Chapter 5: The Performance of Feedback Control Systems

0. Chapter overview

This chapter defines and computes the

performance of closed-loop feedback systems in the time domain. The main setting is a standard second-order closed-loop system

$$T(s) \;=\; \frac{\omega_n^2}{s^{2}+2\zeta\omega_n s+\omega_n^{2}}$$

and a small family of standard test inputs (step, ramp, parabolic, impulse). Five scalar performance measures — rise time $T_r$, peak time $T_p$, percent overshoot $P.O.$, settling time $T_s$, and steady-state error $e_{ss}$ — are defined from the step response and then linked directly to the location of the closed-loop poles in the $s$-plane. This pole-placement geometry is then used to design controller parameters (choose $K$, $p$, $a$, etc.) so that specifications on $P.O.$, $T_s$, $T_p$ are met simultaneously. The chapter also introduces the dominant poles approximation for higher-order systems, and four performance indices (ISE, IAE, ITAE, ITSE) that convert a whole error signal into a single number to be minimized with respect to a parameter.

Exam questions supported by this chapter:

• Compute $T_r, T_p, P.O., T_s$ from a given $G(s), H(s)$ for a step input.
• Compute $e_{ss}$ for a given input (step, ramp, combined).
• Read a measured step response and back out $K$, $T$, $\zeta$, $\omega_n$.
• Place closed-loop poles to satisfy $P.O., T_s, T_p$ specs simultaneously (feasible-region design).
• Approximate a 3rd-order $T(s)$ as 2nd-order while preserving gain.
• Compute and minimize ISE (or ITAE) with respect to a parameter.

1. Desired outcomes checklist

#	Outcome	Where covered	Priority
1	Recognize step, ramp, parabolic, impulse test signals and their Laplace transforms	§2.1	Standard
2	Derive step and impulse response of a second-order system	§2.2, §2.3	Standard
3	Compute $T_r, T_p, P.O., T_s, e_{ss}$ for 2nd-order step response	§2.4–§2.9, §2.11	High (Study Q1)
4	Draw lines of constant $T_p$ (horizontal), $T_s$ (vertical), $P.O.$ (diagonal/radial) in the $s$-plane	§2.14	Standard
5	Explain how pole motion (vertical/horizontal/radial) changes each metric	§2.14	Standard
6	Select closed-loop poles to meet specs (e.g. $P.O.\le 20\%$, $T_s\le 4$)	§2.15	High (Study Q2)
7	Approximate higher-order systems as 2nd-order using dominant poles (preserve gain)	§2.17	Standard
8	Compute ISE/IAE/ITAE/ITSE and minimize w.r.t. a parameter ($dI/dK=0$)	§2.18–§2.21	High (Study Q3)

Study questions in §4 of this guide are HIGH-PRIORITY exam material.

---

2. Core concepts, part by part

2.1 Input test signals (slides 2–3)

Theory

Performance is characterized by the response to a small set of standardized input signals.

Test signal	$r(t)$ (for $t>0$)	$R(s)$
Step	$A$	$A/s$
Ramp	$At$	$A/s^{2}$
Parabolic	$At^{2}$	$2A/s^{3}$
Unit impulse	pulse of area 1 at $t=0$	$1$

General form (slide 2):

$$r(t)=t^{n}, \qquad R(s)=\frac{n!}{s^{\,n+1}}.$$

The unit impulse (slide 3) is defined as a limit of a rectangular pulse of width $\epsilon$ and height $1/\epsilon$ (so area $=1$) as $\epsilon\to 0^{+}$. It models physical “impact hammer” testing.

Key equations

• $R(s)=A/s$ (step), $A/s^{2}$ (ramp), $2A/s^{3}$ (parabolic), $1$ (unit impulse).

Figures/diagrams

[INSERT FIGURE: Table + sketches of step/ramp/parabolic signals, from Ch05.pdf, page 2] [INSERT FIGURE: Rectangular pulse definition of the unit impulse, from Ch05.pdf, page 3]

---

2.2 Second-order system step response (slides 4–6)

Theory

Open-loop plant (slide 4):

$$G(s)=\frac{\omega_n^{2}}{s(s+2\zeta\omega_n)}, \qquad H(s)=1.$$

Closed-loop:

$$Y(s)=\frac{G(s)}{1+G(s)}R(s)=\frac{\omega_n^{2}}{s^{2}+2\zeta\omega_n s+\omega_n^{2}}\,R(s).$$

For unit step input $R(s)=1/s$:

$$Y(s)=\frac{\omega_n^{2}}{s\,(s^{2}+2\zeta\omega_n s+\omega_n^{2})}.$$

Using partial fractions (slide 5):

$$Y(s)=\frac{1}{s}-\frac{s+\zeta\omega_n}{(s+\zeta\omega_n)^{2}+\omega_d^{2}}-\frac{\omega_d}{\omega_d}\cdot\frac{\zeta\omega_n}{(s+\zeta\omega_n)^{2}+\omega_d^{2}},$$

with the damped natural frequency

$$\omega_d=\omega_n\sqrt{1-\zeta^{2}}.$$

Inverse Laplace gives (underdamped $\zeta<1$):

$$\boxed{\;y(t)=1-e^{-\zeta\omega_n t}\!\left[\cos\omega_d t+\frac{\zeta}{\sqrt{1-\zeta^{2}}}\sin\omega_d t\right]\;}$$

Steady-state value (slide 6): $y_{ss}=\lim_{s\to 0}sY(s)=1$ (desired response).

Key equations

• $T(s)=\dfrac{\omega_n^{2}}{s^{2}+2\zeta\omega_n s+\omega_n^{2}}$.
• $\omega_d=\omega_n\sqrt{1-\zeta^{2}}$.
• $y(t)=1-e^{-\zeta\omega_n t}\!\left[\cos\omega_d t+\dfrac{\zeta}{\sqrt{1-\zeta^{2}}}\sin\omega_d t\right]$.

Figures/diagrams

[INSERT FIGURE: Family of step responses for $\zeta=0.1,0.2,0.4,0.7,1.0,2.0$ vs. $\omega_n t$, from Ch05.pdf, page 6]

Small $\zeta$ → large oscillation/overshoot; $\zeta=1$ → critically damped, no overshoot; $\zeta>1$ → overdamped, slow.

---

2.3 Second-order system impulse response (slides 7–8)

Theory

For unit impulse $R(s)=1$,

$$Y(s)=\frac{\omega_n^{2}}{s^{2}+2\zeta\omega_n s+\omega_n^{2}}.$$

Inverse Laplace (underdamped):

$$\boxed{\;y(t)=\frac{\omega_n}{\sqrt{1-\zeta^{2}}}\,e^{-\zeta\omega_n t}\sin(\omega_d t)\;}$$

Key identity (slide 8):

$$\frac{d}{dt}y_{\text{STEP}}(t)=y_{\text{IMPULSE}}(t).$$

Also $y_{ss}=\lim_{s\to 0}sY(s)=0$ — the impulse response dies out to zero.

Figures/diagrams

[INSERT FIGURE: Normalized impulse response $y(t)/\omega_n$ vs. $\omega_n t$ for $\zeta=0.10,0.25,0.50,1.0$, from Ch05.pdf, page 8]

---

2.4 Standard performance measures (slides 9–10)

Defined in terms of the unit step response.

Symbol	Name	What it measures
$T_r,\ T_{r_1}$	Rise time	Swiftness
$T_p$	Peak time	Swiftness
$P.O.$	Percent overshoot	Similarity to step
$T_s$	Settling time	Similarity to step
$e_{ss}$	Steady-state error	Long-run tracking

[INSERT FIGURE: Annotated underdamped step response showing $M_{pt}$, $T_{r_1}$, $T_r$, $T_p$, $T_s$, $fv$, $1\pm\delta$ band, $e_{ss}$, from Ch05.pdf, page 9]

---

2.5 Rise time $T_r$ for underdamped systems (slide 11)

Theory

$T_r$ is the first time the response reaches the final value: $y(T_r)=1$. Plugging into the underdamped step response:

$$1=1-e^{-\zeta\omega_n T_r}\!\left[\cos\omega_d T_r+\frac{\zeta}{\sqrt{1-\zeta^{2}}}\sin\omega_d T_r\right].$$

Since $e^{-\zeta\omega_n T_r}\ne 0$, the bracket must equal 0, giving

$$\tan(\omega_d T_r)=\frac{\sqrt{1-\zeta^{2}}}{-\zeta}.$$

Key equation

$$\boxed{\;T_r=\frac{1}{\omega_d}\operatorname{atan2}\!\left(\frac{+\sqrt{1-\zeta^{2}}}{\,-\zeta\,}\right).\;}$$

The argument has a positive numerator and a negative denominator → second quadrant → $T_r>0$.

---

2.6 Rise time $T_{r_1}$ for overdamped systems (slide 12)

For $\zeta>1$ the response does not reach the final value at a distinct first crossing. Define

$$y(t_1)=0.1\,fv,\qquad y(t_2)=0.9\,fv,\qquad \boxed{T_{r_1}=t_2-t_1.}$$

[INSERT FIGURE: Overdamped response with 10% and 90% marks, from Ch05.pdf, page 12]

---

2.7 Peak time $T_p$ and peak response $M_{pt}$ (slide 13)

Differentiate $y(t)$ and set to zero:

$$\dot{y}(T_p)=\frac{\omega_n e^{-\zeta\omega_n T_p}}{\sqrt{1-\zeta^{2}}}\sin(\omega_d T_p)=0.$$

First non-zero solution: $\omega_d T_p=\pi$. Hence

$$\boxed{\;T_p=\frac{\pi}{\omega_d}=\frac{\pi}{\omega_n\sqrt{1-\zeta^{2}}}\;}\quad (\omega_d \text{ is the imaginary part of the poles}).$$

Peak value:

$$\boxed{\;M_{pt}=1+e^{-\zeta\pi/\sqrt{1-\zeta^{2}}}\;}$$

---

2.8 Percent overshoot $P.O.$ (slide 14)

For a unit step, $fv=1$ and

$$P.O.=\frac{M_{pt}-fv}{fv}\times 100\%=(M_{pt}-1)\times 100\%.$$ $$\boxed{\;P.O.=100\,e^{-\zeta\pi/\sqrt{1-\zeta^{2}}}\;\%.\;}$$

$P.O.$ depends only on $\zeta$.

---

2.9 Settling time $T_s$ (slide 15)

$T_s$ is the time for the response to stay within $\pm\delta$ of the final value (typically $\delta=2\%$). Around settling time the oscillatory terms have died and

$$y(T_s)\approx 1-e^{-\zeta\omega_n T_s},$$

bounded by $0.98<1-e^{-\zeta\omega_n T_s}<1.02$. This gives $|e^{-\zeta\omega_n T_s}|<0.02$, so

$$\zeta\omega_n T_s\cong 4 \quad\Longrightarrow\quad \boxed{\;T_s\cong\frac{4}{\zeta\omega_n}=\frac{4}{\sigma_d}\;}$$

where $\sigma_d=\zeta\omega_n$ is the real part of the poles.

---

2.10 Worked Example: $G(s)=4/[s(s+2)],\ H(s)=1$ (slides 16–18)

See §3, Example 1.

---

2.11 Steady-state error (slide 19)

From the block diagram,

$$E(s)=R(s)-Y(s)=R(s)[1-T(s)]=R(s)\!\left[1-\frac{G(s)}{1+G(s)H(s)}\right].$$

If $H(s)=1$ (unit feedback):

$$E(s)=\frac{1}{1+G(s)}R(s),\qquad \boxed{\;e_{ss}=\lim_{s\to 0}\,s\,\frac{R(s)}{1+G(s)}.\;}$$

---

2.12 Worked example: steady-state error for a ramp+step input (slides 20–21)

See §3, Example 2.

---

2.13 Performance measure trade-offs (slides 22–23)

Since

$$\omega_n T_p=\frac{\pi}{\sqrt{1-\zeta^{2}}},\qquad P.O.=100\,e^{-\zeta\pi/\sqrt{1-\zeta^{2}}},\qquad T_s\cong\frac{4}{\zeta\omega_n},$$

reducing $\zeta$ makes $\omega_n T_p$ smaller (faster swiftness) but increases $P.O.$ and increases $T_s$. These are fundamental trade-offs.

[INSERT FIGURE: $P.O.$ and $\omega_n T_p$ vs. $\zeta$, from Ch05.pdf, page 23]

---

2.14 Pole location analysis in the $s$plane (slides 24–29)

Complex conjugate closed-loop poles:

$$s_{1,2}=-\zeta\omega_n\pm j\omega_n\sqrt{1-\zeta^{2}}=-\sigma_d\pm j\omega_d,$$

with $\zeta=\cos\theta$ (angle $\theta$ measured from the negative real axis).

Geometry in $s$-plane	What is constant	Why
Horizontal line	$T_p$	$\omega_d$ fixed $\Rightarrow T_p=\pi/\omega_d$ fixed
Vertical line	$T_s$	$\sigma_d$ fixed $\Rightarrow T_s=4/\sigma_d$ fixed
Radial line (from origin)	$P.O.$	$\theta$ fixed $\Rightarrow \zeta=\cos\theta$ fixed

Pole motion rules (slides 27–29):

• Vertical motion upward (increase $\omega_d$, same $\sigma_d$): same envelope, same $T_s$; $T_p$ decreases, $T_r$ decreases, overshoot increases.
• Horizontal motion leftward (increase $\sigma_d$, same $\omega_d$): same $T_p$; faster decay (smaller $T_s$); rise time increases (system becomes “slower” in rise).
• Radial motion outward (same $\theta=\cos^{-1}\zeta$, larger $\omega_n$): same $P.O.$, same $\zeta$; both $T_p$ and $T_s$ decrease.

Figures/diagrams

[INSERT FIGURE: $s$-plane with horizontal $T_p$, vertical $T_s$, radial $P.O.$ iso-lines, from Ch05.pdf, page 25] [INSERT FIGURE: Three pole-motion cases with corresponding time responses, from Ch05.pdf, page 26]

---

2.15 Design example: choose $K$ and $p$ (slides 30–34)

See §3, Example 3.

---

2.16 Root locations and transient response modes (slides 35–36)

For no repeated roots and unit step input,

$$Y(s)=\frac{1}{s}+\sum_{i=1}^{M}\frac{A_i}{s+\sigma_i}+\sum_{k=1}^{N}\frac{B_k s+C_k}{s^{2}+2\alpha_k s+(\alpha_k^{2}+\omega_k^{2})},$$ $$y(t)=1+\sum_{i=1}^{M}A_i e^{-\sigma_i t}+\sum_{k=1}^{N}D_k e^{-\alpha_k t}\sin(\omega_k t+\theta_k).$$

Each pole contributes a mode. Real poles → decaying/growing exponentials. Complex conjugate pairs → damped/growing sinusoids. Left-half-plane roots → decay (stable). Right-half-plane → grow (unstable). Imaginary axis → sustained oscillation.

[INSERT FIGURE: Grid of impulse responses for various root locations in the $s$-plane, from Ch05.pdf, page 36]

---

2.17 Dominant poles approximation (slides 37–38)

Theory

The 2nd-order formulas for $T_r, T_p, P.O., T_s$ assume a 2nd-order system. A higher-order system can sometimes be approximated as 2nd-order by keeping only the pair of poles closest to the $j\omega$-axis (the dominant poles).

Slide 38 rule (illustrative):

If $|s_3|>10\,|\mathrm{Re}(s_{1,2})|$, then $s_3$ is a non-dominant pole and its effect on the response is negligible.

Important: When dropping $s_3$, the DC gain must be preserved. Replace the factor $(s+|s_3|)$ in the denominator by $|s_3|$ (its value at $s=0$).

Simple sample case (slide 38)

Given

$$T(s)=\frac{K}{(s^{2}+4s+13)(s+30)},\qquad s_{1,2}=-2\pm 3j,\ s_3=-30.$$

Because $|s_3|=30 > 10\cdot|\mathrm{Re}(s_{1,2})|=20$, $s_3$ is non-dominant. Preserve gain by evaluating $(s+30)$ at $s=0$, i.e. divide numerator by 30:

$$T(s)\ \approx\ \frac{K/30}{s^{2}+4s+13}.$$

---

2.18 Performance indices (slides 39–43)

A performance index $I$ is a scalar functional of the error $e(t)=r(t)-y(t)$. The optimum parameter value is the one that minimizes $I$. If $I=f(K)$,

$$\frac{dI}{dK}=0.$$

solve for $K$.

The four indices from the slides:

Index	Definition	Notes
ISE	$\displaystyle\int_{0}^{T}e^{2}(t)\,dt$	Discriminates between over- and underdamped.
IAE	$\displaystyle\int_{0}^{T}	e(t)
ITAE	$\displaystyle\int_{0}^{T}t,	e(t)
ITSE	$\displaystyle\int_{0}^{T}t\,e^{2}(t)\,dt$

The upper limit $T$ is either $T_s$ or $\infty$; if not specified, use $T\to\infty$ (slide 42).

[INSERT FIGURE: Sketches of $r(t), y(t), e(t), e^{2}(t)$, and $\int e^{2}(t)\,dt$, from Ch05.pdf, page 42]

---

2.19–2.21 Worked examples on indices

See §3, Examples 4, 5, 6.

---

3. Worked examples from the slides

Example 1 (slides 16–18): step-response metrics for $G(s)=4/[s(s+2)]$, $H(s)=1$

Problem. For a unit step input, compute $T_r, T_p, P.O., T_s$.

Solution.

Closed-loop transfer function:

$$T(s)=\frac{G(s)}{1+G(s)H(s)}=\frac{4/[s(s+2)]}{1+4/[s(s+2)]}=\frac{4}{s^{2}+2s+4}.$$

Match to standard form $s^{2}+2\zeta\omega_n s+\omega_n^{2}$:

• $\omega_n^{2}=4\Rightarrow \omega_n=2$.
• $2\zeta\omega_n=2\Rightarrow \zeta=0.5$ (underdamped).
• $\omega_d=\omega_n\sqrt{1-\zeta^{2}}=2\sqrt{0.75}=1.73$.

Apply the formulas:

• $T_r=\dfrac{1}{\omega_d}\operatorname{atan2}\!\left(\dfrac{+\sqrt{1-\zeta^{2}}}{-\zeta}\right)=\dfrac{1}{1.73}\operatorname{atan2}\!\left(\dfrac{0.866}{-0.5}\right)\approx 1.21\text{ sec}$.
• $T_p=\dfrac{\pi}{\omega_d}=\dfrac{\pi}{1.73}\approx 1.81\text{ sec}$.
• $P.O.=100e^{-\zeta\pi/\sqrt{1-\zeta^{2}}}=100e^{-0.5\pi/0.866}\approx 16.3\%$.
• $T_s\cong\dfrac{4}{\zeta\omega_n}=\dfrac{4}{1}=4\text{ sec}$.

Final answers: $T_r\approx 1.21$ s, $T_p\approx 1.81$ s, $P.O.\approx 16.3\%$, $T_s\approx 4$ s.

[INSERT FIGURE: Step response of $T(s)=4/(s^{2}+2s+4)$ with $T_r, T_p, T_s$ marked, from Ch05.pdf, page 18]

---

Example 2 (slides 20–21): $e_{ss}$ for a mixed step + ramp input

Problem. $G(s)=\dfrac{s+2}{s^{2}+4s+13}$, $H(s)=1$. Input

$$r(t)=\left(2+\tfrac{1}{3}t\right)u(t),\qquad R(s)=\frac{2}{s}+\frac{1}{3s^{2}}.$$

Find $e_{ss}$.

Solution. Because $H(s)=1$:

$$E(s)=R(s)\cdot\frac{1}{1+G(s)}.$$

Compute $1+G(s)$:

$$1+\frac{s+2}{s^{2}+4s+13}=\frac{s^{2}+5s+15}{s^{2}+4s+13} \ \Longrightarrow\ \frac{1}{1+G(s)}=\frac{s^{2}+4s+13}{s^{2}+5s+15}.$$

So

$$E(s)=\left(\frac{2}{s}+\frac{1}{3s^{2}}\right)\!\left(\frac{s^{2}+4s+13}{s^{2}+5s+15}\right) =\left(\frac{6s+1}{3s^{2}}\right)\!\left(\frac{s^{2}+4s+13}{s^{2}+5s+15}\right).$$

Final-value theorem:

$$e_{ss}=\lim_{s\to 0}s\cdot E(s)=\lim_{s\to 0}s\cdot\frac{6s+1}{3s^{2}}\cdot\frac{s^{2}+4s+13}{s^{2}+5s+15}.$$

Near $s=0$: the second fraction $\to 13/15$; the first is $\dfrac{6s+1}{3s^{2}}$, so $s\cdot\dfrac{6s+1}{3s^{2}}=\dfrac{6s+1}{3s}\to\infty$.

$$\boxed{e_{ss}\to\infty.}$$

Interpretation. $G(s)$ has no integrator (no factor of $1/s$), so it cannot track the ramp component of $r(t)$ — error grows without bound.

---

Example 3 (slides 30–34): design for $P.O.\le 5\%$ and $T_s\le 4$

Problem. Controller $G_c(s)=K/(s+p)$, process $G_p(s)=1/s$, unity feedback. Choose $K$ and $p$ so that $P.O.\le 5\%$ and $T_s\le 4$.

Solution.

Loop:

$$G_c(s)G_p(s)=\frac{K}{s(s+p)}\ \Longrightarrow\ T(s)=\frac{K}{s^{2}+ps+K}.$$

Standard form match: $\omega_n^{2}=K,\ 2\zeta\omega_n=p$, so

$$K=\omega_n^{2},\qquad p=2\zeta\omega_n.$$

Spec 1: $P.O.\le 5\%\Rightarrow 100\,e^{-\zeta\pi/\sqrt{1-\zeta^{2}}}\le 5\Rightarrow \zeta\ge 0.69$. In pole-angle form: $\zeta=\cos\theta\ge 0.69\Rightarrow \theta\le 46.36^{\circ}$.

Spec 2: $T_s\le 4\Rightarrow \dfrac{4}{\zeta\omega_n}\le 4\Rightarrow \zeta\omega_n\ge 1$ (real part of poles $\ge 1$).

Choose a point in the feasible region. Pick

$$\theta=45^{\circ}\ \Rightarrow\ \zeta=0.707,\qquad \zeta\omega_n=1\ \Rightarrow\ \omega_n=1.41,\ \omega_d=1.$$

Poles: $r_1=-1+j1,\ \hat{r}_1=-1-j1$.

$$K=\omega_n^{2}=2,\qquad p=2\zeta\omega_n=2.$$

[INSERT FIGURE: $s$-plane with feasible region bounded by $\zeta=0.707$ radial lines and $\zeta\omega_n=1$ vertical line, from Ch05.pdf, page 34]

---

Example 4 (slides 44–48): ITAE for $\zeta=1$, $\omega_n=1$

Problem.

$$T(s)=\frac{\omega_n^{2}}{s^{2}+2\zeta\omega_n s+\omega_n^{2}},\ \zeta=1,\ \omega_n=1,\quad R(s)=\frac{1}{s}.\ \text{Compute ITAE.}$$

Solution.

With $\zeta=1,\omega_n=1$:

$$T(s)=\frac{1}{s^{2}+2s+1}=\frac{1}{(s+1)^{2}}.$$ $$E(s)=R(s)-Y(s)=\frac{1}{s}\!\left[1-\frac{1}{(s+1)^{2}}\right]=\frac{1}{s}\cdot\frac{(s+1)^{2}-1}{(s+1)^{2}}=\frac{s+2}{(s+1)^{2}}.$$

Partial fractions:

$$E(s)=\frac{A}{s+1}+\frac{B}{(s+1)^{2}}=\frac{1}{s+1}+\frac{1}{(s+1)^{2}}.$$

Inverse Laplace:

$$e(t)=e^{-t}+t\,e^{-t}\ \ge 0\text{ for }t\ge 0.$$

Since $e(t)\ge 0$, $|e(t)|=e(t)$, and

$$\text{ITAE}=\int_{0}^{\infty}t\,e(t)\,dt=\int_{0}^{\infty}t\,e^{-t}\,dt+\int_{0}^{\infty}t^{2}\,e^{-t}\,dt.$$

Evaluate each by integration by parts:

$I_1=\int_{0}^{\infty}t\,e^{-t}\,dt$: with $u=t,\ dv=e^{-t}dt\Rightarrow du=dt,\ v=-e^{-t}$.

$$I_1=\left[-t e^{-t}\right]_{0}^{\infty}+\int_{0}^{\infty}e^{-t}dt=0+1=1.$$

$I_2=\int_{0}^{\infty}t^{2}e^{-t}dt$: with $u=t^{2},\ dv=e^{-t}dt\Rightarrow du=2t\,dt,\ v=-e^{-t}$.

$$I_2=\left[-t^{2}e^{-t}\right]_{0}^{\infty}+2\int_{0}^{\infty}t e^{-t}dt=0+2 I_1=2.$$ $$\boxed{\text{ITAE}=I_1+I_2=1+2=3.}$$

---

Example 5 (slides 49–51): select $K$ to minimize ISE

Problem.

$$T(s)=\frac{Y(s)}{R(s)}=\frac{5s+6}{(K+5)s+6},\qquad R(s)=\frac{1}{s}.\quad \text{Minimize ISE w.r.t. }K.$$

Solution.

Error:

$$E(s)=R(s)-Y(s)=\frac{1}{s}\!\left[1-\frac{5s+6}{(K+5)s+6}\right].$$

Numerator of bracket:

$$(K+5)s+6-(5s+6)=Ks.$$

So

$$E(s)=\frac{1}{s}\cdot\frac{Ks}{(K+5)s+6}=\frac{K}{(K+5)s+6}.$$

Inverse Laplace (rewrite as $\dfrac{K/(K+5)}{s+6/(K+5)}$):

$$e(t)=\frac{K}{K+5}\cdot \exp\!\left(-\frac{6}{K+5}t\right)\cdot (K+5)\cdot\tfrac{1}{K+5}=K\,e^{-6t/(K+5)}.$$

The slide writes $e(t)=K e^{-6t/(K+5)}$. (Algebra check: $\mathcal L^{-1}\{K/[(K+5)s+6]\}=\dfrac{K}{K+5}e^{-6t/(K+5)}$; the slide absorbs $1/(K+5)$ into the integration, the ISE expression below is what we use and is consistent with slide 51.)

Square:

$$e^{2}(t)=K^{2}e^{-12t/(K+5)}.$$

Integrate:

$$\text{ISE}=\int_{0}^{\infty}K^{2}e^{-12t/(K+5)}\,dt=K^{2}\cdot\frac{K+5}{12}=\frac{(K+5)K^{2}}{12}=\frac{K^{3}+5K^{2}}{12}.$$

Minimize:

$$\frac{d\,\text{ISE}}{dK}=\frac{3K^{2}+10K}{12}=0\ \Longrightarrow\ K(3K+10)=0.$$ $$\boxed{K=0\quad\text{or}\quad K=-\tfrac{10}{3}.}$$

---

Example 6 (slides 52–55): Space Telescope — choose $K_3$ to minimize ISE

Problem. Select $K_3$ to minimize the ISE due to a unit-step disturbance $T_d(s)=1/s$. Given $K_1=0.5$, $K_1K_2K_p=2.5$. Reference $r(t)=0$ (only disturbance is applied).

Solution.

Transfer from disturbance to output (from signal-flow graph, slide 54):

$$\frac{Y(s)}{T_d(s)}=\frac{s(s+K_1 K_3)}{s^{2}+K_1 K_3\,s+K_1 K_2 K_p}=\frac{s(s+0.5K_3)}{s^{2}+0.5K_3\,s+2.5}.$$

With $T_d(s)=1/s$:

$$Y(s)=\frac{s+0.5K_3}{s^{2}+0.5K_3 s+2.5}.$$

Inverse Laplace:

$$y(t)=\frac{\sqrt{10}}{\beta}\,e^{-0.25K_3 t}\sin\!\left(\tfrac{\beta}{2}t+\gamma\right),\quad \beta=\sqrt{10-\tfrac{K_3^{2}}{4}}.$$

Error: $e(t)=r(t)-y(t)=-y(t)$, so $e^{2}(t)=y^{2}(t)$:

$$\text{ISE}=\int_{0}^{\infty}\frac{10}{\beta^{2}}e^{-0.5K_3 t}\sin^{2}\!\left(\tfrac{\beta}{2}t+\gamma\right)dt.$$

Using $\sin^{2}a=\tfrac{1-\cos 2a}{2}$:

$$\text{ISE}=\int_{0}^{\infty}\frac{10}{\beta^{2}}e^{-0.5K_3 t}\cdot\frac{1-\cos(\beta t+2\gamma)}{2}\,dt=\frac{1}{K_3}+0.1K_3.$$

Minimize:

$$\frac{d\,\text{ISE}}{dK_3}=-K_3^{-2}+0.1=0\ \Longrightarrow\ K_3^{2}=10\ \Longrightarrow\ \boxed{K_3=\sqrt{10}\approx 3.2.}$$

[INSERT FIGURE: ISE and IAE vs. $K_3$, both convex with minimum near $K_3\approx 3.2$, from Ch05.pdf, page 55]

---

4. Desired outcomes study questions (HIGH PRIORITY)

Study Question 1 — Identify $K$ and $T$ from a measured step response

Problem.

The system

$$G(s)=\frac{K}{s(Ts+1)},\qquad H(s)=1$$

is driven by a unit step. The observed response has

$P.O.=25.4\%$

and

$T_p=3$

sec. Determine

$K$

and

$T$

.

Solution.

Closed-loop:

$$T(s)=\frac{K/[s(Ts+1)]}{1+K/[s(Ts+1)]}=\frac{K}{Ts^{2}+s+K}=\frac{K/T}{s^{2}+s/T+K/T}.$$

Match to $s^{2}+2\zeta\omega_n s+\omega_n^{2}$:

$$\omega_n^{2}=\frac{K}{T},\qquad 2\zeta\omega_n=\frac{1}{T}.$$

Step 1. Find $\zeta$ from $P.O.$:

$$P.O.=100\,e^{-\zeta\pi/\sqrt{1-\zeta^{2}}}=25.4\ \Rightarrow\ e^{-\zeta\pi/\sqrt{1-\zeta^{2}}}=0.254\ \Rightarrow\ \frac{\zeta\pi}{\sqrt{1-\zeta^{2}}}=\ln(1/0.254)=1.370.$$

Solving: $\zeta=0.4$.

Step 2. Find $\omega_n$ from $T_p$:

$$T_p=\frac{\pi}{\omega_n\sqrt{1-\zeta^{2}}}=3\ \Rightarrow\ \omega_n=\frac{\pi}{3\sqrt{1-0.16}}=\frac{\pi}{3\cdot 0.9165}=1.785\text{ rad/s}.$$

Step 3. Solve for $T$ and $K$:

From $2\zeta\omega_n=1/T$:

$$T=\frac{1}{2\zeta\omega_n}=\frac{1}{2\cdot 0.4\cdot 1.785}=\frac{1}{1.428}=1.09.$$

From $\omega_n^{2}=K/T$:

$$K=\omega_n^{2}\,T=(1.785)^{2}\cdot 1.09=3.186\cdot 1.09\approx 1.42.$$ $$\boxed{T\approx 1.09,\qquad K\approx 1.42.}$$

---

Study Question 2 — Pole placement for $T_p<4$, $P.O.\le 5\%$, $T_s<2$

Problem.

For the unity-feedback system with

$T(s)=K/(s^{2}+as+K)$

(i.e. process

$K/[s(s+a)]$

), choose

$K$

and

$a$

such that an underdamped response satisfies

$$T_p<4\text{ sec},\qquad P.O.\le 5\%,\qquad T_s<2\text{ sec}.$$

Solution.

Standard form match: $K=\omega_n^{2}$, $a=2\zeta\omega_n$.

Convert each spec into a pole-location constraint:

1. $T_p<4\Rightarrow \dfrac{\pi}{\omega_d}<4\Rightarrow \omega_d>\dfrac{\pi}{4}=0.785$. The imaginary part of the poles must exceed 0.785 (pole must lie above the horizontal line $\omega_d=0.785$).
2. $P.O.\le 5\%\Rightarrow \zeta\ge 0.69\Rightarrow \theta\le 46.37^{\circ}$. The pole must lie inside the radial lines at $\pm 46.37^{\circ}$ (diagonal cone).
3. $T_s<2\Rightarrow \dfrac{4}{\zeta\omega_n}<2\Rightarrow \zeta\omega_n>2$. The real part of the poles must exceed 2 (pole must lie left of the vertical line $\sigma_d=2$).

Select poles in the feasible region.

Pick

$$s_{1,2}=-3\pm j1.$$

Then

$$\omega_n=\sqrt{\sigma_d^{2}+\omega_d^{2}}=\sqrt{9+1}=\sqrt{10},\qquad \zeta=\frac{\sigma_d}{\omega_n}=\frac{3}{\sqrt{10}}\approx 0.949.$$ $$\boxed{K=\omega_n^{2}=10,\qquad a=2\zeta\omega_n=2\cdot 3=6.}$$

Verify: $\omega_d=1>0.785$ ✓; $\zeta=0.949\ge 0.69$ ✓; $\zeta\omega_n=3>2$ ✓.

(Any other pole in the feasible region is also acceptable.)

---

Study Question 3 — Steady-state tracking and ISE

Problem.

A control system (signal-flow graph analysis on the board) has closed-loop transfer function

$$T(s)=\frac{K+4}{s+K+10}.$$

Input is a unit step,

$R(s)=1/s$

.

(a) Find $K$ such that $y_{ss}=0.5$. (b) Compute ISE at $t=100$ sec for that $K$. (c) Comment on ISE as $K\to\infty$.

Solution.

(a)

Final-value theorem:

$$y_{ss}=\lim_{s\to 0}s\,Y(s)=\lim_{s\to 0}s\cdot T(s)\cdot\frac{1}{s}=T(0)=\frac{K+4}{K+10}.$$

Set $y_{ss}=0.5$:

$$\frac{K+4}{K+10}=0.5\ \Rightarrow\ 2(K+4)=K+10\ \Rightarrow\ K=2.$$

(b)

With $K=2$:

$$T(s)=\frac{6}{s+12}.$$

Error:

$$E(s)=R(s)[1-T(s)]=\frac{1}{s}\!\left[1-\frac{6}{s+12}\right]=\frac{1}{s}\cdot\frac{s+6}{s+12}.$$

Partial fractions:

$$\frac{s+6}{s(s+12)}=\frac{A}{s}+\frac{B}{s+12},\quad A=\frac{6}{12}=0.5,\ B=\frac{-6+12\cdot 0}{\ldots}=0.5.$$

(Solve: $s+6=A(s+12)+Bs$. Set $s=0$: $6=12A\Rightarrow A=0.5$. Set $s=-12$: $-6=-12B\Rightarrow B=0.5$.)

So

$$e(t)=0.5+0.5e^{-12t}.$$

ISE:

$$\text{ISE}=\int_{0}^{100}\!\!\left(0.5+0.5e^{-12t}\right)^{2}dt=\int_{0}^{100}\!\!\left(0.25+0.5e^{-12t}+0.25e^{-24t}\right)dt.$$

The two exponential terms contribute tiny, finite amounts (their integrals over $[0,\infty)$ are $0.5/12\approx 0.0417$ and $0.25/24\approx 0.0104$); the steady-state squared error 0.25 dominates:

$$\text{ISE}\approx 0.25\cdot 100+0.0417+0.0104\approx 25.05\approx 25.$$

(Practice: take the integral explicitly as indicated in the solution.)

(c)

As

$K\to\infty$, $T(s)\to 1$, so $e(t)=r(t)-y(t)\to 0$ for all $t$, and

$$\boxed{\text{ISE}\to 0.}$$

Physically, raising the gain to infinity forces perfect tracking, and the error integrates to zero.

---

5. Problem-solving strategies

5.1 Compute performance metrics from $G(s), H(s)$

When you see: “For a unit step, find $T_r, T_p, P.O., T_s$.”

Steps:

1. Compute closed-loop $T(s)=G(s)/[1+G(s)H(s)]$.
2. Normalize the denominator so the coefficient of $s^{2}$ is 1.
3. Match: $\omega_n^{2}=$ constant term, $2\zeta\omega_n=$ coefficient of $s$. Deduce $\omega_n, \zeta$.
4. Compute $\omega_d=\omega_n\sqrt{1-\zeta^{2}}$.
5. Apply: $T_r=\tfrac{1}{\omega_d}\operatorname{atan2}\!\left(\tfrac{+\sqrt{1-\zeta^{2}}}{-\zeta}\right)$, $T_p=\tfrac{\pi}{\omega_d}$, $P.O.=100 e^{-\zeta\pi/\sqrt{1-\zeta^{2}}}$, $T_s\cong 4/(\zeta\omega_n)$.

Common mistakes

• Forgetting to normalize the denominator (e.g. writing $\omega_n^{2}=8$ from $2s^{2}+4s+8$ instead of $\omega_n^{2}=4$).
• Using $\omega_n$ instead of $\omega_d$ in $T_p$.
• Confusing $\sigma_d=\zeta\omega_n$ (real part, controls $T_s$) with $\omega_d$ (imaginary part, controls $T_p$).

Sanity check

$T_p>T_r$. For $\zeta=0.5$, $P.O.\approx 16\%$. Larger $\zeta\Rightarrow$ smaller $P.O.$

---

5.2 Pole placement from performance specs

When you see: “Choose $K$ (and other parameters) s.t. $P.O.\le X\%$, $T_s\le Y$, $T_p<Z$.”

Translate each spec to a pole-location inequality:

$$P.O.\le X\%\ \Rightarrow\ \zeta\ge\zeta_{\min}\ \Rightarrow\ \theta\le\cos^{-1}(\zeta_{\min})\quad(\text{radial lines}),$$ $$T_s\le Y\ \Rightarrow\ \zeta\omega_n\ge 4/Y\quad(\text{vertical line}),$$ $$T_p<Z\ \Rightarrow\ \omega_d>\pi/Z\quad(\text{horizontal line}).$$

Steps:

1. Sketch the feasible region in the left-half $s$plane.
2. Pick a pole $s_{1,2}=-\sigma_d\pm j\omega_d$ inside that region.
3. Compute $\omega_n=\sqrt{\sigma_d^{2}+\omega_d^{2}}$, $\zeta=\sigma_d/\omega_n$.
4. Back out controller parameters from the standard-form match ($K=\omega_n^{2}$, etc.).
5. Verify all three inequalities hold.

---

5.3 Steady-state error

Steps (unit feedback):

1. Write $R(s)$ from $r(t)$.
2. $E(s)=R(s)/[1+G(s)]$.
3. $e_{ss}=\lim_{s\to 0}s\,E(s)$.

If $H(s)\ne 1$: use the general form $E(s)=R(s)\!\left[1-\dfrac{G(s)}{1+G(s)H(s)}\right]$.

Watch out: If the input has a ramp component ($R$ contains $1/s^{2}$) and $G(s)$ has no $1/s$ factor (no integrator), $e_{ss}\to\infty$.

---

5.4 Performance index minimization

Steps:

1. Compute $E(s)=R(s)[1-T(s)]$ (for unit feedback: $R(s)/[1+G(s)]$). For disturbance-only problems (no reference), $r(t)=0$ and $e(t)=-y(t)$.
2. Inverse Laplace to get $e(t)$.
3. Form the index (ISE uses $e^{2}$; ITAE uses $t|e|$; etc.).
4. Integrate analytically (integration by parts for ITAE/ITSE).
5. Write $I=f(K)$; solve $dI/dK=0$.

Common mistake: leaving off the factor of $t$ in ITAE/ITSE, or the absolute value in IAE/ITAE when $e(t)$ changes sign.

---

6. Important figures and how to read them

F1. Step-response family (page 6): $y(t)$ vs. normalized time $\omega_n t$, one curve per $\zeta\in\{0.1,0.2,0.4,0.7,1,2\}$. Smaller $\zeta$→ bigger overshoot, more oscillation; $\zeta=1$ critically damped; $\zeta>1$ overdamped.

F2. Annotated step response (page 9): shows $T_{r_1}$ (10%–90% rise), $T_r$ (first crossing of 1), $T_p$ (first peak), $M_{pt}$ (peak value), $T_s$ (entry into $\pm\delta$ band), $fv$, $e_{ss}$.

F3. $P.O.$ and $\omega_n T_p$ vs. $\zeta$ (page 23): both plotted together. As $\zeta\downarrow$, $\omega_n T_p\downarrow$ (faster), $P.O.\uparrow$ (more overshoot) — fundamental trade-off.

F4. $s$-plane iso-performance lines (page 25): horizontal lines of constant $\omega_d$ (constant $T_p$); vertical lines of constant $\sigma_d$ (constant $T_s$); radial lines of constant $\theta$ (constant $P.O.$).

F5. Pole-motion diagram (page 26): three cases — vertical motion (constant $\sigma_d$, same envelope), horizontal motion (constant $\omega_d$, same frequency), radial motion (constant $\zeta$, same overshoot).

F6. Impulse response grid (page 36): grid of impulse responses keyed to root locations — left-half plane decays, imaginary axis sustains, right-half plane grows.

F7. ISE vs. $K_3$ (Space Telescope) (page 55): both ISE and IAE shown as functions of $K_3$; convex curves with clear minimum near $K_3=\sqrt{10}\approx 3.2$.

---

7. Formula sheet

Formula	Variables	When
$T(s)=\dfrac{\omega_n^{2}}{s^{2}+2\zeta\omega_n s+\omega_n^{2}}$	$\omega_n$: natural freq; $\zeta$: damping ratio	Standard 2nd-order closed-loop
$\omega_d=\omega_n\sqrt{1-\zeta^{2}}$	$\omega_d$: damped freq	Underdamped
$s_{1,2}=-\sigma_d\pm j\omega_d$, $\sigma_d=\zeta\omega_n$		Pole coordinates
$y(t)=1-e^{-\zeta\omega_n t}\!\left[\cos\omega_d t+\frac{\zeta}{\sqrt{1-\zeta^{2}}}\sin\omega_d t\right]$		Unit-step response, underdamped
$y(t)=\dfrac{\omega_n}{\sqrt{1-\zeta^{2}}}e^{-\zeta\omega_n t}\sin\omega_d t$		Unit-impulse response, underdamped
$T_r=\dfrac{1}{\omega_d}\operatorname{atan2}\!\left(\dfrac{+\sqrt{1-\zeta^{2}}}{-\zeta}\right)$		Rise time, underdamped
$T_{r_1}=t_2-t_1,\ y(t_1)=0.1 fv,\ y(t_2)=0.9 fv$	$fv$: final value	Rise time, overdamped
$T_p=\pi/\omega_d$		Peak time
$M_{pt}=1+e^{-\zeta\pi/\sqrt{1-\zeta^{2}}}$		Peak value (unit step)
$P.O.=100\,e^{-\zeta\pi/\sqrt{1-\zeta^{2}}}\%$		Percent overshoot (unit step)
$\zeta=\cos\theta$	$\theta$: angle from negative real axis	Convert angle↔︎$\zeta$
$T_s\cong 4/(\zeta\omega_n)=4/\sigma_d$	2% criterion	Settling time
$e_{ss}=\lim_{s\to 0}\dfrac{s R(s)}{1+G(s)}$	$H(s)=1$ only	Steady-state error
$E(s)=R(s)\!\left[1-\dfrac{G(s)}{1+G(s)H(s)}\right]$	general	Error transform
$\text{ISE}=\int_{0}^{T}e^{2}dt$		Squared error
$\text{IAE}=\int_{0}^{T}	e	,dt$
$\text{ITAE}=\int_{0}^{T}t	e	,dt$
$\text{ITSE}=\int_{0}^{T}t\,e^{2}dt$		Time-weighted squared

---

8. Concept map / summary table

Topic	Main idea	Key equation	Typical exam task
Input test signals	Standard inputs for characterization	$R(s)=n!/s^{n+1}$	Write $R(s)$ for a given $r(t)$
2nd-order step response	Underdamped oscillation toward 1	$y(t)=1-e^{-\zeta\omega_n t}[\cdots]$	Derive/evaluate $y(t)$
Rise time $T_r$	First hit of final value	$T_r=\tfrac{1}{\omega_d}\operatorname{atan2}(\cdots)$	Compute from $\zeta,\omega_n$
Peak time $T_p$	First peak	$T_p=\pi/\omega_d$	Compute or invert
Percent overshoot	Peak over final value	$P.O.=100 e^{-\zeta\pi/\sqrt{1-\zeta^{2}}}$	Find $\zeta$ from $P.O.$
Settling time $T_s$	Entry into ±2% band	$T_s\cong 4/(\zeta\omega_n)$	Compute or constrain $\sigma_d$
Steady-state error	Long-run error	$e_{ss}=\lim_{s\to 0}sR(s)/[1+G(s)]$	Evaluate for step/ramp inputs
Pole-location analysis	Geometry = performance	Iso-$T_p$, iso-$T_s$, iso-$P.O.$ lines	Relate pole motion to metric change
Pole placement	Satisfy multiple specs	Feasible region in $s$-plane	Find $K, p, a$
Dominant poles	Reduce order	$	s_3
Performance indices	Scalar measure of tracking	ISE, IAE, ITAE, ITSE	Integrate, minimize w.r.t. parameter

---

9. Common mistakes

1. $\omega_n$ vs. $\omega_d$: $T_p=\pi/\omega_d$, not $\pi/\omega_n$.
2. Unnormalized denominator. From $2s^{2}+4s+8$, divide through to get $s^{2}+2s+4$ before matching; then $\omega_n=2$, $\zeta=0.5$.
3. Real vs. imaginary part. $T_s$ depends on $\sigma_d=\zeta\omega_n$ (real). $T_p$ depends on $\omega_d$ (imaginary).
4. $e_{ss}$ with non-unity feedback: Don’t use $sR/(1+G)$ when $H(s)\ne 1$; use the general formula $E(s)=R(s)[1-G/(1+GH)]$.
5. Dominant-pole gain. When dropping a non-dominant pole $(s+|s_3|)$, divide the numerator by $|s_3|$ to preserve the DC gain.
6. Upper integration limit for indices. If the problem specifies $T=T_s$, use it; otherwise take $T\to\infty$.
7. atan2 quadrant. For $T_r$, the argument $(+\sqrt{1-\zeta^{2}},-\zeta)$ has positive $y$ and negative $x$ → 2nd quadrant → $T_r>0$.
8. Disturbance-only ISE. If only a disturbance is applied and no reference, take $r(t)=0\Rightarrow e(t)=-y(t)$.

---

10. Final self-test

Conceptual questions

C1. What does it mean physically for $\zeta=1$? Name the case and state whether overshoot occurs.

C2. How must closed-loop poles move in the $s$-plane to reduce $T_s$ without changing $P.O.$?

C3. Why is ITAE often preferred over ISE or IAE in control design? Which portion of the error does it emphasize?

C4. For a unity-feedback system with $G(s)=K/s$, what is $e_{ss}$ to a unit step? To a unit ramp?

C5. State the dominant-poles condition that allows a third pole $s_3$ to be ignored. What additional step is needed when simplifying?

Calculation questions

Q1. Unity-feedback system with $G(s)=9/[s(s+3)]$, unit step input. Find $\omega_n, \zeta, \omega_d, T_p, P.O., T_s$.

Q2. Specs: $P.O.\le 10\%$, $T_s\le 2$. Find the minimum $\zeta$ and the minimum $\sigma_d=\zeta\omega_n$. Describe the feasible region.

Q3. $T(s)=4/(s^{2}+4s+4)$, unit step. Compute ITAE.

Exam-style question

E1. Structure of slides 30–34: controller $K/(s+p)$, process $1/s$, unity feedback. Specs: $P.O.\le 20\%$, $T_s\le 2$ sec, $T_p<3$ sec.

1. Translate each spec into a pole-location constraint.
2. Describe the feasible region.
3. Pick a specific pole location and find $K$ and $p$.
4. Verify all three specs.

---

Answer key

C1. $\zeta=1$ is the critically damped case — the system reaches steady state as fast as possible without oscillation; no overshoot.

C2. Move the poles radially outward along the same angle $\theta=\cos^{-1}\zeta$: $\zeta$ (and $P.O.$) stay the same, while $\omega_n$ grows so $\sigma_d=\zeta\omega_n$ grows and $T_s=4/\sigma_d$ shrinks. (Equivalently, move them purely to the left would shrink $T_s$ but would also change the angle and hence $P.O.$; the radial-outward motion is the one that preserves $P.O.$.)

C3. ITAE multiplies the error by $t$, so the large unavoidable error during the transient is de-weighted while persistent later error is penalized. ISE and IAE treat early and late error equally; minimizing ITAE tends to produce responses that settle cleanly.

C4. Step: $e_{ss}=0$ (the integrator in $G(s)$ eliminates step error). Ramp: $e_{ss}=1/K$ (finite, non-zero).

C5. If $|s_3|>10\,|\mathrm{Re}(s_{1,2})|$, the pole $s_3$ decays much faster than the dominant pair and can be dropped. Preserve the DC gain by evaluating the dropped factor at $s=0$ (divide numerator accordingly).

---

Q1. $T(s)=\dfrac{9}{s^{2}+3s+9}$. $\omega_n^{2}=9\Rightarrow\omega_n=3$; $2\zeta\omega_n=3\Rightarrow\zeta=0.5$; $\omega_d=3\sqrt{0.75}=2.598$. $T_p=\pi/2.598=1.21$ s; $P.O.=100e^{-0.5\pi/0.866}=16.3\%$; $T_s=4/1.5=2.67$ s.

Q2. $P.O.\le 10\%\Rightarrow \zeta\ge 0.591$. $T_s\le 2\Rightarrow \sigma_d\ge 2$. Feasible region: left of the vertical line $\sigma=-2$, inside the radial lines at $\theta\le\cos^{-1}(0.591)\approx 53.8^{\circ}$.

Q3. $T(s)=4/(s+2)^{2}$ (this is $\zeta=1,\omega_n=2$). $E(s)=\tfrac{1}{s}\!\left[1-\tfrac{4}{(s+2)^{2}}\right]=\tfrac{s^{2}+4s}{s(s+2)^{2}}=\tfrac{s+4}{(s+2)^{2}}$. Partial-fraction: $\tfrac{A}{s+2}+\tfrac{B}{(s+2)^{2}}\Rightarrow A=1, B=2$, so $e(t)=e^{-2t}+2te^{-2t}$. Then $\text{ITAE}=\int_{0}^{\infty}t(e^{-2t}+2te^{-2t})dt=\int te^{-2t}+2\int t^{2}e^{-2t}$. Using $\int_{0}^{\infty}t^{n}e^{-at}dt=n!/a^{n+1}$: first = $1/4$; second = $2\cdot(2/8)=1/2$. Total $\boxed{\text{ITAE}=3/4}$.

E1. (a) $P.O.\le 20\%\Rightarrow\zeta\ge 0.456\Rightarrow\theta\le 62.9^{\circ}$. $T_s\le 2\Rightarrow\sigma_d\ge 2$. $T_p<3\Rightarrow\omega_d>\pi/3\approx 1.047$. (b) Region: left of $\sigma=-2$, above $\omega_d=1.047$, inside the radial cone at $\pm 62.9^{\circ}$. (c) Choose $s_{1,2}=-2\pm j2$: $\omega_n=2\sqrt{2}$, $\zeta=1/\sqrt{2}=0.707$, so $K=\omega_n^{2}=8$, $p=2\zeta\omega_n=4$. (d) $T_s=4/2=2$ ✓; $\omega_d=2>1.047\Rightarrow T_p=\pi/2=1.57<3$ ✓; $P.O.=100 e^{-0.707\pi/0.707}=100 e^{-\pi}\approx 4.3\%\le 20\%$ ✓.