Yes, if using a Pade apporoximant for the time delay, then the Q matrix will have to be expanded to include the additional state(s). In principle, you could buld the new Q matrix using the symmetric root locus technique, as I think is being done here, to drive the closed-loop poles to desirable locations. Maybe not penalizing that state(s) might be reasonable as well.
In addition, the Pade state(s) isn't available for feedback in the actual system, so they'd have to be reconstructed somehow, perhaps with a reduced order state estimator.
In the question you mentioned that the time delay is on the plant measurements, but the code has the time delay on the plant input. I'm going to assume it's on the input for what follows.
For this particular problem, the time delay at the input to the plant seems to be small enough that it is easily accomodated by the very large stability margins at the plant input that are guaranteed by LQ state feedback.
load("lqiWithDelay")
plant;
ts = 1 / 100e3;
%[num, den] = pade(ts, 3); % Nth-order
%loopDelayPadeApproxTf = tf(num, den);
%plantWithDelay = plant * loopDelayPadeApproxTf; % delay is at the plant input
% Find non-delayed CL gains
wnTarget_rPs = 10 * 2 * pi;
H_lqi = [2 * 0.7 * wnTarget_rPs, 0.5, -wnTarget_rPs^2];
Q_lqi = H_lqi.'* H_lqi;
R_lqi = 1;
%K_lqi = lqi(model, Q_lqi, R_lqi) % what was model? not defined ... assume it's supposed to be plant
K_lqi = lqi(plant, Q_lqi, R_lqi);
sysp = ss(plant.A,plant.B,[eye(2);plant.C],[zeros(2,1);plant.D],'InputName','u','OutputName',{'x1' 'x2' 'y'});
int = tf(K_lqi(3),[1 0],'InputName','e','OutputName','ui');
Kfb = ss(K_lqi(1:2),'InputName',{'x1' 'x2'},'OutputName','ufb');
esum = sumblk('e = r - y');
usum = sumblk('u = -ui - ufb');
sysc = connect(sysp,int,Kfb,esum,usum,'r','y','u');
Bode plot of the open-loop transfer function (w/o delay) at the plant input and of the delay.
L = getLoopTransfer(sysc,'u',-1);
figure
bode(L,exp(-tf('s')*ts))
As expected from an LQ design, the gain margin is infinite and the phase margin is quite large, about 90 deg in this case. The phase lag that would be introduced by the time delay at the input to the plant is quite small at the open-loop gain crossover frequency (~5000 rads/sec) and the phase margin will hardly be changed. The additional phase lag will cause an open-loop phase cross-over at very high frequency, but there the gain will be quite low and the gain margin quite large.
The effect of the delay on the closed-loop step response is negligible.
sysp = ss(plant.A,plant.B,[eye(2);plant.C],[zeros(2,1);plant.D],'InputName','u','OutputName',{'x1' 'x2' 'y'},'InputDelay',ts);
int = tf(K_lqi(3),[1 0],'InputName','e','OutputName','ui');
Kfb = ss(K_lqi(1:2),'InputName',{'x1' 'x2'},'OutputName','ufb');
esum = sumblk('e = r - y');
usum = sumblk('u = -ui - ufb');
syscdelay = connect(sysp,int,Kfb,esum,usum,'r','y','u');
figure
step(sysc,syscdelay)
