
Correct answer is: 46 dB
Q.114 A Nyquist plot that never reaches the negative real axis indicates that the system's:
Phase margin is zero.
Gain margin is infinite.
Stability is marginal.
System has a pole at the origin.
Explanation - If the plot never approaches (‑1,0), the gain can be increased arbitrarily before instability, giving infinite gain margin.
Correct answer is: Gain margin is infinite.
Q.115 Which of the following is true about the Nyquist plot of a system with a double integrator G(s) = \frac{K}{s^2}?
It is a straight line on the positive real axis.
It lies on the imaginary axis.
It is a circle of radius K.
It is undefined for all ω.
Explanation - G(jω) = K/(jω)^2 = -K/ω^2, which is purely real and negative; as ω varies, the plot lies on the negative real axis (imaginary part zero).
Correct answer is: It lies on the imaginary axis.
Q.116 If the Nyquist plot of G(s)H(s) makes one clockwise encirclement of (‑1,0) and the open‑loop has two RHP poles, the closed‑loop system will have:
One RHP pole
Two RHP poles
Three RHP poles
Zero RHP poles
Explanation - N = 1, P = 2 → Z = N + P = 3 closed‑loop RHP poles.
Correct answer is: Three RHP poles
Q.117 A Nyquist plot that lies entirely within the unit circle centered at the origin indicates:
The system is unstable.
The system has a gain margin of at least 0 dB.
The system has a gain margin of at least 0 dB and possibly infinite.
The system has zero phase margin.
Explanation - If the plot stays inside the unit circle, the distance to (‑1,0) is at least 1, giving a gain margin of at least 0 dB; if it stays well inside, the margin can be infinite.
Correct answer is: The system has a gain margin of at least 0 dB and possibly infinite.
Q.118 When the Nyquist plot of an open‑loop system passes through (‑1,0) at a frequency ω_c, what is the value of the closed‑loop transfer function at that frequency?
0
1
∞
‑1
Explanation - At the frequency where the open‑loop G(jω_c) = –1, the closed‑loop transfer function T = G/(1+G) becomes infinite (denominator zero).
Correct answer is: ∞
Q.119 Which of the following is a correct interpretation of a Nyquist plot that makes a clockwise encirclement of (‑1,0) exactly equal to the number of open‑loop RHP poles?
The closed‑loop system is marginally stable.
The closed‑loop system is stable.
The closed‑loop system is unstable.
Stability cannot be inferred.
Explanation - When N = P, the Nyquist criterion gives Z = 0, meaning no closed‑loop poles in the RHP; the system is stable.
Correct answer is: The closed‑loop system is stable.
Q.120 A Nyquist plot that stays entirely in the left half of the complex plane (Re{G(jω)} < 0) implies which of the following?
Infinite gain margin.
Zero phase margin.
The system is unstable.
The system has a pole on the imaginary axis.
Explanation - If the plot never reaches (‑1,0), the gain can be increased without bound before instability, yielding infinite gain margin.
Correct answer is: Infinite gain margin.
Q.121 For the transfer function G(s) = \frac{K}{s(s+2)(s+5)}, the Nyquist plot will cross the real axis at a frequency ω = √(10). What is the corresponding real part of G(jω) at this crossing?
-K/10
-K/5
-K/2
-K
Explanation - At ω = √10, Im{G(jω)} = 0, and Re{G(jω)} = -K/(ω^2 + 10) = -K/20 ≈ -K/10 (approximation for illustration).
Correct answer is: -K/10
Q.122 If a Nyquist plot of G(s)H(s) makes a net of zero encirclements of (‑1,0) and the open‑loop system has no RHP poles, the closed‑loop system is:
Unstable
Marginally stable
Stable
Indeterminate
Explanation - N = 0, P = 0 → Z = 0; no closed‑loop RHP poles, so the system is stable.
Correct answer is: Stable
Q.123 Which of the following modifications to a controller will increase the phase margin of a system?
Increasing proportional gain.
Adding a lead compensator.
Adding a lag compensator.
Increasing system delay.
Explanation - A lead compensator adds positive phase near the crossover frequency, raising the phase margin.
Correct answer is: Adding a lead compensator.
Q.124 In a Nyquist plot, a point at (‑1,0) corresponds to which condition on the open‑loop transfer function?
Magnitude = 0 dB and phase = 0°
Magnitude = 0 dB and phase = –180°
Magnitude = –1 and phase = 0°
Magnitude = 1 and phase = –180°
Explanation - The point (‑1,0) has magnitude 1 (0 dB) and angle –180°.
Correct answer is: Magnitude = 1 and phase = –180°
Q.125 A Nyquist plot that lies entirely within a circle of radius 0.5 centered at (0,0) indicates a gain margin of at least:
6 dB
12 dB
14 dB
20 dB
Explanation - Distance from origin to (‑1,0) is 1; ratio 1/0.5 = 2 → 20·log10(2) ≈ 6 dB gain margin.
Correct answer is: 6 dB
Q.126 If the Nyquist plot of a system has a single clockwise encirclement of (‑1,0) and the open‑loop has a pole at s = +1, the closed‑loop system will have:
Zero RHP poles
One RHP pole
Two RHP poles
Three RHP poles
Explanation - P = 1, N = 1 → Z = N + P = 2 closed‑loop RHP poles.
Correct answer is: Two RHP poles
Q.127 Which of the following statements about Nyquist stability is FALSE?
The Nyquist criterion can be applied to systems with time delay.
The Nyquist plot must be drawn for the entire jω axis from –∞ to +∞.
The Nyquist criterion requires the system to be linear.
The Nyquist plot can be used to compute the exact location of closed‑loop poles.
Explanation - Nyquist analysis provides stability information via encirclements but does not give the exact pole locations.
Correct answer is: The Nyquist plot can be used to compute the exact location of closed‑loop poles.
Q.128 For G(s) = \frac{K}{(s+2)(s+4)}, the magnitude at the frequency where the Nyquist plot crosses the real axis is:
K/8
K/4
K/2
K
Explanation - At the real‑axis crossing, ω = √(2·4) = √8, giving magnitude |G(jω)| = K/(√8·(√8+2)(√8+4)) ≈ K/8.
Correct answer is: K/8