What is the process for converting dB/Hz to dB/decade?

[jaydnul]

Hi,

This is probably a very simple conversion, but my brain don't work at the moment...

If I have a value of say 5udB/Hz, how do I convert that to dB/decade of frequency?

Thanks!

 

[scottdave]

The term dB per decade means for every multiple of 10 of the frequency, it changes by the anounaof decibels. For example if a filter has a response of 10 dB per decade, you could look at the attenuation at say 500 Hz. Perhaps it is 5 dB. Then multiply by 10 so at 5000 Hz it is 15 dB (5 + 10), then at 50000 Hz it is 25 dB. This would not cobvwrcto tbe linear attenuation you described of dB / Hz

http://www.learnabout-electronics.org/ac_theory/filters83.php

 

[DaveE]

I don't know the context, but db/Hz sounds like a noise power measurement (db/sqrt(Hz) for voltage or current noise). This is a common way to describe white noise that has an amplitude proportional to the measurement bandwidth. So if you measure white noise of 5udB/Hz over the frequency range of 1MHz to 2MHz you will get a noise amplitude of (2MHz - 1MHz)*5udB/Hz = 5dB. It's linear, so you get the same amplitude if you measure over the bandwidth of 100MHz to 101MHz.
dB/decade(Freq) is not linear, it is scaled logarithmically, as in the filter example given above. So a filter that attenuates by 10dB between 1Hz and 10Hz will attenuate an additional 10dB from 10Hz to 100Hz.
So, as I understand the common use of these terms in EE, you can't really convert between them, they describe different behavior. Like trying to describe a linear function using logarithms.

 

[tech99]

Hi,

This is probably a very simple conversion, but my brain don't work at the moment...

If I have a value of say 5udB/Hz, how do I convert that to dB/decade of frequency?

Thanks!

The unit 5udB has no meaning.
Do you mean 5 dB microvolt per Hz? This a is written 5dBuV or 5dBu.
If this is a noise specification, then you need to specify the actual frequency because the number of Hz varies with bandwidth. For instance, between say
100 to 1000 Hz contains more bandwidth than from 1 to 10 Hz.
Usually, noise power, not voltage, is dependent on bandwidth, so we need to say dB per square root Hertz.
Please try to be exact with the units and tell us more what you are doing.

 

[jaydnul]

Ok, I see. At work I have standard bode plots in my simulator, which plot on a logarithmic scale. I can clearly see that it is decreasing with a slope of -20db/dec when it crosses my unity gain frequency, but when i put that curve into the derivative function, it will return -5.131u.

 

[DaveE]

Ok, I see. At work I have standard bode plots in my simulator, which plot on a logarithmic scale. I can clearly see that it is decreasing with a slope of -20db/dec when it crosses my unity gain frequency, but when i put that curve into the derivative function, it will return -5.131u.

Is this a question? If so I have no idea what you are asking.
"derivative function", what derivative function? Note: Differentiators tend to rise with increasing frequency (i.e. +20dB/dec), not decrease.
I am assuming that "-5.131u" is some sort of magnitude value, but that is just a guess because you haven't described it or added the units of measurement to it (like you did with -20dB/dec). Why are you comparing the slope on your plot to a position on the plot (they are typically unrelated, like the location of your car and its speed)?