Test: Automatic Control Systems - 3 - Electrical Engineering (EE) MCQ

Total number of roots is 5. Since two sign changes in first column, two roots are in right half plane.

Hence 3 roots in left half plane.

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Two sign changes and hence two poles in right half plane.

For (jω)ⁿ term the slope is (20 n) dB/decade.

It is not always zero.

It changes phase.

Use principle of superposition. Initially assume R(s) = 0 and find C(s).

Then assume U(s) = 0 and find C(s). Add the two responses.

When ω = 0, phase angle is 0.

For high values of phase angle = - 90º - 90º = 180º.

For 5% criterion settling time = 3T = .

Since ξω_n = 4, and 3T = 0.75 secs.

When polar plot passes through (-1, 0) the system is on limit of stability and gain margin is zero.

G₁ and G₂ are cascaded.

Transfer function to inner loop = .

Then .

The characteristic equation is s² + 3s + 6as + 6 = 0.

The error voltage is always very small. A high value of error voltage may cause instability.

A double pole or pair of complex poles result in - 40 dB/decade slope in log magnitude graph.

Use signal flow graph.

For a lead-lag compensator

β > 1 and a < 1.

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Since there are two poles at s = -2 and s = -1 and one zero at s = +3.

If gain is a, gain margin is . In this case gain is |G(jω₁)|.

Coefficient (s + 15) gives the term e^-15t and the coefficient (s + 20) gives the term e^-20t.

These can be neglected as compared to terms corresponding to s and (s + 1).

Compare with standard form.

It is not critical frequency.

Variation in damping factor is small.

jω term in denominator gives a straight line with a slope of - 20 dB/decade and gives a straight line with a slope of - 40 dB/decade.

When ω is high, the output is low.