Question

1. the system has transfer function (10 pt.)

g(s) = \frac{y(s)}{u(s)} = \frac{k}{js^2 + (a + k k_f)s + k}
assume ( j = 1 ) kg-m² and ( a = 1 ) n-m/ rad/sec. if the maximum overshoot of the unit-step input is 0.2 and the peak time is 0.1 sec,
(a) find its damping ratio and natural frequency.
(b) find the gain ( k ) and ( k_f ). also, calculate the settling time.

Explanation:

Part (a)

For a second - order system, the maximum overshoot (OS) is given by the formula \(OS = e^{-\frac{\pi\zeta}{\sqrt{1 - \zeta^{2}}}}\), where \(\zeta\) is the damping ratio. The peak time (\(T_p\)) is given by the formula \(T_p=\frac{\pi}{\omega_d}\), and \(\omega_d=\omega_n\sqrt{1 - \zeta^{2}}\), where \(\omega_n\) is the natural frequency and \(\omega_d\) is the damped natural frequency.

Step 1: Find the damping ratio (\(\zeta\)) from maximum overshoot

We know that \(OS = 0.2=e^{-\frac{\pi\zeta}{\sqrt{1 - \zeta^{2}}}}\)
Take the natural logarithm of both sides: \(\ln(0.2)=-\frac{\pi\zeta}{\sqrt{1 - \zeta^{2}}}\)
\(\ln(0.2)\approx - 1.6094\), so \(1.6094=\frac{\pi\zeta}{\sqrt{1 - \zeta^{2}}}\)
Square both sides: \((1.6094)^{2}=\frac{\pi^{2}\zeta^{2}}{1 - \zeta^{2}}\)
\(2.590=\frac{9.8696\zeta^{2}}{1 - \zeta^{2}}\)
\(2.590(1 - \zeta^{2})=9.8696\zeta^{2}\)
\(2.590-2.590\zeta^{2}=9.8696\zeta^{2}\)
\(2.590 = 9.8696\zeta^{2}+2.590\zeta^{2}\)
\(2.590 = 12.4596\zeta^{2}\)
\(\zeta^{2}=\frac{2.590}{12.4596}\approx0.2079\)
\(\zeta\approx\sqrt{0.2079}\approx0.456\)

Step 2: Find the natural frequency (\(\omega_n\)) from peak time

We know that \(T_p = 0.1=\frac{\pi}{\omega_d}\), so \(\omega_d=\frac{\pi}{0.1}=31.4159\ rad/s\)
Also, \(\omega_d=\omega_n\sqrt{1 - \zeta^{2}}\), so \(\omega_n=\frac{\omega_d}{\sqrt{1 - \zeta^{2}}}\)
We know \(\zeta\approx0.456\), so \(\sqrt{1 - \zeta^{2}}=\sqrt{1 - 0.2079}=\sqrt{0.7921}\approx0.89\)
\(\omega_n=\frac{31.4159}{0.89}\approx35.3\ rad/s\)

Part (b)

The transfer function of the system is \(G(s)=\frac{K}{Js^{2}+(a + Kk_f)s + K}\). For a standard second - order system \(G(s)=\frac{\omega_n^{2}}{s^{2}+2\zeta\omega_ns+\omega_n^{2}}\)

Step 1: Find \(K\)

Compare the denominator \(Js^{2}+(a + Kk_f)s + K\) with \(s^{2}+2\zeta\omega_ns+\omega_n^{2}\) (after dividing by \(J\)). We have \(\omega_n^{2}=\frac{K}{J}\)
Given \(J = 1\ kg - m^{2}\) and \(\omega_n\approx35.3\ rad/s\), then \(K = J\omega_n^{2}=1\times(35.3)^{2}\approx1246.09\)

Step 2: Find \(k_f\)

The coefficient of \(s\) in the denominator: \(2\zeta\omega_n=\frac{a + Kk_f}{J}\)
We know \(J = 1\), \(a = 1\ N - m/rad/sec\), \(\zeta\approx0.456\), \(\omega_n\approx35.3\ rad/s\)
\(2\times0.456\times35.3=\frac{1+1246.09k_f}{1}\)
\(32.1=\ 1+1246.09k_f\)
\(1246.09k_f=32.1 - 1 = 31.1\)
\(k_f=\frac{31.1}{1246.09}\approx0.025\)

Step 3: Find the settling time (\(T_s\))

For a second - order system, the settling time (for \(2\%\) criterion) is given by \(T_s=\frac{4}{\zeta\omega_n}\)
\(T_s=\frac{4}{0.456\times35.3}\approx\frac{4}{16.1}\approx0.248\ s\) (for \(5\%\) criterion, \(T_s=\frac{3}{\zeta\omega_n}\approx\frac{3}{16.1}\approx0.186\ s\))

Final Answers

Part (a)

• Damping ratio \(\zeta\approx\boldsymbol{0.456}\)
• Natural frequency \(\omega_n\approx\boldsymbol{35.3\ rad/s}\)

Part (b)

• Gain \(K\approx\boldsymbol{1246.09}\)
• \(k_f\approx\boldsymbol{0.025}\)
• Settling time (for \(2\%\) criterion) \(T_s\approx\boldsymbol{0.248\ s}\) (for \(5\%\) criterion \(T_s\approx\boldsymbol{0.186\ s}\))

Answer:

For a second - order system, the maximum overshoot (OS) is given by the formula \(OS = e^{-\frac{\pi\zeta}{\sqrt{1 - \zeta^{2}}}}\), where \(\zeta\) is the damping ratio. The peak time (\(T_p\)) is given by the formula \(T_p=\frac{\pi}{\omega_d}\), and \(\omega_d=\omega_n\sqrt{1 - \zeta^{2}}\), where \(\omega_n\) is the natural frequency and \(\omega_d\) is the damped natural frequency.

Step 1: Find the damping ratio (\(\zeta\)) from maximum overshoot

We know that \(OS = 0.2=e^{-\frac{\pi\zeta}{\sqrt{1 - \zeta^{2}}}}\)
Take the natural logarithm of both sides: \(\ln(0.2)=-\frac{\pi\zeta}{\sqrt{1 - \zeta^{2}}}\)
\(\ln(0.2)\approx - 1.6094\), so \(1.6094=\frac{\pi\zeta}{\sqrt{1 - \zeta^{2}}}\)
Square both sides: \((1.6094)^{2}=\frac{\pi^{2}\zeta^{2}}{1 - \zeta^{2}}\)
\(2.590=\frac{9.8696\zeta^{2}}{1 - \zeta^{2}}\)
\(2.590(1 - \zeta^{2})=9.8696\zeta^{2}\)
\(2.590-2.590\zeta^{2}=9.8696\zeta^{2}\)
\(2.590 = 9.8696\zeta^{2}+2.590\zeta^{2}\)
\(2.590 = 12.4596\zeta^{2}\)
\(\zeta^{2}=\frac{2.590}{12.4596}\approx0.2079\)
\(\zeta\approx\sqrt{0.2079}\approx0.456\)

Step 2: Find the natural frequency (\(\omega_n\)) from peak time

We know that \(T_p = 0.1=\frac{\pi}{\omega_d}\), so \(\omega_d=\frac{\pi}{0.1}=31.4159\ rad/s\)
Also, \(\omega_d=\omega_n\sqrt{1 - \zeta^{2}}\), so \(\omega_n=\frac{\omega_d}{\sqrt{1 - \zeta^{2}}}\)
We know \(\zeta\approx0.456\), so \(\sqrt{1 - \zeta^{2}}=\sqrt{1 - 0.2079}=\sqrt{0.7921}\approx0.89\)
\(\omega_n=\frac{31.4159}{0.89}\approx35.3\ rad/s\)

Part (b)

The transfer function of the system is \(G(s)=\frac{K}{Js^{2}+(a + Kk_f)s + K}\). For a standard second - order system \(G(s)=\frac{\omega_n^{2}}{s^{2}+2\zeta\omega_ns+\omega_n^{2}}\)

Step 1: Find \(K\)

Compare the denominator \(Js^{2}+(a + Kk_f)s + K\) with \(s^{2}+2\zeta\omega_ns+\omega_n^{2}\) (after dividing by \(J\)). We have \(\omega_n^{2}=\frac{K}{J}\)
Given \(J = 1\ kg - m^{2}\) and \(\omega_n\approx35.3\ rad/s\), then \(K = J\omega_n^{2}=1\times(35.3)^{2}\approx1246.09\)

Step 2: Find \(k_f\)

The coefficient of \(s\) in the denominator: \(2\zeta\omega_n=\frac{a + Kk_f}{J}\)
We know \(J = 1\), \(a = 1\ N - m/rad/sec\), \(\zeta\approx0.456\), \(\omega_n\approx35.3\ rad/s\)
\(2\times0.456\times35.3=\frac{1+1246.09k_f}{1}\)
\(32.1=\ 1+1246.09k_f\)
\(1246.09k_f=32.1 - 1 = 31.1\)
\(k_f=\frac{31.1}{1246.09}\approx0.025\)

Step 3: Find the settling time (\(T_s\))

For a second - order system, the settling time (for \(2\%\) criterion) is given by \(T_s=\frac{4}{\zeta\omega_n}\)
\(T_s=\frac{4}{0.456\times35.3}\approx\frac{4}{16.1}\approx0.248\ s\) (for \(5\%\) criterion, \(T_s=\frac{3}{\zeta\omega_n}\approx\frac{3}{16.1}\approx0.186\ s\))

Final Answers

Part (a)

• Damping ratio \(\zeta\approx\boldsymbol{0.456}\)
• Natural frequency \(\omega_n\approx\boldsymbol{35.3\ rad/s}\)

Part (b)

• Gain \(K\approx\boldsymbol{1246.09}\)
• \(k_f\approx\boldsymbol{0.025}\)
• Settling time (for \(2\%\) criterion) \(T_s\approx\boldsymbol{0.248\ s}\) (for \(5\%\) criterion \(T_s\approx\boldsymbol{0.186\ s}\))