- #1

arhzz

- 268

- 52

- Homework Statement
- Find the transfer function

- Relevant Equations
- Circuit analysis

Hello!

Consider this filter,and that I have to find the transfer function U2/U1 with the norm Ω= ωRC (also double fractions are not allowed)

Now I can see that that the resistor and capacitor left as well as right are parallel to each other. So simplifying that

## Z_1 = \frac{R}{jωRC +1} ## Where Z1 is the Impedance on the left.

## Z_2 = \frac{R}{jωR199C +1} ## Where Z2 the impedance on the middle, L is irrelevant since no current is flowing there ( the circuit is open)

Now I can simplify with the norm and use the voltage divider to find the transfer function (T)

## T(Ω) = \frac{Z_2}{Z_1+Z_2} ## Inputting this gives me a double fraction

Now since I cannot have a double fraction I multiply everything by jΩ199 +1 and I am left with this

##\frac{R}{\frac{Rj199Ω+R }{jΩ+1 }+R } ## Now I multiply by jΩ +1 and I am left with this

##\frac{R(jΩ +1)}{R(jΩ199+1)+R(jΩ +1)} ##

So the final soultion should be ## \frac{1+jΩ}{jΩ200 +2}##

Now I am not so sure that this is correct.The reason why is because the next part of the problem they want me to find out how the transfer function looks like when Ω = 0,and what type of filter this is(low pass high pass etc..). I get that it should be 1/2 but I am not so sure that this is valid.I think I am susposed to get 1,because a low pass filter has transfer function 1 (ideally) for when the frequency is really low (hence really close to 0).I am not familiar with a filter that has a transfer function 1/2 at Ω = 0. Is it possible that I made a mistake somewhere in the transfer function; I've checked it thorughly and I think its okay,that would mean that I am missing something.

Any insights? Thanks

Consider this filter,and that I have to find the transfer function U2/U1 with the norm Ω= ωRC (also double fractions are not allowed)

Now I can see that that the resistor and capacitor left as well as right are parallel to each other. So simplifying that

## Z_1 = \frac{R}{jωRC +1} ## Where Z1 is the Impedance on the left.

## Z_2 = \frac{R}{jωR199C +1} ## Where Z2 the impedance on the middle, L is irrelevant since no current is flowing there ( the circuit is open)

Now I can simplify with the norm and use the voltage divider to find the transfer function (T)

## T(Ω) = \frac{Z_2}{Z_1+Z_2} ## Inputting this gives me a double fraction

Now since I cannot have a double fraction I multiply everything by jΩ199 +1 and I am left with this

##\frac{R}{\frac{Rj199Ω+R }{jΩ+1 }+R } ## Now I multiply by jΩ +1 and I am left with this

##\frac{R(jΩ +1)}{R(jΩ199+1)+R(jΩ +1)} ##

So the final soultion should be ## \frac{1+jΩ}{jΩ200 +2}##

Now I am not so sure that this is correct.The reason why is because the next part of the problem they want me to find out how the transfer function looks like when Ω = 0,and what type of filter this is(low pass high pass etc..). I get that it should be 1/2 but I am not so sure that this is valid.I think I am susposed to get 1,because a low pass filter has transfer function 1 (ideally) for when the frequency is really low (hence really close to 0).I am not familiar with a filter that has a transfer function 1/2 at Ω = 0. Is it possible that I made a mistake somewhere in the transfer function; I've checked it thorughly and I think its okay,that would mean that I am missing something.

Any insights? Thanks

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