# Resistance and Capacitance effect on bits transfer

I want to transfer a sequence of bits (say 1110) from one end of a wire and then receive at the other end of the wire. Both ends of the wire would be having a microcontroller.

Now I want to know the effect a resistor and a capacitor will have on the bit transfer, compared to them not being present at all.

I know that a wire itself has its own resistance, but I just want to know what effect will a true resistor have.

Thanks

• What about 0 volts? – Andy aka Mar 1 '17 at 14:51

I'm assuming you mean an in series resistor followed by a capacitor to ground. This will act as a low pass filter and will increase the rise/fall times of the digital signal. This will make it less noisy and help reduce reflection effects by impedance matching your input/output to the line.

Too much though and you will no longer have a decipherable digital signal.

• The resistor and capacitor both won't be present together. Once there would be a wire alone, then the second time a resistor and then only a capacitor. And yes, the components would be in series alone. – Samurai Mar 1 '17 at 15:09
• A capacitor alone in series would simply decouple the signal making it a +/- signal. No idea why you would want to do that for a digital signal. – Trevor_G Mar 1 '17 at 15:22
• Here is a useful article on the subject marvintest.com/KnowledgeBase/KBArticle.aspx?ID=196 – Trevor_G Mar 1 '17 at 15:22
• I just want to know what effect a resistor or a capacitor will have on bit transfer. Need to show whether there would be a delay in the bit transfer due to the presence of the components. That's all. – Samurai Mar 1 '17 at 15:51

This would be a model for your situation. R would be the wire resistance, C would be the gate capacitance on the reception side:

These would be the input/output waveforms:

The rising/falling times increase can end up degrading the voltage levels outside of the acceptable limits of a logical 0 or 1.