In the previous post, I indicated that coupled Broadband energy levels are determined by whether the signals are random or coherent. Random/non-coherent signals originate from sources that are not time related. A couple of good examples would be receiver thermal noise and DC motor brush noise. Another would be the RF environment taken as a whole. Because the individual energy components are not on at the same time, they do not combine directly and the total increase in energy as a function of bandwidth is Δe dB = 10 log BW2 / BW1.
Coherent signals on the other hand are time related. This type of noise is frequently called impulsive noise and can be a single impulse like ESD, lightning, or repetitive like a computer clock. The primary difference between single and repetitive signals is that a single transient signal has a continuous RF spectrum. No matter what frequency the receptor circuit responds to, RF energy from the single transient exists at that frequency. The spectral amplitude may be so small that the circuit does not react, but the energy was there for the duration of the transient. For coherent energy, the increase in captured energy levels as a function of bandwidth is Δe dB = 20 log BW2 / BW1. This makes the receptor more susceptible to transients as its bandwidth increases.
For the repetitive transients, energy exists at discrete narrowband frequencies determined by the turn on repetition rate. For example, if a circuit is turned on at a 100 kHz rate, it will produce narrowband harmonic frequency components at 100 kHz intervals starting at 100 kHz. If turned on at a 1 GHz rate, the circuit will produce narrowband harmonic frequency components at 1 GHz intervals starting at 1 GHz. For a 50% duty cycle trapezoidal waveform, the odd harmonics will have the greatest amplitude. If it were possible to generate a perfect square-wave, there would be no even harmonics. That’s not possible because the rise-time must always be greater than zero.
Yes, the signals are coherent, but are they broadband? That all depends on the bandwidth of the receptor. For our 100 kHz example above, if the measurement were made with a 10 kHz receiver (with a perfect rectangular passband) and the bandwidth was increased to 1 MHz, more harmonics would be captured and the increase would be: Δe dB = 20 log 1MHz /10 kHz, or 40 dB. For the 1 GHz example if we were measuring with a 10 kHz band width and changed to 1 MHz or even 1 GHz, no additional harmonics would be captured.
The significance of the comment above regarding a perfect rectangular passband is that no perfectly rectangular filters exist . . . although mechanical, ceramic, and especially the newest digital ones are getting close. Next time we will take another look at broadband bandwidth.
-Ron Brewer
