MarkF Senior Heliman Location: Palo Alto, CA
My Posts This: Topic Forum | Howdy, Gang!
I thought I'd post a few comments about frequencies and bandwidth. I'd mentioned earlier that I'd love to reuse existing crystal frequencies, but as nice as that would be, doing so would probably reduce the performance of this radio. The reason for this is tied to basic material physics: 455 KHz and 10.7 MHz intermediate frequencies have become so popular because they correspond to "natural" resonant frequency ranges for ceramic, and quartz crystals, respectively. While a ceramic filter can do a decent job and is cheap, in and of itself, it can't come close to the performance of a good crystal filter. In other words, a 10.7 MHz I.F. is pretty appealing just for the materials that are available there.
So, at the moment I'm leaning towards the use of a 10.7 MHz single I.F. for both the transmitter and the receiver, with a single four-pole crystal filter used there to provide superior adjacent channel interference elimination. I'll have to learn more before I'll know for sure whether this is the way to go, but it's an appealing possibility. If we do go this route, the first I.F. stage will require a (10.7 - .024) = 10.676 MHz crystal, and the second crystal will bring us directly to the transmit frequency (the final crystal would be the transmit frequency - 10.7 MHz). My guess is that this should work well for both the transmitter and the receiver, but we'll see! [In particular, I'm concerned about the potential for intermodulation products with those four crystals, so this might not be such a good idea, after all. Experts???]
As far as bandwidths go, I wouldn't be surprised if you were confused by all the bandwidth numbers I've been mentioning. To understand them, let's start with the symbol rate, which is the number of output data symbols that are sent per second. QPSK sends two bits for every symbol, so a 12,000 bits/second data rate will yield a symbol rate of 6,000 symbols/second. I'd mentioned previously that this design needs to pick a specific value for "alpha", which is the "excess bandwidth" constant. In this case, it's 0.35. What that means is that the output bandwidth that is required for this signal is 6,000 + 6,000 * 0.35 = 8,100 Hz. Indeed, at the -25 dB point, the signal is right around that bandwidth (slightly better, since the FCC mandates at least -25 dB reduction in signal strength at an 8,000 Hz bandwidth). The way the FCC limits bandwidth is to talk about a +/- range, so they specify the emissions limits at +/- 4000 Hz from the center of the transmit waveform (along with additional limits at +/- 5 KHz, +/- 20 KHz, etc).
Yet another bandwidth figure I've mentioned is 6,000 Hz: this is the nominal bandwidth of the Square-Root Raised Cosine filter that's used in both the transmitter and the receiver, corresponding to a carrier-modulated signal. This is the bandwidth at the -3dB point - in other words, the range of frequencies that the filter will pass. Between 6 KHz (+/- 3,000 Hz) and 8 KHz (+/- 4,000 Hz) the transmitter's output falls off sharply. However, no realistic filter can just cutoff frequencies at one point, so you'll always see a gradual transition from what's called the "passband" to the "stopband". Furthermore, when moving from baseband to carrier modulation, the bandwidth doubles, as you can see in the spectrum plots in this thread.
With all that kept in mind, it is completely correct to refer to this project as needing a bandwidth of 3 KHz (in a spectrum plot showing the baseband of either I or Q, and referring to the -3 dB cutoff point), 4 KHz (in a spectrum plot at baseband referring to the -25 dB cutoff point), 4.5 KHz (baseband, -60 dB), 6 KHz (carrier-modulation, -3 dB), 8 KHz (carrier modulation, -25 dB), 9 KHz (carrier modulation, -60 dB), or 10 KHz (nominal channel bandwidth, as defined by the FCC)!
I'm confused, too!
Cheers!
MarkF |