Question

crosses the imaginary axis at jω, we have How does this (jω)3+3(jω)2+2(jω)+K=0 become the →K−3ω2−jω(ω2−2)=0 Imaginary part implies (ω2−2)=0⇒ω=2​ Real part implies K=3ω2=6

Answer 1

To simplify the equation (jω)3 + 3(jω)2 + 2(jω) + K = 0, we can factor out (jω) as a common factor:

(jω)((jω)2 + 3(jω) + 2) + K = 0

Next, we can factor the quadratic term (jω)2 + 3(jω) + 2:

(jω)((jω + 2)(jω + 1)) + K = 0

Now, we can rearrange the equation to isolate K:

K - 3ω^2 - jω(ω^2 - 2) = 0

The imaginary part implies (ω^2 - 2) = 0, which gives us ω = ±√2.

The real part implies K = 3ω^2 = 3(2) = 6.

Therefore, the simplified equation becomes K - 3ω^2 - jω(ω^2 - 2) = 6 - 3(2) - jω(2 - 2) = 0.