Answer:
a) G(s) = 2(s+2) / (s^2 -1)
First, let's rewrite the transfer function in its factored form:
G(s) = 2(s+2) / [(s-1)(s+1)]
Now we can create the Bode Plot.
Magnitude plot:
For s = 0, |G(jω)| = [(2*2)/(-1)] = 4
For s → ∞, |G(jω)| → 0
For ω = 1, |G(jω)| = 2.83 ≈ -9 dB
For ω → ∞, |G(jω)| → 0
We can plot these points and connect them using asymptotes as shown below:
Gain crossover frequency = 1 rad/s (where the magnitude curve intersects 0 dB line).
Phase plot:
For s = 0, ∠G(jω) = 90°
For s → ∞, ∠G(jω) → 0°
For ω = 1, ∠G(jω) = 164°
For ω → ∞, ∠G(jω) → 0°
We can plot these points and connect them using an asymptote as shown below:
Phase margin can be calculated by finding the difference between the phase angle at the gain crossover frequency and -180°:
PM = -16°
b) G(s) = 2 / (s(2+s)(5+s))
First, let's rewrite the transfer function in its factored form:
G(s) = 2 / [s(2+s)(s+5)]
Now we can create the Bode Plot.
Magnitude plot:
For s → ∞, |G(jω)| → 0
For ω << 1, |G(jω)| ≈ 0 dB (since the s term dominates)
For ω = 1, |G(jω)| = 0.18 ≈ -13.95 dB
For ω = 2, |G(jω)| = 0.10 ≈ -19.97 dB
For ω = 5, |G(jω)| = 0.04 ≈ -28 dB
We can plot these points and connect them using asymptotes as shown below:
Gain crossover frequency = 2 rad/s (where the magnitude curve intersects 0 dB line).
Phase plot:
For s → ∞, ∠G(jω) → 0°
For ω << 1, ∠G(jω) ≈ -90° (since the s term dominates)
For ω = 1, ∠G(jω) = -93°
For ω = 2, ∠G(jω) = -128°
For ω = 5, ∠G(jω) = -160°
We can plot these points and connect them using asymptotes as shown below:
Phase crossover frequency = 1.26 rad/s (where the phase curve intersects -180° line).
Phase margin can be calculated by finding the difference between the phase angle at the gain crossover frequency and -180°:
PM = -49°
Gain margin can be calculated by finding the difference between the 0 dB line and the magnitude at the phase crossover frequency:
GM = 24 dB