# Transfer function Algebra

I have an equation that was derived from a schematic in order to find the transfer function of a filter (electrical engineering) but my algebra is not agreeing with the actual result.

The transfer function is defined as: $\frac{V_o}{V_i} = H(\omega)$

The current equation I have:

$$\frac{(Vi(\frac {R}{(R+R)}) - V_i)}{R} +\frac{(Vi(\frac {R}{(R+R)}) - V_0) }{ Z_f} = 0$$

where $Z_f$ is equal too: $$Z_f = \frac {RZ_c}{(R+Z_c)}$$ and $Z_c = \frac {1}{(j\omega c)}$

My result was $$\frac {V_o}{V_i} =\frac {(1/2 j \omega RC)}{(j \omega Rc +1)}$$

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What you call an equation is just an expression. There's currently no equation in the question to base a solution on. – joriki Oct 12 '12 at 14:43
Sorry forgot to add $= 0$ – Nick Oct 12 '12 at 14:48
I don't see a question. (And yes, your result is wrong.) – joriki Oct 12 '12 at 15:49
I'm very sorry; I'd made a mistake. Your result is actually correct; I've added an answer with a derivation. – joriki Oct 12 '12 at 17:59

We can simplify the equation by cancelling the $R$s and dividing through by $V_i$, which yields
$$\frac{\frac12-1}R+\frac{\frac12-V_0/V_i}{Z_f}=0$$
$$V_0/V_i=\frac12\left(1-\frac{Z_f}R\right)=\frac12\left(1-\frac{Z_c}{R+Z_c}\right)=\frac12\frac R{R+Z_c}\;,$$
which coincides with your result if we substitute $Z_c$ and multiply through by $\mathrm j\omega C$.