Monday, November 3, 2025

If all roots of characteristic equation lie on imaginary axis, system is

 

226. The resonant peak (Mr) of a second-order system is a measure of:

a) Rise time
b) Overshoot
c) Frequency response damping
d) Bandwidth
Answer: c) Frequency response damping
Explanation: Mr is inversely proportional to damping ratio ζ.


227. For a stable second-order system, the damping ratio ζ must be:

a) ζ < 0
b) ζ = 0
c) ζ > 0
d) ζ > 1
Answer: c) ζ > 0
Explanation: Positive damping ensures poles lie in left half-plane.


228. The transfer function of a pure differentiator is:

a) 1/s
b) s
c) 1
d) s + 1
Answer: b) s
Explanation: Differentiator amplifies high frequencies; its transfer function is G(s)=s.


229. The transfer function of a pure integrator is:

a) 1/s
b) s
c) 1/(s+1)
d) s+1
Answer: a) 1/s
Explanation: Integration in Laplace corresponds to division by s.


230. The sensitivity of closed-loop system to parameter variations is:

a) High
b) Low
c) Zero
d) Infinite
Answer: b) Low
Explanation: Feedback reduces sensitivity of system to changes in components.


231. The steady-state error for a unit step in a Type 0 system is:

a) 0
b) 1/(1+Kp)
c) ∞
d) None
Answer: b) 1/(1+Kp)
Explanation: Type 0 → finite steady-state error for step input.


232. The transient response depends mainly on:

a) Zeros
b) Poles
c) Type of input
d) Feedback path
Answer: b) Poles
Explanation: Poles determine natural frequency and damping ratio.


233. A system is said to be critically damped when ζ =

a) 0
b) 0.707
c) 1
d) >1
Answer: c) 1
Explanation: ζ = 1 gives the fastest response without oscillation.


234. The steady-state error for a step input in a Type 1 system is:

a) 0
b) 1/Kp
c) Infinite
d) None
Answer: a) 0
Explanation: Type 1 → one integrator → zero error for step input.


235. A stable system will always have:

a) Positive real poles
b) Poles on left half of s-plane
c) Imaginary poles
d) Zeros on right half
Answer: b) Poles on left half of s-plane
Explanation: Negative real part of poles ensures stability.


236. The rise time of a system decreases with:

a) Decrease in natural frequency
b) Increase in natural frequency
c) Increase in damping
d) Increase in time constant
Answer: b) Increase in natural frequency
Explanation: Higher ωn leads to faster rise time.


237. Phase lag occurs when:

a) Output leads input
b) Output lags input
c) Both in phase
d) Output is zero
Answer: b) Output lags input
Explanation: Phase lag means output occurs later than input in phase.


238. In a stable feedback system, feedback is usually:

a) Positive
b) Negative
c) Zero
d) Infinite
Answer: b) Negative
Explanation: Negative feedback stabilizes and controls gain.


239. The phase margin is defined at:

a) Unity gain frequency
b) Cutoff frequency
c) Resonant frequency
d) Bandwidth frequency
Answer: a) Unity gain frequency
Explanation: Phase margin is measured at the frequency where |G(jω)| = 1.


240. The gain margin is defined at:

a) Phase crossover frequency
b) Gain crossover frequency
c) Resonant frequency
d) None
Answer: a) Phase crossover frequency
Explanation: Gain margin is the amount of gain reduction to reach critical point.


241. The steady-state error constant for parabolic input is:

a) Kp
b) Kv
c) Ka
d) None
Answer: c) Ka
Explanation: Acceleration error constant (Ka) relates to parabolic inputs.


242. A root locus asymptote angle for two branches is given by:

a) ±90°
b) ±45°
c) ±60°
d) ±120°
Answer: a) ±90°
Explanation: Two branches → asymptote angles ±90°.


243. The frequency at which phase = -180° is:

a) Gain crossover frequency
b) Phase crossover frequency
c) Cutoff frequency
d) Resonant frequency
Answer: b) Phase crossover frequency
Explanation: By definition, phase = -180° → phase crossover frequency.


244. The bandwidth of a feedback system is usually:

a) Increased by feedback
b) Decreased by feedback
c) Unchanged
d) None
Answer: a) Increased by feedback
Explanation: Negative feedback broadens frequency response.


245. The steady-state error for ramp input in Type 0 system is:

a) Finite
b) Infinite
c) Zero
d) None
Answer: b) Infinite
Explanation: Type 0 cannot track ramp → infinite error.


246. The Laplace transform of a unit ramp is:

a) 1/s
b) 1/s²
c) s
d) s²
Answer: b) 1/s²
Explanation: L{t} = 1/s².


247. The Laplace transform of δ(t) is:

a) 1
b) 0
c) s
d) 1/s
Answer: a) 1
Explanation: Unit impulse transform is 1.


248. The closed-loop transfer function is G(s)1+G(s)H(s)\frac{G(s)}{1+G(s)H(s)}. This represents:

a) Open-loop system
b) Closed-loop system
c) Feedback path
d) Error function
Answer: b) Closed-loop system
Explanation: Standard closed-loop transfer formula.


249. For a unit feedback system, error signal =

a) R(s) + C(s)
b) R(s) - C(s)
c) R(s)C(s)
d) None
Answer: b) R(s) - C(s)
Explanation: e(s) = input – output (unity feedback).


250. The natural frequency of s2+4s+25=0s^2 + 4s + 25 = 0 is:

a) 5
b) 25
c) 2
d) 4
Answer: a) 5
Explanation: ωn=25=5ω_n = √25 = 5.


251. The damping ratio for s2+4s+25=0s^2 + 4s + 25 = 0:

ζ=42×5=?ζ = \frac{4}{2×5} = ?
a) 0.2
b) 0.4
c) 0.8
d) 1
Answer: b) 0.4
Explanation: ζ = 4/10 = 0.4 → underdamped.


252. The steady-state error for unit step in Type 2 system is:

a) 0
b) ∞
c) 1
d) Depends on gain
Answer: a) 0
Explanation: Type ≥1 → zero step error.


253. A system with gain margin of 10 dB and phase margin of 60° is:

a) Unstable
b) Stable
c) Marginally stable
d) Uncontrollable
Answer: b) Stable
Explanation: Positive margins indicate stability.


254. Increasing derivative gain (Kd) in PID:

a) Increases overshoot
b) Reduces overshoot
c) Slows response
d) Increases steady-state error
Answer: b) Reduces overshoot
Explanation: Derivative adds damping → less overshoot.


255. Increasing integral gain (Ki):

a) Increases steady-state error
b) Reduces steady-state error
c) Increases overshoot
d) b & c
Answer: d) b & c
Explanation: Ki eliminates error but increases oscillation/overshoot.


256. The PID controller equation is:

a) u=Kpe+Kiedt+Kddedtu = Kp e + Ki \int e dt + Kd \frac{de}{dt}
b) u=Kpeu = Kp e
c) u=Kieu = Ki e
d) None
Answer: a) u=Kpe+Kiedt+Kddedtu = Kp e + Ki \int e dt + Kd \frac{de}{dt}
Explanation: Standard PID law.


257. The root locus moves toward infinity when:

a) Zeros < Poles
b) Zeros = Poles
c) Zeros > Poles
d) Gain is 0
Answer: a) Zeros < Poles
Explanation: Unmatched poles → branches go to infinity.


258. Adding poles near origin makes system:

a) Faster
b) Slower
c) More oscillatory
d) Stable
Answer: b) Slower
Explanation: Pole near origin → increased time constant → slower response.


259. In Nyquist plot, encirclement in clockwise direction means:

a) Unstable closed loop
b) Stable
c) Marginally stable
d) None
Answer: a) Unstable closed loop
Explanation: Clockwise encirclement → right-half-plane poles.


260. If all roots of characteristic equation lie on imaginary axis, system is:

a) Stable
b) Unstable
c) Marginally stable
d) Divergent
Answer: c) Marginally stable
Explanation: Oscillations of constant amplitude → marginally stable.

The steady-state error for a unit step input in a type 1 system is

 

201. The damping ratio (ζ) of 0.707 corresponds to:

a) Underdamped
b) Overdamped
c) Critically damped
d) Optimum damping
Answer: d) Optimum damping
Explanation: ζ = 0.707 provides minimal overshoot and fast settling — considered “optimum damping.”


202. The transient response is affected by:

a) Poles only
b) Zeros only
c) Both poles and zeros
d) Input type only
Answer: c) Both poles and zeros
Explanation: Poles determine dynamics; zeros modify transient response.


203. For a stable system, all poles must lie:

a) On imaginary axis
b) In right half-plane
c) In left half-plane
d) Anywhere on s-plane
Answer: c) In left half-plane
Explanation: Stability requires all poles to have negative real parts.


204. The steady-state error for a unit step input in a type 1 system is:

a) 0
b) 1
c) ∞
d) Depends on K
Answer: a) 0
Explanation: Type 1 systems have one integrator → zero steady-state error for step input.


205. A higher gain in closed-loop control generally:

a) Increases stability
b) Reduces steady-state error
c) Increases damping
d) None
Answer: b) Reduces steady-state error
Explanation: Increased gain reduces steady-state error but may affect stability.


206. The number of branches in a root locus equals:

a) Number of poles
b) Number of zeros
c) Difference of poles and zeros
d) Product of poles and zeros
Answer: a) Number of poles
Explanation: Each pole corresponds to one locus branch.


207. For a system G(s)=Ks(s+2)(s+5)G(s) = \frac{K}{s(s+2)(s+5)}, the number of asymptotes is:

a) 1
b) 2
c) 3
d) 0
Answer: b) 2
Explanation: nm=31=2n-m = 3-1 = 2 → two asymptotes (n = poles, m = zeros).


208. A system with pure integrator has:

a) Type 0
b) Type 1
c) Type 2
d) Unstable nature
Answer: b) Type 1
Explanation: One pole at origin → one integrator → Type 1 system.


209. The steady-state error for a ramp input in a Type 1 system is:

a) 0
b) 1/Kv
c) ∞
d) None
Answer: b) 1/Kv
Explanation: For ramp, error constant Kv determines steady-state error.


210. Nyquist plot passes through (-1,0) indicates:

a) Stable
b) Marginally stable
c) Unstable
d) Oscillatory
Answer: b) Marginally stable
Explanation: Passing exactly through critical point indicates marginal stability.


211. Lead compensator adds:

a) Negative phase
b) Positive phase
c) Zero phase
d) None
Answer: b) Positive phase
Explanation: Lead compensation adds phase lead, improving phase margin.


212. Lag compensator is used to:

a) Improve steady-state accuracy
b) Improve phase margin
c) Reduce rise time
d) Increase overshoot
Answer: a) Improve steady-state accuracy
Explanation: Lag compensation adds low-frequency gain, reducing steady-state error.


213. A system is Type 2 if it has:

a) Two zeros
b) Two poles at origin
c) Two feedback loops
d) None
Answer: b) Two poles at origin
Explanation: Each pole at origin represents an integrator → Type number = poles at origin.


214. Root locus branches end at:

a) Poles
b) Zeros
c) Infinity
d) a & b
Answer: d) a & b
Explanation: Locus starts from poles and ends at zeros or infinity.


215. The term “phase margin” refers to:

a) Gain difference
b) Phase difference from -180°
c) Bandwidth
d) Frequency ratio
Answer: b) Phase difference from -180°
Explanation: Phase margin = angle by which phase is above -180° at gain crossover.


216. Gain margin is measured in:

a) Radians
b) Decibels (dB)
c) Hertz
d) Degrees
Answer: b) Decibels (dB)
Explanation: It indicates the gain increase required to reach instability (in dB).


217. A system has poles at -2, -4, and zeros at -3. The centroid is:

(poleszeros)/(nm)(\sum \text{poles} - \sum \text{zeros})/(n-m)
= ((24)(3))/(31)((-2-4) - (-3))/(3-1) = ?
a) -1.5
b) -2.5
c) -3.5
d) -4.5
Answer: b) -2.5
Explanation: Centroid = (-6 + 3)/2 = -1.5? Wait — correction → (-6 - (-3)) / 2 = (-3)/2 = -1.5 actually → Answer: a)
Correction: a) -1.5


218. A system with all real negative poles is:

a) Unstable
b) Stable
c) Marginally stable
d) Oscillatory
Answer: b) Stable
Explanation: All poles in left half-plane → stable.


219. If transfer function numerator order > denominator order, system is:

a) Stable
b) Improper
c) Type 0
d) Causal
Answer: b) Improper
Explanation: Improper systems are physically unrealizable.


220. The “bandwidth” of a system indicates:

a) Speed of response
b) Damping
c) Stability margin
d) Phase shift
Answer: a) Speed of response
Explanation: Higher bandwidth = faster response to input.


221. The Laplace transform of d2y(t)dt2\frac{d^2y(t)}{dt^2} is:

a) sY(s)
b) s²Y(s) – s y(0) – y’(0)
c) s²Y(s)
d) None
Answer: b) s²Y(s) – s y(0) – y’(0)
Explanation: Standard Laplace derivative property.


222. The steady-state error for a parabolic input in a Type 2 system is:

a) 0
b) 1/K
c) Infinite
d) None
Answer: a) 0
Explanation: Type 2 can follow parabolic input with zero steady-state error.


223. The number of poles in a second-order system is:

a) 1
b) 2
c) 3
d) Depends on order
Answer: b) 2
Explanation: Order = number of poles.


224. The breakaway point occurs where:

a) Two loci meet
b) Locus leaves real axis
c) Gain = ∞
d) Both a & b
Answer: d) Both a & b
Explanation: Breakaway occurs when two loci diverge from real axis.


225. Frequency corresponding to maximum magnitude in closed-loop is called:

a) Resonant frequency
b) Crossover frequency
c) Cutoff frequency
d) Gain margin frequency
Answer: a) Resonant frequency
Explanation: Resonant frequency corresponds to the peak of magnitude response.

The number of integrations in open-loop transfer function gives

 

59. For a second-order underdamped system, the peak overshoot MpM_p is given by:

a) eπζ/1ζ2e^{-\pi ζ / \sqrt{1-ζ^2}}
b) 1eζ1 - e^{-ζ}
c) eζπe^{-ζ \pi}
d) 1eπζ1 - e^{-πζ}
Answer: a) eπζ/1ζ2e^{-\pi ζ / \sqrt{1-ζ^2}}
Explanation: Overshoot depends exponentially on the damping ratio ζ.


60. The rise time trt_r for a second-order system is approximately:

a) 1.8ωn\frac{1.8}{ω_n}
b) 3ωn\frac{3}{ω_n}
c) 2.3ωn\frac{2.3}{ω_n}
d) 0.5ωn\frac{0.5}{ω_n}
Answer: a) 1.8ωn\frac{1.8}{ω_n}
Explanation: Approximate relation for 0–100% rise time in an underdamped system.


61. The term “bandwidth” of a control system refers to:

a) Frequency range of zero gain
b) Frequency range of unity phase
c) Frequency range over which system responds satisfactorily
d) Phase crossover frequency
Answer: c) Frequency range over which system responds satisfactorily
Explanation: Bandwidth is the frequency limit of significant response (typically -3 dB).


62. A proportional plus derivative (PD) controller improves:

a) Steady-state error
b) Transient response
c) Steady-state accuracy
d) None
Answer: b) Transient response
Explanation: Derivative term enhances damping, improving transient behavior.


63. The function of a tachogenerator in control systems is:

a) To measure displacement
b) To measure speed
c) To measure position
d) To amplify signal
Answer: b) To measure speed
Explanation: Tachogenerators produce voltage proportional to rotational speed.


64. The transfer function of a DC motor (armature control) is:

a) Ks(Ts+1)\frac{K}{s(Ts+1)}
b) KTs+1\frac{K}{Ts+1}
c) Ks2+sT+1\frac{K}{s^2 + sT + 1}
d) Ks2(Ts+1)\frac{K}{s^2(Ts+1)}
Answer: a) Ks(Ts+1)\frac{K}{s(Ts+1)}
Explanation: One integration from speed to position, and one lag due to motor time constant.


65. The error detector in an automatic control system compares:

a) Output and input
b) Feedback and reference
c) Desired output and actual output
d) All of these
Answer: d) All of these
Explanation: Error is difference between desired (reference) and actual (feedback) signals.


66. The transfer function of a second-order system is:

ωn2s2+2ζωns+ωn2\frac{ω_n^2}{s^2 + 2ζω_n s + ω_n^2}.
Its natural frequency is:
a) ζωnζω_n
b) ωnω_n
c) 2ζωn2ζω_n
d) ωn2ω_n^2
Answer: b) ωnω_n
Explanation: ωnω_n represents natural frequency of oscillation.


67. The steady-state error for a unit parabolic input is zero for:

a) Type 0
b) Type 1
c) Type 2
d) None
Answer: c) Type 2
Explanation: Two integrators eliminate error for parabolic input.


68. The open-loop poles determine:

a) Static error
b) Dynamic response
c) Gain margin
d) Phase margin
Answer: b) Dynamic response
Explanation: Pole locations dictate system speed and oscillation characteristics.


69. The characteristic equation of a unity feedback system is:

a) 1+G(s)=01 + G(s) = 0
b) 1+G(s)H(s)=01 + G(s)H(s) = 0
c) G(s)=0G(s) = 0
d) H(s)=0H(s) = 0
Answer: b) 1+G(s)H(s)=01 + G(s)H(s) = 0
Explanation: Derived from closed-loop transfer function denominator.


70. The steady-state error constants are:

a) Kp,Kv,KaK_p, K_v, K_a
b) Kd,Ki,KpK_d, K_i, K_p
c) K1,K2,K3K_1, K_2, K_3
d) None
Answer: a) Kp,Kv,KaK_p, K_v, K_a
Explanation: They correspond to step, ramp, and parabolic input errors respectively.


71. The Nyquist plot of a Type 1 system passes through:

a) Origin
b) (-1,0)
c) (1,0)
d) (0,1)
Answer: a) Origin
Explanation: Type 1 (one integrator) introduces -90° phase lag at low frequencies, passing through origin.


72. The main advantage of state-space over transfer function is:

a) Easier Laplace computation
b) Handles MIMO systems
c) Requires fewer equations
d) None
Answer: b) Handles MIMO systems
Explanation: State-space can describe multi-input, multi-output systems directly.


73. The transient response of a second-order system is primarily determined by:

a) Zeros
b) Damping ratio and natural frequency
c) Gain
d) Feedback
Answer: b) Damping ratio and natural frequency
Explanation: ζ and ωₙ define rise time, overshoot, and settling time.


74. A system with characteristic equation s2+6s+25=0s^2 + 6s + 25 = 0 is:

a) Overdamped
b) Underdamped
c) Critically damped
d) Unstable
Answer: b) Underdamped
Explanation: ζ=6/(225)=0.6<1ζ = 6 / (2√25) = 0.6 < 1 → underdamped.


75. A phase lead compensator improves:

a) Stability margin
b) Steady-state error
c) Low-frequency response
d) All
Answer: a) Stability margin
Explanation: Lead compensators add positive phase → improved phase margin.


76. The polar plot of a first-order system is a:

a) Circle
b) Straight line
c) Parabola
d) Spiral
Answer: a) Circle
Explanation: For G(jω)=11+jωTG(jω) = \frac{1}{1+jωT}, the locus is a circle in complex plane.


77. The Bode magnitude of 1s\frac{1}{s} is:

a) +20 dB/dec
b) -20 dB/dec
c) 0 dB
d) None
Answer: b) -20 dB/dec
Explanation: Integrator causes -20 dB per decade slope.


78. For stability using Routh-Hurwitz, all elements of the first column must be:

a) Positive
b) Negative
c) Alternate sign
d) Zero
Answer: a) Positive
Explanation: All positive → no sign changes → all poles in LHP.


79. If any row of Routh array becomes zero, the system has:

a) Imaginary roots
b) Repeated real roots
c) Complex roots
d) None
Answer: a) Imaginary roots
Explanation: Zero row indicates symmetric root pairs on imaginary axis → marginal stability.


80. The condition for controllability is:

a) Rank [B AB A²B … Aⁿ⁻¹B] = n
b) det(A) ≠ 0
c) All poles distinct
d) Trace(A) ≠ 0
Answer: a) Rank [B AB A²B … Aⁿ⁻¹B] = n
Explanation: Kalman controllability criterion.


81. For observability, the matrix used is:

a) [C; CA; CA²; …; CAⁿ⁻¹]
b) [B AB …]
c) [A B C D]
d) [A C D]
Answer: a) [C; CA; CA²; …; CAⁿ⁻¹]
Explanation: Kalman observability test.


82. The open-loop transfer function K(s+3)s(s+2)(s+5)\frac{K(s+3)}{s(s+2)(s+5)} has how many asymptotes?

a) 1
b) 2
c) 3
d) None
Answer: b) 2
Explanation: 3 poles − 1 zero = 2 asymptotes.


83. For the same system, centroid of asymptotes =

(25+0)(3)2\frac{(-2-5+0)-(−3)}{2} = ?
a) -2
b) -1
c) -2
d) -2
Answer: a) -2
Explanation: Centroid = (sum of poles − sum of zeros)/(p−z) = (-7+3)/2 = -2.


84. When feedback gain is increased:

a) Stability increases
b) Bandwidth increases
c) Sensitivity increases
d) Gain margin increases
Answer: b) Bandwidth increases
Explanation: Higher feedback improves response speed but may reduce stability.


85. The number of encirclements in Nyquist plot equals:

a) Number of open-loop poles in LHP
b) Number of closed-loop poles in RHP
c) Difference between right-half open and closed-loop poles
d) None
Answer: c) Difference between right-half open and closed-loop poles
Explanation: N=ZPN = Z - P, where N = encirclements, Z = RHP closed-loop poles, P = RHP open-loop poles.


86. Servo systems are primarily used for:

a) Speed control
b) Position control
c) Current control
d) Temperature control
Answer: b) Position control
Explanation: Servo systems precisely control angular/linear position.


87. A stepper motor is used in control systems for:

a) Continuous control
b) Discrete position control
c) Speed control
d) None
Answer: b) Discrete position control
Explanation: Stepper moves in fixed angular steps → discrete control.


88. In frequency response, the phase lag increases with:

a) Frequency
b) Gain
c) Damping ratio
d) Time constant decrease
Answer: a) Frequency
Explanation: Lag increases as system’s phase angle drops at higher frequencies.


89. The final value theorem is used to find:

a) Initial value
b) Steady-state value
c) Transient response
d) Time constant
Answer: b) Steady-state value
Explanation: limtf(t)=lims0sF(s)\lim_{t→∞} f(t) = \lim_{s→0} sF(s).


90. The initial value theorem is used to find:

a) Steady-state response
b) Starting value of response
c) Average value
d) None
Answer: b) Starting value of response
Explanation: limt0f(t)=limssF(s)\lim_{t→0} f(t) = \lim_{s→∞} sF(s).


91. In a unity feedback system, steady-state error for step input is given by:

a) 1/(1+Kp)1/(1+K_p)
b) 1/Kp1/K_p
c) Kp/(1+Kp)K_p/(1+K_p)
d) 1+Kp1+K_p
Answer: a) 1/(1+Kp)1/(1+K_p)
Explanation: Derived from position error constant Kp=lims0G(s)K_p = \lim_{s→0} G(s).


92. If damping ratio ζ = 1, the system is:

a) Overdamped
b) Underdamped
c) Critically damped
d) Unstable
Answer: c) Critically damped
Explanation: ζ = 1 → fastest response without oscillation.


93. For a second-order system, the damping ratio ζ = 0 gives:

a) Overdamped
b) Critically damped
c) Undamped
d) Oscillatory stable
Answer: c) Undamped
Explanation: ζ = 0 → no damping → continuous oscillations.


94. A unity feedback system has G(s)=10s(s+2)G(s) = \frac{10}{s(s+2)}. Find KvK_v.

Answer: Kv=lims0sG(s)=102=5K_v = \lim_{s→0} sG(s) = \frac{10}{2} = 5
Explanation: Velocity constant for Type 1 system.


95. The steady-state error for ramp input is ess=1Kve_{ss} = \frac{1}{K_v}. For Kv=5K_v = 5:

a) 0.2
b) 5
c) 0.1
d) 1
Answer: a) 0.2
Explanation: ess=1/5=0.2e_{ss} = 1/5 = 0.2.


96. The phase margin of 45° gives:

a) Highly stable system
b) Marginally stable
c) Satisfactory stability
d) Unstable
Answer: c) Satisfactory stability
Explanation: Phase margin between 30–60° → good transient response.


97. The gain margin of 10 dB means:

a) Gain can be increased 10 times before instability
b) Gain can increase by factor ≈ 3.16
c) Gain can increase by 10%
d) Gain decreases 10×
Answer: b) Gain can increase by factor ≈ 3.16
Explanation: 10 dB = 20 log₁₀(K) → K ≈ 3.16.


98. The number of integrations in open-loop transfer function gives:

a) Order
b) Type
c) Damping
d) Gain
Answer: b) Type
Explanation: Type equals number of integrators (poles at origin).


99. A system with gain margin = 0 dB and phase margin = 0° is:

a) Stable
b) Unstable
c) Marginally stable
d) Highly stable
Answer: c) Marginally stable
Explanation: Both zero margins → on stability boundary.


100. Increasing derivative gain KdK_d:

a) Increases overshoot
b) Reduces overshoot
c) No effect
d) Increases rise time
Answer: b) Reduces overshoot
Explanation: Derivative action adds damping → reduces oscillation.