3.2 HW 2
Date due and handed in Feb. 18,2010
3.2.1 Problem 3.8
Find the impulse response of the following systems defined by the following differential equations.
Verify your answer
3.2.1.1 Part a
\(\left ( D^{2}+7D+12\right ) y\left ( t\right ) =u\left ( t\right ) \)
Answer
The impulse reponse \(h\left ( t\right ) \) satisfies the homogenouse part of the differential equation under the initial
conditions \(h\left ( 0\right ) =0,h^{\prime }\left ( 0\right ) =1\).
Hence we solve the following
\begin{equation} \left ( D^{2}+7D+12\right ) h\left ( t\right ) =0 \tag {1}\end{equation}
The charateristic equation is \(r^{2}+7r+12=0\,\ \)or \(\left ( r+4\right ) \left ( r+3\right ) =0\), hence
\begin{equation} h\left ( t\right ) =\left ( c_{1}e^{-3t}+c_{2}e^{-4t}\right ) \xi \left ( t\right ) \tag {2}\end{equation}
Where \(\xi \left ( t\right ) \) is the unit step
function. Now find \(c_{1}\,\)and \(c_{2}\) from initial conditions
\begin{equation} h\left ( 0\right ) =0=c+c_{2} \tag {3}\end{equation}
and
\begin{align} h^{\prime }\left ( t\right ) & =\left ( -3c_{1}e^{-3t}-4c_{2}e^{-4t}\right ) \xi \left ( t\right ) +\left ( c_{1}e^{-3t}+c_{2}e^{-4t}\right ) \delta \left ( t\right ) \nonumber \\ h^{\prime }\left ( 0\right ) & =1=\left ( -3c_{1}-4c_{2}\right ) +\left ( c_{1}+c_{2}\right ) \nonumber \\ 1 & =-2c_{1}-3c_{2} \tag {4}\end{align}
From (3) and (4), we solve for \(c_{1},c_{2}\)
\begin{align*} c_{1} & =1\\ c_{2} & =-1 \end{align*}
Hence \(h\left ( t\right ) \) from (2) becomes
\begin{equation} h\left ( t\right ) =\left ( e^{-}3t-e^{-}4t\right ) \xi \left ( t\right ) \tag {5}\end{equation}
Now we verify this solution (note that \(\xi ^{\prime }\left ( t\right ) =\delta \left ( t\right ) \))
\begin{align} h^{\prime }\left ( t\right ) & =\left ( -3e^{-3t}+4e^{-4t}\right ) \xi \left ( t\right ) +\left ( e^{-3t}-e^{-4t}\right ) \delta \left ( t\right ) \nonumber \\ h^{\prime }\left ( t\right ) & =\left ( -3e^{-3t}+4e^{-4t}\right ) \xi \left ( t\right ) \tag {6}\end{align}
And
\begin{align} h^{\prime \prime }\left ( t\right ) & =\left ( 9e^{-3t}-16e^{-4t}\right ) \xi \left ( t\right ) +\left ( -3e^{-3t}+4e^{-4t}\right ) \delta \left ( t\right ) \nonumber \\ & =\left ( 9e^{-3t}-16e^{-4t}\right ) \xi \left ( t\right ) +\left ( -3+4\right ) \delta \left ( t\right ) \nonumber \\ & =\left ( 9e^{-3t}-16e^{-4t}\right ) \xi \left ( t\right ) +\delta \left ( t\right ) \tag {7}\end{align}
Substitute (5),(6) and (7) into LHS of (1) we obtain
\begin{align*} \left ( D^{2}+7D+12\right ) h\left ( t\right ) & =h^{\prime \prime }\left ( t\right ) +7h^{\prime }\left ( t\right ) +12h\left ( t\right ) \\ & \\ & =\left ( 9e^{-3t}-16e^{-4t}\right ) \xi \left ( t\right ) +\delta \left ( t\right ) +\\ & 7\left ( -3e^{-3t}+4e^{-4t}\right ) \xi \left ( t\right ) +\\ & 12\left ( e^{-3t}-e^{-4t}\right ) \xi \left ( t\right ) \\ & \\ & =\left ( 9e^{-3t}-16e^{-4t}-21e^{-3t}+28e^{-4t}+12e^{-3t}-12e^{-4t}\right ) \xi \left ( t\right ) +\delta \left ( t\right ) \\ & =\left [ \left ( 9-21+12\right ) e^{-3t}+\left ( -16+28-12\right ) e^{-4t}\right ] \xi \left ( t\right ) +\delta \left ( t\right ) \\ & =\delta \left ( t\right ) \end{align*}
Hence we see that when the input is \(\delta \left ( t\right ) \), then the solution is \(h\left ( t\right ) \), which is the definition of \(h\left ( t\right ) \). Hence the
solution is verified.
3.2.1.2 Part d
\(\left ( D^{3}+6D^{2}+12D+8\right ) y\left ( t\right ) =u\left ( t\right ) \)
Answer
The impulse reponse \(h\left ( t\right ) \) satisfies the homogenouse part of the differential equation under the initial
conditions \(h\left ( 0\right ) =0,h^{\prime }\left ( 0\right ) =0,h^{\prime \prime }\left ( 0\right ) =1\)
Hence we solve the following
\begin{equation} \left ( D^{3}+6D^{2}+12D+8\right ) h\left ( t\right ) =0\tag {1}\end{equation}
The charateristic equation is \(r^{3}+6r^{2}+12r+8=0\,\)or \(\left ( r+2\right ) \left ( r+2\right ) (r+2)=0\), hence
\begin{equation} h\left ( t\right ) =\left ( c_{1}e^{-2t}+c_{2}te^{-2t}+c_{3}t^{2}e^{-2t}\right ) \xi \left ( t\right ) \tag {2}\end{equation}
Now we find unknown \(c^{\prime }s\). We
start from \(h\left ( 0\right ) =0\) and obtain
\[ h\left ( 0\right ) =0=c_{1}\]
Hence the solution becomes
\begin{align*} h\left ( t\right ) & =\left ( c_{2}te^{-2t}+c_{3}t^{2}e^{-2t}\right ) \xi \left ( t\right ) \\ h^{\prime }\left ( t\right ) & =\left ( c_{2}t\left ( -2e^{-2t}\right ) +c_{2}e^{-2t}+c_{3}t^{2}\left ( -2e^{-2t}\right ) +2c_{3}te^{-2t}\right ) \xi \left ( t\right ) +\left ( c_{2}te^{-2t}+c_{3}t^{2}e^{-2t}\right ) \delta \left ( t\right ) \\ & =\left ( -2c_{2}te^{-2t}+c_{2}e^{-2t}-2c_{3}t^{2}e^{-2t}+2c_{3}te^{-2t}\right ) \xi \left ( t\right ) \end{align*}
And from \(h^{\prime }\left ( 0\right ) =0\) we obtain
\[ 0=c_{2}\]
Hence the solution becomes
\begin{align*} h\left ( t\right ) & =\left ( c_{3}t^{2}e^{-2t}\right ) \xi \left ( t\right ) \\ h^{\prime }\left ( t\right ) & =\left ( 2c_{3}te^{-2t}-2c_{3}t^{2}e^{-2t}\right ) \xi \left ( t\right ) +\left ( c_{3}t^{2}e^{-2t}\right ) \delta \left ( t\right ) \\ & =\left ( 2c_{3}te^{-2t}-2c_{3}t^{2}e^{-2t}\right ) \xi \left ( t\right ) \\ h^{^{\prime \prime }}\left ( t\right ) & =\left ( 2c_{3}e^{-2t}-4c_{3}te^{-2t}-4c_{3}te^{-2t}+4c_{3}t^{2}e^{-2t}\right ) \xi \left ( t\right ) +\left ( 2c_{3}te^{-2t}-2c_{3}t^{2}e^{-2t}\right ) \delta \left ( t\right ) \\ & =\left ( 2c_{3}e^{-2t}-4c_{3}te^{-2t}-4c_{3}te^{-2t}+4c_{3}t^{2}e^{-2t}\right ) \xi \left ( t\right ) \end{align*}
And from \(h^{\prime \prime }\left ( 0\right ) =1\) we find that
\begin{align*} h^{\prime \prime } & =1=2c_{3}\\ c_{3} & =\frac {1}{2}\end{align*}
Hence the final solution is
\[ h\left ( t\right ) =\left ( \frac {1}{2}t^{2}e^{-}2t\right ) \xi \left ( t\right ) \]
To verify, we need to evaluate \(h^{\prime \prime \prime }\left ( t\right ) +6h^{\prime \prime }\left ( t\right ) +12h^{\prime }\left ( t\right ) +8h\left ( t\right ) \) and see if we obtain \(\delta \left ( t\right ) \) as the
result.
\begin{align*} h^{\prime }\left ( t\right ) & =\left ( te^{-2t}-t^{2}e^{-2t}\right ) \xi \left ( t\right ) +\left ( \frac {1}{2}t^{2}e^{-2t}\right ) \delta \left ( t\right ) \\ & =\left ( te^{-2t}-t^{2}e^{-2t}\right ) \xi \left ( t\right ) \end{align*}
And
\begin{align*} h^{\prime \prime }\left ( t\right ) & =\left ( e^{-2t}-2te^{-2t}-2te^{-2t}+2t^{2}e^{-2t}\right ) \xi \left ( t\right ) +\left ( te^{-2t}-t^{2}e^{-2t}\right ) \delta \left ( t\right ) \\ & =\left ( e^{-2t}-4te^{-2t}+2t^{2}e^{-2t}\right ) \xi \left ( t\right ) \end{align*}
And
\begin{align*} h^{\prime \prime \prime }\left ( t\right ) & =\left ( -2e^{-2t}-4e^{-2t}+8te^{-2t}+4te^{-2t}-4t^{2}e^{-2t}\right ) \xi \left ( t\right ) +\left ( e^{-2t}-4te^{-2t}+2t^{2}e^{-2t}\right ) \delta \left ( t\right ) \\ & =\left ( -6e^{-2t}+12te^{-2t}-4t^{2}e^{-2t}\right ) \xi \left ( t\right ) +\delta \left ( t\right ) \end{align*}
Therefore, \(LHS=h^{\prime \prime \prime }\left ( t\right ) +6h^{\prime \prime }\left ( t\right ) +12h^{\prime }\left ( t\right ) +8h\left ( t\right ) \) becomes
\begin{align*} LHS & =\left ( -6e^{-2t}+12te^{-2t}-4t^{2}e^{-2t}\right ) \xi \left ( t\right ) +\delta \left ( t\right ) \\ & +6\left ( \left ( e^{-2t}-4te^{-2t}+2t^{2}e^{-2t}\right ) \xi \left ( t\right ) \right ) \\ & +12\left ( \left ( te^{-2t}-t^{2}e^{-2t}\right ) \xi \left ( t\right ) \right ) \\ & +8\left ( \left ( \frac {1}{2}t^{2}e^{-2t}\right ) \xi \left ( t\right ) \right ) \\ & \\ & =e^{-2t}\left ( -6+6\right ) +te^{-2t}\left ( 12-24+12\right ) +t^{2}e^{-2t}\left ( -4+12-12+4\right ) +\delta \left ( t\right ) \\ & =\delta \left ( t\right ) \end{align*}
Hence we see that when the input is \(\delta \left ( t\right ) \), then the solution is \(h\left ( t\right ) \), which is the definition of \(h\left ( t\right ) \). Hence the
solution is verified
3.2.1.3 Part e
\[ \left ( D^{3}+6D^{2}+12D+8\right ) y\left ( t\right ) =\left ( D-1\right ) u\left ( t\right ) \]
Note: There is a typo in the textbook. The problem as shown in the text had the number \(4\) in the
above equation when it should be \(6\). I confirmend this with our course instructor. I am sloving the
correct version of the problem statment as shown above.
We start by finding the impluse response for the system \(\left ( D^{3}+6D^{2}+12D+8\right ) y\left ( t\right ) =u\left ( t\right ) \), which we call \(\hat {h}\left ( t\right ) \), then find the required
impulse reponse using
\[ h\left ( t\right ) =\left ( D-1\right ) \hat {h}\left ( t\right ) \]
However, the impulse reponse of the above was found in part (d), and it is
\[ \hat {h}\left ( t\right ) =\left ( \frac {1}{2}t^{2}e^{-2t}\right ) \xi \left ( t\right ) \]
Therefore the required reponse reponse is
\begin{align*} h\left ( t\right ) & =\left ( D-1\right ) \left ( \frac {1}{2}t^{2}e^{-2t}\right ) \xi \left ( t\right ) \\ & =\left ( te^{-2t}-t^{2}e^{-2t}\right ) \xi \left ( t\right ) +\left ( \frac {1}{2}t^{2}e^{-2t}\right ) \delta \left ( t\right ) -\left ( \frac {1}{2}t^{2}e^{-2t}\right ) \xi \left ( t\right ) \\ & =\left ( te^{-2t}-\frac {3}{2}t^{2}e^{-2t}\right ) \xi \left ( t\right ) \end{align*}
Therefore
\[ h\left ( t\right ) =\left ( te^{-}2t-\frac {3}{2}t^{2}e^{-}2t\right ) \xi \left ( t\right ) \]
Now we need to verify this solution.
\begin{align*} h^{\prime }\left ( t\right ) & =\left ( e^{-2t}-2te^{-2t}-3te^{-2t}+3t^{2}e^{-2t}\right ) \xi \left ( t\right ) +\left ( te^{-2t}-\frac {3}{2}t^{2}e^{-2t}\right ) \delta \left ( t\right ) \\ & =\left ( e^{-2t}-5te^{-2t}+3t^{2}e^{-2t}\right ) \xi \left ( t\right ) \end{align*}
And
\begin{align*} h^{\prime \prime }\left ( t\right ) & =\left ( -2e^{-2t}-5e^{-2t}+10te^{-2t}+6te^{-2t}-6t^{2}e^{-2t}\right ) \xi \left ( t\right ) +\left ( e^{-2t}-5te^{-2t}+3t^{2}e^{-2t}\right ) \delta \left ( t\right ) \\ & =\left ( -7e^{-2t}+16te^{-2t}-6t^{2}e^{-2t}\right ) \xi \left ( t\right ) +\delta \left ( t\right ) \end{align*}
And
\begin{align*} h^{\prime \prime \prime }\left ( t\right ) & =\left ( 14e^{-2t}+16e^{-2t}-32te^{-2t}-12te^{-2t}+12t^{2}e^{-2t}\right ) \xi \left ( t\right ) +\left ( -7e^{-2t}+16te^{-2t}-6t^{2}e^{-2t}\right ) \delta \left ( t\right ) +\delta ^{\prime }\left ( t\right ) \\ & =\left ( 30e^{-2t}-44te^{-2t}+12t^{2}e^{-2t}\right ) \xi \left ( t\right ) -7\delta \left ( t\right ) +\delta ^{\prime }\left ( t\right ) \end{align*}
Now using the above, we evaluate the LHS of the ODE, we obtain
\begin{align*} LHS & =\left ( D^{3}+6D^{2}+12D+8\right ) h\left ( t\right ) \\ & =h^{\prime \prime \prime }\left ( t\right ) +6h^{\prime \prime }\left ( t\right ) +12h^{\prime }\left ( t\right ) +8h\left ( t\right ) \\ & \\ & =\left ( 30e^{-2t}-44te^{-2t}+12t^{2}e^{-2t}\right ) \xi \left ( t\right ) -7\delta \left ( t\right ) +\delta ^{\prime }\left ( t\right ) \\ & +6\left [ \left ( -7e^{-2t}+16te^{-2t}-6t^{2}e^{-2t}\right ) \xi \left ( t\right ) +\delta \left ( t\right ) \right ] \\ & +12\left [ \left ( e^{-2t}-5te^{-2t}+3t^{2}e^{-2t}\right ) \xi \left ( t\right ) \right ] \\ & +8\left [ \left ( te^{-2t}-\frac {3}{2}t^{2}e^{-2t}\right ) \xi \left ( t\right ) \right ] \\ & \\ & =e^{-2t}\left ( 30-42+12\right ) \xi \left ( t\right ) \\ & +te^{-2t}\left ( -44+96-60+8\right ) \xi \left ( t\right ) \\ & +t^{2}e^{-2t}\left ( 12-36+36-12\right ) \xi \left ( t\right ) \\ & -\delta \left ( t\right ) +\delta ^{\prime }\left ( t\right ) \\ & \\ & =e^{-2t}\left ( 0\right ) +te^{-2t}\left ( 0\right ) +t^{2}e^{-2t}\left ( 0\right ) -\delta \left ( t\right ) +\delta ^{\prime }\left ( t\right ) \\ & \\ & =\delta ^{\prime }\left ( t\right ) -\delta \left ( t\right ) \end{align*}
But the RHS is \(\left ( D-1\right ) \delta \left ( t\right ) \) which is \(\delta ^{\prime }\left ( t\right ) -\delta \left ( t\right ) \). Hence LHS=RHS, hence verified.