puzzled with waveguide filter

I design a filter,but the measured response is that,
in the pass band,the s11 is about -10dB,and the s21 is also -10db.
it doesnt follow the power conservation theory.
who can give me some instruction?

There is also the internal power loss.

I can't see a problem. You've a filter with not very good S11 and poor S21, only.

hi,borich,flatuent,
if the s11 is very poor,it will be about 0db which means full reflection,but not -10dB.-10dB is poor but not very poor,and most of the power should go through the filter and the s21 should be close to 0 db. I dont think there will be a very large internal loss in a passive component which absorb almost all of the power.
thanks

by the way,any one can introduce me some good methods to optimise a waveguide filter?HFSS is too timeconsuming and I dont have a software of mode matching method.Who have the experience?

I know what is your question:
if the input power is 0 dBm, and the s11 is -10, then the reflected power is -10dBm,so the output power is 0-10=-10dBm, tht means the S21 is -10, that is all!

springf,
you are quite wrong
db(s21)=10log10(s21)
and |s21|**2+|s11|**2≈1 if it is a lossless system.

Powers use 10 Log x, voltages use 20 Log x.

dB[s11] = 20 Log (s11)
dB[s21] = 20 Log (s21)

S11 amd S21 are VOLTAGE travelling waves.

In a network, power is conserved. That is the power going into a black box is equal exactly to the power comming out of the black box's ports PLUS the power turned into heat inside of the black box.

Power is proporitonal to voltage squared, so conservation of power says, if the input traveling voltage wave is of unity amplitude, the normallized input power is 1 squared:

(1^2) = 1 = (s11^2) + (s21*2) + power dissipated inside of the black box

S11 = -10 dB = 0.316 so: S11^2 = 0.1
S21 = -10 dB = 0.316 so: S21^2 = 0.1

1 - (0.316^2) - (0.316^2) = power dissipated in the black box

so

1 - 0.1 - 0.1 = 0.8 = (input power -reflected power -transmitted power) = power dissipated inside of the black box.

the power dissipated inside of your box is 0.8, or 80% of the input power in watts. Sounds high-loss to me.

Another way to visualize this: If your input power was 0 dBm (1 mW), as someone suggested, you could use a power meter/directional coupler on the input port and measure that the reflected power was 0.1 mW (ie s11 was 10 dB down from the input power of 0 dBm). You could hook the power meter directly to the output port, and also see 0.1 mW coming out of the output.

SOOOOO, if 1 mw comes in, 0.1 mw reflects back, and 0.1 mW goes into the output load, there must be 0.8 mW of heat being generated inside of the black box (barring things like black holes and dark matter, etc).

amicloud said:

hi,borich,flatuent,
if the s11 is very poor,it will be about 0db which means full reflection,but not -10dB.-10dB is poor but not very poor,and most of the power should go through the filter and the s21 should be close to 0 db. I dont think there will be a very large internal loss in a passive component which absorb almost all of the power.
thanks
by the way,any one can introduce me some good methods to optimise a waveguide filter?HFSS is too timeconsuming and I dont have a software of mode matching method.Who have the experience?

Click to expand...

There is not problem.
S11=-10dB means that the power reflected is 12%. And other part of energy you lost inside the filter. If your design is not correct, you've mismatching and loss, between different parts of the filter. You need to optimize the structure or repeat a calculation.
For optimizing you can use CST-soft. For waveguide filters, with a AR-setting, the programme is more faster than HFSS.

A 10 dB return loss is ok, industry standard is more like 12 to 15 dB for return loss.

A 10 dB insertion loss is NOT ok. Unless this is a very narrowband filter, something is probably broken inside of the filter.