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How does the "unit" per Hertz in behave in the exponent? My specific question is about the calculation of the modulation index \$\eta\$ from a measured spectral density of a phase modulation on a spectrum analyser. For a read value of P in dBc/Hz (for example a noise floor), the modulation index can be calculated by $$\eta=2\cdot10^{\frac{P~[\text{dBc/Hz}]}{20}}$$ What happens with the per Hertz in the exponent? Is the result of the unit rad/Hz? I have read several threats about units in exponents not being possible because an exponent can be developed into a power series, just to give two examples: Exponentation and Units-Exponential.

My guess is the following: Per Hertz is not a unit anyway but a normalisation to the equivalent noise bandwidth. Thus, the outcome would be rad/Hz.

To make the question clearer: Units in exponents cannot exist because exponents can be developed into power series leading to an addition in the from of ()+()^2+()^3+... And Units of different power cannot be added. So how can I deduce a modulation index per Hertz from a noise power per Hertz?

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  • \$\begingroup\$ I'm unsure what it is that is giving you a problem here. \$\endgroup\$ Commented Feb 19, 2018 at 17:48
  • \$\begingroup\$ I will try to rephrase: I can read a value in dBc from the spectrum analyser and calculate the modulation index. But if I use a noise marker, I read a value in dBc/Hz, and then I am not sure if the equation above still holds because then the exponent has a unit. \$\endgroup\$ Commented Feb 19, 2018 at 17:53
  • \$\begingroup\$ The noise will be set to produce a certain noise per Hz in the bandwidth it occupies. If the bandwidth it occupies is 100 Hz then the dBc power for the whole noise will be 100x higher. Does this help? Not sure it will but maybe we still have crossed-wires here? \$\endgroup\$ Commented Feb 19, 2018 at 17:59
  • \$\begingroup\$ This does not help to answer my question. Also, I do not understand the 100 Hz to factor 100. \$\endgroup\$ Commented Feb 26, 2018 at 11:37
  • \$\begingroup\$ Then you need to be clearer and you need to take much less than a week to respond if you want decent help. \$\endgroup\$ Commented Feb 26, 2018 at 11:39

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(Translated to english):

Floor noise has a PSD proportional to 1/f. That means, in measure units, your PSD of floor noise depends both parameters, the dBc measured at certain bandwidth and the bandwith. So, when you want to calculate the modulation index, you must consider the noise contains this frequency dependant component.

https://en.wikipedia.org/wiki/Noise_spectral_density

Hope it helps.

Fast answer in Spanish (tried to translate the best I can, sorry for my english):

El ruido de fondo tiene una densidad de potencia espectral proporcional a 1/f. En unidades, eso quiere decir que la densidad de potencia espectral del ruido de fondo dependerá de los decibelios en dBc, y también, del ancho de banda utilizado para la medición de dBc. Por lo que al hallar el factor de modulación o índice de modulación, tendrás que tener en cuenta que el ruido de fondo tiene esa componente dependiente de la frecuencia.

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  • \$\begingroup\$ I understand, why a measurement gives dBc/Hz. It is the output from typical noise analyzers such as for example a R&S FWSP. But, my questions is more of a physical/mathematical nature. "Units" in general do not make sense in the exponent because of the power series as mentioned above. But, if a noise analyzer measures dBc/Hz, that goes in to exponent. dBc is only a ratio and no unit, but /Hz is kind of a unit. Development in a power serios leads to an addition of 1/Hz+1/Hz^2+1/Hz^3+... So I can not get rad/Hz because it contradicts mathematics. \$\endgroup\$ Commented Feb 26, 2018 at 11:42
  • \$\begingroup\$ You can't measure noise without a bandwidth nor a exposition time. Spectrum analyzer specifications include a parameter called displayed average noise level (DANL), which is the amplitude of the analyzer’s noise floor over a given frequency range with the input terminated in 50 ohms and the internal attenuator set to 0 dB. DANL values are normalized to a bandwidth of 1 Hz, so it is necessary to compensate for the resolution bandwidth (RBW) setting of the analyzer. More information about measuring noise: reeve.com/Documents/Noise/Reeve_Noise_6_NFMeasSpecAnalyz.pdf \$\endgroup\$ Commented Mar 7, 2018 at 8:27
  • \$\begingroup\$ That is understood, but does not answer the question. It is about the calculation of the modulation index and a limit of detection like quantity. Modulation index refers to amplitudes, but I would like a modulation index like quantity that refers to noise (that is measured within a bandwidth) that can act as a limit of detection in rad/Hz. \$\endgroup\$ Commented Mar 7, 2018 at 8:42
  • \$\begingroup\$ Using the dimensional analysis, use the 1/hz as a unit. $$ \left[x \right ] = \left[\frac{1}{Hz}\right] $$ Eta would be, now, a variable that depends on 1/Hz. $$ \eta\left(\frac{1}{Hz}\right) = 2 \cdot 10 ^{ \cfrac{ P {\left(dBc\cdot\frac{1}{Hz}\right) } }{20}} $$ The similar way leads to $$ \eta\left(x\right) = 2 \cdot 10 ^{ \cfrac{ P {\left(dBc\cdot x\right) } }{20}} $$ \$\endgroup\$ Commented Mar 7, 2018 at 10:36
  • \$\begingroup\$ I thought a lot about this problem and read some more and I conclude that the per Hz cannot be seen as a unit in this case, but it can be omitted for this calculation leading to the caluclation that @José Manuel Ramos stated. In literature, it seems to be common to ignore this. \$\endgroup\$ Commented Oct 17, 2018 at 11:26

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