Problem with cascading two microstrip circuits

Hi,

Could anyone tell me what if I have resonance at one frequency when I cascade two microstrip bandpass filter? All other frequencies are Okay except one point(see attachment).

Obviously I can remove it by using an ideal isolator in-between the two filters. Is there any simple but feasible way to remove it? I tried to use the single stub but seemed useless.

kevin

Two cascaded bandpass filters will disturb the stopband performance, as each filter is designed using a wideband 50 ohm load (assumed). The result may be spurious response or degraded attenuation in the stopbands.

When the filters are cascaded, the 1st filter will see a high mismatch at its stobands, as the load consists of the 2nd filter input, which has low return loss at in the stopbands of the 1st filter. If you cannot use an isolator (with sufficient bandwidth) between the two filters, maybe you can use a buffer amplifier (with sufficient bandwidth). Or an attenuator, if the increased overall insertion loss is acceptable. An attenuator located between the two filters will improve the return loss of the load the 1st filter will see by an amount of 2*A, where A is the attenuator attenuation in dB. So a 3 dB attenuator will improve the return loss by 6 dB, which in many cases will be sufficient for nice stopband performance.

ok, I see. Thank you very much for your fascinating explanation!!

Yes, very true. The simple explanation is that a bandpass filter is designed to work with a 50 ohm source, and a 50 ohm load impedance. If you give it something like a short circuit as a load (like a filter in its stopband might be), you can not really expect it to have the same dB of rejection.

Sometimes, you only need something like a 3 dB pad between the two. You do not need too much loss, because the signal goes thru it twice--from the output of filter one to the input to filter 2 is one pass, and the reflection off of the input of filter 2 back to the output of filter one is the 2nd pass. So a 3 dB attenuator will improve the load return loss by 6 dB.

Also, without any attenuation or isolator between the two filters you could have a lot of passband ripple--especially if the two filters have some length between them.

Also, do not forget that you can negate most of the filter rejection if a big standing wave is set up between the two filters. Suppose the output impedance of filter one is near a short circuit. Suppose the input impedance of filter two is near a short circuit. Now, suppose the length of line between the two filters is λ/2 at that stopband frequency. Well...a whopping big standing wave will set up between the two filters, the 2nd filter will think the input signal is 40 dB higher than you though it was, and it will output a much higher signal, even though you are well into the stopband of both filters.

If you really needed two filters in series (maybe one filter in one box, and the second filter in a second enclosure for EMI reasons), but can not take the loss of an attenuator or the size/cost of a ferrite isolator, then I suppose you could design a custom filter to handle it without a weird standing wave between the two. I would design the input filter as a straight 50 to 50 ohm filter. I would design the 2nd filter as a diplexer, with a couple of poles in the stopband to a 50 ohm load, and with the number of poles desired in the passband to the real output connector.

Thank you guys, I really learned it lot from your discussion.

The main reason I employed the second wider-band BPF is that I have to remove the unwanted passband at the lower frequency in the first BPF(you can see it in my uploaded pdf file.). Donno if there is a better way to do this.

biff44, could you pls tell me more details or some material about how to design this diplexer? I read from Microwaves101.com about the diplexer. It says "A diplexer can also be used to create a "matched" filter that is non-reflective outside of the intended passband." I still don't know how it works... thx.

Best Regards,
kevin

it is a tricky job to do, you must maesue the first filter's out impedance, at the frequency of interst for the second filter(Fo2). then try to adjust the second filter's output impedance ( at its frequency of operation Fo2) so that it equals to be conjugate of the out impedance of fist filter(at same freq). I have cascaded some filter at 900MHz using this method and it works!!!!

if it was helpfull , please press helped me botton,

Mehdi

It is a little complicated, but not too bad. You normally design a filter to be 50 ohm matched at both ends.

But in a diplexer filter you design them differently, by using what is called a "singly terminated" lowpass prototype.

Why, well the output of the bandpass in diplexer filter is your 50 ohm system load. The output of the stopband in your diplexer filter is the 50 ohm chip resistor.
BUT, where they both combine (in shunt lets say) to a single input, the stopband has the following input NORMALIZED Admitance:

1 mho real in the two stop bands
0 mhos real + -jY(f) in the passband

And the bandpass has the following input Normalized admitance:
0 mhos real + +jY(f) in the stopband
1 mho real in the passband.

SO, when you comine the two in shunt at the input port, at any frequency, the input normalized admitance is 1 -jY(f) + jY(f) =1 or matched to 50 ohms.

You can find some examples in Mathaie Young and Jones under singley terminated filters. There are probably some more modern versions.

As a simplification, you might not need to use the same number of poles in the stopband as the passband. for instance, if you need 5 poles in the bandpass, you might be able to get away with 2 or 3 poles in the stopband, as you can stand some out of band VSWR mismatch, just not a lot.