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I'm trying to simulate a RF low pass filter in Pspice. The result don't match with my calculations and a filter simulator I found online.

Does anyone know what I'm doing wrong in Pspice?

My source has a 50 ohm resistance. It seems strange that I already have so much attenuation at the input.

The filter is a 5th order Chebyshev filter with maximum 3dB in the passband at 500MHz and 30dB attenuation at 750MHz.

How can I solve my Pspice settings to get a correct result?

EDIT: I tried some different things from the feedback I got here. It's still not showing the right result( see last image.)

Why is there a 6-12dB attenuation in the passband?

Pspice results

Simulator result

enter image description here

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  • \$\begingroup\$ Where’s your signal source resistor? \$\endgroup\$
    – Andy aka
    May 21, 2020 at 18:38
  • \$\begingroup\$ @Andyaka He specified the source's Rser, but it's obscured by the .AC command. OP, your simulation uses linear frequency axis, but you need it logarithmic. Try .ac dec 1001 100meg 10G. Also, the result that you show from the calculator shows a 0 dB peak, which will never happen with passive elements. \$\endgroup\$ May 21, 2020 at 19:38
  • \$\begingroup\$ @aconcernedcitizen I did indeed specified the internal resistance of the source. Although when i define it as an external resistance the results are different. Why i there so much attenuation at the input? I'm not planning to make this filter, this is just an exercise for school. But i don't get it why there is such a difference. Pspice should just give me the simulation as i calculated with a lumped method. \$\endgroup\$ May 21, 2020 at 20:14

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You should add a label at least for the plotted waveforms because nobody can tell which node is N002 & co. Also, using passive elements means there is no gain, and thus the output will always be less than the input. What you have there is a voltage divider, followed by another, etc. Even if those are ideal LC elements, the resistive divider formed by the input and output cannot be avoided.

If you mean that you should see the 6 dB at the lowest frequency, then you should simulate to include DC (or closest to it). And, since your filter goes into hundreds of MHz, then a linear scale is better (in the comments I proposed log scale to match the one in the example): .ac lin 1001 1m 1g will allow you to see what you've missed. Using dec (log scale) means every decade there are N points. Using lin (linear) means from start to finish there are N points. Since for this case you want to see pretty much from DC to light, then lin makes more sense, but .AC analysis go very fast and, if you're really in dire need of points, you can choose dec with 10k points; it's overkill for this case, but it's not forbidden or impossible.

filter

If you're expecting the response to be as in the picture you provided, as I said in the comments, they are showing the response of the mathematical formula, which does not include the attenuation given by the I/O impedances:

math

Not lastly, there are some builtin parasitics that may affect the response, but not in this case. If you really want to go pedantic about it, then you can set Rser, Rpar and Cpar for all elements to zero, and for capacitors Lser and Rshunt, too. Also, this is LTspice, not PSpice, and the results are accurate for the given input (same in other SPICE programs).

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