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Author Topic: Behavior of filters placed in series  (Read 13042 times)
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Tom WA3KLR
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« Reply #25 on: July 05, 2014, 07:56:48 AM »

I originally threw in all 3 coaxes as that is what really would be in a bench setup.  I rationalized at the time that the output coax at least would probably not make a difference since the load resistor equals the transmission line impedance; just what we want from a transmission line - it puts the load right at the end of the second filter.  That's why I first stated that there were 3 coaxes but effectively only two.

I did a series of simulations this morning to prove what is what and the input and output coaxes have no effect. It is the presence of the coax between the two LPFs that is the sole cause of the response upset.

Enjoy the weekend weather.
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73 de Tom WA3KLR  AMI # 77   Amplitude Modulation - a force Now and for the Future!
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"Season's Greetings" looks okay to me...


« Reply #26 on: July 05, 2014, 11:34:20 AM »

Tom

I used the 10MHz cut-off frequency filter LTspiceIV file that you posted to do some further simulations:

A. I tweaked the 1-way delay of the center coaxial cable... until the first zero-loss (cavity resonance) frequency was 20MHz. This turned out to correspond to a one-way delay of 8.25ns.

B. Then I added 25ns of additional delay (total of 33.25ns) to make the cable 0.5 wavelengths longer (at 20MHz). I.e. 1/20MHz = 50ns. Therefore, a half wavelength at 20MHz corresponds to 25ns of delay.

As expected, the extra 360 degrees of round trip phase change (again) produced a resonance at 20MHz, with zero loss from the left side of filter A to the right side of filter B. (see attached slide 1).

I.e. the input-output loss at 20MHz was 6dB; which is the same as the 6dB low frequency loss that occurs when the source (which has a 50 ohm source impedance) feeds into a 50 ohm load. (see attached slide 2)

The 2-sided 3dB bandwidth of this resonance (i.e. measuring from -9dB to -9dB) is 58kHz.

Note: at 20MHz, the simulated transmission loss of a single filter, driven by a source with a 50 ohm output impedance, and with a 50 ohm load, is: 24dB. (see attached slide 2). As stated above, 6dB of this corresponds to the loss associated with the 50 ohm source output impedance. Therefore the loss from input-to-output of the single filter, itself, is: 18dB.  This means that the amplitude of the transmitted wave, passing through a single filter, is 1/(7.94) x the amplitude of the incoming wave = 0.126 x the amplitude of the incoming wave. To build up the field in the cavity to steady state amplitude (i.e. the amplitude of the circulating wave in the cavity has to be 7.94 x the amplitude of the wave that leaves the cavity from the right side of filter B), we need 63 round trips (i.e. 0.126 x 63 x 0.126 =1). Therefore, we would expect the turn on time constant of the filter (i.e. how long it takes the field in the cavity to build up, after the source is turned on) to be 63 round trips x (2 x 33.25)ns per round trip = 4196ns. The corresponding 2-sided 3dB bandwidth is (roughly): 2 x (1/4196ns) / (2 x pi) = 76kHz. This is roughly consistent with the results of the simulation.

Since 8.25 ns of 1-way coaxial cable delay corresponds to -180 degrees x (8.25/25) = -59.4 degrees of phase change at 20MHz... the phase shift that occurs (at 20MHz), each time a wave is reflected by one of the filters, must be -180 degrees + 59.4 degrees = -120.6 degrees.

C. From slide 2, we see that the phase shift, on transmission, caused by a single filter feeding into a 50 ohm load is: -210 degrees at 20MHz. Therefore the filter, by itself, introduces -210 degrees of phase shift when a 20MHz wave passes through it (in addition to 18dB of attenuation). The phase shift that a wave that is reflected from the filter experiences must be the phase shift that a transmitted wave experiences + or - 90 degrees. Therefore (using +90 degrees), the phase shift that a wave that is reflected from the filter experiences must be -210 degrees + 90 degrees = -120 degrees. Therefore, from this analysis, the phase shift experienced by a wave that is reflected by one of the filters is -120 degrees. This is (within round off errors) equal to the result calculated above.

The simulation is fun to play with.

Stu



I did a simulation using Linear Technology's LTspice IV.  The LPFs are just a 3 element Butterworth LPF, 18 db/octave roll-off.  There is a 3 foot section of coax between the generator and the first filter, a 5 foot section between filter1 and filter2.  Then a 6 foot section to the 50 Ohm load.

Attached is a pdf of the response plot output showing the 2 LPFs with no transmission lines and a plot with the 3 coaxes.

In review, I see that Stu stated 5 MHz cut-off and my filters are 10 MHZ cut-off.  But the effect of the coaxes is shown never the less.

The .doc file is not a Word document but is the LT spice schematic file with the extension renamed so it can be attached here.  For those with LTspice IV, save the .doc file and rename it with the .asc extension and then you can open it in LTspice and play.


* Tandem filter simulations.jpg (82.35 KB, 960x720 - viewed 320 times.)

* Slide2.JPG (89.47 KB, 960x720 - viewed 322 times.)

* Slide3.JPG (89.95 KB, 960x720 - viewed 327 times.)
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Stewart ("Stu") Personick. Pictured: (from The New Yorker) "Season's Greetings" looks OK to me. Let's run it by the legal department
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