Consider the signal \(x\left ( t\right ) =e^{-5t}u\left ( t\right ) +e^{-\beta t}u\left ( t\right ) \) and denote its Laplace transform by \(X\left ( s\right ) \) What are the constraints placed on the
real and imaginary parts of \(\beta \) if the region of convergence of \(X\left ( s\right ) \) is \(\operatorname{Re}\left ( s\right ) >-3\) ?
solution
The Laplace transform is \begin{align*} X\left ( s\right ) & =\int _{-\infty }^{\infty }x\left ( t\right ) e^{-st}dt\\ & =\int _{0}^{\infty }\left ( e^{-5t}+e^{-\beta t}\right ) e^{-st}dt\\ & =\int _{0}^{\infty }e^{-5t}e^{-st}dt+\int _{0}^{\infty }e^{-\beta t}e^{-st}dt\\ & =\int _{0}^{\infty }e^{-t\left ( 5+s\right ) }dt+\int _{0}^{\infty }e^{-t\left ( \beta +s\right ) }dt\\ & =\frac{1}{-\left ( s+5\right ) }\left [ e^{-t\left ( 5+s\right ) }\right ] _{0}^{\infty }-\frac{1}{\beta +s}\left [ e^{-t\left ( \beta +s\right ) }\right ] _{0}^{\infty } \end{align*}
For the first term \(\frac{1}{-\left ( s+5\right ) }\left [ e^{-t\left ( 5+s\right ) }\right ] _{0}^{\infty }=\frac{1}{-\left ( s+5\right ) }\left [ e^{-\infty \left ( 5+s\right ) }-1\right ] \). For this term to converge we need \(5+\operatorname{Re}\left ( s\right ) >0\) or \[ \operatorname{Re}\left ( s\right ) >-5 \] For the second term, let \(\beta =a+ib\) and let \(s=\sigma +j\omega \), hence the second term becomes \begin{align*} \frac{1}{\beta +s}\left [ e^{-t\left ( \beta +s\right ) }\right ] _{0}^{\infty } & =\frac{1}{\beta +s}\left [ e^{-t\left ( \left ( a+ib\right ) +\left ( \sigma +j\omega \right ) \right ) }\right ] _{0}^{\infty }\\ & =\frac{1}{\beta +s}\left [ e^{-t\left ( a+\sigma +j\left ( b+\omega \right ) \right ) }\right ] _{0}^{\infty }\\ & =\frac{1}{\beta +s}\left [ e^{-t\left ( a+\sigma \right ) }e^{-jt\left ( b+\omega \right ) }\right ] _{0}^{\infty }\\ & =\frac{1}{\beta +s}\left [ e^{-\infty \left ( a+\sigma \right ) }e^{-j\infty \left ( b+\omega \right ) }-1\right ] \end{align*}
The complex exponential terms always converges since its norm is bounded by \(1\). For the real exponential term, we need \(a+\sigma >0\) or \(a+\operatorname{Re}\left ( s\right ) >0\) or \(\operatorname{Re}\left ( s\right ) >-a\). Since we are told that \(\operatorname{Re}\left ( s\right ) >-3\), then \[ a=3 \] Is the requirement on real part of \(\beta \). There is no restriction on the imaginary part of \(\beta \).
Given that \(e^{-at}u\left ( t\right ) \Longleftrightarrow \frac{1}{s+a}\) for \(\operatorname{Re}\left ( s\right ) >\operatorname{Re}\left ( -a\right ) \), determine the inverse Laplace transform of\[ X\left ( s\right ) =\frac{2\left ( s+2\right ) }{s^{2}+7s+12}\qquad \operatorname{Re}\left ( s\right ) >-3 \] solution
Writing \(X\left ( s\right ) \) as\begin{align*} X\left ( s\right ) & =\frac{2\left ( s+2\right ) }{\left ( s+4\right ) \left ( s+3\right ) }\\ & =\frac{A}{\left ( s+4\right ) }+\frac{B}{\left ( s+3\right ) } \end{align*}
Hence \(A=\left . \frac{2\left ( s+2\right ) }{\left ( s+3\right ) }\right \vert _{s=-4}=\frac{2\left ( -4+2\right ) }{\left ( -4+3\right ) }=\allowbreak 4\) and \(B=\left . \frac{2\left ( s+2\right ) }{\left ( s+4\right ) }\right \vert _{s=-3}=\frac{2\left ( -3+2\right ) }{\left ( -3+4\right ) }=-2\), therefore the above becomes\[ X\left ( s\right ) =\frac{4}{\left ( s+4\right ) }-\frac{2}{\left ( s+3\right ) }\] Using \(e^{-at}u\left ( t\right ) \Longleftrightarrow \frac{1}{s+a}\) gives the inverse Laplace transform as\begin{align*} x\left ( t\right ) & =4e^{-4t}u\left ( t\right ) -2e^{-3t}u\left ( t\right ) \\ & =\left ( 4e^{-4t}-2e^{-3t}\right ) u\left ( t\right ) \end{align*}
With \(\operatorname{Re}\left ( s\right ) >-4\) and also \(\operatorname{Re}\left ( s\right ) >-3\). Therefore the ROC for both is \(\operatorname{Re}\left ( s\right ) >-3\).
Consider the two right-sides signals \(x\left ( t\right ) ,y\left ( t\right ) \) related through the differential equations \begin{align*} \frac{dx\left ( t\right ) }{dt} & =-2y\left ( t\right ) +\delta \left ( t\right ) \\ \frac{dy\left ( t\right ) }{dt} & =2x\left ( t\right ) \end{align*}
Determine \(Y\left ( s\right ) ,X\left ( s\right ) \) along with their ROC.
solution
The Laplace transform of \(\delta \left ( t\right ) \) is \(1\). Taking the Laplace transform of both the ODE’s above, and assuming zero initial conditions gives\begin{align} sX\left ( s\right ) & =-2Y\left ( s\right ) +1\tag{1}\\ sY\left ( s\right ) & =2X\left ( s\right ) \tag{2} \end{align}
Using the second equation in the first gives\begin{align*} sX\left ( s\right ) & =-2\left ( \frac{2X\left ( s\right ) }{s}\right ) +1\\ & =\frac{-4X\left ( s\right ) +s}{s}\\ s^{2}X\left ( s\right ) & =-4X\left ( s\right ) +s\\ \left ( s^{2}+4\right ) X\left ( s\right ) & =s\\ X\left ( s\right ) & =\frac{s}{\left ( s^{2}+4\right ) } \end{align*}
Using the above in (2) gives \(Y\left ( s\right ) \)\begin{align*} sY\left ( s\right ) & =2\frac{s}{\left ( s^{2}+4\right ) }\\ Y\left ( s\right ) & =\frac{2}{\left ( s^{2}+4\right ) } \end{align*}
Considering \(X\left ( s\right ) \) to find its ROC, let us write it as\[ X\left ( s\right ) =\frac{s}{\left ( s^{2}+4\right ) }=\frac{s}{\left ( s+2j\right ) \left ( s-2j\right ) }=\frac{A}{\left ( s+2j\right ) }+\frac{B}{\left ( s-2j\right ) }\] We see that the ROC for first term is \(\operatorname{Re}\left ( s\right ) >-\operatorname{Re}\left ( 2j\right ) \) which means \(\operatorname{Re}\left ( s\right ) >0\) since real part is zero. Same for the second term. Hence we see that for \(X\left ( s\right ) \) the ROC is \(\operatorname{Re}\left ( s\right ) >0\). Similarly for \(Y\left ( s\right ) \). Therefore the overall ROC is \[ \operatorname{Re}\left ( s\right ) >0 \]
A causal LTI system with impulse response\(\ h(t)\) has the following properties: (1) When the input to the system is \(x\left ( t\right ) =e^{2t}\) for all \(t\), the output is \(y\left ( t\right ) =\frac{1}{6}e^{2t}\) for all \(t\). (2) The impulse response \(h(t)\) satisfies the differential equation \[ \frac{dh\left ( t\right ) }{dt}+2h\left ( t\right ) =e^{-4t}u\left ( t\right ) +bu\left ( t\right ) \] Where \(b\) is unknown constant. Determine the system function \(H(s)\) of the system, consistent with the information above. There should be no unknown constants in your answer; that is, the constant \(b\) should not appear in the answer
solution
First \(H\left ( s\right ) \) is found from the differential equation. Taking Laplace transform gives (assuming zero initial conditions)\begin{align} sH\left ( s\right ) +2H\left ( s\right ) & =\frac{1}{s+4}+\frac{b}{s}\nonumber \\ H\left ( s\right ) \left ( 2+s\right ) & =\frac{1}{s+4}+\frac{b}{s}\nonumber \\ H\left ( s\right ) & =\frac{1}{\left ( s+4\right ) \left ( s+2\right ) }+\frac{b}{s\left ( s+2\right ) }\nonumber \\ & =\frac{s+b\left ( s+4\right ) }{s\left ( s+4\right ) \left ( s+2\right ) }\tag{1} \end{align}
Now we are told when the input is \(e^{2t}\) then the output is \(\frac{1}{6}e^{2t}\). In Laplace domain this means \(Y\left ( s\right ) =X\left ( s\right ) H\left ( s\right ) \,\). Therefore\begin{align*} Y\left ( s\right ) & =\frac{1}{6}\frac{1}{s-2}\qquad \operatorname{Re}\left ( s\right ) >2\\ X\left ( s\right ) & =\frac{1}{s-2}\qquad \operatorname{Re}\left ( s\right ) >2 \end{align*}
Hence \begin{align} H\left ( s\right ) & =\frac{Y\left ( s\right ) }{X\left ( s\right ) }\nonumber \\ H\left ( s\right ) & =\frac{\frac{1}{6}\frac{1}{s-2}}{\frac{1}{s-2}}\nonumber \\ & =\frac{1}{6}\tag{2} \end{align}
Comparing (1,2) then\[ \frac{1}{6}=\frac{s+b\left ( s+4\right ) }{s\left ( s+4\right ) \left ( s+2\right ) }\] Solving for \(b\) gives\begin{align*} \frac{s\left ( s+4\right ) \left ( s+2\right ) }{6} & =s+b\left ( s+4\right ) \\ \frac{s\left ( s+4\right ) \left ( s+2\right ) }{6\left ( s+4\right ) }-\frac{s}{\left ( s+4\right ) } & =b\\ b & =\frac{s\left ( s+2\right ) }{6}-\frac{s}{\left ( s+4\right ) }\\ & =\frac{s\left ( s+2\right ) \left ( s+4\right ) -6s}{6\left ( s+4\right ) }\\ & =\frac{s\left ( s^{2}+6s+2\right ) }{6\left ( s+4\right ) } \end{align*}
This is true for \(\operatorname{Re}\left ( s\right ) >2\). Hence for \(s=2\) the above reduces to\begin{align*} b & =\frac{2\left ( 4+12+2\right ) }{6\left ( 2+4\right ) }\\ & =1 \end{align*}
Therefore (1) becomes\begin{align*} H\left ( s\right ) & =\frac{s+\left ( s+4\right ) }{s\left ( s+4\right ) \left ( s+2\right ) }\\ & =\frac{2s+4}{s\left ( s+4\right ) \left ( s+2\right ) }\\ & =\frac{2\left ( s+2\right ) }{s\left ( s+4\right ) \left ( s+2\right ) }\\ & =\frac{2}{s\left ( s+4\right ) } \end{align*}
Consider the system \(S\) characterized by the differential equation\[ y^{\prime \prime \prime }\left ( t\right ) +6y^{\prime \prime }\left ( t\right ) +11y^{\prime }\left ( t\right ) +6y\left ( t\right ) =x\left ( t\right ) \] (a) Determine the zero-state response of this system for the input \(x\left ( t\right ) =e^{-4t}u\left ( t\right ) \) (b) Determine the zero-input response of the system for \(t>0^{-}\) given the initial conditions \(y\left ( 0^{-}\right ) =1,\left . \frac{dy}{dt}\right \vert _{t=0^{-}}=-1,\left . \frac{d^{2}y}{dt^{2}}\right \vert _{t=0^{-}}=1\). (c) Determine the output of \(S\) when the input is \(x\left ( t\right ) =e^{-4t}u\left ( t\right ) \) and the initial conditions are the same as those specified in part (b).
Solution
Applying Laplace transform on the ODE and using zero initial conditions gives\begin{align} s^{3}Y\left ( s\right ) +6s^{2}Y\left ( s\right ) +11sY\left ( s\right ) +6Y\left ( s\right ) & =\frac{1}{s+4}\nonumber \\ Y\left ( s\right ) \left ( s^{3}+6s^{2}+11s+6\right ) & =\frac{1}{s+4}\nonumber \\ Y\left ( s\right ) & =\frac{1}{\left ( s+4\right ) \left ( s^{3}+6s^{2}+11s+6\right ) }\nonumber \\ & =\frac{1}{\left ( s+4\right ) \left ( s+1\right ) \left ( s+2\right ) \left ( s+3\right ) }\tag{1} \end{align}
Using partial fractions\[ \frac{1}{\left ( s+4\right ) \left ( s+1\right ) \left ( s+2\right ) \left ( s+3\right ) }=\frac{A}{\left ( s+4\right ) }+\frac{B}{\left ( s+1\right ) }+\frac{C}{\left ( s+2\right ) }+\frac{D}{\left ( s+3\right ) }\] Hence \begin{align*} A & =\left . \frac{1}{\left ( s+1\right ) \left ( s+2\right ) \left ( s+3\right ) }\right \vert _{s=-4}=\frac{1}{\left ( -4+1\right ) \left ( -4+2\right ) \left ( -4+3\right ) }=\allowbreak -\frac{1}{6}\\ B & =\left . \frac{1}{\left ( s+4\right ) \left ( s+2\right ) \left ( s+3\right ) }\right \vert _{s=-1}=\frac{1}{\left ( -1+4\right ) \left ( -1+2\right ) \left ( -1+3\right ) }=\allowbreak \frac{1}{6}\\ C & =\left . \frac{1}{\left ( s+4\right ) \left ( s+1\right ) \left ( s+3\right ) }\right \vert _{s=-2}=\frac{1}{\left ( -2+4\right ) \left ( -2+1\right ) \left ( -2+3\right ) }=\allowbreak \frac{-1}{2}\\ D & =\left . \frac{1}{\left ( s+4\right ) \left ( s+1\right ) \left ( s+2\right ) }\right \vert _{s=-3}=\frac{1}{\left ( -3+4\right ) \left ( -3+1\right ) \left ( -3+2\right ) }=\frac{1}{2} \end{align*}
Hence (1) becomes\begin{align*} Y\left ( s\right ) & =\frac{A}{\left ( s+4\right ) }+\frac{B}{\left ( s+1\right ) }+\frac{C}{\left ( s+2\right ) }+\frac{D}{\left ( s+3\right ) }\\ & =-\frac{1}{6}\frac{1}{\left ( s+4\right ) }+\frac{1}{6}\frac{1}{\left ( s+1\right ) }-\frac{1}{2}\frac{1}{\left ( s+2\right ) }+\frac{1}{2}\frac{1}{\left ( s+3\right ) }\qquad \operatorname{Re}\left ( s\right ) >-1 \end{align*}
From tables, the inverse Laplace transform is (one sided) is\[ y\left ( t\right ) =-\frac{1}{6}e^{-4t}u\left ( t\right ) +\frac{1}{6}e^{-t}u\left ( t\right ) -\frac{1}{2}e^{-2t}u\left ( t\right ) +\frac{1}{2}e^{-3t}u\left ( t\right ) \]
Applying Laplace transform on the ODE \(y^{\prime \prime \prime }\left ( t\right ) +6y^{\prime \prime }\left ( t\right ) +11y^{\prime }\left ( t\right ) +6y\left ( t\right ) =0\) and using the non-zero initial conditions given above gives\begin{align} \left ( s^{3}Y\left ( s\right ) -s^{2}y\left ( 0\right ) -sy^{\prime }\left ( 0\right ) -y^{\prime \prime }\left ( 0\right ) \right ) +6\left ( s^{2}Y\left ( s\right ) -sy\left ( 0\right ) -y^{\prime }\left ( 0\right ) \right ) +11\left ( sY\left ( s\right ) -y\left ( 0\right ) \right ) +6Y\left ( s\right ) & =0\nonumber \\ \left ( s^{3}Y\left ( s\right ) -s^{2}+s-1\right ) +6\left ( s^{2}Y\left ( s\right ) -s+1\right ) +11\left ( sY\left ( s\right ) -1\right ) +6Y\left ( s\right ) & =0\nonumber \\ s^{3}Y\left ( s\right ) -s^{2}+s-1+6s^{2}Y\left ( s\right ) -6s+6+11sY\left ( s\right ) -11+6Y\left ( s\right ) & =0\nonumber \\ Y\left ( s\right ) \left ( s^{3}+6s^{2}+11s+6\right ) -s^{2}+s-1-6s+6-11 & =0\tag{1} \end{align}
Hence\begin{align*} Y\left ( s\right ) \left ( s^{3}+6s^{2}+11s+6\right ) & =s^{2}-s+1+6s-6+11\\ Y\left ( s\right ) & =\frac{s^{2}+5s+6}{s^{3}+6s^{2}+11s+6}\\ & =\frac{\left ( s+3\right ) \left ( s+2\right ) }{\left ( s+1\right ) \left ( s+2\right ) \left ( s+3\right ) }\\ & =\frac{1}{s+1}\qquad \operatorname{Re}\left ( s\right ) >-1 \end{align*}
Hence the inverse Laplace transform (one sided) gives\[ y\left ( t\right ) =e^{-t}u\left ( t\right ) \]
This is the sum of the response of part(a) and part(b) since the system is linear ODE. Hence\begin{align*} y\left ( t\right ) & =-\frac{1}{6}e^{-4t}u\left ( t\right ) +\frac{1}{6}e^{-t}u\left ( t\right ) -\frac{1}{2}e^{-2t}u\left ( t\right ) +\frac{1}{2}e^{-3t}u\left ( t\right ) +e^{-t}u\left ( t\right ) \\ & =\left ( -\frac{1}{6}e^{-4t}+\frac{1}{6}e^{-t}-\frac{1}{2}e^{-2t}+\frac{1}{2}e^{-3t}+e^{-t}\right ) u\left ( t\right ) \\ & =\left ( -\frac{1}{6}e^{-4t}+\frac{7}{6}e^{-t}-\frac{1}{2}e^{-2t}+\frac{1}{2}e^{-3t}\right ) u\left ( t\right ) \end{align*}