Recent content by Lucille

Thank you so so so much! It makes so much more sense to me now.

Whoops - I got 0.109

It is for a low pass filter -- and so I used the value for C = 1.45 * 10^-8 F and subbed it into the equation for the gain of a low pass filter

if you don't mind could you help me with this question? https://www.physicsforums.com/threads/bandpass-filter.799647/

** i meant 5 E-12

Sorry, I meant that the new capacitance that I was solving for is 4.99*10^-10 F

Wait in part a : L = 1.24 * 10 ^-6 H So that changes the capacitance to 4.99*10^-10 F

Alright: so for b then I would have to divide by 4 to get their individual capacitance so 3.125*10^-17 F

So then: for a) Z = 497 ohms and finding the inductance: Z = jwL so L = 1.24 H b) L = 1.24 H so Z = jwL = 2Pi(128 MHz)(1.24) = 1/(jwC) --> C = 1.25*10^-16 F

Z_c = 1/(jwC) But we're asked to find the capacitance, so it would be 2 unknowns?

Homework Statement A simple way to build a bandpass filter is to filter the output of an RC highpass filter with an RC lowpass filter, as shown on the left in the diagram below. The isolation buffer is a circuit element that keeps the two circuits isolated so they behave as they would on their...

how am i to find that?

Oh, okay. So then for b: Z = 497 = 4/[C(128MHz)(2pi)] ---> C = 4/[(497)(2pi)(128MHz)] = 10 pF

Okay, so: a) Z = 497 H b) I'm a little confused as to how to solve for the capacitance if the impedance is different?

Wait for b: would it be C = 4/[(3125 ohms)(128 Mhz) = 1*10^-11 C