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# after convert the pulse waves to the sine waves with RC filters, how can I remove DC offset?

#### goatmxj666

##### Member level 2
Hello,

I made a 4 RC filter stages to convert pulse waves to sine waves.

pulse source: 0V ~ 1.8V, f= 2.45GHz, duty cycle= 50%

r= 600ohm, c= 100fF, load= port(resistance= 1M ohm)

to get -Vout ~ +Vout, I add AC coupling capacitor(C17) between 4th capacitor and load.

However, the voltage at both ends of C17 shows the same result...

I think it's because the output voltage is like a rectified voltage.

How can I get a sine wave with the DC offset removed?

Solution
The circuit should work but as Klaus pointed out your simulation scale is wrong. C17 will take 5 seconds to charge through your 1M load. Only once C17 is charged will you see a voltage drop across it since it is a short to the high frequencies that you are measuring. The 0.9V that you see is the voltage across the last stage of your filter. In order for C17 to block that DC voltage it needs time to charge. And the time constant is 1uF X 1Mohm.
Hi,

ever heard of tau? or cut off frequency?
If not, you need to learn about it. Do an internet search for "basic analog filter tutorial", or "basicRC filter tutorial" or similar

Tau = R x C
fc = 1 / (2 x Pi x R x C)

If your tau = 1M x 1uF= 1s but your diagram also should show 1s (or in this magnitude) but it shows only 6ns.
So it´s a factor of 160.000.000 off!

Btw: your schematic is inclomplete, and it doesn´t even show the values of the parts.
Also "r= 600ohm, c= 100fF, load= port(resistance= 1M ohm)" is meaningless as long as you don´t tell us what "r" and what "c" you refer to.

Klaus

How can I get a sine wave with the DC offset removed?
There's the tactic of applying a DC voltage to the opposite end of the load.

Your sine wave appears to be riding 0.9 VDC component. Therefore apply 0.9 VDC to the other end of the load.

Or if your power supply is 1.8V, stack two capacitors in series between the supply rails. This creates 0.9V at the join.

Question, is this only a problem of unsuitable simulation startup, or a real circuit design problem. I guess the former. If it's however a real circuit design problem, we should know the conditions and requirements.

You could DC-block the square wave with a much bigger
cap (than is used in the filter) beforehand.

Simulation problem is that initial transient solution is performed with 0 V voltage source but average jumps to 0.9 V after start. To avoid the discontinuity, the pulse generator should have an initial voltage of 0.9 V which isn't possible with a simple square wave generator unfortunately.

Often in real circuits you will have to apply some initialization "preamble" to get to realistic steady state. That chews time but is just "what it takes".

The circuit should work but as Klaus pointed out your simulation scale is wrong. C17 will take 5 seconds to charge through your 1M load. Only once C17 is charged will you see a voltage drop across it since it is a short to the high frequencies that you are measuring. The 0.9V that you see is the voltage across the last stage of your filter. In order for C17 to block that DC voltage it needs time to charge. And the time constant is 1uF X 1Mohm.