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The polar plot of the transfer function G(s) = 10(s+1)/s+10 will be in the :
  • a)
    First quadrant
  • b)
    Second quadrant
  • c)
    Third quadrant
  • d)
    Fourth quadrant
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
The polar plot of the transfer function G(s) = 10(s+1)/s+10 will be in...
Answer: d
Explanation: Polar plot of the given transfer function lies in the fourth quadrant.
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The polar plot of the transfer function G(s) = 10(s+1)/s+10 will be in...
Polar plot of transfer function G(s)

The polar plot of a transfer function G(s) is a graph that represents the magnitude and phase angle of G(s) as a function of the frequency.

The transfer function G(s) = 10(s+1)/(s+10) can be rewritten as:

G(s) = 10 * (1 + s/1) / (1 + s/10)

From this expression, we can see that the DC gain of G(s) is 10, meaning that the magnitude of G(s) is 10 at zero frequency.

To plot the polar plot of G(s), we need to evaluate its magnitude and phase angle for various values of frequency.

Magnitude and phase angle of G(s)

The magnitude of G(s) is given by:

|G(jω)| = 10 * |(1 + jω/1) / (1 + jω/10)|

To evaluate this expression, we need to find the frequency at which the magnitude of the denominator is equal to the magnitude of the numerator. This frequency is called the crossover frequency.

At the crossover frequency, we have:

|1 + jω/10| = |1 + jω/1|

Squaring both sides and simplifying, we get:

ω2 = 10

Therefore, the crossover frequency is ω = √10.

At frequencies below the crossover frequency, the magnitude of the denominator is larger than the magnitude of the numerator, and hence |G(jω)| is smaller than 10. At frequencies above the crossover frequency, the magnitude of the numerator is larger than the magnitude of the denominator, and hence |G(jω)| is larger than 10.

The phase angle of G(s) is given by:

∠G(jω) = atan(ω/1) - atan(ω/10)

This expression represents the difference between the phase angles of the numerator and denominator. At low frequencies, the phase angle of the numerator is close to zero, while the phase angle of the denominator is negative and close to -90 degrees. Therefore, the phase angle of G(s) is close to +90 degrees. At high frequencies, the phase angle of the numerator is close to +90 degrees, while the phase angle of the denominator is close to zero. Therefore, the phase angle of G(s) is close to zero degrees.

Polar plot of G(s)

Using the magnitude and phase angle expressions for G(s), we can plot its polar plot as shown below:

At low frequencies, the magnitude of G(s) is small and its phase angle is close to +90 degrees. Therefore, the polar plot of G(s) starts in the fourth quadrant (i.e., at 90 degrees).

As the frequency increases, the magnitude of G(s) increases and its phase angle decreases. Therefore, the polar plot moves towards the origin along the positive real axis (i.e., towards the first quadrant).

At the crossover frequency, the magnitude of G(s) is equal to 10 and its phase angle is zero degrees. Therefore, the polar plot passes through the positive real axis at a magnitude of 10.

At high frequencies, the magnitude of G(s) is large and its phase angle is close to zero degrees. Therefore, the polar plot moves towards the origin along the negative real axis (i.e., towards the fourth quadrant).

Therefore, the polar plot of G(s) starts in the fourth quadrant
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The polar plot of the transfer function G(s) = 10(s+1)/s+10 will be in the :a)First quadrantb)Second quadrantc)Third quadrantd)Fourth quadrantCorrect answer is option 'D'. Can you explain this answer?
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