Electronic – Why simulink step response differs from matlab’s step function

A not-at-all elegant way to do it is:

zeta=[...]; %your zeta values
wn = ... % calculate your wn values according to your zeta values
figure;
hold('on');
for idx = 1:length(zeta)
    % sys = tf([wn(idx)],[1 2*wn(idx)*zeta(idx) wn(idx)^2]); %system's transfer function
    % EDIT : numerator corrected
    sys = tf([wn(idx)^2],[1 2*wn(idx)*zeta(idx) wn(idx)^2]); %system's transfer function
    step(alpharef*sys);
end

The step is a stimulus to the system. The system is defined by the pole zero diagram irrespective of the stimulus and it looks like a 2nd order response so here's the math behind the poles: -

You have poles at co-ordinates -5 on the real part of the s-plane and +/-9 on the \$j\omega\$ of the s-plane. So you can say 5 = \$\zeta\omega_n\$ and 9 = \$\omega_n\sqrt{1-\zeta^2}\$. From this you can work out what \$\omega_n\$ and \$\zeta\$ are.

Because it looks like a 2nd order system you can use this to define the shape from the above values for \$\omega_n\$ and \$\zeta\$: -