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What happens when I have a low pass filter with R = 1kΩ C = 0.1 µF and I input a square signal with different frequencies? Will the time constant be the same for all frequencies?

τ = 100 µs for 1 kHz but when I increase the frequency τ starts to decrease on my oscilloscope. I thought τ was constant regardless of the frequency.

oscilloscope screenshot

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  • \$\begingroup\$ What do you mean, "τ starts to decrease on my oscilloscope"? How can you tell? \$\endgroup\$ – Dave Tweed Nov 10 '18 at 22:06
  • \$\begingroup\$ How do you measure \$\tau\$ on your scope? \$\endgroup\$ – Arsenal Nov 10 '18 at 22:06
  • \$\begingroup\$ To calculate the time constant I do this: Delta C = C2-C1=487.67- (-481.99)=969.66 0.63(969.66)=610.88 610.88-487.67=123.21 the I placed C2 on approximately 123.21 and read dT which is 100us \$\endgroup\$ – guasabito Nov 10 '18 at 22:20
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A 1 kHz square wave has a total period of 1000 us. With a low pass filter having a time constant of 100 us, there is enough time for the output capacitor to charge to the full value of the peak of the square wave. In general 4 time constants (or 400 us in this case) is sufficient. If you increase the frequency of the input square wave, eventually there will not be enough time for the output capacitor to charge to the full value. For example, if you input 5 kHz, which has a total period of only 200 us, the output capacitor will not be fully charged before the square wave changes state. Thus the output level will decrease as the input frequency increases. The filter time constant has not changed but the spectrum of the input signal compared to the cutoff frequency of the filter has changed. A 5 kHz square wave has most of its spectral components (every odd harmonic) above the filter cutoff frequency. Thus the filter is doing what it is supposed to do. The output wave shape will always have the shape of a 100 us time constant exponential rise, but as the frequency increases you will see less and less of the total rise.

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