Adding RF Filters in Series - Odd Results

[killingtime]

Hi,

I'm designing a low pass filter for the 2m (144-146MHz) band. It's going on the (low impedance) collector of a BJT (amplifier - class E). To save any surprises, I thought I'd  run a simulation first in LTSpiceIV. Using an online freeware filter calculator for the L match and chebyshev filters. Screenshots below.

https://images2.imgbox.com/82/b9/8UmZgw3E_o.jpg

Two filters. An 'L' match for impedance conversion (1 Ohm to 50 Ohms, BJT collector output, 145MHz) and a chebyshev (50 Ohms to 50 Ohms, 170MHz to keep the operating frequency (145MHz) away from large phase changes near the cut off point).

Here's the thing; if I model each filter on its own in LTSpice, I see the expected S21 response. When I add them in series, I see anomalies in the stop band.

Image below: Single 8th order chebyshev 50 Ohm ip and op. Filter designed for 50 Ohms. Expected response.
https://images2.imgbox.com/cb/cc/LMjpTfY6_o.jpg

Image below: two of ther same chebyshev. Spike at 250MHz.
https://images2.imgbox.com/a5/47/qjoXqw5S_o.jpg

Image below; an L match and 2 chebyshev. 2spikes in the stop band. L match set for 50 Ohms output.
https://images2.imgbox.com/33/aa/NRQPRkxY_o.jpg

I always understood that provided you match the input and output impedances to the filter design, you should get the designed S21 response, and that appears to be the case for individual filters, but not for multiple series filters...I'm missing something obvious here.
Can anyone explain why these anomalies appear? I thought filter responses just summed together (provided the design impedances match). The online calculators could be wrong, but that's unlikely. I could be using LTSpice incorrectly.

It's not a transmission line issue, this isn't real world (yet) as it all in LTSpice so you can treat the circuit as a lumped element.

Thanks.

[Joel_Dunsmore]

your final circuit has a resistor in parallel with a voltage source which means the impedance looking back into the voltage source from the filter is zero (not the value of the resistor).  In general you should only have series elements with a voltage source and only shunt elements with a current source.   If you are emulating a 2 ohm output impedance, put the resistor in series and see what you see.

But: remember your filter is only matched in the pass band, in the stop band it will be effectively a full reflection (open or short or in-between but 0 dB return loss).  so the transfer function will be effected in the concatenation of filters in the stop band.

I guess the spikes will go away when you put a proper resistor in series with the voltage source.

[killingtime]

your final circuit has a resistor in parallel with a voltage source which means the impedance looking back into the voltage source from the filter is zero (not the value of the resistor).  ......
I guess the spikes will go away when you put a proper resistor in series with the voltage source.

Corrected schematic and simulation below. The spikes are still there but the shape is different. I added the parallel resistor to see if it made any difference (racking my brain on this one...).

https://images2.imgbox.com/34/a6/ZOOmOGLv_o.jpg

The whole point of a stop band is that it stops the signals from getting through. If there's a spike of 'gain' then that defeats the point of the filter. Admittedly, the 2nd anomaly is at -30dB, which I can live with becuase it's not on a harmonic, but why is it even there to begin with. I hope this is a simulation error.

[T3sl4co1l]

Yeah, lossless filters don't have resistive port impedances.  That's how they filter, after all!  Incident waves are either passed or reflected.

To cascade filters, you need to match the conjugate impedances at each point, or use a diplexing filter so that the stopband is resistive (a termination resistor goes on the extra port, unless you happen to have another use for it).

The design can be simplified a bit, when multiple resistors can be used in random places; a constant-resistance filter usually has some RC, RL or RLC which acts to cancel out the bumps in the filter's stopband impedance.  This is limited to certain types, I think (as anything sharper than a Butterworth necessarily (I think?) has an inversion in its impedance, so cannot be dealt with by just shunt or just series elements alone).

Example: http://jeroen.web.cern.ch/jeroen/reports/crfilter.pdf

Note that the output impedance of a class E stage need not be anything in particular.  Hmm, let me see.  There is a mechanism to absorb power (rectification -- assuming an antiparallel diode on the, bipolar collector you say?, or sufficient base drive to achieve synchronous rectification).  There is a mechanism for reactance (obviously, the inductors around the amp), that is to say, the exact shape of the flyback wave can vary.  The flux is constrained by driven pulse width, frequency and supply voltage.  It should be pretty low impedance, then?

BTW, you can make asymmetrical impedance (matching) filters, though I don't know of any online calculators that use them.  Electronic Filter Design Handbook (Williams, Taylor) comes to mind, or the classic Handbook of Filter Synthesis (Zverev).  May be able to find copies online.

Obviously, you need to know the source impedance for that to apply.  If it is in fact a low resistance (and some reactance but that doesn't matter, we can always tune that out), then a low resistance or one-port-shorted (L-input) filter would apply.

Same thing that works with class D amplifiers for example; which, for applications like audio power and mains inverters, the resistance might be very poorly defined over the filter's transition band, so you must add your own resistance -- shunt R+C and series R||L ensure reasonable behavior into any impedance load.  This is kind of the reciprocal case of a constant-resistance filter, where it needs to provide behavior within a certain envelope (the response will still vary with load impedance by some arbitrary amount) and the output impedance isn't necessarily an important factor (but, both can be done, as for the case of the CR filter proper).

Tim

[G0HZU]

I hope this is a simulation error.

It looks like you used infinite unloaded Q for the inductors and caps so in this sense this is a simulation error. The stopband resonances are meant to be there but if you were to put a realistic Q for the inductors into your simulation the spikes wouldn't look so pointy and significant.

Putting two even order filters (8th order)  in between 50R ports probably makes things slightly worse than they would be compared to using odd order filters but there should still be stopband resonances even if you do this.  Normally, you should always choose an odd order filter design if both ports are the same. An even order filter is normally used between ports of differing impedance.

Note: your linear simulation shows the steady state case after equilibrium is reached. One way to understand what is happening is to think in the time domain and how quickly the filter reaches steady state when a single filter is tested between 50R ports. You can get some idea if you plot the group delay of an even order filter (like yours) and an odd order filter. The even order filter will typically have a much bigger spike of group delay near the cutoff compared to the odd order filter.

When you double up on this poor practice by stacking two even order filters the group delay spikes in the stopband region will become HUGE and this is a key clue that the system has to spend ages storing energy inside the filter sections in order to reach steady state at the stopband spike frequency. This will inevitably cause a stopband spike somewhere. There will be a lot of energy stored in the filter where those stopband spikes are if you simulate with infinite Q components. If you use inductors with realistic Q then this naturally damps out the stopband spike resonances and reduces the amount of unwanted energy stored in the filter.

Try giving the inductors an unloaded Q of 150 and try odd order filters and maybe it won't look so bad. However, I still don't recommend stacking filters like this between equal 50R ports. Note it is possible to stack narrow bandpass filters in series if you put a 90degree delay between sections.

[G0HZU]

The other thing to consider is the amount of signal loss there will be in your filter. If you want to preserve efficiency you might want to look at ways to minimise the loaded Q of your matching section. Maybe absorb this into a ladder matching network that will have lowpass properties. You can then use high Q inductors without the risk of having those nasty stopband spikes you see when you combine two even order filters in series.

[killingtime]

Putting two even order filters (8th order)  in between 50R ports probably makes things slightly worse than they would be compared to using odd order filters but there should still be stopband resonances even if you do this.  Normally, you should always choose an odd order filter design if both ports are the same. An even order filter is normally used between ports of differing impedance.
.....
Try giving the inductors an unloaded Q of 150 and try odd order filters and maybe it won't look so bad. However, I still don't recommend stacking filters like this between equal 50R ports. Note it is possible to stack narrow bandpass filters in series if you put a 90degree delay between sections.

I've been playing around with this all morning, and figured out the problem while you were posting, but I believe you're correct.
I tried 3 different filter calculator programs with exact values instead of standard component series values (3 decimal places) - same result - so it's not the programs (wait for it...).
Then I stumbled on a calculator that only allowed odd order chebyshev filters. I gave that a try and the anomalies disappeared.

https://images2.imgbox.com/27/62/3qQyTRme_o.jpg

T3sl4co1l was right with his statement 'To cascade filters, you need to match the conjugate impedances at each point', that's always been my understanding as well. Even if you're using infinite Q inductors and capacitors, the simulation should still work (real world is a different story).

This lead me to think that there can only be one reason for the simulation anomalies  - the input or  output filter impedances aren't 50 Ohms resistive - as specified, but how to check as LTSpiceIV doesn't offer smith charts. Well, if we're only looking at one frequency, you don't need a complex impedance calculator, you can just look at voltage & current phase relationship on the input and output. Anything purely resistive will be in phase. Anything with a reactive component will have a phase shift. Take a look at the simulation below.

https://images2.imgbox.com/bc/36/morN77z8_o.jpg

The filter output is resistive (as specified), but the input is reactive. This is where the anomalies are coming from. So the freeware online calculator is just doing a 'best guess' at matching to 50 Ohms resistive, and not warning you when it's wide of the mark. All the online calculators did this.

This raises another interesting point. You have to know something about an RF amplifier design and its output filter before you add another one to reduce harmonics (say). Not all filter designs play well together. It's interesting that the anomalies only show up when you add a second filter.  Learned something new today.

Thanks for the replies.

EDIT
-----

In reply to G0HZU, here's the bode plot of the (2x) 8 element chebyshev with a Q of around 100, and 1000. The anomaly spike drops by about 30dB.

https://images2.imgbox.com/a5/ca/EXRXK408_o.jpg

[G0HZU]

If it helps there is a much easier way to understand why your dual 8th order filter stores lots of energy at the ringing resonance at 250MHz. Because you have used 8th order filters and stacked them end to end in the way you have it can be visualised to have created an LC resonance at the centre of the filter at 250MHz.

Here's how to visualise what is happening at 250MHz. If you were to split the filter back into two sections as in the image below and look into each side with a VNA then the filter on the left will look like a (lossy) variable capacitor (vs frequency) above the filter cutoff frequency of 160MHz. If you look with a VNA into the other half of the filter it will look like a lossy inductor where the inductance changes in frequency. At 250MHz there will be a special case where the reactance of the left side exactly cancels the reactance at the right side.

In this case the whole network can be visualised (at 250MHz) as a series capacitor feeding a series inductor and they are series resonant at 250MHz. This is why the stopband suddenly misbehaves and starts to look like an extra passband in the filter response at 250MHz.

[G0HZU]

I suppose the other test you could do (on a simulator) would be to hit the 8th order + 8th order filter with a fast and very narrow pulse and do a transient analysis. The pulse would have to be maybe 100 picoseconds wide with very fast edges. Then look at the centre node where the two filters join together and it should ring with stored energy and the stored energy will be ringing at 250MHz. A simple FFT at the middle of the filter will show the frequency it wants to store energy at. It should be somewhere around 250MHz.

[coppercone2]

that paper is excellent, the reflective/dissipative filter thing.. I tried to approach it a few times and usually its not so clear. I think that document makes things very much more clear.

Ah time to dust off the series inductors to try to make some, its very interesting. The amount of dust on that inductor book makes me feel like Allan Quartermaine.

[T3sl4co1l]

If it helps there is a much easier way to understand why your dual 8th order filter stores lots of energy at the ringing resonance at 250MHz. Because you have used 8th order filters and stacked them end to end in the way you have it can be visualised to have created an LC resonance at the centre of the filter at 250MHz.

And just to wrap this up even tighter -- note that a filter isn't a brick wall in the physical sense.  It's a superposition of waves, at all times.  We would like to think of a filter as stopping unwanted frequencies unconditionally, but the fact is, it is utterly dependent on its environment, and it is always passing all frequencies -- it just so happens that, in the right environment, those frequencies interfere destructively in the stopband.

So it's perfectly possible, indeed natural, to have spikes in the stopband, even in the absence of parasitic modes -- when under conditions such as this.

Hmm, which -- I'm not sure I'd go so far as to ascribe a mechanical nature to this -- I can see one coming up with the association that, based on the LC resonant circuit argument above, perhaps the phase shift induced by that mode causes those frequencies to fail to interfere -- perhaps this is an adequate explanation, but I would prefer sticking to structural and numeric justifications, namely that this topology and set of values happens to give an overall filter with, well, *gestures at the screenshot* -- it looks like that.  Mainly because, the same can be said for the bisection of any filter -- it's not that you're doing something structurally wrong here, it's still just another ladder network; it's the values that are doing it.  But the values are non-obvious, you have to calculate the port impedances to know if there's any weird response expected from any particular junction.

And the values are ultimately given by some gross high-order polynomial roots, so it's not obvious how you could adjust values to compensate; you have to start over with a whole filter of the desired total order.

Which, even if you have two identical filters cascaded, with the correct match, you can still get better performance with a properly designed (equal order) filter -- though at the cost of needing much higher Q components, assuming it's the same type of course.  (Or, you may instead find it's more economical to increase the order modestly, while using a more gentle type -- Butterworth and Bessel don't require as high Q factor as Chebyshev.  Uh, for the same order at least; I'm not sure offhand if the minimum component Q is actually lower for the same cutoff steepness, actually!)

Tim