filter bandwidth losses
I simulate several kinds of microwave filters and find a very interesting point.
That is when a filter possesses a larger bandwidth and it usually accompanies with a lower insertion loss. I think it may be interpreted by the Dishal's method. However, how do we explain this phenomenon by physical viewpoint?
Could someone kindly give me some replies?
Thank you!
Hi Ycchiou,
It is not necessary. Insersion loss is designed by the component values in lumped element filter design. Bandwidth depends upon the order of the filter and also the design. Also, in maximally flat design it gives fixed insersion loss of 3 dB. So, it can be really hard to generalized the relationship between insersion loss and bandwidth.
This is the general rule that bandwidth and insertion loss are inversely proportional for the filters. General equation for filter insertion loss is: IL(dB)=4.34*(Qloaded/Qunloaded)*SUM(gn). For given unloaded Q insertion loss is directly proportional to loaded Q, which is inversely proportional to the bandwidth. I hope it help.
Thanks for RF-OM.
I appreciate your response; however how to physically interpret this phenomenon, but not depend on equation.
This is what I want to know.
Could you kindly give me responses?
The dissipation loss level in filters depends primarily on three factors: unloaded Q of filter elements, the filter type and order (this is related to the number of lossy elements) and degree of filtering. For example, for lowpass filter of given type only unloaded Q affects the loss. But for bandpass filter we need additionally count the bandwidth too. As more narrow the filter is as higher the degree of filtering and more dissipation loss involved. For physical analogy we may say that narrowband filter is doing more work than wideband and therefore there is higher loss. Another way is to compare the reactance slopes. For narrowband filter loaded Q is higher, therefore reactance slopes are steeper. This means higher speed of energy transfer, which requires more loss.
Thanks for RF-OM's valuable comments.
However, I do not understand what the following sentence means.
As more narrow the filter is as higher the degree of filtering and more dissipation loss involved. For physical analogy we may say that narrowband filter is doing more work than wideband and therefore there is higher loss.
Could you give me more explicit explanation? I appreciate it.
Thanks a lot again!
Well, let's try this way.
Filtering is the process of energy change and energy exchange. As more narrow the bandwidth as more changes are necessary, but energy change is the lossy process. Therefore as narrower the filter as higher the loss.
It is very general explanation. If you need more details you may think about filter as an electrical network that have some number of series and parallel elements. You can write a differential equation that counts all the reactance slopes for series and susceptance slopes for parallel elements. This will visualize your problem and may help to understand the underlaying physics. You can also use the theory of coupling circuits that may help to understand well the interaction processes inside the filter. This processes also contribute a lot into filter's insertion loss. I know you do not like the math, but here math can help you a lot. It is not necessary to solve all the equations, but just looking on them may help to visualize the process of energy transfer.
Best regards,
RF-OM
Thanks RF-OM's kindly reply.
:D
You are welcome! I am glad when I was able to help.
Best regards,
RF-OM