Well, it is just converting the numerator and denominator from complex numbers to magnitude-phase (phasor) notation, which is really short-hand of imaginary exponential.
Numerator
j6
is a vector with no real component (only has imaginary component). So in phasor terms, the magnitude is 6
, and the angle is 90°
(if it had been negative (-j6
), the angle would have been -90°
). The angle is measured positive in the anti-clockwise direction, with respect to the positive real axis.
$$ j6 = 6\angle{90°} $$
Denominator
-1-j
has real and imaginary components equal to -1
. So from the origin it looks like an arrow pointing to the bottom-left in the complex plane. So the angle is -135°
, and the magnitude is \$(1^2+1^2)^{1/2} = \sqrt{2} \$
$$ -1-j = \sqrt{2} \angle{-135°} $$
To answer the question in the comments of why \$-j=\frac{1}{j} \$:
When you divide phasors, the resulting magnitude is the quotient of the magnitudes, and the resulting angle is the difference between the angles.
$$ -j = 1\angle{-90} = \frac{1\angle{0}}{1\angle{90}} = \frac{1}{j} $$
Just redraw the diagram like this: -
I've used Y instead of theta for my own convenience.
You get Y = I.G(M - Y.D) which reduces to \$\dfrac{Y}{M} = \dfrac{1}{\frac{1}{G.I} + D}\$
I can see that if you plug the expressions for G, I and D in you get the correct answer.
Best Answer
Here is how you proceed :
Reduce the system to either unity feedback system or a Single block representing the closed loop transfer function. In your case, reducing to a single block should be easy.
For a unity feedback system , as shown
Then from the diagram
Or
If you reduce it to a single block, Then
If you have non unity feedback system, then
will reduce it to the form in figure.
For steady state error, you need to specify the input. 3 inputs are used :
Substitute the transform of input into the equation for your error obtained in step 2.
Use the final value theorem acc to which
E(infinity) = lim s-> 0 [ s.E(s)]