Chapter B Problem Sheet 2 (Lectures 4–5)

• 1. Calculate the step response of \(\dot {x}=2x+5u\), \(y=x\).

  • Solution. We obtain the step response by solving \(\dot {x}=2x+5\), \(x(0)=0\). The homogeneous equation \(\dot {x}=2x\) has general solution \(a\e ^{2t}\) and a particular solution of \(\dot {x}=2x+5u\) is \(x=\frac {-5}{2}\). Therefore the general solution of \(\dot {x}=2x+5u\) is \(x(t)=\frac {-5}{2}+a\e ^{2t}\). From the initial condition we obtain \(a=\frac {5}{2}\) so that \(x(t)=\frac {-5}{2}+\frac {5}{2}\e ^{2t}\). Therefore \(H(t)=\frac {-5}{2}+\frac {5}{2}\e ^{2t}\) is the step response. □

• 2. Calculate the step response of \(\ddot {q}+4\dot {q}+3q=6u\), \(y=5q\).

  • Solution. We obtain the step response by solving \(\ddot {q}+4\dot {q}+3q=6\), \(q(0)=\dot {q}(0)=0\). The homogeneous equation \(\ddot {q}+4\dot {q}+3q=0\) has general solution \(a\e ^{-t}+b\e ^{-3t}\) and a particular solution of \(\ddot {q}+4\dot {q}+3q=6\) is \(q=2\). Therefore the general solution of \(\ddot {q}+4\dot {q}+3q=6\) is \(q(t)=2+a\e ^{-t}+b\e ^{-3t}\). From the initial condition we obtain

    \[ 0=q(0)=2+a+b,\qquad 0=\dot {q}(0)=-a-3b. \]

    Solving this gives \(a=-3\), \(b=1\) so that \(q(t)=2-3\e ^{-t}+\e ^{-3t}\). We then have \(y(t)=5q(t)=10-15\e ^{-t}+5\e ^{-3t}\), so that \(H(t)=10-15\e ^{-t}+5\e ^{-3t}\) is the step response. □

• 3. Calculate the step response of \(\ddot {q}+4\dot {q}+3q=6u\), \(y=\bbm {q\\\dot {q}}\).

  • Solution. In the previous problem we already solved \(\ddot {q}+4\dot {q}+3q=6\), \(q(0)=\dot {q}(0)=0\) and obtained \(q(t)=2-3\e ^{-t}+\e ^{-3t}\). We therefore have

    \[ y(t)=\bbm {2-3\e ^{-t}+\e ^{-3t}\\3\e ^{-t}-3\e ^{-3t}}, \]

    so that

    \[ H(t)=\bbm {2-3\e ^{-t}+\e ^{-3t}\\3\e ^{-t}-3\e ^{-3t}}, \]

    is the step response. □

• 4. Calculate the step response of

  \[ \dot {x}=\bbm {1&0\\0&2}x+\bbm {1&1\\0&1}u,\qquad y=x_1+x_2. \]

  • Solution. With \(u_1=1\) and \(u_2=0\) we obtain the equations \(\dot {x}_1=x_1+1\), \(\dot {x}_2=2x_2\). Solving this with zero initial conditions gives \(x_1(t)=-1+\e ^t\) and \(x_2=0\). Therefore \(y(t)=-1+\e ^t\) which therefore is the first column of the step response.

    With \(u_1=0\) and \(u_2=1\) we obtain the equations \(\dot {x}_1=x_1+1\), \(\dot {x}_2=2x_2+1\). Solving this with zero initial conditions gives \(x_1(t)=-1+\e ^t\) and \(x_2=\frac {-1}{2}+\frac {1}{2}\e ^{2t}\). Therefore \(y(t)=\frac {-3}{2}+\e ^t+\frac {1}{2}\e ^{2t}\) which therefore is the second column of the step response. Therefore

    \[ H(t)=\bbm {-1+\e ^t&\frac {-3}{2}+\e ^t+\frac {1}{2}\e ^{2t}}, \]

    is the step response. □

• 5. Calculate the transfer function of \(\dot {x}=2x+5u\), \(y=x\).

  • Solution. Laplace transforming with zero initial conditions gives \(s\hat {x}=2\hat {x}+5\hat {u}\), so that \(\hat {x}=\frac {5}{s-2}\hat {u}\) and therefore

    \[ G(s)=\frac {5}{s-2}, \]

    is the transfer function. □

• 6. Calculate the transfer function of \(\ddot {q}+4\dot {q}+3q=6u\), \(y=5q\).

  • Solution. Laplace transforming with zero initial conditions gives \((s^2+4s+3)\hat {q}=6\hat {u}\), so that \(\hat {y}=5\hat {q}=\frac {30}{s^2+4s+3}\hat {u}\) and therefore

    \[ G(s)=\frac {30}{s^2+4s+3}, \]

    is the transfer function. □

• 7. Calculate the transfer function of \(\ddot {q}+4\dot {q}+3q=6u\), \(y=\bbm {q\\\dot {q}}\).

  • Solution. In the previous problem we saw that \(\hat {q}=\frac {6}{s^2+4s+3}\hat {u}\), so that \(\hat {y}_1=\hat {q}=\frac {6}{s^2+4s+3}\hat {u}\) and \(\hat {y}_2=s\hat {q}=\frac {6s}{s^2+4s+3}\hat {u}\). Therefore

    \[ G(s)=\bbm {\frac {6}{s^2+4s+3}\\\frac {6s}{s^2+4s+3}}, \]

    is the transfer function. □

• 8. Calculate the transfer function of

  \[ \dot {x}=\bbm {1&0\\0&2}x+\bbm {1&1\\0&1}u,\qquad y=x_1+x_2. \]

  • Solution. We write this in standard form \(\dot {x}=Ax+Bu\), \(y=Cx+Du\) with

    \[ A=\bbm {1&0\\0&2},\qquad B=\bbm {1&1\\0&1},\qquad C=\bbm {1&1},\qquad D=0, \]

    and calculate the transfer function as

    \[ G(s)=C(sI-A)^{-1}B+D=\bbm {1&1}\bbm {\frac {1}{s-1}&0\\0&\frac {1}{s-2}}\bbm {1&1\\0&1} =\bbm {\frac {1}{s-1}&\frac {1}{s-1}+\frac {1}{s-2}}. \]

    □