Correct Answer - Option 1 : ∞, 35
Concept:
For a negative unity feedback system, if CLTF is given, the Open-loop transfer function is evaluated as:
OLTF \( = \frac{G}{{1 + GH - G}}\)
The phase margin is defined as:
\(PM=180^\circ - ϕ \)
where ϕ is the phase of the OLTF at ωgc
ωgc (Gain crossover frequency) = frequency at which system gain is unity.
If the polar plot is not crossing the negative real axis, then GM = ∞
For a negative unity feedback system, the phase at ω = 0 is 0 degrees.
And phase at ω = ∞ is:
(p-z) × 90°
p = number of open-loop poles and z = no. of open-loop zeros.
Analysis:
\( TF= \frac{{500000}}{{{s^2} + 700s + 250000}}\)
Hence OLTF will be:
\(GH = \frac{{500000}}{{{s^2} + 700s + 250000 - 500000}}\)
\( = \frac{{500000}}{{{s^2} + 700s - 250000}}\)
Replacing s with jω:
\(G(jω )H(jω ) = \frac{{500000}}{{ - {ω ^2} + 700jω - 250000}}\)
For ωgc:
\(\left| {\frac{{500000}}{{ - {ω ^2} + 700jω - 250000}}} \right|=1\)
\(\frac{{500000}}{{\sqrt {\left( { - {ω ^2} - {{250000}}} \right)^2 + {{\left( {700ω } \right)}^2}} }}=1\)
\( \frac{{{{\left( {500000} \right)}^2}}}{{{{\left( { - {ω ^2} - 250000} \right)}^2} + {{\left( {700ω } \right)}^2}}}=1\)
\(\frac{{25 × {{10}^{10}}}}{{{ω ^4} + 500000{ω ^2} + 625 × {{10}^8} + 490000{ω ^2}}}=1\)
\(\frac{{25 × {{10}^{10}}}}{{{ω ^4} + 990000{ω ^2} + 625 × {{10}^8}}}=1\)
25× 1010 = ω4 + 990000ω2 + 625× 108
On solving the above, we get:
gc = 403.32 rad/sec
Now we will calculate the phase of the system at ωgc
GH(jω) \(=\frac{{500000}}{{ - 412667 + j282324}}\)
ϕ = -180o + tan-1\([\frac{{282324}}{{412667}}]\) = ( -180o + 34.37) degrees
PM = \(180^\circ \) + ϕ
= 180o + (- 180o + 34.37o) = 34.37o
If we draw the polar plot of the above system, then we get to see that its polar plot will not cut the negative real axis, in that case:
GM = ∞
