4.8 HW 8

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4.8.1 Problem 9.3

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 $\mathrm{Re}\left(s\right)>-3$ ?

solution

The Laplace transform is $$\begin{array}{rl}X\left(s\right)& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x\left(t\right)e^{-st}dt\\ & ={\int }_0^{\mathrm{\infty }}(e^{-5t}+e^{-\beta t})e^{-st}dt\\ & ={\int }_0^{\mathrm{\infty }}{e}^{-5t}{e}^{-st}dt+{\int }_0^{\mathrm{\infty }}{e}^{-\beta t}{e}^{-st}dt\\ & ={\int }_0^{\mathrm{\infty }}{e}^{-t(5+s)}dt+{\int }_0^{\mathrm{\infty }}{e}^{-t(\beta +s)}dt\\ & =\frac{1}{-(s+5)}{\left[e^{-t(5+s)}\right]}_0^{\mathrm{\infty }}-\frac{1}{\beta +s}{\left[e^{-t(\beta +s)}\right]}_0^{\mathrm{\infty }}\end{array}$$

For the first term $\frac{1}{-(s+5)}{\left[e^{-t(5+s)}\right]}_0^{\mathrm{\infty }}=\frac{1}{-(s+5)}[e^{-\mathrm{\infty }(5+s)}-1]$. For this term to converge we need $5+\mathrm{Re}\left(s\right)>0$ or $$\mathrm{Re}\left(s\right)>-5$$ For the second term, let $\beta =a+ib$ and let $s=\sigma +j\omega$, hence the second term becomes $$\begin{array}{rl}\frac{1}{\beta +s}{\left[e^{-t(\beta +s)}\right]}_0^{\mathrm{\infty }}& =\frac{1}{\beta +s}{\left[e^{-t((a+ib)+(\sigma +j\omega ))}\right]}_0^{\mathrm{\infty }}\\ & =\frac{1}{\beta +s}{\left[e^{-t(a+\sigma +j(b+\omega ))}\right]}_0^{\mathrm{\infty }}\\ & =\frac{1}{\beta +s}{\left[e^{-t(a+\sigma )}{e}^{-jt(b+\omega )}\right]}_0^{\mathrm{\infty }}\\ & =\frac{1}{\beta +s}[e^{-\mathrm{\infty }(a+\sigma )}{e}^{-j\mathrm{\infty }(b+\omega )}-1]\end{array}$$

The complex exponential terms always converges since its norm is bounded by $1$. For the real exponential term, we need $a+\sigma >0$ or $a+\mathrm{Re}\left(s\right)>0$ or $\mathrm{Re}\left(s\right)>-a$. Since we are told that $\mathrm{Re}\left(s\right)>-3$, then $$a=3$$ Is the requirement on real part of $\beta$. There is no restriction on the imaginary part of $\beta$.

4.8.2 Problem 9.9

Given that $e^{-at}u\left(t\right)⟺\frac{1}{s+a}$ for $\mathrm{Re}\left(s\right)>\mathrm{Re}(-a)$, determine the inverse Laplace transform of$$X\left(s\right)=\frac{2(s+2)}{s^2+7s+12}\mathrm{Re}\left(s\right)>-3$$ solution

Writing $X\left(s\right)$ as$$\begin{array}{rl}X\left(s\right)& =\frac{2(s+2)}{(s+4)(s+3)}\\ & =\frac{A}{(s+4)}+\frac{B}{(s+3)}\end{array}$$

Hence $A={\frac{2(s+2)}{(s+3)}|}_{s=-4}=\frac{2(-4+2)}{(-4+3)}=4$ and $B={\frac{2(s+2)}{(s+4)}|}_{s=-3}=\frac{2(-3+2)}{(-3+4)}=-2$, therefore the above becomes$$X\left(s\right)=\frac{4}{(s+4)}-\frac{2}{(s+3)}$$ Using $e^{-at}u\left(t\right)⟺\frac{1}{s+a}$ gives the inverse Laplace transform as$$\begin{array}{rl}x\left(t\right)& =4e^{-4t}u\left(t\right)-2e^{-3t}u\left(t\right)\\ & =(4e^{-4t}-2e^{-3t})u\left(t\right)\end{array}$$

With $\mathrm{Re}\left(s\right)>-4$ and also $\mathrm{Re}\left(s\right)>-3$. Therefore the ROC for both is $\mathrm{Re}\left(s\right)>-3$.

4.8.3 Problem 9.15

Consider the two right-sides signals $x\left(t\right),y\left(t\right)$ related through the differential equations $$\begin{array}{rl}\frac{dx\left(t\right)}{dt}& =-2y\left(t\right)+\delta \left(t\right)\\ \frac{dy\left(t\right)}{dt}& =2x\left(t\right)\end{array}$$

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$$\begin{array}{}\text{(1)}& sX\left(s\right)& =-2Y\left(s\right)+1\text{(2)}& sY\left(s\right)& =2X\left(s\right)\end{array}$$

Using the second equation in the first gives$$\begin{array}{rl}sX\left(s\right)& =-2\left(\frac{2X\left(s\right)}{s}\right)+1\\ & =\frac{-4X\left(s\right)+s}{s}\\ s^{2}X\left(s\right)& =-4X\left(s\right)+s\\ (s^2+4)X\left(s\right)& =s\\ X\left(s\right)& =\frac{s}{(s^2+4)}\end{array}$$

Using the above in (2) gives $Y\left(s\right)$$$\begin{array}{rl}sY\left(s\right)& =2\frac{s}{(s^2+4)}\\ Y\left(s\right)& =\frac{2}{(s^2+4)}\end{array}$$

Considering $X\left(s\right)$ to find its ROC, let us write it as$$X\left(s\right)=\frac{s}{(s^2+4)}=\frac{s}{(s+2j)(s-2j)}=\frac{A}{(s+2j)}+\frac{B}{(s-2j)}$$ We see that the ROC for first term is $\mathrm{Re}\left(s\right)>-\mathrm{Re}\left(2j\right)$ which means $\mathrm{Re}\left(s\right)>0$ since real part is zero. Same for the second term. Hence we see that for $X\left(s\right)$ the ROC is $\mathrm{Re}\left(s\right)>0$. Similarly for $Y\left(s\right)$. Therefore the overall ROC is $$\mathrm{Re}\left(s\right)>0$$

4.8.4 Problem 9.32

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)$$\begin{array}{rl}sH\left(s\right)+2H\left(s\right)& =\frac{1}{s+4}+\frac{b}{s}\\ H\left(s\right)(2+s)& =\frac{1}{s+4}+\frac{b}{s}\\ H\left(s\right)& =\frac{1}{(s+4)(s+2)}+\frac{b}{s(s+2)}\\ \text{(1)}& & =\frac{s+b(s+4)}{s(s+4)(s+2)}\end{array}$$

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$$\begin{array}{rl}Y\left(s\right)& =\frac{1}{6}\frac{1}{s-2}\mathrm{Re}\left(s\right)>2\\ X\left(s\right)& =\frac{1}{s-2}\mathrm{Re}\left(s\right)>2\end{array}$$

Hence $$\begin{array}{rl}H\left(s\right)& =\frac{Y\left(s\right)}{X\left(s\right)}\\ H\left(s\right)& =\frac{\frac{1}{6}\frac{1}{s-2}}{\frac{1}{s-2}}\\ \text{(2)}& & =\frac{1}{6}\end{array}$$

Comparing (1,2) then$$\frac{1}{6}=\frac{s+b(s+4)}{s(s+4)(s+2)}$$ Solving for $b$ gives$$\begin{array}{rl}\frac{s(s+4)(s+2)}{6}& =s+b(s+4)\\ \frac{s(s+4)(s+2)}{6(s+4)}-\frac{s}{(s+4)}& =b\\ b& =\frac{s(s+2)}{6}-\frac{s}{(s+4)}\\ & =\frac{s(s+2)(s+4)-6s}{6(s+4)}\\ & =\frac{s(s^2+6s+2)}{6(s+4)}\end{array}$$

This is true for $\mathrm{Re}\left(s\right)>2$. Hence for $s=2$ the above reduces to$$\begin{array}{rl}b& =\frac{2(4+12+2)}{6(2+4)}\\ & =1\end{array}$$

Therefore (1) becomes$$\begin{array}{rl}H\left(s\right)& =\frac{s+(s+4)}{s(s+4)(s+2)}\\ & =\frac{2s+4}{s(s+4)(s+2)}\\ & =\frac{2(s+2)}{s(s+4)(s+2)}\\ & =\frac{2}{s(s+4)}\end{array}$$

4.8.5 Problem 9.40

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 $y\left(0^{-}\right)=1,{\frac{dy}{dt}|}_{t=0^{-}}=-1,{\frac{d^{2}y}{d{t}^2}|}_{t=0^{-}}=1$. (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

4.8.5.1 Part a

Applying Laplace transform on the ODE and using zero initial conditions gives$$\begin{array}{rl}{s}^{3}Y\left(s\right)+6s^{2}Y\left(s\right)+11sY\left(s\right)+6Y\left(s\right)& =\frac{1}{s+4}\\ Y\left(s\right)(s^3+6s^2+11s+6)& =\frac{1}{s+4}\\ Y\left(s\right)& =\frac{1}{(s+4)(s^3+6s^2+11s+6)}\\ \text{(1)}& & =\frac{1}{(s+4)(s+1)(s+2)(s+3)}\end{array}$$

Using partial fractions$$\frac{1}{(s+4)(s+1)(s+2)(s+3)}=\frac{A}{(s+4)}+\frac{B}{(s+1)}+\frac{C}{(s+2)}+\frac{D}{(s+3)}$$ Hence $$\begin{array}{rl}A& ={\frac{1}{(s+1)(s+2)(s+3)}|}_{s=-4}=\frac{1}{(-4+1)(-4+2)(-4+3)}=-\frac{1}{6}\\ B& ={\frac{1}{(s+4)(s+2)(s+3)}|}_{s=-1}=\frac{1}{(-1+4)(-1+2)(-1+3)}=\frac{1}{6}\\ C& ={\frac{1}{(s+4)(s+1)(s+3)}|}_{s=-2}=\frac{1}{(-2+4)(-2+1)(-2+3)}=\frac{-1}{2}\\ D& ={\frac{1}{(s+4)(s+1)(s+2)}|}_{s=-3}=\frac{1}{(-3+4)(-3+1)(-3+2)}=\frac{1}{2}\end{array}$$

Hence (1) becomes$$\begin{array}{rl}Y\left(s\right)& =\frac{A}{(s+4)}+\frac{B}{(s+1)}+\frac{C}{(s+2)}+\frac{D}{(s+3)}\\ & =-\frac{1}{6}\frac{1}{(s+4)}+\frac{1}{6}\frac{1}{(s+1)}-\frac{1}{2}\frac{1}{(s+2)}+\frac{1}{2}\frac{1}{(s+3)}\mathrm{Re}\left(s\right)>-1\end{array}$$

From tables, the inverse Laplace transform is (one sided) is$$y\left(t\right)=-\frac{1}{6}{e}^{-4t}u\left(t\right)+\frac{1}{6}{e}^{-t}u\left(t\right)-\frac{1}{2}{e}^{-2t}u\left(t\right)+\frac{1}{2}{e}^{-3t}u\left(t\right)$$

4.8.5.2 Part b

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$$\begin{array}{rl}(s^{3}Y\left(s\right)-s^{2}y\left(0\right)-s{y}^{\prime }\left(0\right)-y^{\prime \prime }\left(0\right))+6(s^{2}Y\left(s\right)-sy\left(0\right)-y^{\prime }\left(0\right))+11(sY\left(s\right)-y\left(0\right))+6Y\left(s\right)& =0\\ (s^{3}Y\left(s\right)-s^2+s-1)+6(s^{2}Y\left(s\right)-s+1)+11(sY\left(s\right)-1)+6Y\left(s\right)& =0\\ s^{3}Y\left(s\right)-s^2+s-1+6s^{2}Y\left(s\right)-6s+6+11sY\left(s\right)-11+6Y\left(s\right)& =0\\ \text{(1)}& Y\left(s\right)(s^3+6s^2+11s+6)-s^2+s-1-6s+6-11& =0\end{array}$$

Hence$$\begin{array}{rl}Y\left(s\right)(s^3+6s^2+11s+6)& =s^2-s+1+6s-6+11\\ Y\left(s\right)& =\frac{s^2+5s+6}{s^3+6s^2+11s+6}\\ & =\frac{(s+3)(s+2)}{(s+1)(s+2)(s+3)}\\ & =\frac{1}{s+1}\mathrm{Re}\left(s\right)>-1\end{array}$$

Hence the inverse Laplace transform (one sided) gives$$y\left(t\right)=e^{-t}u\left(t\right)$$

4.8.5.3 Part c

This is the sum of the response of part(a) and part(b) since the system is linear ODE. Hence$$\begin{array}{rl}y\left(t\right)& =-\frac{1}{6}{e}^{-4t}u\left(t\right)+\frac{1}{6}{e}^{-t}u\left(t\right)-\frac{1}{2}{e}^{-2t}u\left(t\right)+\frac{1}{2}{e}^{-3t}u\left(t\right)+e^{-t}u\left(t\right)\\ & =(-\frac{1}{6}{e}^{-4t}+\frac{1}{6}{e}^{-t}-\frac{1}{2}{e}^{-2t}+\frac{1}{2}{e}^{-3t}+e^{-t})u\left(t\right)\\ & =(-\frac{1}{6}{e}^{-4t}+\frac{7}{6}{e}^{-t}-\frac{1}{2}{e}^{-2t}+\frac{1}{2}{e}^{-3t})u\left(t\right)\end{array}$$

4.8.6 key solution

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