Node Volt Analysis: Find \frac{V_o}{V_i}

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The discussion focuses on finding the ratio \(\frac{V_o}{V_i}\) in a node voltage analysis problem. The proposed solution is \(\frac{V_o}{V_i}=\frac{1}{s+2}\), but there is confusion regarding the derivation, particularly the presence of two 's' terms in the equation provided by the solution manual. Participants clarify that the variable \(V_o\) should not be treated as ground, and the bottom node voltage should be set to zero for consistency with standard practices. Correcting these misunderstandings is essential for achieving the accurate result. The thread emphasizes the importance of proper node voltage referencing in circuit analysis.
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Homework Statement



Find \frac{V_o}{V_i}

for

http://img685.imageshack.us/img685/8658/nodevoltage.png

Homework Equations



s=σ+jω, where σ=real component & jω=imaginary component

The Attempt at a Solution



Solution is
\frac{V_o}{V_i}=\frac{1}{s+2}

Solution manual says:
\frac{V_o-V_i}{s}+\frac{V_o}{s}+V_o=0

I don't know where that comes from. I don't understand why there are 2 s's
Shouldn't it be;

\frac{V_L-V_i}{R}+\frac{V_L-0}{Ls}+\frac{V_L-V_o}{R}=0

where V_L=V_o

so

\frac{V_o-V_i}{R}+\frac{V_o-0}{Ls}+\frac{V_o-V_o}{R}=0
\frac{V_o-V_i}{R}+\frac{V_o-0}{Ls}=0

That won't give me an s+2 in the denominator...
 
Last edited by a moderator:
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Cursed said:

Homework Statement



Find \frac{V_o}{V_i}

for

http://img685.imageshack.us/img685/8658/nodevoltage.png

Homework Equations



s=σ+jω, where σ=real component & jω=imaginary component

The Attempt at a Solution



Solution is
\frac{V_o}{V_i}=\frac{1}{s+2}

Solution manual says:
\frac{V_o-V_i}{s}+\frac{V_o}{s}+V_o=0

I don't know where that comes from. I don't understand why there are 2 s's
Shouldn't it be;

\frac{V_L-V_i}{R}+\frac{V_L-0}{Ls}+\frac{V_L-V_o}{R}=0

where V_L=V_o

...

You seem to be using V_o for ground, whereas the diagram shows V_o = V_L.
You're close, fix that up, & shoot again.
You should call ground (the bottom node voltage) zero like everybody else does! :smile:
 
Last edited by a moderator:

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