0
\$\begingroup\$

For t > 0 find v0(t)

enter image description here

Using nodal analysis I was found v1 and v2, take difference and find correct result

But I can’t make same with direct KCL, KVL approach

enter image description here

May anyone help with that?

\$\endgroup\$
11
  • \$\begingroup\$ That was fixed but result still incorrect \$\endgroup\$
    – MaxMil
    Dec 29, 2016 at 9:00
  • \$\begingroup\$ Is \$8(e^{-t}-e^{-6t})\$ incorrect? \$\endgroup\$
    – Chu
    Dec 29, 2016 at 9:09
  • \$\begingroup\$ That is correct! But my in MathCAD "v0(s) invlaplace -> ..." is not equal to your calculations, but why? \$\endgroup\$
    – MaxMil
    Dec 29, 2016 at 9:14
  • \$\begingroup\$ Ask MathCAD!!!!! \$\endgroup\$
    – Chu
    Dec 29, 2016 at 9:15
  • \$\begingroup\$ Are you find result handy? \$\endgroup\$
    – MaxMil
    Dec 29, 2016 at 9:16

1 Answer 1

0
\$\begingroup\$

Your equation for \$V_o(s)\$ may be factorised to: $$\frac{40}{(s+1)(s+6)}=\frac{8}{s+1}-\frac{8}{s+6}$$

Giving: $$v_o(t)=8(e^{-t}-e^{-6t})$$

The MathCAD answer is also correct (expand it by hand to see), but not in as compact/useful a form.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.