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Lowpass filter with no inductance, and voltage output over a resistor 
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#1
Apr2212, 01:28 AM

P: 7

1. The problem statement, all variables and given/known data
I am supposed to find the circuit design for a firstorder lowpass filter. The filter must be made of resistors and capacitors only (no inductances). Furthermore, the output voltage must be measured over a resistor or something in parallel with a resistor. 2. Relevant equations Impedance for a capacitor is 1/[jwC], for a resistor is R Voltage divider for impedances 3. The attempt at a solution i) Attempt at following those constraints: A resistor R_{1} in series with the signal source and in series with (a resistor R_{2} in parallel with a capacitor C). Note: I realize this is not a strictly lowpass filter. Impedance over the series resistor z_{1} = R_{1} Impedance over the parallel component z_{2} = R_{2}/[jwCR_{1} + 1] Equivalent impedance z = (R_{1} + R_{2})[jwCR_{1}R_{2}(1/[R_{1}+R_{2}) + 1]/[jwCR_{2}+1] As far as I can tell, this is actually a bandpass (or reject) filter with cutoff frequencies at (R1 + R2)/(C R1 R2) and 1/(C R2). ii) I realize I can make an RC lowpass filter (but then the output voltage is not measured over a resistor) and an RL lowpass filter (but then there is an inductor). Please help. Or, if you know definitively that this is an impossible task, please tell me so. Thanks. 


#2
Apr2212, 05:19 AM

HW Helper
Thanks
PF Gold
P: 4,757




#3
Apr2212, 06:49 AM

HW Helper
Thanks
P: 5,141

You can take the output as the voltage across that R∥C combo, or alternatively you could break that R into 2 and take output from across its lower portion. 


#4
Apr2212, 07:25 AM

P: 7

Lowpass filter with no inductance, and voltage output over a resistor
That was a typo:
z2 should have been R2/(jwcR2 + 1) But really, I forgot to take: z2/(z1+z2) Now I have R1/(R1 + R2)/[jwcR1R2(1/[R1+R2]) + 1] with cutoff frequency (R1 + R2)/(R1 R2) Thanks! 


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