De Wikilibros, la colección de libros de texto de contenido libre.
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{\displaystyle {\mathcal {L}}[x'']=s^{2}X(s)-sx(t_{0})-x'(t_{0})=s^{2}X(s)-s-1}
L
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{\displaystyle {\mathcal {L}}[x']=sX(s)-x(t_{0})=sX(s)-1}
Reemplazando en la ecuación
L
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4
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3
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L
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e
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{\displaystyle {\mathcal {L}}[x''+4x'+3x]={\mathcal {L}}[e^{-t}]}
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1
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X
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{\displaystyle s^{2}X(s)-s-1+4sX(s)-4+3X(s)={\frac {1}{s+1}}}
X
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{\displaystyle X(s)(s^{2}+4s+3)-s-5={\frac {1}{s+1}}}
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{\displaystyle X(s)(s^{2}+4s+3)={\frac {1+(s+5)(s+1)}{s+1}}}
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{\displaystyle X(s)={\frac {1+(s+5)(s+1)}{(s+1)(s^{2}+4s+3)}}}
X
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5
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s
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{\displaystyle X(s)={\frac {1+(s+5)(s+1)}{(s+1)(s+1)(s+3)}}}
X
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1
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s
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5
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s
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2
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3
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=
A
2
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s
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1
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2
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A
1
s
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1
+
B
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3
{\displaystyle X(s)={\frac {1+(s+5)(s+1)}{(s+1)^{2}(s+3)}}={\frac {A_{2}}{(s+1)^{2}}}+{\frac {A_{1}}{s+1}}+{\frac {B}{s+3}}}
Los valores de
A
2
{\displaystyle A_{2}}
y de
B
{\displaystyle B}
se pueden obtener directamente:
A
2
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1
+
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s
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5
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1
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3
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1
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1
2
{\displaystyle A_{2}=\left[{\frac {1+(s+5)(s+1)}{s+3}}\right]_{s=-1}={\frac {1}{2}}}
B
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2
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4
{\displaystyle B=\left[{\frac {1+(s+5)(s+1)}{(s+1)^{2}}}\right]_{s=-3}=-{\frac {3}{4}}}
Para obtener el valor de
A
1
{\displaystyle A_{1}}
pueden utilizarse estos valores obtenidos y resolver el sistema, o puede utilizarse la derivada:
A
1
=
d
d
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1
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1
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s
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s
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1
=
[
1
+
3
(
s
+
3
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2
]
s
=
−
1
=
7
4
{\displaystyle A_{1}={\frac {d}{ds}}\left[{\frac {1+(s+5)(s+1)}{s+3}}\right]_{s=-1}=\left[1+{\frac {3}{(s+3)^{2}}}\right]_{s=-1}={\frac {7}{4}}}
X
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1
2
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s
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2
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7
4
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{\displaystyle X(s)={\frac {\frac {1}{2}}{(s+1)^{2}}}+{\frac {\frac {7}{4}}{s+1}}+{\frac {\frac {3}{4}}{s+3}}}
L
−
1
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X
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x
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t
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1
2
t
e
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7
4
e
t
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3
4
e
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t
{\displaystyle {\mathcal {L}}^{-1}[X(s)]=x(t)={\frac {1}{2}}te^{-t}+{\frac {7}{4}}e^{t}-{\frac {3}{4}}e^{-3t}}