Chapter A Problem Sheet 1 (Lectures 1–3)
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1. Solve the initial value problem \(\dot {x}+x=0\), \(x(0)=1\).
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2. Solve the initial value problem \(\ddot {q}+q=0\), \(q(0)=1\), \(\dot {q}(0)=0\).
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3. Solve the initial value problem \(\ddot {q}+4\dot {q}+3q=0\), \(q(0)=1\), \(\dot {q}(0)=0\) and determine the output \(y=\dot {q}\).
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Solution. The characteristic polynomial is \(s^2+4s+3\), which factors as \((s+1)(s+3)\). Therefore the general solution is \(q(t)=a\e ^{-t}+b\e ^{-3t}\) and \(\dot {q}(t)=-a\e ^{-t}-3b\e ^{-3t}\). From the initial conditions we then obtain
\[ a+b=q(0)=1,\quad -a-3b=\dot {q}(0)=0, \]
from which we obtain \(a=\frac {3}{2}\), \(b=\frac {-1}{2}\) so that
\[ y(t)=\dot {q}(t)=-\frac {3}{2}\e ^{-t}+\frac {3}{2}\e ^{-3t}. \]
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4. Write the second order scalar differential equation \(\ddot {q}+4\dot {q}+3q=u\), \(y=\dot {q}\) as a first order system of differential equations with state \(x:=\sbm {q\\\dot {q}}\).
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Solution. In terms of \(x_1\) and \(x_2\) we can rewrite the given second order scalar differential equation as \(\dot {x}_2+4x_2+3x_1=u\), \(y=x_2\). We encaptulate that \(x:=\sbm {q\\\dot {q}}\) by \(\dot {x}_1=x_2\). Combining these equations gives
\[ \bbm {\dot {x}_1\\\dot {x}_2}=\bbm {x_2\\-3x_1-4x_2}+\bbm {0\\u},\qquad y=x_2. \]
We can write this in matrix form as
\[ \dot {x}=\bbm {0&1\\-3&-4}x+\bbm {0\\1}u,\qquad y=\bbm {0&1}x, \]
so that
\[ A=\bbm {0&1\\-3&-4},\qquad B=\bbm {0\\1},\qquad C=\bbm {0&1},\qquad D=0. \]
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5. Write \(\dot {q}=2q+3r+v\), \(\dot {r}=-q+v+2w\) as a first order system of differential equations with state \(x:=\sbm {q\\r}\) and input \(u:=\sbm {v\\w}\).
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Solution. In terms of \(x_1\), \(x_2\), \(u_1\) and \(u_2\) we can rewrite the given equations as
\[ \dot {x}_1=2x_1+3x_2+u_1,\qquad \dot {x}_2=-x_1+u_1+2u_2. \]
Combining these equations gives
\[ \bbm {\dot {x}_1\\\dot {x}_2}=\bbm {2x_1+3x_2\\-x_1}+\bbm {u_1\\u_1+2u_2}. \]
We can write this in matrix form as
\[ \dot {x}=\bbm {2&3\\-1&0}x+\bbm {1&0\\1&2}u. \]
so that
\[ A=\bbm {2&3\\-1&0},\qquad B=\bbm {1&0\\1&2}. \]
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6. Show using only the definition of asymptotic stability that
\[ A=\bbm {-1&0\\0&-5}, \]
is asymptotically stable.
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Solution. Since \(A\) is diagonal, we can easily solve \(\dot {x}=Ax\) (it amounts to solving two uncoupled scalar first order differential equations):
\[ x_1(t)=\e ^{-t}x^0_1,\qquad x_2(t)=\e ^{-5t}x^0_2. \]
We have \(\lim _{t\to \infty }x(t)=0\) for all initial conditions \(x^0\in \mR ^2\) (since the solution consists of decaying exponentials). Hence all solutions of \(\dot {x}=Ax\) satisfy \(\lim _{t\to \infty }x(t)=0\) and therefore \(A\) is asymptotically stable. □
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7. Show using only the definition of asymptotic stability that
\[ A=\bbm {-1&0\\0&3}, \]
is not asymptotically stable.
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Solution. Since \(A\) is diagonal, we can easily solve \(\dot {x}=Ax\) (it amounts to solving two uncoupled scalar first order differential equations):
\[ x_1(t)=\e ^{-t}x^0_1,\qquad x_2(t)=\e ^{3t}x^0_2. \]
Choosing \(x^0=\sbm {0\\1}\) we therefore have the solution \(x(t)=\sbm {0\\\e ^{3t}}\) which does not satisfy \(\lim _{t\to \infty }x(t)=0\) (it contains an increasing exponential). Therefore \(\dot {x}=Ax\) has a solution which does not satisfy \(\lim _{t\to \infty }x(t)=0\) and therefore \(A\) is not asymptotically stable. □
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8. Determine whether or not the following polynomials are stable:
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(a) \(s^2+1\);
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(b) \(s^2+s+1\);
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(c) \(s^3+s+1\);
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(d) \(s^3-s^2+s+1\);
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(e) \(s^3+2s^2+4s+5\).
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Solution. (a) Since the coefficient of \(s\) is zero, not all coefficients have the same sign, so this polynomial is not stable. In this case we can also compute the roots as \(s=\pm i\) which have zero real part and are therefore not stable.
(b) Since all coefficients are positive (which for a second degree polynomial is necessary and sufficient), this polynomial is stable.
(c) Since the coefficient of \(s^2\) is zero, not all coefficients have the same sign, so this polynomial is not stable.
(d) Since the coefficient of \(s^2\) is negative (and the other coefficients are positive), not all coefficients have the same sign, so this polynomial is not stable.
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(e) We use Routh–Hurwitz for third degree polynomials. We have that all coefficients are positive, so the only condition left to check is that \(a_2a_1-a_0a_3>0\). We have \(a_2a_1-a_0a_3=8-5=3>0\). Therefore, this polynomial is stable. □
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9. Determine whether or not the following matrix is asymptotically stable
\[ A=\bbm {1&0&-1\\2&1&3\\-5&-1&-2}. \]