large bandwidth white noise leads to higher Vrms noise, why??

From the noise power perspective, the power of white noise is equal to KTB. Boltzmann's constant, T is the temperature, and B is the noise bandwidth.

To normalize it 50 ohms resistor, p=(Vrms)^2/R=KTB, so Vrms=sqrt(KTBR).

From the above equation, a larger bandwidth leads to a larger RMS noise.

However, from the bellowing figure, white noise is distributed among the frequency range. No matter within 1KHz or 1MHz band, the noise does not change a lot. Noise is equal regardless of the bandwith.

Could you please help to figure out? Thanks.

shanmei said:

From the noise power perspective, the power of white noise is equal to KTB. Boltzmann's constant, T is the temperature, and B is the noise bandwidth.

To normalize it 50 ohms resistor, p=(Vrms)^2/R=KTB, so Vrms=sqrt(KTBR).

From the above equation, a larger bandwidth leads to a larger RMS noise.

However, from the bellowing figure, white noise is distributed among the frequency range. No matter within 1KHz or 1MHz band, the noise does not change a lot. Noise is equal regardless of the bandwith.

Could you please help to figure out? Thanks.

View attachment 140660

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I hope I understand your question well..
kTB is the lowest White Noise Power and if you pay attention there is no noise source at all in that equation.
If you add a resistor, this component will add extra noise onto it.Therefore RMS_Noise is higher than kTB..

Let's compare the two cases:

1. A 50 ohms resistor, with 1kHz bandwidth, the Vrms=14nV; with 1MHz, the Vrms=450nV. The Vrms values can be calculated from the above equations.

2. White noise is randomly distributed among 1K-1MHz, so the noise should be approximately the same.

Why are the two cases different? 1. Noise changes with the different bandwidth. 2. Noise is almost the same along the bandwidth since it is white noise.

Thanks.

Let me try to put it in a qualitative manner.
A rectangular wave can be broken down by analysis into a combination of a fundamental sine wave and an infinite number of harmonic frequencies. The fundamental and the harmonics are all present at the same time. If the wave is passed through a low-pass filter, a wide bandwidth filter will pass more harmonics without attenuation than a narrower bandwidth.

White noise contains all frequencies at the same time. The wider the bandwidth, the more the energy contained within the frequency band.
Imagine a large collection of people, each given a microphone and assigned to make sound of a particular frequency into the mike. The higher the number of mikes switched on, the greater the total volume of sound produced.

That's a good analogy example.

Although the white noise is "equally" distributed among the frequency range, but when they are mapped back to the time domain, the higher the bandwidth, the higher energy of the noise. Thanks.