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I have designed this basic RC high pass filter circuit in LtSpice IV:

enter image description here

It should have cutoff frequency around 10 kHz. When I perform the ac small signal analysis the output is as expected.

enter image description here

But if I give as input tree AC sinusoidal signals each with different frequency(400 Hz, 4 kHz and 18 kHz), and perform transient analysis, it seems that filter doesn't work.

Here's the procedure...

I made desired input using summing amplifier. Circuit is shown belove.

enter image description here

Then I perform transient analysis. The output and input signals are:

enter image description here

Output signal seems to have all tree frequency components.

Then I have done FFT on both input and output signals and it confirmed that both signals contain same frequencies, just like the high pass filter didn't worked.

FFT of input signal

FFT of output signal

By inspecting the magnitude axes (the one in dB) I noticed that the signals have been suppressed, but as it seems not significantly. If you look at output signal in time domain this becomes apparent.

What I'm doing wrong?

How could I design a filter that will keep only 18 kHz signal?

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    \$\begingroup\$ Are you aware that you compare two C-R highpass sections which have DIFFERENT corner frequencies? \$\endgroup\$
    – LvW
    May 10, 2014 at 15:58
  • \$\begingroup\$ I uploaded the wrong picture of circuit. All analysis are done on circuit with correct RC values. I'm now editing question. Thank you. \$\endgroup\$
    – balboa
    May 10, 2014 at 16:00
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    \$\begingroup\$ It looks like it worked perfectly to me. The 4 kHz signal is down by about 8 dB, and the 400 Hz signal is down by about 28 dB, which is a 20 dB/decade slope, exactly what I'd expect for a single-pole filter. What did you expect? \$\endgroup\$
    – Dave Tweed
    May 10, 2014 at 16:06
  • \$\begingroup\$ If you want to select only 18kHz signals, you will need a bandpass filter. Also, getting a frequency response from time domain data is very hard, you would need to run the simulation for a very looong time to allow all the start up transients to die out. These start up transients are still easily visible in your time domain plot, so the FFT will be inaccurate. \$\endgroup\$
    – gsills
    May 10, 2014 at 16:07

2 Answers 2

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If you want a filter that passes 18 kHz but blocks 4 kHz, you need to specify the minimum amount of attenuation that you require at 4 kHz. This will determine how complex (e.g., number of "poles") your filter needs to have.

For example, if you expect to have, say, 60 dB of attenuation over that ~2-octave span, you'll need something like a 5-pole filter, which will give you 5 × 6 dB/octave = 30 dB/octave.

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  • \$\begingroup\$ I want to check weather there is or not the component of 18 kHz in my input signal. All other components of signal have frequencies belove 10 kHz. What I planned to do is to pass input signal through high pass filter, then through full wave rectifier, then through integrator and then through analog to digital comparator. And at output I should have a digital signal that tells me whether the 18 kHz component is present or not. Which circuit would be appropriate for high pass filter? If you provide answer I will also edit the question. \$\endgroup\$
    – balboa
    May 10, 2014 at 16:44
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    \$\begingroup\$ It sounds like you're looking for something like the pilot tone in a stereo audio broadcast system. Rather than brute-forcing it with filters and envelope detectors, it might be better to use something like a PLL-based tone-detector chip. The NE567 is one example that's been around for a long time. \$\endgroup\$
    – Dave Tweed
    May 10, 2014 at 16:51
  • \$\begingroup\$ This could be just what I need, I'll research it. If it works, you saved me. Thank you! \$\endgroup\$
    – balboa
    May 10, 2014 at 17:05
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I don´t know how exact your FFT operation is - however, for my opinion both frequencies below 10 kHz appear at the output with an attenuation that approximately looks as expected. Don´t forget that you have the most simple C-R highpass with a slope of 20 dB/Dec only.

As to your second question - there is no filter which can "keep only 18kHz". You only can expect an attenuation of lower frequencies up to a certain level. For stronger attenuation as shown you need, of course, a more complex filter of higher order.

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