Chapter C Problem Sheet 3 (Lectures 6–7)
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1.
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(a) Determine the frequency response \(F\) of \(\dot {x}=2x+5u\), \(y=x\).
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(b) Calculate \(|F(\omega )|\). Your answer should only include real numbers.
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Solution. (a) On the previous problem sheet we calculated the transfer function
\[ G(s)=\frac {5}{s-2}, \]
so that the frequency response is
\[ F(\omega )=G(i\omega )=\frac {5}{i\omega -2}. \]
(b) We have
\[ |F(\omega )|^2=\left |\frac {5}{i\omega -2}\right |^2=\frac {25}{\omega ^2+4}, \]
so that
\[ |F(\omega )|=\frac {5}{\sqrt {\omega ^2+4}}. \]
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2.
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(a) Determine the frequency response \(F\) of \(\ddot {q}+4\dot {q}+3q=6u\), \(y=5q\).
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(b) Calculate \(|F(\omega )|\). Your answer should only include real numbers.
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Solution. (a) On the previous problem sheet we calculated the transfer function
\[ G(s)=\frac {30}{s^2+4s+3}, \]
so that the frequency response is
\[ F(\omega )=G(i\omega )=\frac {30}{-\omega ^2+4i\omega +3}. \]
(b) We have
\[ |F(\omega )|^2=\left |\frac {30}{-\omega ^2+4i\omega +3}\right |^2=\frac {30^2}{16\omega ^2+(3-\omega ^2)^2}, \]
so that
\[ |F(\omega )|=\frac {30}{\sqrt {\omega ^4+10\omega ^2+9}}. \]
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3.
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(a) Determine the frequency response \(F\) of \(\ddot {q}+4\dot {q}+3q=6u\), \(y=\bbm {q\\\dot {q}}\).
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(b) Calculate \(\|F(\omega )\|\) (your answer should only include real numbers); here the norm is the usual Euclidean norm.
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Solution. (a) On the previous problem sheet we calculated the transfer function
\[ G(s)=\bbm {\frac {6}{s^2+4s+3}\\\frac {6s}{s^2+4s+3}}, \]
so that the frequency response is
\[ F(\omega )=G(i\omega )=\bbm {\frac {6}{-\omega ^2+4i\omega +3}\\\frac {6i\omega }{-\omega ^2+4i\omega +3}}. \]
(b) We have
\(\seteqnumber{0}{C.}{0}\)\begin{align*} \|F(\omega )\|^2&=\left |\frac {6}{-\omega ^2+4i\omega +3}\right |^2+\left |\frac {6i\omega }{-\omega ^2+4i\omega +3}\right |^2 \\&=\frac {36}{16\omega ^2+(3-\omega ^2)^2}+\frac {36\omega ^2}{16\omega ^2+(3-\omega ^2)^2} \\&=\frac {36+36\omega ^2}{\omega ^4+10\omega ^2+9} \end{align*} so that
\[ \|F(\omega )\|=\frac {6\sqrt {1+\omega ^2}}{\sqrt {\omega ^4+10\omega ^2+9}}. \]
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4. Consider the frequency response
\[ F(\omega )=\frac {1}{-\omega ^2+2\zeta i\omega +1}, \]
of the second order scalar differential equation \(\ddot {q}+2\zeta \dot {q}+q=u\), \(y=q\).
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(a) Show that
\[ |F(\omega )|^2=\frac {1}{(1-\omega ^2)^2+4\zeta ^2\omega ^2}. \]
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(b) Show that when \(\zeta \in [0,\frac {1}{\sqrt {2}})\phantom {]}\), \(|F(\omega )|\) has a maximum at \(\omega _{\rm max}:=\sqrt {1-2\zeta ^2}\) and that when \(\zeta \geq \frac {1}{\sqrt {2}}\), \(|F(\omega )|\) is monotonically decreasing for \(\omega >0\).
[Hint: it suffices to show these statements for \(|F(\omega )|^2\) rather than for \(|F(\omega )|\) since the squareroot function is strictly increasing. Also note that one can work only with the denominator and subsequently draw conclusions about the fraction.]
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Solution. (a) The formula for \(|F(\omega )|^2\) is immediate.
(b) We consider the denominator \(g(\omega ):=(1-\omega ^2)^2+4\zeta ^2\omega ^2\). We have
\[ g'(\omega )=-4\omega (1-\omega ^2)+8\zeta ^2\omega . \]
Setting \(g'\) equal to zero gives \(\omega =0\) or \(1-\omega ^2=2\zeta ^2\), i.e. \(\omega =\pm \sqrt {1-2\zeta ^2}\). For \(\pm \sqrt {1-2\zeta ^2}\) to be real and nonzero we require \(\zeta \in [0,\frac {1}{\sqrt {2}}\!\phantom {]})\). We see that at \(\omega _{\max }:=\sqrt {1-2\zeta ^2}\) the function \(g'\) changes sign from negative to positive, so \(g\) has a minimum at \(\omega _{\max }\). It follows that \(|F|^2\) and therefore \(|F|\) has a maximum at \(\omega _{\max }\). If \(\zeta \geq \frac {1}{\sqrt {2}}\), then \(g'>0\) on \((0,\infty )\) so that \(g\) is strictly increasing and therefore \(|F|^2\) and \(|F|\) are strictly decreasing. □
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5. Consider the second order scalar differential equation
\[ \ddot {q}+4\dot {q}+3q=3u,\qquad y=q. \]
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(a) Write this in standard form \(\dot {x}=Ax+Bu\), \(y=Cx+Du\) with \(x:=\sbm {q\\\dot {q}}\).
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(b) Calculate the impulse response by determining \(\e ^{At}\) and forming \(h(t)=C\e ^{At}B\).
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Solution. The matrix \(A\) has characteristic polynomial \(s^2+4s+3=(s+3)(s+1)\), so has eigenvalues \(-3\) and \(-1\). Corresponding eigenvectors are
\[ \bbm {1\\-3} \quad \bbm {1\\-1}. \]
Therefore a fundamental matrix function is
\[ \Phi (t)=\bbm {\e ^{-3t}&\e ^{-t}\\-3\e ^{-3t}&-\e ^{-t}}. \]
We have
\[ \Phi (0)=\bbm {1&1\\-3&-1},\qquad \Phi (0)^{-1}=\bbm {-1&-1\\3&1}\frac {1}{2}, \]
so that
\[ \e ^{At}=\Phi (t)\Phi (0)^{-1} =\bbm {-\e ^{-3t}+3\e ^{-t}&-\e ^{-3t}+\e ^{-t}\\ 3\e ^{-3t}-3\e ^{-t}&3\e ^{-3t}-\e ^{-t}}\frac {1}{2}. \]
Therefore
\[ h(t)=C\e ^{At}B=\left (-\e ^{-3t}+\e ^{-t}\right )\frac {3}{2} =-\frac {3}{2}\e ^{-3t}+\frac {3}{2}\e ^{-t}. \]
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(c) Calculate the step response by solving \(\ddot {q}+4\dot {q}+3q=3\), \(q(0)=\dot {q}(0)=0\), \(y=q\) and then differentiate this to obtain the impulse response.
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Solution. A particular solution is \(q=1\). The homogeneous equation is \(\ddot {q}+4\dot {q}+3q=0\). It characteristic polynomial is \(s^2+4s+3=(s+3)(s+1)\). So the general solution of the homogeneous equation is \(a\e ^{-3t}+b\e ^{-t}\). Therefore the general solution of the original problem is \(q(t)=1+a\e ^{-3t}+b\e ^{-t}\). The initial conditions give \(1+a+b=0\), \(-3a-b=0\). Solving this gives \(a=\frac {1}{2}\), \(b=-\frac {3}{2}\). So the step response is
\[ H(t)=1+\frac {1}{2}\e ^{-3t}-\frac {3}{2}\e ^{-t}. \]
It follows that the impulse response is
\[ h(t)=H'(t)=-\frac {3}{2}\e ^{-3t}+\frac {3}{2}\e ^{-t}. \]
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(d) Determine the transfer function and find the inverse Laplace transform of this to determine the impulse response.
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Solution. The transfer function is
\[ G(s)=\frac {3}{s^2+4s+3}. \]
The partial fraction expansion of the transfer function is
\[ \frac {-3/2}{s+3}+\frac {3/2}{s+1}. \]
The inverse Laplace transform of this is
\[ \frac {-3}{2}\e ^{-3t}+\frac {3}{2}\e ^{-t}, \]
which therefore is the impulse response. □
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(e) Use the initial value problem from Remark 7.2 to determine the impulse response.
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Solution. Since \(B=\sbm {0\\3}\), this amounts to the initial condition \(q(0)=0\), \(\dot {q}(0)=3\). So we solve
\[ \ddot {q}+4\dot {q}+3q=0,\qquad q(0)=0,\quad \dot {q}(0)=3. \]
The general solution of the differential equation is \(q(t)=a\e ^{-3t}+b\e ^{-t}\). The initial conditions give \(a+b=0\), \(-3a-b=3\). Solving this gives \(a=-\frac {3}{2}\), \(b=\frac {3}{2}\). Therefore we obtain
\[ h(t)=-\frac {3}{2}\e ^{-3t}+\frac {3}{2}\e ^{-t}. \]
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6. For a system with impulse response \(\e ^{-t}\), determine the output for initial condition zero and input \(\e ^{-2t}\).
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Solution. We have
\(\seteqnumber{0}{C.}{0}\)\begin{align*} y(t)&=\int _0^t h(t-\theta )u(\theta )\,d\theta =\int _0^t \e ^{-(t-\theta )}\e ^{-2\theta }\,d\theta =\e ^{-t}\int _0^t \e ^{-\theta }\,d\theta =\e ^{-t}\left [-\e ^{-\theta }\right ]_0^t \\&=\e ^{-t}(1-\e ^{-t})=\e ^{-t}-\e ^{-2t}. \end{align*} □
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