Where do the units of SFDR “dB·Hz^(2/3)” come from?

The units of spurious-free dynamic range (SFDR) are dB·Hz^(2/3). The units can be a source of confusion. The short answer is that it is a product of ratios between power levels (dBm) and noise power spectral density (dBm/Hz). The units of dBHz^(2/3) are for SFDR normalized to a 1Hz bandwidth. For the real SFDR of a system, the units are in dB.

If we look at a plot of the equivalent input noise (EIN), the fundamental tone, OIP3 (output intercept point of the third order distortion), and IMD3 (intermodulation distortion of the third order), a ratio of 2/3 exists between OIP3 and SFDR. This can be recognized from the basic geometry, given that the slop of the fundamental is 1 and the slope of IMD3 is 3.

Now, we need to look at the units of both OIP3 and EIN. The units of OIP3 are dBm and the units of the equivalent input noise (a noise power spectral density) are dBm/Hz.

SFDR = (2/3)*(OIP3 – EIN)

[SFDR] = (2/3) * ( [dBm] – [dBm/Hz] )

Now, remember that in logarithmic operations, division is equal to subtracting the denominator from the numerator. and therefore:

[dBm/Hz] = [dBm] – 10*log_10([Hz])

Note that the [Hz] term is still in logarithmic scale. We can use dBHz to denote the logarithmic scale in Hertz.

[dBm/Hz] = [dBm] – [dBHz]

Substituting this into the SFDR unit calculation:

[SFDR] = (2/3) * ( [dBm] – ( [dBm] – [dBHz] )

This simplifies to:

[SFDR] = (2/3) * ( [dBm] – [dBm] + [dBHz] )

Remember that the difference between two power levels is [dB].

[SFDR] = (2/3) * ( [dB] + [dBHz] )_

The units of [dB] + [dBHz] is [dBHz], as we know from the same logarithmic relation used above for [dBm] and [dB].

[SFDR] = (2/3) * [dBHz]

Now, remember that this is a lkogarithmic operation, and a number multiplying a logarithm can be taken as an exponent in the inside of the logarithm.Therefore, we can express Hz again explicitly in logarithm scale, and move the (2/3) into the logarithm.

(2/3) * [dBHz] = (2/3) * 10*log_10([Hz]) = 10*log_10([Hz]^(2/3))

We can return our units back to the dB scale now, giving us the true units for SFDR: dBHz^(2/3):

[SFDR] = [dBHz^(2/3)]

1 thought on “Where do the units of SFDR “dB·Hz^(2/3)” come from?

  1. Pingback: Converting from normalized SFDR (dBHz^(2/3)) to real SFDR (dB) | RF Photonics Lab

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