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When I was trying to see the effect of load at the output of an RC low-pass filer I found myself in some sort of confusion. Below is an RC low-pass filter circuit with another R as a load and its Thevenin equivalent:

enter image description here

LTspice simulation in time domain shows that these circuits are giving same outputs at any frequency. Therefore I thought the cut-off frequency of the Original circuit must be the same as its Thevenin equivalent.

To verify it I first plotted the frequency response of the Original circuit as below:

enter image description here

As you see in the above plot, the Vout starts from -6dB and there is no -3dB in the entire plot for a cut-off frequency.

Then I plotted the frequency response of the Thevenized equivalent as below:

enter image description here

Above plot shows the -3dB point corresponds to 2.3Hz cut-off frequancy.

And finally I used the cut-off formula fc = 1/(2 * pi* R * C) for the Thevenized equivalent circuit and found fc = 31.8Hz.

What is going on wrong here?

All three ways give different results. Which one is the right one?

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    \$\begingroup\$ I think you're looking at the phase not the gain in the second plot. Use the same horizontal axis settings as the first plot. The first plot has 6dB loss because of the 2:1 attenuator - which will be the same voltage as the V source in the second. You're interested in the -3dB point relative to that loss, i.e. -9dB. \$\endgroup\$ – Brian Drummond Jul 15 '16 at 11:48
  • \$\begingroup\$ I see, but I still dont get why the first plot starts from -6dB. \$\endgroup\$ – user16307 Jul 15 '16 at 11:55
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    \$\begingroup\$ Quote: "As you see in the above plot, the Vout starts from -6dB and there is no -3dB in the entire plot for a cut-off frequency." REMEMBER: The cut-off frequency is defined as a frequency where the ouput is 3 dB less than the output at DC. \$\endgroup\$ – LvW Jul 15 '16 at 11:56
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    \$\begingroup\$ Because the two resistors halve the input voltage, which is 6dB loss. In the Thevenin circuit you have no similar loss, you turn the source voltage down by 6dB instead. \$\endgroup\$ – Brian Drummond Jul 15 '16 at 12:07
  • \$\begingroup\$ @LvW your definition of cut-off relieved my pain i was blindly looking for -3dB \$\endgroup\$ – user16307 Jul 15 '16 at 12:14
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All three methods give the same answer.

The filter with the load resistor starts from -6dB, so the '-3dB point' is -9dB on that graph. Reading the first graph by eye looks consistent with just over 30Hz. Zoom into it for better accuracy.

In the second graph, for some reason you have taken the phase curve (the dotted line), which is the same height for 4 degrees shift on the right y axis and -3dB on the left y axis. You don't see the gain curve at this frequency because the y axis scale starts at -1dB (or maybe you've not plotted the gain curve). Replot your second graph with frequency going up to 100Hz, and gain axis from 0dB to -10dB. When you do, you'll see it agrees with the theoretical 31.8Hz figure you've got from calculation.

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  • \$\begingroup\$ Oh thanks alright I plot the wrong one in the Thevenin equivalent. But I still dont get "The filter with the load resistor starts from -6dB," Whay is that? Isnt the original circuit's freq. response should be exactly the same as the Thevenin's? Could you explain it in a simple manner? \$\endgroup\$ – user16307 Jul 15 '16 at 11:55
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    \$\begingroup\$ Remove the capacitor, what is the voltage division ratio for the two 10k resistors you've drawn? \$\endgroup\$ – Neil_UK Jul 15 '16 at 12:57

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