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Chapter 9: Problem 7

At \(0.1 \mathrm{~Hz}\), a low-frequency measurement has a noise value of \(-60 \mathrm{dBm}\) when a resolution bandwidth of \(1 \mathrm{mHz}\) is used. Assuming \(1 / \mathrm{f}\) noise dominates, what would be the expected noise value \((\) in \(\mathrm{dBm})\) over the band from \(1 \mathrm{mHz}\) to \(1 \mathrm{~Hz}\) ?

Short Answer

Expert verified
The expected noise value is approximately -51.6 dBm.

Step by step solution

01

Understand Noise Power Scaling

Noise power scales linearly with bandwidth. If you know the noise power ( ext{in} ext{dBm}) for a certain bandwidth, the noise power for a different bandwidth can be calculated, assuming a linear relation if it's white noise. However, here we have \(1 / f\) noise, indicating that the power spectral density (PSD) decreases with frequency. To find the change in total noise power from one bandwidth to another, we typically need to integrate the PSD over the new bandwidth.
02

Recognize Bandwidth Contribution

In the given problem, the initial condition is set for a very small bandwidth of \(1\,\text{mHz}\). The total noise power over a frequency band should be calculated by considering all contributions from \(1 \, \text{mHz}\) to \(0.1 \, \text{Hz}\). For \(1/f\) noise, since PSD \(\propto \frac{1}{f}\), you'll integrate over the frequency band to determine the total power.
03

Integrate the Power Spectral Density

The PSD \(S(f)\) is given by the noise in the \(1 \text{mHz}\) bandwidth, \(-60 \text{dBm/mHz}\), meaning each millihertz contributes \(10^{-6} \text{mW/mHz}\). Convert dBm to mW: 1 dBm = 0.001 mW, so \(-60 \text{dBm} = 10^{-6} \text{mW}\). For \(1/f\) noise, the power \(P(f)\) will integrate the inverse of frequency over the desired band.
04

Calculate the Integral of \(1/f\) from 1mHz to 1Hz

The total noise over the band is calculated by integrating \(S(f) = \frac{10^{-6}}{f}\, ext{mW/mHz}\) from \(1mHz\) to \(1Hz\). This results in \(\int_{0.001}^{1} \frac{10^{-6}}{f} df = 10^{-6} \times (\ln(1) - \ln(0.001)) = 10^{-6} \times 6.9077 \approx 6.9 \times 10^{-6} \text{ mW}\).
05

Convert the Result to dBm

Convert the power from mW to dBm. The formula for conversion is: \(\text{dBm} = 10 \log_{10}(\text{mW})\). Plug in the values: \(10\log_{10}(6.9 \times 10^{-6}) \approx 10\times(-5.161) \approx -51.61 \text{ dBm}\).
06

Final Step: Present the Result

Thus, the expected noise value over the band from \(1 ext{mHz}\) to \(1 ext{Hz}\) is approximately \(-51.6 \text{ dBm}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Spectral Density
In noise analysis, one major aspect to consider is the Power Spectral Density (PSD).
PSD provides a representation of how power is distributed over frequency.
For noise sources, it helps us understand how much power falls within a specific bandwidth.
  • Why is PSD Important? Understanding PSD is crucial for managing and anticipating noise behavior in systems.
  • How to Calculate PSD: It typically involves measuring noise power over a small bandwidth, then using these data points to assess the noise at other frequencies.
  • Example: In the given problem, we know the noise at a bandwidth of 1 mHz is 0 dBm. This implies a PSD value that diminishes as you move to higher frequencies.
By understanding how PSD works, you can make more informed decisions about managing noise across different frequency ranges.
1/f Noise
1/f Noise, also known as "flicker noise," is a type of noise that distinctly varies with frequency.
This means that its power spectral density decreases with increasing frequency.
  • Understanding 1/f Behavior: In our scenario, the 1/f noise indicates a non-uniform distribution of noise across frequencies, where lower frequencies experience higher noise intensities.
  • Impact on Systems: Systems dominated by 1/f noise often have a significant amount of noise at lower frequencies. This makes careful bandwidth selection pivotal.
  • Expression of 1/f Noise: The relationship can be depicted mathematically by \(S(f) = \frac{C}{f}\) where \(C\) is a constant derived from specific conditions.
It's imperative to consider 1/f characteristics in designing systems to ensure performance isn't compromised by unexpected noise spikes at lower frequencies.
Bandwidth Integration
When analyzing noise across a range of frequencies, bandwidth integration becomes vital.
This process sums the contributions of power across the desired frequency range to compute total power.
  • Significance: Bandwidth integration allows us to move from understanding noise at a narrow segment (like 1mHz) to a broader spectrum (like 1Hz).
  • Application: Integration involves mathematical techniques to handle continuous variations, particularly useful in calculating total 1/f noise output.
  • Example from Exercise: The calculation involved integrating the expression \(S(f) = \frac{10^{-6}}{f}\) from 1mHz to 1Hz to obtain a total noise power in mW.
Mastering bandwidth integration is essential for precise noise analysis, enabling the conversion of point-specific data to whole system insights.

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Most popular questions from this chapter

A sinusoidal signal of \(150 \mathrm{mV} \mathrm{rms}\) is in series with a noise voltage. a. If the noise voltage is \(1 \mathrm{mV}\) rms, what is the signal to noise ratio expressed in \(\mathrm{dB}\) ? b. If a SNR of \(70 \mathrm{~dB}\) is desired, how small must the noise voltage be made?

A noise current has a power of \(-30 \mathrm{dBm}\) referenced to a \(50 \Omega\) load. What is its rms value in Amperes?

What signal level is required, in \(\mathrm{V}_{(\mathrm{rms})}\), to achieve a SNR of \(62 \mathrm{~dB}\) in the presence of \(-35 \mathrm{dBm}\) noise?

Calculate the percentage THD value of a signal in which the fundamental component is at \(1 \mathrm{MHz}\) and has a voltage level of \(1 \mathrm{~V}_{\text {rms }}\), and the only significant harmonics are the first three above the fundamental. These harmonics, at \(2 \mathrm{MHz}, 3 \mathrm{MHz}\), and \(4 \mathrm{MHz}\), are measured to have voltage levels of \(1 \mathrm{mV}_{\mathrm{rms}}, 0.5 \mathrm{mV}_{\mathrm{rms}}\), and \(0.3 \mathrm{mV}_{\mathrm{rms}}\), respectively.

We saw in equation (9.41) that the noise bandwidth of a first-order, low-pass filter is \((\pi / 2) f_{0}\). Show that the noise bandwidth of a second-order, low-pass filter given by $$ A(s)=\frac{A_{o}}{\left(1+\frac{s}{\left(2 \pi f_{0}\right)}\right)^{2}} $$ is equal to \((\pi / 4) \mathrm{f}_{\mathrm{o}}\).
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