02-25-2006 03:45 PM
02-27-2006 03:10 PM
Hey,
Those are some great questions! Let me provide a little background on PAM modulation and maybe we can go from there. First, although PAM is generally though of as a way to modulate analog signals, it really is a form of digital modulation in that it is represented by discrete voltage levels. For example, consider 2-PAM modulation. In this circumstance, four different bits are used to represent 4 (2^2) different levels of analog voltage (unique symbols). Thus, by sending digital data as 0’s and 1’s. we can represent analog voltage levels.
First, you will need to determine what type of PAM modulation you are using. For example, with 2-PAM, you can represent 4 voltage levels. With 4-PAM, you can represent 16 levels. Second, you will derive the digital data based on these levels. With 4-PAM modulation, the resulting bitstream can be thought of as being generated in 4-bit chunks. Each “chunk” can be thought of as an individual symbol which represents a unique analog value.
I hope this helps and let me know if you have any further questions.
Regards,
David Hall | Applications Engineering | National Instruments
02-27-2006 05:03 PM
02-28-2006 08:21 AM
03-01-2006 11:11 AM
Hey,
I guess the best answer to your question is yes and no. The "Modulate PAM" VI does expect a bit stream of 0's and 1's as bytes and not words. However, in your code, I noticed that you had an array of 16-bit words wired to this input. That why there is a green "coersion dot" right at the input of this VI. The coersion dot tells us that LabIVEW is converting your array of 16-bit words to an array of 8-bit bytes. Looking at your array, this is all 0's and 1's anyway, so there is no difference between representing the bitsream in 8 bits or 16 bits.
I did notice, however, that with the "generate system parameters" VI, you have kept the default setting of 2-PAM. This means that you are generating PAM data at 2 bits per symbol, for a total of 4 symbols. I think the actual polar coordinates are: (-1 + 0i),( -0.33 + 0i),( 0.33 + 0i), and (1 + 0i). That means that every two bits of your input bit stream will be used to represent a different symbol. I looked at your example, and it seemed like you were useing every 8 bits to represent a differnet symbol. In that case, you would want to select 8-PAM. Otherwise, you will need to figure out how to represent your sine wave with only 4 symbols (which is a little tough to do). I hope this helps and let me know if you have any more questions.
Regards,
David Hall | Applications Engineering | National Instruments
03-02-2006 07:59 PM
03-03-2006 05:49 PM
Hey,
I looked at your code, and I'm not completly sure how you're evaluating how "good" your modulated signal is. In general, the best way to evaluate this would be to de-modulate the signal, and then see if the sine wave can be sucessfully demodulated. If you are looking to improve the bit resolution of the sine wave, you could try representing it with more bits. Currently, it looks like you are representing your sine wave with 32-bits. This is actually probably higher than you need. In general, 8-bits (8-PAM) is much more typical. Again, let me know what about your modulated signal that you are trying to improve.
Also, just a word on symbol rate. This affects, how often the characteristics of the modulated wave are changing. For a high symbol rate, the characteristics of the modulated signal change very quickly. Wtih low symbol rates, the characteristics of the modulated signal are changing more slowly. Thus, demodulating the signal will be more accurate. The actual numeric value of symbol rate represents the number of symbols per second. With 8-PAM, there are 2^8 or 256 different symbols. Thus, bit-rate = symbol rate * 8 for this type of modulation. I hope this helps and let me know if you have more information.
Regards,
David Hall | Applications Engineering | National Instruments
03-04-2006 08:38 AM
03-04-2006 08:42 AM
03-06-2006 09:50 PM
Hey,
First of all, you are absolutely right in your analysis of what 8-PAM really means. As you mentioned, there are 256 individual symbols associated with 8-PAM, each of these distinguished by amplitude. In terms of seeing noisy signals, the data in your VI's didn't appeary THAT abnormal. First, your FFT of the modulated frequency spectrum seems relativley OK. In general, it is not uncommon to see several frequency spikes around the modulated carrier frequency. In some ways, I think some of the filtering VI's have actually made this appear worse (apply complex filter). Also, the gaussian noise VI also adds nose to some of the IQ data plotted on the screen. I were you, I might take a look at some of the PAM examples again, to see what kind of filters are used there.
Regards,
David Hall | Applications Engineering | National Instruments