Simple Servo System Block Diagram: Finding $a$, $b$ from Overshoot and Time Constant, then $t_r$, $t_p$, $\omega_d$, $t_{ss}$

A standard “simple servo system” under unity feedback is typically modeled as a second-order closed-loop system whose step response is governed by the characteristic polynomial $s^2 + a s + b = 0.$

For a standard second-order form $\frac{Y(s)}{R(s)}=\frac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2},$ matching coefficients gives the key relations:

• $a = 2\zeta\omega_n$
• $b = \omega_n^2$

We will use the standard closed-loop step-response relationships for a dominant underdamped second-order system ($0<\zeta<1$) to connect time-domain specs (overshoot, time constant/decay) to $(\zeta,\omega_n,\omega_d)$.

Key system-step metrics used

Assumption (typical in such problems): “time constant $=0.1$ s” refers to the exponential decay constant of the underdamped response, i.e. $\tau=\frac{1}{\zeta\omega_n}.$

So, $\zeta\omega_n=\frac{1}{0.1}=10.$

If your specific course notes define “time constant” differently (e.g., based on $1/\alpha$ where $\alpha$ is the real part of the poles), it still maps to the same exponential decay rate $\alpha=\zeta\omega_n$ for the standard second-order model.

We will also interpret overshoot $16\%$ as percent maximum overshoot: $M_p = 16\% = 0.16.$

(i) For $k=10$: find $a$ and $b$ to achieve $16\%$ overshoot and time constant $0.1$ s

Step 1: Use the maximum overshoot formula to get $\zeta$

For an underdamped second-order step response, percent overshoot is $M_p = e^{-\frac{\zeta\pi}{\sqrt{1-\zeta^2}}}.$

Set $M_p=0.16$: $0.16 = e^{-\frac{\zeta\pi}{\sqrt{1-\zeta^2}}}.$

Take natural log: $\ln(0.16)= -\frac{\zeta\pi}{\sqrt{1-\zeta^2}}.$

So $\frac{\zeta\pi}{\sqrt{1-\zeta^2}} = -\ln(0.16).$

Compute the RHS constant: $-\ln(0.16)=\ln\left(\frac{1}{0.16}\right)=\ln(6.25).$

Numerically, $\ln(6.25)\approx 1.8326.$

Thus $\frac{\zeta\pi}{\sqrt{1-\zeta^2}} \approx 1.8326.$

Rearrange: $\frac{\zeta}{\sqrt{1-\zeta^2}} \approx \frac{1.8326}{\pi}.$ $\frac{\zeta}{\sqrt{1-\zeta^2}} \approx \frac{1.8326}{3.1416}\approx 0.5835.$

Square both sides: $\frac{\zeta^2}{1-\zeta^2} \approx 0.5835^2\approx 0.3405.$

So $\zeta^2 \approx 0.3405(1-\zeta^2)=0.3405-0.3405\zeta^2.$ $\zeta^2(1+0.3405)\approx 0.3405.$ $\zeta^2 \approx \frac{0.3405}{1.3405}\approx 0.2538.$

Hence $\zeta \approx \sqrt{0.2538}\approx 0.5038.$

So the damping ratio required for $16\%$ overshoot is approximately: $\boxed{\zeta \approx 0.504}.$

---

Step 2: Use the time constant condition $\tau=0.1$ to get $\omega_n$

Given $\tau=\frac{1}{\zeta\omega_n},\quad \tau=0.1,$ then $\zeta\omega_n = 10.$

So $\omega_n = \frac{10}{\zeta} \approx \frac{10}{0.5038}\approx 19.85 \text{ rad/s}.$

Therefore $\boxed{\omega_n \approx 19.85\ \text{rad/s}}.$

---

Step 3: Match coefficients to obtain $a$ and $b$

Coefficient matching for $s^2 + a s + b$ vs. $s^2+2\zeta\omega_n s+\omega_n^2$ gives: $a=2\zeta\omega_n,\quad b=\omega_n^2.$

Compute $a$: $a=2\zeta\omega_n \approx 2(0.5038)(19.85)\approx 20.03.$

Compute $b$: $b=\omega_n^2 \approx (19.85)^2\approx 394.0.$

So $\boxed{a \approx 20.0},\qquad \boxed{b \approx 394}.$

Note: In many servo-system problems, the given plant parameters $a,b$ are inserted directly into the closed-loop characteristic equation. If your diagram uses $k$ in building that polynomial, then $k=10$ may affect how $a,b$ map into the denominator. However, once the closed-loop characteristic polynomial is explicitly of the form $s^2+as+b$, the above matching is the correct second-order spec-matching method.

(ii) For $k=40$: determine $t_r$, $t_p$, $\omega_d$, and $t_{ss}$

We now use the second-order parameters obtained above (or re-derived under the assumption that the same design specs apply). Since the question statement changes only $k$ to $40$, the only consistent way to compute time-domain response quantities is that the closed-loop dynamics correspond to the second-order model with the damping ratio and $\omega_n$ determined by the design conditions.

Thus we proceed with: $\zeta \approx 0.504,\quad \omega_n \approx 19.85\ \text{rad/s}.$

Step 1: Damped natural frequency $\omega_d$

For underdamped second-order systems, $\omega_d=\omega_n\sqrt{1-\zeta^2}.$

Compute $\sqrt{1-\zeta^2}$: $1-\zeta^2 \approx 1-0.2538=0.7462.$ $\sqrt{1-\zeta^2}\approx \sqrt{0.7462}\approx 0.8639.$

So $\omega_d \approx 19.85(0.8639)\approx 17.14\ \text{rad/s}.$

Therefore $\boxed{\omega_d \approx 17.1\ \text{rad/s}}.$

---

Step 2: Peak time $t_p$

Peak time for an underdamped second-order step response is $t_p=\frac{\pi}{\omega_d}.$

So $t_p \approx \frac{\pi}{17.14}\approx \frac{3.1416}{17.14}\approx 0.1833\ \text{s}.$

Hence $\boxed{t_p \approx 0.183\ \text{s}}.$

---

Step 3: Rise time $t_r$

A common textbook approximation for rise time (for $0<\zeta<0.7$) is $t_r \approx \frac{\pi - \theta}{\omega_d},\quad \theta=\cos^{-1}(\zeta).$

Compute $\theta$: $\theta=\cos^{-1}(0.5038)\approx 1.045\ \text{rad}.$

Then $t_r \approx \frac{\pi-1.045}{17.14}=\frac{2.0966}{17.14}\approx 0.1224\ \text{s}.$

So $\boxed{t_r \approx 0.122\ \text{s}}.$

---

Step 4: Settling time $t_{ss}$

For the standard $2\%$ settling-time criterion, the approximation is $t_{ss}\approx \frac{4}{\zeta\omega_n}.$

But $\zeta\omega_n=10$ from part (i), so $t_{ss}\approx \frac{4}{10}=0.4\ \text{s}.$

Thus $\boxed{t_{ss}\approx 0.40\ \text{s}\ \text{(2\% criterion)}}.$