and HW3 combinedand HW3 combinedand HW3 combined and HW3 combined
Date due and handed in Feb. 18,2010
Find the impulse response of the following systems defined by the following differential equations. Verify your answer
\(\left ( D^{2}+7D+12\right ) y\left ( t\right ) =u\left ( t\right ) \)
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.
\(\left ( D^{3}+6D^{2}+12D+8\right ) y\left ( t\right ) =u\left ( t\right ) \)
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
\[ \left ( D^{3}+6D^{2}+12D+8\right ) y\left ( t\right ) =\left ( D-1\right ) u\left ( t\right ) \] 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.
and HW3 combined