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For the transfer function
G(s) H(s) = 1 / s(s+1) (s+0.5), the phase cross-over frequency is
- a)0.5 rad/sec
- b)0.707 rad/sec
- c)1.732 rad/sec
- d)2 rad/sec
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
For the transfer functionG(s) H(s) = 1 / s(s+1) (s+0.5), the phase cro...
Answer: b
Explanation: Phase cross over frequency is calculated at the point where magnitude of the polar plot is 1.
Explanation: Phase cross over frequency is calculated at the point where magnitude of the polar plot is 1.
Most Upvoted Answer
For the transfer functionG(s) H(s) = 1 / s(s+1) (s+0.5), the phase cro...
Given transfer function: G(s) H(s) = 1 / s(s+1) (s+0.5)
To find phase cross-over frequency, we follow these steps:
Step 1: Convert the transfer function into polar form
G(s) H(s) = K / s(s+1) (s+0.5)
where K = 1
G(s) H(s) = K / (1+s)(0.5+s)
G(s) H(s) = K / (s+1) (s+0.5)^(1/2) (s+0.5)^(-1/2)
Step 2: Find the angle of the transfer function at the phase cross-over frequency
At the phase cross-over frequency, the magnitude of the transfer function is 1 (in dB), i.e., |G(jω) H(jω)| = 1
∴ 1 = K / [(jω+1) ((jω+0.5)^(1/2) (jω+0.5)^(-1/2))]
1 = K / [(jω+1) (jω+0.5)^0.5 (jω+0.5)^(-0.5)]
1 = K / [(jω+1) (jω+0.5)^0.5 / (jω+0.5)]
1 = K / [(jω+1) (jω+0.5)^(1/2) / (jω+0.5)^(1/2)]
(jω+0.5)^(1/2) = (jω+1)
(jω+0.5) = (jω+1)^2
jω+0.5 = -ω^2 + 2jω + 1
Equating real and imaginary parts:
ω^2 - 0.5 = 1
ω^2 = 1.5
ω = ± 1.225
We take only the positive value of ω as phase cross-over frequency.
Step 3: Find the phase at the phase cross-over frequency
At ω = 1.225, the transfer function becomes:
G(j1.225) H(j1.225) = K / (1+j1.225) (1.725)^(1/2)
G(j1.225) H(j1.225) = K / (1+j1.225) (1.313)
G(j1.225) H(j1.225) = 0.606 / (1+j1.225)
∠G(j1.225) H(j1.225) = -56.31°
Therefore, the phase cross-over frequency is 0.707 rad/sec (approximately).
Hence, option B is the correct answer.
To find phase cross-over frequency, we follow these steps:
Step 1: Convert the transfer function into polar form
G(s) H(s) = K / s(s+1) (s+0.5)
where K = 1
G(s) H(s) = K / (1+s)(0.5+s)
G(s) H(s) = K / (s+1) (s+0.5)^(1/2) (s+0.5)^(-1/2)
Step 2: Find the angle of the transfer function at the phase cross-over frequency
At the phase cross-over frequency, the magnitude of the transfer function is 1 (in dB), i.e., |G(jω) H(jω)| = 1
∴ 1 = K / [(jω+1) ((jω+0.5)^(1/2) (jω+0.5)^(-1/2))]
1 = K / [(jω+1) (jω+0.5)^0.5 (jω+0.5)^(-0.5)]
1 = K / [(jω+1) (jω+0.5)^0.5 / (jω+0.5)]
1 = K / [(jω+1) (jω+0.5)^(1/2) / (jω+0.5)^(1/2)]
(jω+0.5)^(1/2) = (jω+1)
(jω+0.5) = (jω+1)^2
jω+0.5 = -ω^2 + 2jω + 1
Equating real and imaginary parts:
ω^2 - 0.5 = 1
ω^2 = 1.5
ω = ± 1.225
We take only the positive value of ω as phase cross-over frequency.
Step 3: Find the phase at the phase cross-over frequency
At ω = 1.225, the transfer function becomes:
G(j1.225) H(j1.225) = K / (1+j1.225) (1.725)^(1/2)
G(j1.225) H(j1.225) = K / (1+j1.225) (1.313)
G(j1.225) H(j1.225) = 0.606 / (1+j1.225)
∠G(j1.225) H(j1.225) = -56.31°
Therefore, the phase cross-over frequency is 0.707 rad/sec (approximately).
Hence, option B is the correct answer.
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