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Basic Elementry Circuit

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Wond3rboy

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Hi,

I am simulating some circuits from The Art of Electronics Student Guide.I simulated the LPF shown in figure,the problem is that it does not remove all high frequenies and i get a wiggle in the output. Can any one tell me why?
 

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Hi,

I am simulating some circuits from The Art of Electronics Student Guide.I simulated the LPF shown in figure,the problem is that it does not remove all high frequenies and i get a wiggle in the output. Can any one tell me why?

hi,
Its a simple filter, have you plotted the graph for your filter, and where down the slope of the graph does the 16KHz freq appear.??:)

Look here:
Analog Devices : Analog Dialogue : Phase Relations
 

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Last edited:
Hi,

Yes, the filter isnt sharp enough to cut the higher frequency out
altogether. You'll need to lower the cutoff frequency to get that
or else go to a better filter like with two resistors and two caps.
 
16Khz doesnt appear at the ouput as such.Its the distortion i get which i shouldnt be getting.I am uploading the file.



PS:Didnt get the link by Eric and also AL's Reply i will work at it and tell you thanks.
 

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16Khz doesnt appear at the ouput as such.Its the distortion i get which i shouldnt be getting.I am uploading the file.
PS:Didnt get the link by Eric and also AL's Reply i will work at it and tell you thanks.

I dont agree ref the 16KHz, count the number of 'distortion' blips you get over 1 cycle of the 1KHz, there are 16!!!.

Its the 16KHz thats getting thru your single pole filter and modulating the output.

The 'wiggle' is 16KHz.
 
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Thanks Two stages make it much better.Can i know why did i require two stages whereas in the book it is said to be done with a single stage?And also when i reduce the R of the first stage to 1k it doesnt help.But when i change iit to 10k(2 stages having same R) then i get a better output wavaform.
 
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Thanks Two stages make it much better.Can i know why did i require two stages whereas in the book it is said to be done with a single stage?

It really depends how much 'hf component' you can tolerate in the circuitry following the LPF.

Do you know the attenuation in db/octave for a single pole/element LPF compared to a two pole LPF.???

Please let me know what you calculate it to be.

Also once you know the attenuation of the single pole LPF you should be able to work out the attenuation of the 16KHz freq compared to the LPF and 1KHz freq.???:)
 
Here is how i calculate the attenuation to be:

There are 3 octaves between the f3db freq of my LPF.So the 16KHz signal should be attenuated to .125 its original value.For a 2 pole LPF i dont know how they add up but if i take it as voltage in series(two seperate stages) its going to be .125/8=.015. I know i got that wrong.My other question is that why is the signal of 1Khz being attenuated?It should pass unscathed shouldnt it?It decreases by nearly 4 times every stage.

PS: If i count decades than how does a signal attenuates per decade?
 
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My other question is that why is the signal of 1Khz being attenuated?It should pass unscathed shouldnt it?It decreases by nearly 4 times every stage.

hi,

The impedance of a capacitor is Xc= 1/(2 * pi * f *C)

So at 1KHz, Xc = 15.9KΩ [16KΩ]

As you have a 10K resistor connected to this 16K [Xc] it is acting as a potential divider between the input and output.
So, Vout~ = Vinp~ * (16K/(10K+16K)) = 0.61

I have ignored the effect of the 100K Rload !, it should be included for accuracy.The ideal Rload should be infinity.

Do you follow.?:)
 
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Hi again,

For a single stage RC low pass filter the
-3db point is at w=1/RC. At any frequency above this point the
attenuation is going to be even more, so if you tune 1/RC even lower
by making R bigger you effectively increase the attenuation to the
signal to be rejected so you'll see less of it in the output. Unfortunately
this means the -3db point also gets closer to the pass frequency so
you'll get some more attenuation there too.
A double filter does twice the work and has a steeper slope too,
so it's sharper and this means it doesnt affect the pass frequency as
much for the same amount of rejection of the noice frequency.

In the book they may have assumed that this would be good enough,
or that the input noise was not equal in amplitude to the signal as
you have shown in your circuit. Usually the noise is lower than the
signal and if that is really the case then maybe one stage would work
ok. It also depends how clean the output signal really has to be...
noise 1/10 th of the pass frequency output is ok or do you need 1/100 th
of the pass frequency output. This specification affects the requirements
of the filter greatly.

Also, how much attenuation of the signal frequency is acceptable.
This affects component values as well.

A very strong filter can be built with three stages, where the resistance
of the first stage starts at 1k and the second stage has 10k and the last
stage has 100k, and the caps go 0.1uf, 0.01uf, and 0.001uf.
This has lots of cut at 16kHz and the output at 1kHz is around 0.6 of
the input (not too bad) and you'll get a very clean 1kHz signal.
 
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Integrator and Differentiatior

Hi its me again and this time its the integrator and differentiator. They work fine although there is a wierd phase shift in the integrator.I want to know how do you fix values for R and C.And why is there not an exact integration of the sine wave?

Thanks.
 

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hi,

The impedance of a capacitor is Xc= 1/(2 * pi * f *C)

So at 1KHz, Xc = 15.9KΩ [16KΩ]

As you have a 10K resistor connected to this 16K [Xc] it is acting as a potential divider between the input and output.
So, Vout~ = Vinp~ * (16K/(10K+16K)) = 0.61

That's not completely true you forgot that Z is the length of the hypotenuse with a an angel of 45°.

Z = √(R²+Xc²)

Xc = 1/(2πFC)

Vout = Vin - R*Vin/Z

@16kHz
Vout = Vin * 0.707
 
We've already answered the question many times over, you just don't understand use.

I'll try again.

A filter doesn't stop all frequencies above the cut-off, a first order low pass filter, reduces the frequencies by 6dB per decade above the cut-off.
 
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