Engineering Control Systems Engineering - Circuits

[GreenPrint]

Homework Statement

Find the transfer function, G(s) = \frac{V_{0}(s)}{V_{i}(s)} for each network shown in figure P2.3. [Section 2.4]

http://imagizer.imageshack.us/v2/800x600q90/20/1ocq.png

Homework Equations

The Attempt at a Solution

When I try and solve this problem I Kirchhoff's Current Law and get

\frac{V_{O}(t) - V_{i}(t)}{R} + \frac{1}{L}∫_{0}^{t}V_{0}(\tau)d\tau + \frac{V_{0}(t)}{R} = 0

This doesn't really seem to help.

The solutions manual writes the node equation as

\frac{V_{0} - V_{i}}{s} + \frac{V_{0}}{s} + V_{0} = 0

and then solves this for

\frac{V_{0}}{V_{i}} = \frac{1}{s + 2}

I don't exactly see how they get the node equation they do. I know that the impedance of a resistor is just the resistance and that the impedance of the inductor is Ls. So to me it looks like it should be

\frac{V_{0} - V_{i}}{R} + \frac{V_{0}}{Ls} + \frac{V_{0}}{R} = 0
\frac{V_{0} - V_{i}}{1} + \frac{V_{0}}{1s} + \frac{V_{0}}{1} = 0
V_{0} - V_{i} + \frac{V_{0}}{s} + V_{0} = 0

I don't see were the \frac{V_{0} - V_{i}}{s} term comes from.

Thanks for any help.

 

[gneill]

It looks like a mistake in their solution; They have taken the first 1 Ω resistor (connected to Vs) and treated it as a 1H inductor. So either the circuit diagram is incorrect or their equation is.