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s-plane region for ζ>0.707 and ts<2 s

Short answer

The region is left of the vertical line Re(s) = -2 and inside the sector bounded by lines at ±45° to the negative real axis (ζ > 0.707).

Draw the region in s-plane that satisfies following requirements: i) ζ > 0.707 ii) ts < 2 s.

Each specification puts a boundary on where the closed-loop poles s=−ζωn±jωn1−ζ2s = -\zeta\omega_n \pm j\omega_n\sqrt{1-\zeta^2} may lie.

i) ζ>0.707\zeta > 0.707

A pole makes angle θ\theta with the negative real axis, where cos⁡θ=ζ\cos\theta = \zeta.

ζ>0.707  ⇒  θ<cos⁡−1(0.707)=45∘\zeta > 0.707 \;\Rightarrow\; \theta < \cos^{-1}(0.707) = 45^\circ

So the poles must lie inside the wedge between the two lines at ±45∘\pm 45^\circ to the negative real axis.

ii) ts<2t_s < 2 s

Using the 2% criterion, ts=4ζωn=4σt_s = \dfrac{4}{\zeta\omega_n} = \dfrac{4}{\sigma}:

4σ<2  ⇒  σ=ζωn>2\frac{4}{\sigma} < 2 \;\Rightarrow\; \sigma = \zeta\omega_n > 2

So the poles must lie to the left of the vertical line Re(s)=−2\text{Re}(s) = -2.

Required region

Both conditions together: the shaded region (marked #), left of σ=−2\sigma = -2 and inside the ±45∘\pm 45^\circ lines.

                          jw
  ##############\     :   |
  ################\   :   |
  ##################\ :   |
  ####################\   |
  ####################: \ |
----------------------+---+------ sigma
  ####################: / |
  ####################/   |
  ##################/ :   |
  ################/   :   |
  ##############/     :   |
                     -2   0
  # = allowed region; \ / = 45 deg lines

The region is bounded by the lines ω=±σ\omega = \pm\sigma (45 degrees) and the line σ=−2\sigma = -2; it extends to the left without limit. (With the 5% criterion, ts=3/σt_s = 3/\sigma, the vertical line is at σ=−1.5\sigma = -1.5.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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