Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a very simple low-pass filter with an AC source a capacitor and a resistance in series. Now I set C and R so that R*C=100 but when I plot the diagram whit db(out), it looses 3dB at 156Hz instead of 100Hz.

Vin  in    0  DC 0.0  AC 1.0 0.0
C1   out 0   1uF
R1   in     out   1kOhm

ac dec 50 0.1Hz 100kHz

I'm just starting with this things, can someone help me?

share|improve this question
RC is a time constant, not a frequency. In your case: tau = RC = (1 uF)(1 kohm) = 1 ms. This time constant doesn't correspond to the cutoff frequency directly, only to the point where the signal reaches 1-1/e or 63% in magnitude. The cutoff frequency is where your signal is at 50%. This is at 1/(2π*tau). – tyblu Dec 4 '10 at 2:41
yeah I meant 1/RC thanks, was missing the conversion w=2*pi*f anyway – gurghet Dec 4 '10 at 7:19
up vote 6 down vote accepted

The 3db cut-off frequency is \$\dfrac{1}{2 \cdot \pi \cdot RC}\$ = 159 Hz, which is quite close to your 156 Hz.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.