A family of constant N circles has the centre as

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ESE Electronics 2011 Paper 1: Official Paper
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  1. X = 1 and Y = 2 N
  2. \(X=-\dfrac{1}{4} \ \text{and} \ Y = 4 \ N\)
  3. \(X=-\dfrac{1}{2} \ \text{and} \ Y=\dfrac{1}{4N}\)
  4. \(X=-\dfrac{1}{2} \ \text{and} \ Y=\dfrac{1}{2N}\)

Answer (Detailed Solution Below)

Option 4 : \(X=-\dfrac{1}{2} \ \text{and} \ Y=\dfrac{1}{2N}\)
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Detailed Solution

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Explanation:

  • Constant magnitude loci that are M-circles and constant phase angle loci that are N-circles are the fundamental components in designing the Nichols chart. The constant M and constant N circles in G (jω) plane can be used for the analysis and design of control systems.
  • The M and N circles of G (jω) in the gain phase plane are transformed into M and N contours in rectangular coordinates.
  • The critical point in G (jω), plane corresponds to the point of zero decibels and -1800 in the gain phase plane. The plot of M and N circles in the gain phase plane is known as the Nichols chart /plot.

 

Constant N circle equation is:

\({\left( {x + \frac{1}{2}} \right)^2} + {\left( {y - \frac{1}{{2N}}} \right)^2} = \frac{{{N^2} + 1}}{{4{N^2}}}\)

Where,

N = tan α 

1) The Centre of circles is at \((\frac{-1}{2},\frac{1}{2N})\)

2) Radius is \(\frac{{\sqrt {{N^2} + 1} }}{{2N}}\)

3) Constant N circles always pass through (-1,0) and (0,0).

So, the answer is an option (4).

Important Points

Constant M circle equation is:

\({\left( {x + \frac{{{M^2}}}{{{M^2} - 1}}} \right)^2} + {\left( y \right)^2} = \frac{{{M^2}}}{{({M^2} - 1)^2}}\)

1) The centre of circles is at \((\frac{-M^2}{M^2-1},0)\)

2) Radius is \(\space (\frac{M}{M^2-1})\)

3) The Constant M circle is a straight line at x = -1/2.

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