There are several comments to be made:
* The passband is in the 100 kHz range (not MHz)
* Passband and stopband attenuation can be determined only if you know (and tell us) about the passband and stopband frequencies.
* passband ripple (if any!) cannot be detected because of bad frequency resolution. You need more calculated points of the transfer curve.
* You should use logarithmic scales for both axes (dB) and display alos the phase .
*At first, I suppose your bandpass is of 2nd order.
In this case, the passband is mostly defined between the two frequencies for which the gain is 3dB less than the maximum gain.
In your case this would be app. a level of -3.1 dB on both sides of the maximum. However, this region is not shown in your diagram.
*According to your diagram the passband attenuation is 0.1 dB.
*Passband ripple exist only for bandpass functions of higher order (4, 6,...).
*The task of a bandpass is to attenuate some freqency components outside the passband. For these frequencies the stopband attenuation is defined. Do you have any requirenents in this respect?
---------- Post added at 15:20 ---------- Previous post was at 15:17 ----------
By the way: Is it the same filter as in your 1st posting?
(1)Thank you for telling me that passband ripple only exists for higher order. Yes my filter is 2-order, so there is no ripple, thank you for this instruction!
If you want you can see the gain variation within the passband as ripple (some people do that). See (2) below.
(2) You said -3.1db, do you mean from the top of the plot, it is around -0.15? Is it OK for a filter?
It is the normal case to specify the -3 dB points (below the maximum) as passband. However, depending on the application and the requitements, respectively, you also can define the passband for any other values.
Example: If you allow only a 0.1 dB variation within the passband (0.25-0.15 dB) , one can derive the following edge frequencies from your drawing: 20 kHz.....10 MHz. (By the way, that is very very broad!).
For other attenuation values you must extend the displayed frequency range.
(3) Now I just want to present this filter as a project work (not have a specification already and tried to meet it), so I need to define a passband range (like from f1 to f2). How can I define the range from the plot and present that in this range the filter works well?
See above. The answer, if it works "well" can be given only in case the requirements are known. But, as mentioned above, the bandpass seems to exhibit a very small quality factor Q (very large bandwidth).
(4) After you define the range from the plot, could you tell me how to calculate the stopband rejection accroding to its db?
As mentioned earlier: What are the stopband frequencies?
Some general remarks (perhaps helpful):
Normally, a bandpass and its parameters are defined as follows:
1.) The bandpass center frequency is fo=...Hz. The gain or attenuation at fo shall be ...dB
2.) The filter must transfer frequencies between f1 and f2 (passband) ; the gain/attenuation in this region must not vary by more than x dB (example 1 or 3 dB).
3.) The following frequencies f<f3 and f>f4 must be attenuated at least by y dB (example 60 dB). This defines the stopband frequencies.
Regards, LvW
It's a 1st order lowpass rather than a bandpass transfer function. It has a real pole, for which no Q can be determined.The transfer function is (Gm/C)/{s+(Gm/C)}, could you tell me how to calculate Q from this equation?
Some additional comments:
It's a 1st order lowpass rather than a bandpass transfer function. It has a real pole, for which no Q can be determined.
Your original frequency response is showing a second order bandpass, as said. But for high relative bandwidth/low Q, it can be better represented as a lowpass/highpass combination. If you want to improve the stop band attenuation, you can increase the lowpass and/or highpass order without changing the bandwidth.
P.S.: It's difficult to guess about the purpose of the bandpass following the charge amplifier without knowing about the involved signal. It can be e.g. a means to improve the signal-noise ratio or just a special preprocessing required by the succeeding signal chain. A possible application would be e.g. a nuclear particle detector with pulse height discrimination or energy spectrum measurement.
Strictly spoken, the charge amplifier output can't be a step function, because a real amplifier needs a time constant to reset the integrator. Otherwise, it can't process repeated input events.
(1)I am terribly sorry that I made the mistake.The transfer function is [(Gm/C)/{s+(Gm/C)}]×[1/{s+(Gm/C)}]
I am afraid, you again have made a mistake. The 2nd part of the function equals a 1st order highpass only when you replace the "1" by "s". OK?
(2)You said that it can be better represented as a lowpass/highpass combination, and I am curious that is it any differences between lowpass/highpass combination and bandpass? Because from the textbook I saw that bandpass is lowpass/highpass combination, please advise!
To realize a 2nd order bandpass you generally have two options:
(a) A combined frequency-selective LC or active RC combination that can produce a conjugate-complex pole pair (for high Qp and good selectivity), or
(b) A series combination of a 1st order lowpass with a 1st order highpass. In this case - as mentioned by FvM - you realize only two real poles (instead of complex) with rather poor selectivity (rather large bandwidth).
I don't know, which text book you are referring to. In my view, a passive RLC circuit or an equivalent active "resonator" circuit would be considered as the common 2nd order bandpass prototype. For low Q, the complex pole pair turns however into two real poles and the bandpass can be represented by separate low- and highpasses.
I fear, the CSA topic tends to go too far beyond the present thread's scope. The "output can't be a step function" point is simply referring to the fact, that the integrator must be reset either periodically by a switch or "continuously" by a resistor in parallel to the capacitor.
Hello LvW, thank you for your kindest help. I just found that you also asked me if the two plots I put are for the same filter. The answer is YES!
*The bandpass transfer curve in your 1st posting #1 exhibits a 3-dB bandwidth (70.7% of the maximum level) between
f1=100 kHz and f2=1.5 MHz.
* Question: Which expression is plottet against the frequency in posting#3 ? Up to now, I was of the opinion it is the magnitude in dB. But this is not the case if both plots originate for the same filter.
Please clarify.
Hello LvW, thank you for pointing out this problem! I am sorry that I just found this reply from you! There are some problems of the plots in #3. And I want to clarify with something and ask some more questions.
Let us focus on plot in #1.
(1)The input of the bandpass filter is 100mV. And the peak value of the output is around 400mV. So the magnitude was amplified. Is it OK for a filter to amplified the input?
(2) Is the peak db of the filter MUST be 0 or less than 0?
Thank you very much!!!!!!!!!!!!!!!!!!
OK, it is good to know if a diagram like #1 shows an output voltage or the transfer function. Please note, that it is really important to explain what a diagram shows.
To your question (1): In many cases it is desired or even a requirement to have passband gain. In your case the maximum gain is 12 dB. Please note, that normally such a diagram should be scaled in dB on a logarithmic frequency axis.
Question (2): I don't understand. I think it is answered by (1), is it not?
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