The AM Forum

Here is an interesting thought experiment, whose outcome may surprise you.

Suppose you have two low pass filters: filter A and filter B. Each is designed for 50 ohms (unbalanced) input and 50 ohms (unbalanced) output. Each of them has an SO-239 connector on the input side and on the output side. Each of them has a cutoff (i.e. 3dB of input-to-output attenuation) frequency of 5MHz. Each of them has a rolloff of 24dB per octave beyond the low pass region. Each filter is symmetrical... so it doesn't matter which side you use as the input side and which side you use as the output side.

You connect a length L (meters) of very low loss 50 ohm coaxial jumper cable between the right side of filter A and left side of filter B (i.e. the filters are now in series, with L meters of very low loss coaxial cable between them).

The question:

What does the above series combination behave like in terms of the "input-output"  response at 10MHz (i.e. well into the stop band of each of the individual filters) from the left side of filter A to the right side of filter B? You should assume that this input-output response will be measured with a 50 ohm output impedance source driving the left side of filter A, and a 50 ohm load on the right side of filter B.

Stu

Off the top of my head, it depends on the length of the connecting coax.

Stu, you have piqued my curiosity, and I will make an uneducated guess at the result. 

First of all, you specify a very low-loss coax between the two filters, so I would assume that many meters of coax would likely not make a measurable difference in the attenuation at ten megahertz.  Probably very similar to connecting the two filters directly without any coax.

So the filter in the passband has 3dB attenuation.  10 megahertz is one octave above the 5 megahertz passband, so I assume we are looking at 24dB plus 3dB or 27dB for one filter at 10 megahertz.  Two filters in series would then yield 56dB attenuation at 10 megahertz, while only 6dB attenuation within the 5 megahertz passband.  Seems too good to be true, so I assume this is an incorrect answer.  Anxiously awaiting the actual results and explanation.

I'm gonna make a wild guess and say the cutoff is now at 10 MHz.

Shelby

I'm thinking this is a trick question since a 4th order filter (24dB/oct) usually consists of two 2nd order filters in cascade (series) with the values adjusted for filter shape and frequency. They simply are not symmetrical, you'd have L ---> C to ground --->L and another C to ground.

So, if for example you turned one around while cascading you'd have either two C's in parallel (changing the value) or two L's in series (changing the value again).

Dunno if this is where it was headed or not.

The coax might be irrelevant, and a magicians trick to distract?

Otoh, IF the coax was the equivalent of a series resistor, that might change something.

                        _-_-

The response of two identical filters in series is much steeper than a single filter.

I'm not sure what answer you are fishing for.

Hint

Not a trick question... but the answer may be surprising.

Add the following to your thought process.

1. What happens to the power carried by a 10 MHz wave, traveling down the coaxial cable from left-to-right, if it arrives at filter B? Most of the power carried by that wave doesn't pass through to the output of the filter B. Where does the portion of the power, that is carried by that wave, but that doesn't pass through filter B, go?

Remember: The filter is made of capacitors and inductors, which do not dissipate power. Averaged over time (e.g. one RF cycle), the power flowing into the filter must equal the power flowing out of the filter.

A wave traveling along the coaxial cable, in either direction, carries power flowing in that direction.

Stu

Maybe too obvious, but heat?    Could it be circulating eddy currents turned into heat, or the series resistance of the coax or the series resistance of the L/C parts producing heat or maybe RF radiation - or a combination of them all?

The reason I axe, is that I remember an 80M  1 KW low pass filter I built with a cutoff at 4.5 Mhz got quite warm.

T

Hint #2 (a similar experimental set-up)

Consider the following situation:

You have a tuner in your shack, feeding a length of low loss transmission line, feeding an antenna that is very poorly matched to the impedance of the transmission line at the frequency of operation.

A wave traveling along the transmission line, from the tuner to the antenna, will mostly be reflected at the junction of the transmission line and the antenna.

A wave traveling along the transmission line, from the antenna to the tuner, will mostly be reflected at the junction of the transmission line and the tuner.

Considered individually, the tuner and the antenna both act as filters...which will attenuate an RF wave trying to pass though in either direction.  The power in the incoming wave, arriving at one side of the filter, will be much larger than the power in the wave that comes out of the other side of the filter.

Nevertheless, when the two filters (the ATU and the antenna) are connected by a coaxial cable... and with proper adjustment of the ATU... the end-to-end system passes the RF injected into the ATU to the antenna's radiated output... with no attenuation!

The proper adjustment of the ATU will depend upon the length of the transmission line.

How does all of this carry over to the solution of the original problem that I posed?

Stu

The rejected wave will reflected between the filter and source until dissipated as heat by the coax losses.

(This is a Walt, W2DU-sk question!)  :)

I'm sorry, but this is not the correct solution.

Also, no correct solution so far in any of the above posts.

In a real situation, there will, of course, be power dissipation in every component, including the coaxial cable.

But, for the purpose of solving this problem (the solution will apply to real implementations... it is not just "hypothetical"), you should ignore all losses, except the loss of the 50 ohm termination at the output of the second filter. You should assume that heating/dissipative losses in the coaxial cable and in the filter components, at 10MHz, are low enough so as to not have a significant effect on the in-output behavior of the series combination of these two filters interconnected by a very low loss coaxial cable.

Stu

24dB down at the coax
27 dB down at B output

???????????????????

Sorry

Not correct.

Check out hint #2 (posted earlier today)

Stu

So...

Are you saying that the output at the end of filter B is more or less dependent upon the length of said coax - IF it looks like the proper length (50 ohms?) then the filters act as advertised, if not, then not?

Of course if a given passive filter is not terminated with the proper impedance, its characteristics will be "non-ideal"...


                           _-_-

Bear

Yes, the behavior of the series combination of the two filters is dependent upon the length of the coaxial cable between the two filters.

Hint #3

Suppose the phase change (at f=10MHz) experienced by a wave traveling from the right side of filter A to the left side of filter B (via the coaxial cable) + the phase shift resulting from the reflection of the wave arriving at the left side of filter B + the phase change experienced by the reflected wave traveling back from the left side of filter B to the right side of filter A + the phase shift resulting from the reflection of the (reflected) wave arriving at the right side of filter A = a multiple of 360 degrees.

I.e. suppose the phase change experienced by a wave making a complete round trip between the filters is a multiple of 360 degrees.

How does that affect the build up of the energy stored in the standing wave that exists on the coaxial cable that is between the two filters?

Stu  

...well that would mean that the energy that is reflected from the "B" side is going to be in phase at the exit of the "A" side. So, I'd expect a voltage increase that would be substantial at that point, assuming little loss on the returning wave. In practice this sounds like an ideal way to obtain a supply of "magic smoke".

Bear

You're getting warm!

The wave that "leaks out" of (emerges from) the resonant cavity from the left side of filter A (traveling back toward the source) will be 180 degrees out of phase with the wave that is reflected from the left side of filter A (I.e the reflection of the incoming wave, from the source, by the left side of filter A).

The 180 degree phase shift arises from the 90 degree phase shift between the reflected wave and the transmitted wave that occurs at an interface that is partially reflecting and partially transmitting. When you go through the phase shift bookkeeping... you end up with the 180 degree phase shift (above).

Stu

Hi Stu,

The significance of the coax is not in the loss but the transmission line effects (propagation delay and impedance transformation).  Also the low-pass filters will look like a pass-thru device at frequencies well below the cut-off frequency.  Near and just below the cut-off frequency and higher frequencies, the LPF’s will have an input impedance and output impedance that is complex and not 50 Ohms – the coax impedance.  Additionally. these LPFs are passive devices; without isolation amps. They will interact with each other even with no transmission lines present and may produce an altered response.

There are 3 different lengths of coax in the system (effectively only 2).  The most influence on modifying the cascaded filters is when each section of coax is excited at its quarter wave and half-wave resonances and also the x/4 wave and x/2 wave resonances out to infinity frequency.    The result is that this should add a large comb of peaks and notches to the end-to-end response.  Mostly this alteration would be in the region say 80 % of the cut-off frequency on up.  

I have not simulated this.  I just joined this thread.  I have run into a similar response problem in the past when measuring noise across a wide ratio frequency spectrum from high voltage switching power supplies, with very high voltage divider towers and a scope connected to the divider tower with coax (by someone else) rather than using a 10X scope probe which eliminated the transmission line problem.

Tom

Correct!

The input-to-output attenuation of the combination... of the two filters, in series, with the coaxial cable in between... will be that of a band pass filter, with Q of 100, having 0dB of minimum attenuation, and 46dB of maximum attenuation*

*This assumes that the attenuation of either filter, by itself, at 10MHz, is 20dB.

The way I would describe the solution is as follows:

Recall, that the attenuation of a single filter, considered separately, (filter A or filter B) is around 20 dB at 10MHz. That means the portion of an incoming 10MHz wave's power that will pass through a single filter (in either direction) is about 1%. That implies that if a 10MHz wave of specific amplitude, V volts, arrives, via a 50 ohm coaxial cable on one side of a single filter, then the wave that emerges, via the 50 ohm coaxial cable, from the other side of that filter, will have an amplitude 0.1V volts (i.e. 10% of the amplitude of the incoming wave).

The remaining 99% of the power will be reflected. The amplitude of the reflected wave will be: [the square root of 0.99] x V volts =  0.995V volts.

Case 1: The round trip phase shift experienced by a 10MHz circulating wave, in the resonant cavity formed by filters A & B + the coax between filters A&B is 360 degrees. I.e. the phase change passing through the coax from the right side of filter A to the left side of filter B + the phase shift associated with the reflection of the circulating wave from the left side of filter B + the phase change passing through the coax from the left side of filter B to the right side of filter A + the phase shift associated with the reflection of the circulating wave from the right side of filter A = 360 degrees.

In this case, the amplitude of the circulating wave (a.k.a. the standing wave) will build up...after the time that the source is turned on... with a time constant equal to the time its takes for the circulating wave to make 100 trips around the resonant cavity:

A. The amplitude of the circulating wave inside the resonant cavity will build up to 10 x the amplitude of the wave arriving on the left side of filter A (from the source)... i.e. 10 x V volts.

B. The portion of this circulating wave the passes through filter B, and continues on (as a travelling wave) to the 50 ohm load, will have amplitude 10 x V volts x 0.1 = V volts. Therefore, the wave that exits from the right side of filter B (traveling toward the 50 ohm load) will have the same amplitude are the wave arriving at the left side of filter A (from the source).

C. The portion of the circulating wave that passes through filter A, and travels back toward the source, will also have amplitude V volts. However, because it is 180 degrees out of phase with the reflection of the incoming wave arriving from the source (that is reflected by the left side of filter A)...and has the same amplitude... the reflection from the left side of filter A will be cancelled.

Therefore, in this case, the two highly reflecting (at 10MHz) filters in series... with the coaxial cable between them (of an electrical length that results in a 360 degree round trip phase shift)... act is a transparent filter with no reflection at the input side (the left side of filter A), and 100% transmission to the output side (the right side of filter B).

This is, of course, a non-intuitive result when first presented.

The time it takes for the circulating wave inside the resonant cavity to build up (or decay, if the source is turned off) is roughly 100x the round trip propagation delay around the cavity. [This is actually the time constant of the buildup or the decay of the field in the resonant cavity]. Because of the 90 degree phase shift that is associated with each reflection (a total of 2 reflections per round trip), the round trip electrical length of the coaxial cable has to be an odd multiple of 1/2 wavelength... i.e. 1/2 wavelength, or 3/2 wavelength, or 5/2 wavelength. If the electrical round trip length of the coaxial cable is 1/2 wavelength (1/4 wavelength in each direction)... then the 2-way delay through the cable is 0.5/10MHz = 50ns. Therefore the time required for the circulating wave in the cavity to build up to its steady state amplitude (or to decay, if the source is turned off) is roughly 100 x 50ns = 5 microseconds.

Case 2: The round trip phase shift experienced by a 10MHz circulating wave, in the (non resonant) cavity formed by filters A & B + the coax between filters A&B is 180 degrees. I.e. the phase change passing through the coax from the right side of filter A to the left side of filter B + the phase shift associated with the reflection of the circulating wave from the left side of filter B + the phase change passing through the coax from the left side of filter B to the right side of filter A + the phase shift associated with the reflection of the circulating wave from the right side of filter A = 180 degrees.

In this case, the amplitude of the circulating wave will be 0.05 x V (volts). I.e. the amplitude of the circulating wave in the (non-resonant) cavity will be half as big as the amplitude of the portion of the incoming (from the source) wave that passes through filter A (i.e. 0.5 x 0.1V volts).

The amplitude of a wave exiting from the right side of filter B (toward the load) will be 0.1 x 0.05V = 0.005V volts. The power carried by the wave exiting filter B toward the 50 ohm load will be 0.005 x 0.005 x the power arriving at the left side on filter A (from the source). Therefore, for Case 2, the attenuation (input-to-output) of the series combination of 2 filters with coaxial cable between them,  will be 10log(0.005 x 0.005) = 46dB.

Note that this is 6dB greater than the stand-alone attenuation of filter A + the stand-alone attenuation of filter B. 46dB = 20dB + 20dB + 6dB

Case 3: The round trip phase shift experienced by a 10MHz circulating wave, in the cavity formed by filters A & B + the coax between filters A&B is something between 360 degrees and 180 degrees. I.e. the phase change passing through the coax from the right side of filter A to the left side of filter B + the phase shift associated with the reflection of the circulating wave from the left side of filter B + the phase change passing through the coax from the left side of filter B to the right side of filter A + the phase shift associated with the reflection of the circulating wave from the right side of filter A = something between 360 degrees and 180 degrees.

In this (general) case... the attenuation from the input of filter A to the output of filter B will increase smoothly and monotonically from 0dB to 46dB. Most of this increase in the attenuation will occur when the round trip phase shift (in the cavity) decreases from 360 degrees to (roughly): [360 - 360/100] degrees = 356.4 degrees. I.e. the end-to-end response is that of a high Q (i.e. Q=100) band pass filter. Starting at 360 degrees, if you change the frequency by 0.5 %... to cause a 1.8 degree change in the round trip phase shift...  then the attenuation of the end-to-end filter will increase by around 3dB. Phase changes beyond that will result in a rapid increase in the attenuation... toward a maximum of 46dB.

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.

Hmmm... interesting effect.

Now I was just thinking - is that still legal?
The output of a typical boatanchor rig is a filter, connected by coax (in some stations) to a resonant antenna, which acts like a high Q bandpass filter.

What about that?

                    _-_-

My mistake was bothering me, so here is a plot with 5 MHz LPFs as the given by Stu.  The first peak in the final response has moved down to 15 MHz from 21 MHz.

A pdf of the schematic input file is attached too.

Tom

If it is not too much work:

Please make the transmission line between the filters about 50% longer.

Also, in your simulation, is the length in feet ( as you state in your earlier post) or is it in meters?

Thanks
Stu

The coax length is by specifying the propagation delay time.  The speed of light is 11.8 inches per nanosecond and the cable commonly used for these lab experiments is RG-58 which has a velocity factor of 0.66.  So I use 1.5 nanoseconds per foot in the ideal transmission line model definitions.

Attached is the results of the simulation with the coax lengths increased to 4.5 feet for the generator to LPF1, 7.5 feet for the cable between LPF1 and LPF2, and 9 feet for the cable to the 50 Ohm load where the response is measured.

The first peak has moved down to 12 MHz and the second peak is at 47 MHz.

Tom

Thank you for doing and posting these simulations.

Since the output load is a 50 ohm resistor, and the coaxial cable has a 50 ohm impedance, I don't think that the length of the output cable will make any difference.

Since the source impedance is 50 ohms (i.e. a voltage source in series with a 50 ohm resistor), I don't think that the length of the input cable will make any difference.

Since the electrical length of the cable that connects the two filters (assuming the wave propagation speed in the cable is 66.7% of the speed of light in free space) is much less than 0.25 wavelengths at the frequency where the first transmission peak occurs (12MHz) ... this seems to imply that the phase shift associated with a reflection from either filter (at that frequency) is more than 90 degrees.

I.e. at 12MHz, 0.25 wavelengths corresponds to a physical distance of 0.25 x 200,000,000 meters per second / 12,000,000 Hz = 4.167 meters = 13.67 feet. But the distance between the filters in your latest simulation is only 7.5 feet.

So the one way phase change experienced by a 12MHz wave propagating through the cable is less than 90 degrees. It is actually around 49 degrees.

If the round trip phase shift is 360 degrees, then the reflection at each filter must be producing a phase shift of 180 degrees - 49 degrees = 131 degrees.

Stu

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.

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