Problems

1. Taking the Laplace transform of the I/O ODE with zero initial conditions, \[\begin{aligned} (s^3 + 6s^2 + 11s + 6) Y(s) &= (2s + 3) U(s). \end{aligned}\] Therefore, \[\begin{aligned} H(s) = \frac{Y(s)}{U(s)} = \frac{2s + 3}{s^3 + 6s^2 + 11s + 6}. \end{aligned}\]

2. The numerator has degree 1 and the denominator has degree 3.

3. The DC gain is \[\begin{aligned} H(0) = \frac{3}{6} = \frac{1}{2}. \end{aligned}\]

4. The zero is found from \(2s + 3 = 0\), giving \(s = -3/2\). The denominator factors as \((s+1)(s+2)(s+3) = s^3 + 6s^2 + 11s + 6\), so the poles are at \(s = -1, -2, -3\).

5. All poles have negative real parts (\(-1\), \(-2\), and \(-3\)), so the system is asymptotically stable.

1. From \(H(s) = \frac{Y(s)}{U(s)} = \frac{5s + 2}{s^2 + 4s + 13}\), cross-multiplying gives \[\begin{aligned} (s^2 + 4s + 13) Y = (5s + 2) U. \end{aligned}\] Taking the inverse Laplace transform: \[\begin{aligned} \ddot{y} + 4\dot{y} + 13 y = 5\dot{u} + 2u. \end{aligned}\]

2. The standard second-order denominator form is \(s^2 + 2\zeta\omega_n s + \omega_n^2\). Comparing with \(s^2 + 4s + 13\): \[\begin{aligned} \omega_n^2 &= 13 \implies \omega_n = \sqrt{13} \approx 3.61 \text{ rad/s}, \\ 2\zeta\omega_n &= 4 \implies \zeta = \frac{2}{\sqrt{13}} \approx 0.555. \end{aligned}\]

3. Since \(0 < \zeta \approx 0.555 < 1\), the system is underdamped.

4. The zero is at \(s = -2/5\) (from \(5s + 2 = 0\)). The poles are the roots of \(s^2 + 4s + 13 = 0\): \[\begin{aligned} s = \frac{-4 \pm \sqrt{16 - 52}}{2} = \frac{-4 \pm \sqrt{-36}}{2} = -2 \pm 3j. \end{aligned}\] So \(\sigma = -2\) and \(\omega_d = 3\) rad/s.

5. The pole-zero plot has poles (×) at \(-2 \pm 3j\) and a zero (○) at \(-2/5\) on the real axis. The distance from the origin to either pole is \(\omega_n = \sqrt{(-2)^2 + 3^2} = \sqrt{13}\). The angle from the negative real axis to the line from the origin to the upper pole is \(\cos^{-1}(\zeta) = \cos^{-1}(2/\sqrt{13}) \approx 56.3°\).

1. We compute \(H(s) = C(sI - A)^{-1}B + D\). First, \[\begin{aligned} sI - A = \begin{bmatrix} s & -1 \\ 5 & s+4 \end{bmatrix}. \end{aligned}\] The determinant is \(\det(sI - A) = s(s+4) + 5 = s^2 + 4s + 5\). The inverse is \[\begin{aligned} (sI - A)^{-1} = \frac{1}{s^2 + 4s + 5} \begin{bmatrix} s+4 & 1 \\ -5 & s \end{bmatrix}. \end{aligned}\] Then \[\begin{aligned} (sI - A)^{-1} B = \frac{1}{s^2 + 4s + 5} \begin{bmatrix} s+4 & 1 \\ -5 & s \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \frac{1}{s^2 + 4s + 5} \begin{bmatrix} 1 \\ s \end{bmatrix}. \end{aligned}\] So \[\begin{aligned} H(s) = C(sI - A)^{-1}B + D = \begin{bmatrix} 3 & 0 \end{bmatrix} \frac{1}{s^2 + 4s + 5} \begin{bmatrix} 1 \\ s \end{bmatrix} + 0 = \frac{3}{s^2 + 4s + 5}. \end{aligned}\]

2. There are no finite zeros (the numerator is the constant 3). The poles are the roots of \(s^2 + 4s + 5 = 0\): \[\begin{aligned} s = \frac{-4 \pm \sqrt{16 - 20}}{2} = -2 \pm j. \end{aligned}\]

3. The characteristic polynomial of \(A\) is \(\det(\lambda I - A) = \lambda^2 + 4\lambda + 5\), giving eigenvalues \(\lambda = -2 \pm j\). These match the poles.

4. Both poles have real part \(-2 < 0\), so the system is asymptotically stable.

1. \(H_1(s) = \frac{2}{s^2 + 3s + 2}\): The denominator factors as \((s+1)(s+2)\). Poles: \(s = -1, -2\). No finite zeros.

\(H_2(s) = \frac{s+1}{s^2 + 4}\): The denominator gives \(s^2 = -4\), so \(s = \pm 2j\). Poles: \(s = \pm 2j\). Zero: \(s = -1\).

\(H_3(s) = \frac{1}{s^2 - s - 6}\): The denominator factors as \((s-3)(s+2)\). Poles: \(s = 3, -2\). No finite zeros.

2.

• \(H_1\): two real poles at \(-1\) and \(-2\) on the negative real axis; no zeros.
• \(H_2\): two purely imaginary poles at \(\pm 2j\) on the imaginary axis; one real zero at \(-1\).
• \(H_3\): two real poles at \(-2\) and \(3\) (one in each half-plane); no zeros.

3.

• \(H_1\): Asymptotically stable. Both poles (\(s = -1, -2\)) have strictly negative real parts.
• \(H_2\): Marginally stable. The poles (\(s = \pm 2j\)) lie on the imaginary axis with \(\operatorname{Re}(s) = 0\).
• \(H_3\): Unstable. The pole at \(s = 3\) has a positive real part.

4. \(H_1\) is the only asymptotically stable system. Its poles are both real and negative, so the free response is a sum of two decaying exponentials \(c_1 e^{-t} + c_2 e^{-2t}\) with no oscillation.

1. Series connection: \[\begin{aligned} G(s)H(s) = \frac{1}{s+1} \cdot \frac{s+3}{s+5} = \frac{s+3}{(s+1)(s+5)} = \frac{s+3}{s^2 + 6s + 5}. \end{aligned}\]

2. Parallel connection: \[\begin{aligned} G(s) + H(s) &= \frac{1}{s+1} + \frac{s+3}{s+5} = \frac{(s+5) + (s+3)(s+1)}{(s+1)(s+5)} \\ &= \frac{s + 5 + s^2 + 4s + 3}{s^2 + 6s + 5} = \frac{s^2 + 5s + 8}{s^2 + 6s + 5}. \end{aligned}\]

3. Negative unity feedback: \[\begin{aligned} \frac{G}{1 + GH} &= \frac{\frac{1}{s+1}}{1 + \frac{1}{s+1}\cdot\frac{s+3}{s+5}} = \frac{\frac{1}{s+1}}{\frac{(s+1)(s+5) + (s+3)}{(s+1)(s+5)}} = \frac{s+5}{(s+1)(s+5) + (s+3)} \\ &= \frac{s+5}{s^2 + 7s + 8}. \end{aligned}\]

4. Poles and zeros:

• Series: poles at \(s = -1, -5\); zero at \(s = -3\). All poles in LHP \(\Rightarrow\) stable.
• Parallel: poles at \(s = -1, -5\); zeros at \(s = \frac{-5 \pm j\sqrt{7}}{2} \approx -2.5 \pm 1.32j\). Stable.
• Feedback: zero at \(s = -5\); poles at \(s = \frac{-7 \pm \sqrt{17}}{2} \approx -1.44, -5.56\). Stable.

All three connections are asymptotically stable. The feedback connection shifts the pole locations relative to the open-loop poles.

1. The Laplace transform of \(u(t) = 5\,u_s(t)\) is \(U(s) = \frac{5}{s}\).

2. The output in the Laplace domain is \[\begin{aligned} Y(s) = H(s) U(s) = \frac{3}{s+2} \cdot \frac{5}{s} = \frac{15}{s(s+2)}. \end{aligned}\]

3. Partial fraction expansion: we write \[\begin{aligned} \frac{15}{s(s+2)} = \frac{A}{s} + \frac{B}{s+2}. \end{aligned}\] Multiplying both sides by \(s(s+2)\): \[\begin{aligned} 15 = A(s+2) + Bs. \end{aligned}\] Setting \(s = 0\): \(15 = 2A\), so \(A = 15/2\). Setting \(s = -2\): \(15 = -2B\), so \(B = -15/2\). Therefore, \[\begin{aligned} Y(s) = \frac{15/2}{s} - \frac{15/2}{s+2}. \end{aligned}\]

4. Using the standard pairs \(\mathcal{L}^{-1}\{1/s\} = u_s(t)\) and \(\mathcal{L}^{-1}\{1/(s+a)\} = e^{-at} u_s(t)\): \[\begin{aligned} y(t) = \frac{15}{2}\left(1 - e^{-2t}\right), \quad t \ge 0. \end{aligned}\]

5. As \(t \to \infty\), \(e^{-2t} \to 0\), so \(\lim_{t\to\infty} y(t) = 15/2 = 7.5\). The DC gain is \(H(0) = 3/2\), and the input amplitude is \(5\), so \(H(0) \times 5 = (3/2)(5) = 15/2 = 7.5\). Consistent. \(\checkmark\)