Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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 $\mathrm j\omega C$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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