# Z-domain root locus advise

#### tech84

Hello all,

Sorry for inconvenience done, there were some misprints on the thread posted before.

I need your kind help answering coursework questions(fig.1) with the following Z-transform open loop transfer function
Wo () = (0.04(z+ 0.675)/ (z− 1) (z− 0.584) (z − 0.563))

My attempt is as follows:

For desired poles: T = 0.1s, = wn=3rad/s, ζ=0.7, P1,2= e^(ζwnT)e^(±jwnT√(1−ζ^2))
P1,2 = 0.8105∠0.214= 0.792-+0.172j

Desired poles = 0.792+0.172j, Actual poles: 1, 0.584,0.563 Actual zeros: 0, -0.675

For angle theta, I plugged in 0.792+0.172j , Wo (z) = (0.04(z+ 0.675)/ (z− 1) (z− 0.584) (z − 0.563))

(12.5+6.687) -(140+39.58+36.9) =-197deg, 197-180= 17.56=theta degrees angle deficiency,
please refer root locus fig2
Digital compensator attempt answer is D(z)=K(z-0.7586)/(z-0.6983)

then I introduced compensator D(z) found before in the open loop TF to find gain K
plug in z=0.792+0.172j
Open loop (with compensator D(z)) Tf (z) = (K (0.04) (z-0.7586) (z + 0.675)/(z-0.6985) (z− 1) (z− 0.584) (z− 0.563)) =1
K=0.4458

And finally, I did Matlab simulation fig3 which is nearly to specs, but I am not totally sure

Kindly could you comment if my working(method) of the root locus(fig2) and Matlab simulation(fig3) if is it correct?

TY

#### Attachments

• 45.2 KB Views: 2
• 61.5 KB Views: 3
• 154.9 KB Views: 3