Finding the value (ZERO) on real axis (Control System)

Hello all,

I have attached the image below



Kindly tell me how does 8/4.77 come from? (RED BOX)
How does 2.38 come from? (BLUE BOX)

Thank you.

hi
first he replaced
s=-2.5+j*4.33;
in
abs(40/(s*(s+0.5)))=
matlab spits out =1.6773 which is =8/4.77
further se exemple 6-8 how that 55' was obtained

Here is 55° calculations from example 6-8
This is our original control system below

You can see we have two roots from this i.e. s(s+0.5) at s1=0 and s2=-0.5 such that s(s+0.5)=0.

See below how does 55° calculated.

-235° is equal to 55°

Now tell me how does 2.38 come in the post#1 (BLUE BOX)

Thanks.

Eshal said:

-235° is equal to 55°

Click to expand...

i don't get it
anyway for now i can tell that the 2.38 comes like the 8.34

180-235=-55
This is angle deficiency. It means +55 must be compensate by ZERO. I hope you know what are poles and zeros in control system

anyway for now i can tell that the 2.38 comes like the 8.34

Click to expand...

sorry, didn't get it.

are you modelling an automatic control system ? Is this about using laplace transform to get it in discrete system f(t)-> f(s)?

no no.. it is just a lead-lag compensator using root locus approach. I just want to know how does 2.38 come? Because in the attached image in post#1, they say by using simple trigonometry. But how?

Eshal said:

180-235=-55
This is angle deficiency. It means +55 must be compensate by ZERO. I hope you know what are poles and zeros in control system


sorry, didn't get it.

Click to expand...

this is new for me
hehe, I wanted to say that the 2.38 segment was calculated like the 8.34 segment which you didn't put in a box so i assume that you know how it was calculated

No, I don't know how does both calculate. you tell.

Kind regards,
Princess

Eshal said:

no no.. it is just a lead-lag compensator using root locus approach. I just want to know how does 2.38 come? Because in the attached image in post#1, they say by using simple trigonometry. But how?

Click to expand...

oh "root locus approach " .. first year mathematics , i don't remember much since it was not my pleasure . I think that it could use trigonometry by changing representation from cartezian to polar ... no clue for now . I have to get over memories hehe

ok fine :'(

Eshal said:

ok fine :'(

Click to expand...

heeey , you'r smart . i bet you find answer somehow . Trigonometry is about knowing angles and segments which to use with trigonometric functions to get other segments and angles + it's still a coordinate system . It can't be hard ... If you find answer post it . You made me curious also

PS: do know the coordinates of P ? (cartesian or polar )

yes. P is at -2.5+4.33i. x-axis is a real axis and y-axis is a imaginary axis.

The value of P on real axis is -2.5 and on imaginary axis is 4.33i

α+β=125°
β=125°-α
sin(β)=sin(125°-α)

sin(125°-α)=sin(125°)cos(α)-cos(125°)sin(α)
since, sin(β)=sin(125°-α)
so,
sin(β)=sin(125°)cos(α)-cos(125°)sin(α)
dividing both sides by sin(α)
so,
sin(β)/sin(α)=sin(125°)cos(α)/sin(α)-cos(125°)sin(α)/sin(α)
sin(β)/sin(α)=sin(125°)cos(α)/sin(α)-cos(125°)
sin(β)/sin(α)=0.8192*cos(α)/sin(α)-(-0.5736)
4.77/8=0.8192*cot(α)+0.5736
0.596=0.8192*cot(α)+0.5736
0.0227=0.8192*cot(α)
cot(α)=0.0227/0.8192
cot(α)=0.028
α=88° and α+β=125° => β=37°
and
AB=sin(P)*PA/sin(β)=6.5

Then,
Draw the line from P perpendicular to the horizontal axis. It will cross the horizontal axis in a point Q, just to the left of A.
PAQ is a right triangle and you know the coordinates of P, the coordinates of Q and the angle α. That should lead you to the coordinates of A.

In triangle APQ:
cot(α)=|AQ||PQ|. The only unknown here is |AQ|.
Thus:
|AQ|=cot(α)∗|PQ|=0.028∗4.33=0.12
So the coordinates of A are: (−2.5+0.12,0)=(−2.38,0)

For B (same as above, but now in the right triangle BPQ):
β=π−55∗π180−1.543=0.638 (in radians).
|BQ|=cot(β)∗|PQ|=cot(0.638)∗4.33=1.349∗4.33=5.84
So the coordinates of B are: (−2.5−5.84,0)=(−8.34,0)

ok smarty ,

Thanks