Question

Consider a unity feedback system having an open loop transfer function

\(G(j\omega ) = \frac{k}{{j\omega \left( {j0.2\omega + 1} \right)\left( {j0.05\omega + 1} \right)}}\)

Find open loop gain (k) with gain margin of 20 dB

1. 5.2
2. 2.5
3. 0.1
4. 2.25

Answer 1

Correct Answer - Option 2 : 2.5

Concept:

Gain Margin:

• The gain margin is a factor by which the gain of a stable system is allowed to increase before the system reaches the verge of instability.
• It is also defined as the reciprocal of the gain at which the phase angle becomes 1800.
• The frequency at which the phase angle is 1800 is called the phase crossover frequency (ωpc). 

Mathematically,

∠G(jω)H(jω) = -1800

If  |G(jω)H(jω)| = a, at ω = ωpc 

Then Gain margin,

\(G.M.=20log\frac{1}{a} db\)    

Calculation:

\(G\left( {jω } \right) = \frac{{{k}}}{{jω \left( {j0.2\;ω \; + \;1} \right)\left( {j0.05\;ω \; + \;1} \right)}}\) 

\(20dB = 20\log \left( {\frac{1}{a}} \right)\) 

Gain = a = 0.1

∠h(jω) = -90 - tan^-1 (0.2 ω) - tan^-1(0.05 ω)

-180 + 90 = -tan^-1(0.2 ω) - tan^-1(0.05 ω)

\(90 = {\tan ^{ - 1}}\left( {\frac{{0.2\;ω \; + \;0.05\;ω }}{{1\; - \;0.2\; \times \;0.05\;{ω ^2}}}} \right)\) 

∴ 1 - 0.2 × 0.05 ω² = 0

\({ω ^2} \times \frac{2}{{10}} \times \frac{5}{{100}} = 1\) 

ω² = 10 rad/sec

at 

ω = 10

|G(jω)| = 0.1 

\({\left| {G\left( {jω } \right)} \right|} = \frac{K}{{\omega \sqrt {{{\left( {0.2ω } \right)}^2} + 1} \sqrt {{{\left( {0.05\;ω } \right)}^{2}} + 1} }}\) 

\(0.1 = \frac{K}{{10\sqrt {{{\left ( {0.2\; \times \;10} \right)}^2} + 1} \sqrt {{{\left( {0.05\; \times \;10} \right)}^2} + 1} }}\) 

\(0.1 = \frac{K}{{10\sqrt 5 \;\sqrt {1.25} }}\) 

K = 2.5