Superheterodyne Receiver Using RF Budget Analyzer App

Hello, I am trying to use RF Budget Analyzer App to do some budget analysis, before that I try to rerealize the example shown in the link below:

https://www.mathworks.com/help/rf/examples/superheterodyne-receiver-using-rf-budget-analyzer-app.html

However, I got couple questions about that, hope anyone got ideas or intuitions can help me out, Thanks!

1). The input signal setup signal bandwidth = 20MHz, then RF filter bandpass = 20MHz (which makes sense), but bandpass for IF filter is 5MHz, how to understand that? Is the parameters wrong or some points I didn't get.

2). I tried to change the bandpass for both RF filter and IF filter, but these changes didn't change the budget result at all. Only signal bandwidth input at the beginning of budget link affacts the results (mainly SNR). I can undersand the SNR changes because of reduce of noise power bandwidth, but how could bandpass of filter not affecting the results?

Again, hope anyone having ideas can give me some clues.

Thank you!

ZHX

Answer 1

[Mehmed Saad]

The paper implemented in example is

https://ieeexplore.ieee.org/document/5688819

see Table II of paper (you can also view it here)

[Hanxiang Zhang]

Thanks Mehmed. I have checked that paper. I can understand the 3-dB bandwidth is 5MHz, but shounld't the passband for both RF & IF filter be the same (might not exactly same)? Also, it dosen't make sense that nothing happens by changing the bandpass parameter of filters. Would you like to give more explanations?

Thanks!!

[Mehmed Saad]

I have not studied that example personally. As far as i remember, practically it is not efficient to get that narrow passband at such a high RF frequency (maybe i am wrong for this scenario).

because signal is of limited bandwidth (<5MHz) so RF filter has no effect or very slight effect (on the corners).

[Hanxiang Zhang]

Yes, I just found it today that paper states the occupied BW is < 5MHz. And I was misled by the explanation on Mathswork that the note "Signal bandwidth = 20MHz". Thanks again!

By the way, how should understand the saying that "receiving Bandwidth" in the paper? Is this the band width that going through the RF filter and only used to calculate the noise source power at the beginning of the link?

[Mehmed Saad]

I think so.