- #1

yungman

- 5,618

- 225

[tex]\frac {s+a}{(s+a)^2+b^2}[/tex]

But here is my work, I can't get rid of the -ve sign. I must be too blind to see the obvious, please help:

[tex]\overline{y(s)}=\int^∞_0 e^{-(s+a)t}\cos (bt) dt\;=\; \frac 1 b [e^{-(s+a)t}\sin(bt)]^∞_0\;+\frac {s+a}{b}\int^∞_0\;e^{-(s+a)t}\sin (bt) dt[/tex]

[tex][e^{-(s+a)t}\sin(bt)]^∞_0 =0 \;\Rightarrow\; \overline{y(s)}= \frac {s+a}{b}\int^∞_0\;e^{-(s+a)t}\sin (bt) dt\;=\;-\frac {s+a}{b^2} [e^{-(s+a)t}\cos(bt)]^∞_0\;-\frac {(s+a)^2}{b^2}\int^∞_0\;e^{-(s+a)t}\cos (bt) dt[/tex]

[tex][e^{-(s+a)t}\cos(bt)]^∞_0 =1 \;\Rightarrow\; \overline{y(s)}= -\frac {s+a}{b^2}-\frac {(s+a)^2}{b^2}\int^∞_0\;e^{-(s+a)t}\cos (bt) dt\;=\;-\frac {s+a}{b^2}-\frac {(s+a)^2}{b^2}\overline{y(s)}[/tex]

[tex]\Rightarrow\;\left(1+\frac {(s+a)^2}{b^2}\right)\overline{y(s)}\;=\;-\frac {s+a}{b^2}\;\Rightarrow\; \overline{y(s)}\;=\; -\frac {s+a}{(s+a)^2+b^2}[/tex]

You see the -ve sign? Please tell me what I did wrong.

Thanks