- #1
photonsquared
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Find v(t) at t=800ms for the circuit in Figure 1.
Ans: 802mV
Writing a single node equation we have
[tex]\frac{v(t)-2tu(t)}{5}+0.1\frac{dv}{dt}=0.[/tex]
Taking the Laplace transform we have
[tex]L\left\{ \frac{v(t)-2tu(t)}{5}+0.1\frac{dv}{dt}=0\right\} .[/tex]
[tex]tu(t)\Rightarrow\frac{1}{s^{2}}[/tex]
[tex]\frac{df}{dt}\Rightarrow sF(s)-f(0^{-})[/tex]
[tex]\frac{V(s)}{5}-\frac{2}{5s^{2}}+0.1sV(s)-0.1v(0^{-})=0[/tex]
Assume that [tex]v(0^{-})=0[/tex] we have
[tex]\frac{V(s)}{5}-\frac{2}{5s^{2}}+0.1sV(s)=0[/tex]
multiplying through by 10 and combing like terms
[tex]V(s)\left(2+s\right)=\frac{4}{s^{2}}[/tex]
Solving for V(s) we have
[tex]V(s)=\frac{4}{s^{2}(s+2)}[/tex]
Applying the method of residues we have
[tex]\frac{4}{s^{2}(s+2)}=\frac{A}{s^{2}}+\frac{B}{s+2}[/tex]
Multiplying throught by [tex]s^{2}[/tex] we have
[tex]\frac{4}{s+2}=A+\frac{Bs^{2}}{s+2}[/tex]
[tex]A=\frac{4-Bs^{2}}{s+2}\mid_{s=0}=2[/tex]
Multiplying through by s+2 we have
[tex]\frac{4}{s^{2}}=\frac{A(s+2)}{s^{2}}+B[/tex]
[tex]B=\frac{4-A(s+2)}{s^{2}}\mid_{s=-2}=1[/tex]
Substituting A and B back into the equation we have
[tex]V(s)=\frac{2}{s^{2}}+\frac{1}{s+2}[/tex]
Applying the known Laplace transform pairs we have
[tex]v(t)=2t+e^{-2t}[/tex]
[tex]v(800ms)=2(0.8)+e^{-2(0.8s)}=1.802V[/tex]
Ans: 802mV
Writing a single node equation we have
[tex]\frac{v(t)-2tu(t)}{5}+0.1\frac{dv}{dt}=0.[/tex]
Taking the Laplace transform we have
[tex]L\left\{ \frac{v(t)-2tu(t)}{5}+0.1\frac{dv}{dt}=0\right\} .[/tex]
[tex]tu(t)\Rightarrow\frac{1}{s^{2}}[/tex]
[tex]\frac{df}{dt}\Rightarrow sF(s)-f(0^{-})[/tex]
[tex]\frac{V(s)}{5}-\frac{2}{5s^{2}}+0.1sV(s)-0.1v(0^{-})=0[/tex]
Assume that [tex]v(0^{-})=0[/tex] we have
[tex]\frac{V(s)}{5}-\frac{2}{5s^{2}}+0.1sV(s)=0[/tex]
multiplying through by 10 and combing like terms
[tex]V(s)\left(2+s\right)=\frac{4}{s^{2}}[/tex]
Solving for V(s) we have
[tex]V(s)=\frac{4}{s^{2}(s+2)}[/tex]
Applying the method of residues we have
[tex]\frac{4}{s^{2}(s+2)}=\frac{A}{s^{2}}+\frac{B}{s+2}[/tex]
Multiplying throught by [tex]s^{2}[/tex] we have
[tex]\frac{4}{s+2}=A+\frac{Bs^{2}}{s+2}[/tex]
[tex]A=\frac{4-Bs^{2}}{s+2}\mid_{s=0}=2[/tex]
Multiplying through by s+2 we have
[tex]\frac{4}{s^{2}}=\frac{A(s+2)}{s^{2}}+B[/tex]
[tex]B=\frac{4-A(s+2)}{s^{2}}\mid_{s=-2}=1[/tex]
Substituting A and B back into the equation we have
[tex]V(s)=\frac{2}{s^{2}}+\frac{1}{s+2}[/tex]
Applying the known Laplace transform pairs we have
[tex]v(t)=2t+e^{-2t}[/tex]
[tex]v(800ms)=2(0.8)+e^{-2(0.8s)}=1.802V[/tex]