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Design a low-pass RC filter

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Design a low-pass RC filter with a break frequency of 30MHz. limit the value of the resistor to less than 1MΩ, and the value of the capacitor in the range of µF to pF. Draw the ideal bode plot (Av (dB) vs f)
this is the question, I kind of didn't get the question.
I tried to use the formula f2=1/2piRC
how would I use the formula to figure out the numbers...like use 1Mohm for R and rang of µF to pF ofr capacitor...and for y-axis it shud be gain in dB...so shud I increase by 20 db each time...very confused...
plz if any2 can help it will be great..thanks
will the graph on 4th quadrant...going from 0 db to -20 db, -40 db, -60db
 
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f=1/(2piRC), R=1/(2piCf), C=1/(2piRf). R in ohms and C in Farads.
It is a very simple RC filter. Its response drops 3dB per octave or 20dB per decade.
 
The bode plot is a flat line then a downward curve at its cutoff frequency into a straight line that drops at -6dB per octave which is -20dB per decade.
 
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The bode plot is a flat line then a downward curve at its cutoff frequency into a straight line that drops at -6dB per octave which is -20dB per decade.
thanks...I was thinking sumthing like tht for low-pass..goes from 0 to negative on the 4th quadrant...
 
thanks for answerring the question...really appricicate..
This is another similar question...
Build the graph Av vs frequency if C1 = C = 0.002 μF, R1 = 22 kΩ and R = 2.2 k Ω. first calculate the break frequencies for the two low-pass and high-pass filters and then plot an approximate graph.
I calculated the break frequency which is also called the cut-off frequency FC=1/2πRC=1/ (2π) (2.2kΩ)(0.002µF)=36.17kHz
FC=1/2πR1C1=1/ (2π) (22kΩ)(0.002µF)=3.61kHz
I am not sure which one is high-pass and which one is the low-pass filter break frequency..
 
You forgot to attach a schematic so we are just guessing.
A lowpass filter has a series resistor feeding a capacitor to ground.
A highpass filter has a series capacitor feeding a resistor to ground.

LOWPASS means it passes low frequencies and attenuates high frequencies. The 36.17kHz is probably the cutoff for a lowpass filter.
HIGHPASS means it passes high frequencies and attenuates low frequencies. The 3.61kHz is probably the cutoff for a highpass filter.
 
Please attach your schematic to your post HERE instead of posting it over at tinypic.

You have the highpass filter first so the total circuit is not a bandpass filter.

The 2nd filter loads down the 1st filter so the 1st filter's cutoff frequency is wrong.

The lowpass and highpass filters are very simple so they both reduce the output near 10kHz.
 
Please attach your schematic to your post HERE instead of posting it over at tinypic.

You have the highpass filter first so the total circuit is not a bandpass filter.

The 2nd filter loads down the 1st filter so the 1st filter's cutoff frequency is wrong.

The lowpass and highpass filters are very simple so they both reduce the output near 10kHz.

thanks man...so how I find the cuttoff frequency for the 1st filter which is the high pass filter? dnt I have to use the formula fc=1/2piRC
 
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The formula doesn't work properly because the lowpass filter loads down the output of the highpass filter.
Usually there is a buffer (an opamp) in between so there is no loading.
 
ok..got it...
ok lets sayIf we decrease the value of R1 to abt 1kΩ, so this will make the high-pass filter break frequency to increase...or wont affect the filter?
 
If you decrease R1 to 1k ohms then the cutoff frequency of the lowpass filter increases.
Look at the load on the output of the highpass filter that would change its frequency. Its load is nothing in my circuit.

The signal source must be able to drive 1k ohms.

The calculation of the cutoff frequency uses R1 plus the output impedance of the signal source.
 
I think the simple filter is schoolwork where it is correct or it is wrong. Maybe the teacher does not know that a buffer is required for it to be correct.
 
Hi,


If you want to make a two stage totally passive filter behave similar to a two stage with a unity gain amplifier in between (to achieve impedance isolation) then the usual idea is to make the impedance of the second stage 10 times the impedance of the first stage, but keep the dual single stage break frequency the same.

This is very easy to achieve because the break frequency is proportional to R and C, and to increase the impedance of the second stage we increase R. This also leads to decreasing the second cap value since we increased the second R.

So for a passive circuit with R1, C1 as the first stage and R2, C2 as the second stage, the values for the second stage are simply:
R2=10*R1, and
C2=C1/10

This makes the break frequencies of both sections the same but keeps the second stage impedance 10 times higher than the first.

This really works for any ratio not just 10, but 10 is considered to be a good minimum and anything higher than that is great.

For a bandpass however, there is also the thought about the bandwidth and that means we might want to shift the high pass section break frequency a little higher so we get wider bandwidth (if that's what we really want). This would simply mean changing the values of the second stage (HP) so that the break frequency is a little higher, and the best way to do this would be to lower the value of C2.

Of course we also have to think about the input driving impedance and the output load impedance. The input drive impedance acts as a series impedance that would be in series with R1 (R1 and C1 make up the LP section), and the output load impedance acts in parallel with R2.
 
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