IF Filtering vs RF Filtering

[robismyname]

rf vs if

Can someone explain simply to me why it is much easier to perform filtering at IF rather than RF? THe way its explained to me in this book by "Dobkin" is not all that clear to me.

Dobkins says, " The task of filtering is much easier if it is performed at the IF rather than the original RF frequency. For example, a 20MHz channel represents 0.83% of the RF frequency of 2.4GHz but 5% of the 374-MHz IF, it is both plausible and true that distinguishing two frequencies 5% apart is much easier than the same exerfcise for a 1% difference."

I mean I understand how he did the math to get .83% and 5% but I dont understand how a 20MHz channel at 2.4GHz is harder to filter than a 20MHz channel at 374MHz.

 

[biff44]

if vs rf

There are physical limits to the Quality Factor of resonators used in making bandpass filters. Without using something pretty exotic, common Q factors range from 50 to 300 for most microwave resonators (transmission line, L-C, etc). For specific circumstances, you can get higher Q's (maybe up to a few thousands) using Surface Acoustic Wave (SAW), Bulk Acoustic Wave (BAW), or Crystal resonators.

One definition of Q for a resonator is:

Q = (center frequency)/(3 dB bandwidth)

So, if you have a given bandwidth to pass with sharp roll off, the lower the center frequency you use, the lower a Q you can have in your resonators that make up the filter.

Rich

 

[robismyname]

if vs. rf

One definition of Q for a resonator is:
Q = (center frequency)/(3 dB bandwidth)
So, if you have a given bandwidth to pass with sharp roll off, the lower the center frequency you use, the lower a Q you can have in your resonators that make up the filter.

So in order to obtain a step rolloff you need a lower Q? Isn't higher Q better?

 

[biff44]

To make a bandpass filter, you need a number of resonators. The Q I was talking about was the performance of a single one of those resonators.

If you simulate a bandpass filter with ideal components, and then change the inductors to have a low Q, you will see an increase in insertion loss (especially at the bandedges), and the rejection will not be as steep.

 

[robismyname]

quote]

filtering is easier at lower frequency(IF) than higher frequency(RF) all because of the value of Q?

As an example:
BW = 20MHz,f = 2400MHz,Q=120
BW = 20MHz,f = 374MHz,Q=19

so how does this correlate to filtering is easier at lower frequencies than at higher frequencies?



Another example:
BW = 1MHz,f = 2400MHz,Q=2400
BW = 1MHz,f = 374MHz,Q=374


again I ask how does this correlate to filtering is easier at lower frequencies than at higher frequencies?

 

[Phytech]

quote]

filtering is easier at lower frequency(IF) than higher frequency(RF) all because of the value of Q?

As an example:
BW = 20MHz,f = 2400MHz,Q=120
BW = 20MHz,f = 374MHz,Q=19

so how does this correlate to filtering is easier at lower frequencies than at higher frequencies?



Another example:
BW = 1MHz,f = 2400MHz,Q=2400
BW = 1MHz,f = 374MHz,Q=374


again I ask how does this correlate to filtering is easier at lower frequencies than at higher frequencies?

In order to build the filter of your interest, you need a minimum unloaded resonator Q (Qu)to build your filter. Depending on the number of poles (resonators) you use, Qu needs to be a few times or more bigger than the filter Q (which is defined by the center frequency divided by 3-dB band width) so that you can obtain a good filter response.

In your first example, you would need minimum Qu of about 120 X 5 = 600 to build your filter centered at 2400 MHz, while you only need 19 x 5 = 95 to build a filter with similar response at 374 MHz.