Automatic Control Systems - Section 1

Directions to Solve

Automatic Control Systems - Section 1

For the system of the given figure the transfer function is

27-245

A.
B.
C.
D.

Answer: Option D

Explanation:

Transfer function = 43-245

In a minimum phase system

A.	all poles lie in the left half plane
B.	all zeros lie in the left half plane
C.	all poles lie in the right half plane
D.	all except one pole or zero lie in the left half plane

Answer: Option B

Explanation:

For stability poles must be in left half plane. Zeros in left half plane lead to minimum phase shift.

For very low frequencies, v0/vi in the given figure equals

32-346

A.	1
B.	0
C.	∞
D.	0.5

Answer: Option A

Explanation:

For low frequencies X_c is very high as compared to R. Hence 44-346

The given figure shows a pole zero diagram. The transfer function G(j1) is

21-174

A.	0.5 ∠0
B.	2.7 ∠- 31°
C.	2 ∠45°
D.	2 ∠- 67.4°

Answer: Option A

Explanation:

Join j₁ point with the zero and poles. Find magnitude and phase angles of all these lines.

The system in the given figure, has

36-395

A.	good gain and phase margin
B.	poor gain and phase margin
C.	good gain margin but poor phase margin
D.	poor gain margin but good phase margin

Answer: Option D

Explanation:

Since the plot is very near (-1 + j0) point, gain margin is poor.

For the circuit in the given figure, V₂(s)/V₁(s) =

27-258

A.
B.
C.
D.

Answer: Option B

Explanation:

43-258

In the given figure, if R = X_C, voltage gain is

32-346

A.	0 dB
B.	3 dB
C.	-3 dB
D.	20 dB

Answer: Option C

Explanation:

If R = X_C, V₀ = 0.5 V_i Hence gain is -3 dB.

For the system in the given figure,

25-225u

A.
B.
C.
D.	none of the above

Answer: Option B

Explanation:

For inner loop we have transfer function = 43-225

43-225-1 = 43-225-2

Automatic Control Systems - Section 1

Automatic Control Systems - Section 1

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