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

Which I don't believe is right. Can someone please help me find the correct transfer function.

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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
    
Your comments are constructive and helpful –  Nick Oct 12 '12 at 17:42
    
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

1 Answer 1

up vote 1 down vote accepted

Your result is correct.

We can simplify the equation by cancelling the $R$s and dividing through by $V_i$, which yields

$$ \frac{\frac12-1}R+\frac{\frac12-V_0/V_i}{Z_f}=0 $$

with solution

$$ V_0/V_i=\frac12\left(1-\frac{Z_f}R\right)=\frac12\left(1-\frac{Z_c}{R+Z_c}\right)=\frac12\frac R{R+Z_c}\;, $$

which coincides with your result if we substitute $Z_c$ and multiply through by $\mathrm j\omega C$.

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