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Polar plot of G (jw) = 1/jw(jw+t) is
- a)Crosses the negative real axis
- b)Crosses the negative imaginary axis
- c)Crosses the positive imaginary axis
- d)None of the mentioned
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
Polar plot of G (jw) = 1/jw(jw+t) isa)Crosses the negative real axisb)...
Explanation: Polar plot can be made of the following function by following appropriate steps and thus the plot neither crosses the real axis nor imaginary axis.
Most Upvoted Answer
Polar plot of G (jw) = 1/jw(jw+t) isa)Crosses the negative real axisb)...
Explanation:
Given transfer function is G(jω) = 1/(jω)(jωt)
To plot the polar plot of G(jω), we need to find the magnitude and phase angle of G(jω) for different values of ω.
Magnitude of G(jω):
|G(jω)| = 1/|jω||jωt| = 1/ω|ωt|
Phase angle of G(jω):
∠G(jω) = -90° - arctan(ωt/ω)
Now, let's analyze the polar plot of G(jω):
- As ω → 0, |G(jω)| → ∞ and ∠G(jω) → -90°. Therefore, the polar plot starts from the positive imaginary axis.
- As ω → ∞, |G(jω)| → 0 and ∠G(jω) → -180°. Therefore, the polar plot approaches the negative real axis.
- When ω = 1/t, |G(jω)| = 1/(1/t)^2 = t^2 and ∠G(jω) = -90° - arctan(t). Therefore, the polar plot passes through the point (t^2, -90° - arctan(t)).
- Therefore, the polar plot does not cross the negative real or imaginary axis or the positive imaginary axis.
Hence, the correct answer is option D - None of the mentioned.
Given transfer function is G(jω) = 1/(jω)(jωt)
To plot the polar plot of G(jω), we need to find the magnitude and phase angle of G(jω) for different values of ω.
Magnitude of G(jω):
|G(jω)| = 1/|jω||jωt| = 1/ω|ωt|
Phase angle of G(jω):
∠G(jω) = -90° - arctan(ωt/ω)
Now, let's analyze the polar plot of G(jω):
- As ω → 0, |G(jω)| → ∞ and ∠G(jω) → -90°. Therefore, the polar plot starts from the positive imaginary axis.
- As ω → ∞, |G(jω)| → 0 and ∠G(jω) → -180°. Therefore, the polar plot approaches the negative real axis.
- When ω = 1/t, |G(jω)| = 1/(1/t)^2 = t^2 and ∠G(jω) = -90° - arctan(t). Therefore, the polar plot passes through the point (t^2, -90° - arctan(t)).
- Therefore, the polar plot does not cross the negative real or imaginary axis or the positive imaginary axis.
Hence, the correct answer is option D - None of the mentioned.
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