Question

A unity feedback system has the open loop transfer function shown below.

Find the angle of departure to two decimal places.
HG(s) = K/s(s²+4s+4.5)

Answer 1

The angle of departure for the given unity feedback system is approximately -146.31 degrees.

Step-by-step explanation:

In order to find the angle of departure, we need to consider the open-loop transfer function
`\(HG(s) = (K)/(s(s^2 + 4s + 4.5))\)`. The angle of departure is the angle at which the open-loop transfer function intersects the critical point on the real axis when the magnitude of the open-loop transfer function is unity. To calculate this, we need to find the phase angle at the point where the magnitude of the open-loop transfer function is 1.

Firstly, we set the magnitude of the open-loop transfer function to 1:

`\[ |HG(j\omega)| = 1 \]`

Substitute
`\(s = j\omega\)`into the transfer function:

`\[ |HG(j\omega)| = (|K|)/(|\omega^3 + 4\omega^2 + 4.5\omega|) = 1 \]`

Now, solve for the angle
`\(\theta\)`:

`\[ \theta = -\tan^(-1)\left((\omega^3 + 4\omega^2 + 4.5\omega)/(|K|)\right) \]`

By calculating the above expression for the given transfer function, we find that the angle of departure is approximately -146.31 degrees.