Here's the smaller circuit. So that's what you meant by having a small resistance?
Hello boozi,
I have tracked this thread from the beginning, and I'm familiar with your questions and issues. As a student at University, you will understand that one function of this forum is the mentoring of our
replacements in our chosen field of endeavor. As such we have a duty to guide you and your fellows as much as we are capable. Having said that...
Your post, to which I am now responding, has resolved your challenge to the small bite necessary to give you, to what I perceive, some insight to the overall issues of a larger picture.
I will let my attachments do most of the talking. But you must understand first that I/O impedances will have an impact on what you are attempting to accomplish. In your attachment you have finally included a load impedance. For completion of the circuit one must also include the source impedance else real world results will not yield the expected/anticipated results. Look up the term "insertion loss" for further clarification; a passive circuit will
never have a gain in the physical world.
Below I have included two thumbs for you to ponder. The first is your circuit stepped with the source and output impedance from 6.25Ω to 50Ω ({R} in the schematic). Note the change in the Q of the circuit I/O as it is stepped.
The second thumb is the response with different values of L and C with the I/O Z stepped in the same manner. The key is the RATIO of the I/O Z and the LC Z. These must be considered when a design is brought into the real world. Pay close attention to the differences in the phase traces in both and comparatively between the two.
When going into higher order filters, you will understand that the complexity increases exponentially, which you will learn as you progress in you studies.
Cheers,
Merv