# Tag Info

4

Yes, it is possible. The approach that I would probably take would be to use a small-to-medium size FPGA to do the DSP (basically 100 DDS generators), a high-speed (250 MHz) DAC, and a reasonably good LC bandpass filter at the output of the DAC. This could be interesting: I did some digging, and it is now possible to do general-purpose programming on the ...

3

An "arbitrary waveform generator" (look it up) in its simplest form is an oscillator that drives a counter that drives a memory that drives a DAC: simulate this circuit – Schematic created using CircuitLab The number of address bits you need is determined by the number of steps in your waveform. The number of data bits is determined by how much ...

3

The reason why OFDM is confusing is mainly because it is never fully presented. There are always gaps and holes in the details. For old-school OFDM, you see details of multiple orthogonal carriers (eg. a fundamental sinusoid and harmonics of it are all orthogonal), where each orthogonal carrier amplitude may be altered (eg. either make the amplitude equal to ...

2

There are commercial chips that will digitize an NTSC (or PAL) signal for you a 8 to 10 bits of precision, but they typically do it at the standard rate of 27 MHz, giving a raw bit rate of 216 - 270 Mbps. But after you have this digital signal, it is relatively straightforward to throw away the parts you don't care about (the chroma channels, the audio, etc....

1

What happens when you increase the amount of distinct states in QAM is you get a higher BER (bit error rate). source To achieve the higher data rate with the same BER, you need to use error correction coding. You see, the BER rises, but your ability to correct the errors increases at a greater pace, as Shannon discovered.

1

You're getting nothing wrong. This just looks like your receiver isn't perfectly using the same frequency as the transmitter. A frequency offset is just a phase that is a linear function over time – hence the rotation in the constellation diagram. Also, you've got some I and Q offset. This should be rotationally symmetrical to 0+0j.

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The bandwidth is defined by the pulse shape, not the modulation type. So, this depends on how you define bandwidth (there's different definitions, and which one you use depends on the purpose of defining the bandwidth) which transmit pulse shaping filter you use.

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