RF Journey

01 / 12

Noise sets the weak-signal limit

A −88.5 dBm available signal reaches a named receiver input at 2.45 GHz. Is it detectable?

Not enough information. The power level and carrier frequency do not determine a decision. We still need the source or antenna noise temperature, noise-density convention, equivalent noise bandwidth, receiver-added noise, in-channel interference and blockers, oscillator phase noise, a detection criterion, and assurance that every quantity uses the same reference plane.

−88.5 dBm signal
source N₀
Bₑₙᵦw
receiver F or Tₑ
I and criterion

The −88.5 dBm value is a new supplied receiver-input value. It is deliberately notderived from the +10 dBm EIRP example in Fields & Waves. Connecting those two values would require the missing propagation, antenna, polarization, mismatch, and loss terms.

Detectability is a conditional engineering statement

A signal can be below the integrated noise yet recoverable by coherent processing, or above it yet unusable because of interference, fading, distortion, or an unsuitable detector. “Above the noise floor” is not a complete requirement.

02 / 12

Random signals have measurable structure

If a noise waveform cannot be predicted sample by sample, what can an engineer measure repeatably?

A realization is one finite record from a random process. Its sample mean, RMS, variance, amplitude distribution, autocorrelation, and power spectral density are estimators—not promises that the next record will look the same. Longer records improve frequency resolution; independent averaging reduces estimator variation.

$x(t) = s(t) + n(t) \qquad \mu _{n} = E[n]$ $\begin{aligned}\sigma_n^2&=\mathrm E[(n-\mu_n)^2]\\n_{\mathrm{rms}}&=\sqrt{\mathrm E[n^2]}\end{aligned}$ $\text{uncorrelated sources}: \sigma _{\mathrm{total}}^{2} = \sum \sigma _{i}^{2}$ For zero-mean noise, RMS equals standard deviation—not zero. Correlated sources require covariance or cross-spectral terms.

Power spectral density carries squared-amplitude per hertz, such as V²/Hz. Amplitude spectral density is its square root, such as V/√Hz. Integrate PSD—not ASD—over the weighted bandwidth to obtain mean-square noise, then take a square root if an RMS amplitude is required.

Interactive · realization, distribution, spectrum

Random-noise laboratory

Step between repeatable finite records. The waveform changes; its estimated statistics and spectrum become more stable only with more samples and independent averages.

Prediction: what happens to a noisy PSD estimate when more independent records are averaged?

Noise model

Record length

PSD averages

Realization seed: 2718
Sample mean: -0.0118
Sample RMS: 0.9571
Sample variance: 0.9159

μ̂ = (1/N)Σxᵢ · RMS = √[(1/N)Σxᵢ²] · σ̂² = (1/N)Σ(xᵢ − μ̂)²Normalized amplitude units. For zero-mean noise, RMS equals standard deviation. Uncorrelated source variances add; correlated sources require covariance or cross-spectral terms.

Model boundary: synthetic, finite, discrete-time, Nyquist-band-limited model. “White” describes the modeled spectrum, while “Gaussian” describes the amplitude distribution; neither implies physically infinite bandwidth or power.

Static takeaway: one random record does not reveal a deterministic waveform. Mean, RMS, variance, histogram, and PSD are estimates whose variability falls as record length and independent averaging increase.

Common misconceptionWhite noise means Gaussian noise.

White describes a flat spectral density over a stated band. Gaussian describes an amplitude probability distribution. A model may be one without the other, and no physical source is white over infinite bandwidth.

03 / 12

From noise density to noise power

Why does a flat −174 dBm/Hz source not produce −174 dBm on a 20 MHz power measurement?

Density is power per unit bandwidth. A linear filter weights that density, and total noise is its area after the filter. For a flat one-sided input density N₀ and a response normalized to unity peak gain, equivalent noise bandwidth replaces the real filter with an equal-area rectangle.

$N=\int_0^\infty S_n(f)|H(f)|^2\,\mathrm df$ $B_{\mathrm{ENBW}}=\frac{\int_0^\infty|H(f)|^2\,\mathrm df}{|H|_{\max}^2}$ $\begin{aligned}\text{Flat }N_0:\quad N&=N_0B_{\mathrm{ENBW}}\\N[\mathrm{dBm}]&=N_0[\mathrm{dBm/Hz}]+10\log_{10}\left(\frac{B_{\mathrm{ENBW}}}{1\,\mathrm{Hz}}\right)\end{aligned}$ This module consistently uses one-sided, positive-frequency RF noise density.

Think about itIf flat-noise bandwidth doubles while density stays fixed, what happens to noise power and RMS noise voltage?

Answer

Power doubles, increasing 3.0103 dB. RMS voltage rises by √2. Increasing bandwidth tenfold raises noise power by 10 dB.

Common misconceptionA filter’s 3 dB bandwidth is always its noise bandwidth.

Not generally. For a one-pole low-pass, the one-sided ENBW is (π/2)f₃dB. Filter shape matters because noise power depends on the whole squared-magnitude response.

04 / 12

Thermal noise and the −174 dBm/Hz reference

Which temperature and bandwidth are hidden inside the familiar “−174 dBm” shortcut?

In the classical matched-source regime, available thermal-noise density is kT W/Hz. The Boltzmann constant is exact: k = 1.380649 × 10⁻²³ J/K. At the standard noise temperature T₀ = 290 K, kT₀ = 4.0038821 × 10⁻²¹ W/Hz = −173.975187 dBm/Hz. At 300 K it is approximately −173.828 dBm/Hz. Neither should silently be called “room temperature” without naming T.

Interactive · kT, resistance, and ENBW

Thermal-noise and bandwidth explorer

Start with available noise-power density, integrate it through a filter, then choose the correct resistor voltage reference: open circuit or matched load.

Prediction: with flat density, what does 10× more ENBW do to integrated noise power?

Temperature 290 K

Resistance 50 Ω

Equivalent noise bandwidth 20.0 MHz

Filter model

Voltage reference

Available density kT: 4.003882e-21 W/Hz-173.975 dBm/Hz
Integrated available noise: 80.08 fW-100.965 dBm
Open-circuit voltage density: 894.9 pV/√Hz
Matched-load RMS voltage: 2.001 µV
Rectangular width: 20.00 MHz

N = ∫₀∞ Sₙ(f)|H(f)|²df = kTBₑₙᵦwFor the one-pole low-pass model, Bₑₙᵦw = (π/2)f₃dB. The plot uses a one-sided, positive-frequency density.

Reference matters: a resistor’s open-circuit density is √(4kTR) V/√Hz. A matched load receives half that voltage, one quarter the mean-square voltage, and exactly kT W/Hz of available power.

Static 290 K, 50 Ω, 20 MHz result: −173.975 dBm/Hz, −100.965 dBm integrated; 0.8949 nV/√Hz open circuit and 2.001 µV RMS across a matched load.

$\begin{aligned}S_v&=4kTR\quad[\mathrm{V^2/Hz}]\\e_n&=\sqrt{4kTR}\quad[\mathrm{V/\sqrt{Hz}}]\end{aligned}$ $v_{n,\mathrm{oc}}=\sqrt{4kTRB}\qquad v_{n,\mathrm{matched}}=\sqrt{kTRB}$ $R = 50 \Omega , T = 290 K, B = 20 \mathrm{MHz}: P_{n} = 80.0776 \mathrm{fW} = -100.9649 \mathrm{dBm}$ Open-circuit density is 0.894861 nV/√Hz; matched-load density is 0.447431 nV/√Hz. The corresponding RMS values are 4.00194 µV and 2.00097 µV.

The factor of four is not missing power. A Thevenin source and equal load form a 2:1 voltage divider, so the load sees half the open-circuit voltage and one quarter the mean-square voltage. That voltage across R delivers kTB of available noise power.

Go deeperWhere the classical kT approximation stops being universal

Planck’s law replaces the classical approximation when hf is no longer much smaller than kT. At 2.45 GHz, hf/k ≈ 0.1176 K and hf/(kT) ≈ 4.05 × 10⁻⁴ at 290 K, so kTB is excellent here. It is not universal at cryogenic temperature or sufficiently high frequency.

05 / 12

Noise sources leave different fingerprints

Does every irregular spectrum belong in the same kTB calculation?

No. Source physics shapes spectrum, amplitude distribution, correlation, direction, time behavior, and mitigation. Thermal and ideal shot noise can be broadband; flicker grows close to low offset; burst noise is intermittent; the antenna collects a changing brightness scene; spurs and other transmitters can be deterministic.

Classify the mechanism before selecting a model
Source	Physical or system origin	Fingerprint and caveat
Thermal	Random carrier motion in resistive loss	Nearly white across ordinary RF bandwidths; scales with temperature
Shot	Discrete charge crossing a barrier	One-sided Sᵢ = 2qI A²/Hz for the ideal Poisson model
Flicker	Device and material processes	Rises toward low offset frequency; often described as 1/f-like
Burst / popcorn	Discrete device-state changes	Time-domain steps or pulses; non-Gaussian and nonstationary
Atmosphere	Oxygen, water vapor, rain, and other propagation media	Frequency-, elevation-, weather-, and path-dependent brightness
Galactic / celestial	Sky brightness and discrete radio sources	Strongly depends on frequency, direction, beam, and polarization
Ground / objects	Thermal radiation coupled through the antenna pattern	Often warm, directional, and installation-dependent
Man-made	Transmitters, electronics, machinery, and infrastructure	May be impulsive, narrowband, broadband, periodic, or intermittent
Quantization	Mapping a continuous input to discrete converter codes	Can resemble white noise only under stated signal and decorrelation conditions
Spurs / interference	Deterministic leakage or other emitters	Lines or occupied signals—not automatically random noise

$\begin{aligned}\text{One-sided shot noise:}\quad S_i&=2qI\quad[\mathrm{A^2/Hz}]\\i_n&=\sqrt{2qI}\quad[\mathrm{A/\sqrt{Hz}}]\end{aligned}$ q = 1.602176634 × 10⁻¹⁹ C exactly. Device bandwidth, excess noise, correlations, operating point, and a non-unity Fano factor may change the practical model.

Common misconceptionQuantization error is always independent white noise with Δ²/12 variance.

That approximation needs sufficient code activity and decorrelation. A coherent or low-level input can produce deterministic, signal-correlated error and spurs; clipping and nonlinearity are different failure modes again.

06 / 12

Phase noise is not an additive broadband floor

Why can a strong signal outside the channel raise in-channel noise even when the RF filter rejects its carrier?

A real oscillator’s phase fluctuates. Mixing a blocker with the local oscillator’s noise sidebands can translate energy into the wanted channel: reciprocal mixing. This impairment depends on blocker power, frequency offset, local-oscillator phase-noise density, filtering, and mixer behavior—not only on the receiver’s additive noise figure.

Interactive · additive noise, modulation noise, and spurs

Additive and phase-noise explorer

Scrub a slowed, normalized carrier phasor and compare mechanisms that can look similar on a spectrum display but require different models.

Prediction: which mechanism lets a strong adjacent blocker raise in-channel noise through reciprocal mixing?

Disturbance model

Paused carrier phase 18°

Carrier power -20 dBm

Offset frequency 100 kHz

Phase noise moves the phasor tangentially; close-in sidebands are normalized to the carrier.

Example carrier: -20 dBm at 2.45 GHz
Example SSB phase-noise density: -100 dBc/Hzat 100 kHz offset, in a 1 Hz bandwidth
Small-noise sideband estimate: -120 dBm/Hz-20 dBm + (-100 dBc/Hz)

v(t) = A[1 + a(t)]cos(2πfᶜt + φₙ(t)) + nₐ(t)Illustrative normalized geometry, not a calibrated oscillator simulation. L(fₒffₛₑₜ) is a single-sideband density ratio to a named carrier power, carrier frequency, offset, and 1 Hz bandwidth.

Static takeaway: additive noise is an absolute floor; AM and phase noise modulate a carrier; flicker is close-in and colored; spurs are discrete. Phase noise from a strong nearby signal can be mixed into a receiver channel as reciprocal-mixing noise.

Single-sideband phase noise L(f_offset) in dBc/Hz is incomplete unless the carrier frequency and power, offset frequency, 1 Hz normalization, sideband convention, and measurement conditions are stated. In the small-noise direct-spectrum approximation, an absolute sideband density is the carrier level in dBm plus L in dBc/Hz.

Common misconceptionA low receiver noise figure guarantees good blocker performance.

NF describes small-signal SNR degradation under stated source conditions. Reciprocal mixing, compression, intermodulation, desensitization, and ADC range can dominate in the presence of strong signals.

07 / 12

What a spectrum analyzer actually displays

When the displayed trace falls after you narrow RBW or VBW, did the source become quieter?

The analyzer passes input through a resolution filter, detects the result, and estimates a display value. RBW sets RF resolution and the noise collected per bin; its true ENBW depends on filter shape. VBW smooths detected display values. Trace averaging reduces estimator variation. Sample, RMS, average, and peak detectors answer different statistical questions, while linear-power and logarithmic averaging are not interchangeable. Neither VBW nor averaging changes the source PSD.

Interactive · density, bins, and display statistics

Spectrum-analyzer noise explorer

Separate the input density from noise collected in one resolution-filter bin, the analyzer’s own input-referred noise, and the final displayed estimate.

Prediction: when RBW increases 10× for flat input noise, what rises by about 10 dB?

Resolution bandwidth

Video bandwidth

Trace averages

Resolution-filter model

Detector definition in this model

Input attenuation 10 dB

Illustrative 15 dB preamplifier

Displayed estimate
Input source per bin
Instrument floor per bin

Input noise density: -160 dBm/Hz
Filter ENBW: 10.60 kHz
Source noise / bin: -119.75 dBm
Instrument floor / bin: -119.75 dBm
Approximate observation time: 64.8 ms

Pbin = PSD + 10log₁₀(Bₑₙᵦw/1 Hz); powers from source and instrument add in wattsRMS = mean of 8 linear-power samples, sample = 1 sample, and positive peak = maximum of exactly 8 samples. Trace averaging is then performed in linear power; VBW is modeled as a moving average.

Educational model, not a calibrated instrument: the assumed input-referred instrument density is −170 dBm/Hz at 0 dB attenuation, and the illustrative preamp lowers it 15 dB. Real DANL, ENBW, detector behavior, and sweep time are model- and setting-specific. VBW and averaging reduce display variation; they do not reduce the source PSD or change RF resolution.

Static takeaway: increasing RBW by 10× collects about 10 dB more flat-noise power per bin. Narrow RBW lowers noise per bin but usually lengthens acquisition. Attenuation protects the front end while raising input-referred instrument noise; a preamp improves measurement sensitivity but reduces overload margin.

Displayed average noise level is a settings-specific instrument quantity, not a universal physical floor. More input attenuation protects against overload but normally raises the input-referred analyzer floor. A preamplifier can lower that floor while reducing headroom. Narrower RBW collects less flat noise per bin and generally requires more acquisition time.

Go deeperSubtracting signal and noise in linear power

If a signal-plus-noise measurement is only 1 dB above a separate equal-bandwidth noise measurement, linear subtraction gives signal/noise = 10^(1/10) − 1 = 0.2589. The signal alone is therefore about 5.87 dB below the noise—not 1 dB above it. This requires stable, comparable measurements and uncorrelated added powers.

Common misconceptionEvery noise trace needs a universal 2.51 dB correction.

Corrections depend on the analyzer architecture, detector, averaging domain, logarithmic processing, filter shape, and the quantity being reported. Use the instrument’s documented noise-marker method and uncertainty, not a memorized universal offset.

08 / 12

SNR, C/N, C/N₀, and SINR

Which ratio still makes sense before a receiver bandwidth has been chosen?

C/N₀ compares carrier power to noise-power density and therefore carries the logarithmic unit dB-Hz. It does not become the dimensionless in-band C/N until a bandwidth is specified. Use “carrier” only for a clearly identified carrier; otherwise SNR is the safer general label.

$\mathrm{SNR} = \frac{P_{s}}{P_{n}} \qquad \mathrm{SINR} = \frac{P_{s}}{P_{n} + P_{i}}$ $\frac{C}{N_{0}} [\mathrm{dB}-\mathrm{Hz}] = C[\mathrm{dBm}] - N_{0}[\frac{\mathrm{dBm}}{\mathrm{Hz}}]$ $\frac CN\ [\mathrm{dB}]=\frac C{N_0}\ [\mathrm{dB\!\cdot\!Hz}]-10\log_{10}\left(\frac B{1\,\mathrm{Hz}}\right)$ dB-Hz is not an ordinary dimensionless dB ratio; the “Hz” records the density normalization.

Each numerator and denominator must share a reference plane, bandwidth, and averaging definition. Add independent noise and interference in watts before taking a ratio. As a later digital-link preview, E_b/N₀ = (C/N₀)/R_b only after the information-bit rate and consistent reference definitions are known.

Common misconceptionC/N and C/N₀ are interchangeable because both are reported in decibels.

C/N is bandwidth-dependent. C/N₀ is a density ratio in dB-Hz. Confusing them silently discards the 10log₁₀(B/1 Hz) term.

09 / 12

Noise factor, noise figure, and equivalent temperature

How can one receiver be described by either a dimensionless factor or a temperature in kelvin?

Noise factor is the degradation of available signal-to-noise ratio through a two-port under stated source and reference conditions. Noise figure is that same factor in decibels. Equivalent input noise temperature represents the receiver’s added noise at its input plane.

$F = \frac{\mathrm{SNR}_{\mathrm{in}}}{\mathrm{SNR}_{\mathrm{out}}} \qquad \mathrm{NF} = 10\log _{10}F$ $T_{e} = T_{0}(F - 1) \qquad F = 1 + \frac{T_{e}}{T_{0}} \qquad T_{0} = 290 K$ $N_{\mathrm{out}} = G k B(T_{s} + T_{e})$ G is available power gain and every temperature is referred to the same input plane.

A matched passive loss L at physical temperature T_p has G = 1/L, Tₑ = (L − 1)T_p, and F = 1 + (L − 1)T_p/T₀. Only when T_p = T₀ does F = L and noise figure equal loss in decibels.

Go deeperY-factor measurement in one line

With calibrated hot and cold source temperatures at the same input plane, Y = Phot/Pcold = (Thot + Tₑ)/(Tcold + Tₑ), so Tₑ = (Thot − YTcold)/(Y − 1). Real measurements must account for source ENR, mismatch, loss, bandwidth, gain stability, linearity, and uncertainty.

Common misconceptionA 2 dB noise figure means the receiver adds a fixed −172 dBm/Hz floor.

NF is a ratio tied to T₀ and reference conditions, not an absolute output floor. Gain, source temperature, bandwidth, losses, impedance conditions, and reference plane still matter.

10 / 12

Why receiver stage order matters

Can the same LNA, filter, and IF amplifier have different total noise figure when their order changes?

Yes. Friis’ equation discounts a stage’s excess noise by all preceding power gain. That makes early low-noise gain powerful and early passive loss expensive. Convert every dB value to linear G and F before applying Friis; never add stage noise figures in decibels.

Interactive · Friis equation in linear units

Cascade-noise explorer

Reorder the same three stages and watch early gain suppress later noise contributions. Add cable loss only at the named input reference plane.

Prediction: which matters more here—removing 1 dB before the LNA or improving NF after substantial gain?

Matched cable loss ahead of the chain, 290 K 0.0 dB

All Friis calculations use power gain G and noise factor F in linear units
Order / stage	Gain	NF / F	Friis contribution	Cumulative gain
1. LNA	15.0 dB / 31.623	1.5 dB / 1.4125	1.412538	15.0 dB
2. Passive filter (290 K)	-3.0 dB / 0.501	3.0 dB / 1.9953	0.031473	12.0 dB
3. IF amplifier	30.0 dB / 1000.000	5.0 dB / 3.1623	0.136430	42.0 dB

Total power gain: 42.000 dB1.5849e+4 linear
Total noise factor: 1.580441
Total noise figure: 1.9878 dB
Equivalent input noise temperature: 168.328 K

Ftotal = F₁ + (F₂ − 1)/G₁ + (F₃ − 1)/(G₁G₂) + …At 290 K, matched passive loss L has G = 1/L and F = L. The initial LNA–filter–IF order gives F = 1.580441, NF = 1.987783 dB, and Tₑ = 168.328 K.

Noise is not the only ordering constraint. A preselector ahead of the LNA may be essential for blocker rejection, stability, linearity, image rejection, or protection even when it worsens small-signal NF. The quietest Friis order is not automatically a buildable receiver.

Static takeaway: with +15 dB/NF 1.5 dB LNA, −3 dB/NF 3 dB filter, and +30 dB/NF 5 dB IF amplifier, total gain is 42 dB and total NF is 1.9878 dB. Moving the filter first raises NF to 4.7053 dB.

For the initial +15 dB/NF 1.5 dB LNA, −3 dB/NF 3 dB filter at 290 K, and +30 dB/NF 5 dB IF amplifier, the three Friis contributions are 1.412538, 0.031473, and 0.136430. Total gain is 42 dB, F = 1.580441, NF = 1.987783 dB, and Tₑ = 168.328 K. Moving the filter ahead of the LNA gives F = 2.954813, NF = 4.705301 dB, and Tₑ = 566.896 K.

11 / 12

Antenna temperature and receiver sensitivity

Is antenna temperature the temperature of the antenna metal, the weather, or the sky?

It is an equivalent noise temperature derived from the brightness-temperature scene weighted by the receiving antenna’s directional, polarization-sensitive pattern, plus applicable antenna losses. It varies with frequency, pointing, environment, atmosphere, ground, nearby objects, and polarization. It is not automatically 290 K, 3 K, or 4 K.

$T_a=\frac{\iint T_b(\theta,\phi)G_r(\theta,\phi)\,\mathrm d\Omega}{\iint G_r(\theta,\phi)\,\mathrm d\Omega}$ $T_{\mathrm{sys}} = T_{a} + T_{e}$ $290 K \mathrm{convention}: \text{sensitivity} = -173.9752 + 10\log _{10}(\frac{B}{1 \mathrm{Hz}}) + \mathrm{NF} + \mathrm{SNR}_{\mathrm{required}}$ Tsys addition is valid only when Tₐ and Tₑ are referred to the same plane. The antenna integral shown is a simplified single-polarization form.

Interactive · common-plane system noise and threshold

Receiver margin explorer

Choose either the conventional 290 K noise-figure method or an actual antenna temperature plus equivalent receiver temperature. Never add NF in decibels directly to a non-290 K source.

Temperature model at the named receiver input plane

Available signal -88.5 dBm

Carrier frequency 2.45 GHz

Equivalent noise bandwidth 20.0 MHz

Receiver noise figure 1.988 dB

Required output ratio 10.0 dB

Matched pre-receiver loss, 290 K 0.0 dB

In-channel interference at this same reference plane

Decision at the receiver input

Passes the entered criterion

The 10.0 dB criterion is an explicit hypothetical requirement, not a universal definition of detectability. Coding, integration time, false-alarm probability, fading, and implementation loss can move the real threshold.

Reference-plane model

Standard 290 K source + 168.3 K receiver

All source and receiver temperatures are referred to the plane ahead of the selected cable loss.

Source noise density: -173.9752 dBm/Hz
Receiver NF / Tₑ at input plane: 1.9878 dB168.328 K
System temperature: 458.328 K
Total noise density: -171.9874 dBm/Hz
Integrated noise: -98.9771 dBm
C/N₀: 83.4874 dB-Hz
C/N: 10.4771 dB
Conditional sensitivity: -88.9771 dBm
Margin: +0.4771 dB
Frequency check: 2.45 GHzclassical kT model selected
Noise power: 126.6 fW

N = kT₀BF; sensitivity = N[dBm] + required C/NC/N₀ = C − N₀ in dB-Hz. C/N = C/N₀ − 10log₁₀(B/1 Hz). Powers, not decibel values, are added when interference is included.

Explore a deliberately simple sky-and-ground antenna-temperature model

This is a hypothetical two-region weighting exercise, not a sky map or antenna simulator. It assumes a lossless normalized pattern, 0.65 of pattern weight on a 30 K sky region and 0.35 on a 290 K ground region, giving Tₐ = 121.0 K.

Hypothetical sky brightness 30 K

Hypothetical ground brightness 290 K

Pattern weight on sky 0.65

Real Tₐ is a pattern-weighted brightness-temperature integral over direction, polarization, frequency, atmosphere, ground, objects, antenna loss, and pointing. It is not the metal temperature and is not automatically 290 K or the cosmic-background temperature.

Static journey result: −88.5 dBm in 20 MHz with Tₐ = 290 K, NF = 1.9878 dB, and a 10 dB requirement has +0.4771 dB margin. A matched 1 dB cable at 290 K ahead of the receiver changes the referenced NF to 2.9878 dB and the margin to −0.5229 dB.

When Tₐ differs from 290 K, convert receiver NF to Tₑ and use kB(Tₐ + Tₑ). The conventional sensitivity equation is otherwise liable to hide a warm ground contribution, a cold-sky advantage, or loss between the antenna and receiver. A quoted “sensitivity” must also state its bandwidth, criterion, probability or error target, and implementation assumptions.

12 / 12

Putting it together: one 2.45 GHz noise journey

Under one explicit set of assumptions, does the supplied −88.5 dBm signal meet a 10 dB in-band requirement?

Assume the available signal and antenna noise are referred to the receiver input; Tₐ = 290 K; one-sided density and Bₑₙᵦw = 20 MHz are used; the LNA–filter–IF chain is in its initial order; and there is no in-channel interference, reciprocal mixing, compression, fading, or implementation loss. The 10 dB requirement is hypothetical but explicit.

Common-plane calculation from source density to conditional margin
Step	Value	Meaning
Available signal	−88.5000 dBm	Supplied at the receiver input; not derived from the prior module’s +10 dBm EIRP
Source	Tₐ = 290 K	N₀ = −173.9752 dBm/Hz at the same reference plane
Channel	2.45 GHz; Bₑₙᵦw = 20 MHz	Source noise = −100.9649 dBm
Receiver	G = 42 dB; F = 1.580441	NF = 1.9878 dB; Tₑ = 168.328 K
System	Tsys = 458.328 K	Input-referred density = −171.9874 dBm/Hz; noise = −98.9771 dBm
Ratios	C/N₀ = 83.4874 dB-Hz	C/N = 10.4771 dB in 20 MHz
10 dB criterion	Sensitivity = −88.9771 dBm	Margin = +0.4771 dB: conditional pass
Receiver output	Signal = −46.5000 dBm	Noise = −56.9771 dBm; SNR remains 10.4771 dB

$C/N=-88.5000-(-98.9771)=10.4771\,\mathrm{dB}$ $\mathrm{margin} = -88.5000 - (-88.9771) = +0.4771 \mathrm{dB}$ Conditional answer: yes, narrowly, under only the assumptions above.

Now place a matched 1 dB cable at 290 K ahead of the LNA and move the reference plane ahead of that cable. The complete network NF becomes 2.9878 dB—exactly 1 dB worse—while receiver Tₑ becomes 287.001 K and Tsys becomes 577.001 K. Input-referred noise becomes −97.9771 dBm, C/N becomes 9.4771 dB, sensitivity becomes −87.9771 dBm, and margin becomes −0.5229 dB: a conditional fail.

The connected receiver-noise model

Classify the source. State the density convention and reference plane. Integrate with ENBW. Refer source and receiver temperatures to one plane. Cascade linear factors in order. Add independent powers in watts. Then compare the resulting SNR or SINR with a named criterion.

Ungraded review

Check your understanding

Answer each question in your own words, then reveal the model answer.

01What thermal-noise power does −173.9752 dBm/Hz produce across a 20 MHz equivalent noise bandwidth?
Model answer
−100.9649 dBm at 290 K. Doubling the bandwidth would raise total noise by 3.0103 dB and RMS noise voltage by √2.
02Why does placing a passive filter before an LNA usually worsen receiver noise performance?
Model answer
Friis’ equation discounts later excess noise by preceding gain. Early passive loss receives no such protection: in the lesson’s chain, filter-first ordering raises NF from about 1.99 dB to 4.71 dB.
03A −88.5 dBm signal faces −98.9771 dBm of input-referred noise and requires 10 dB C/N. Does it pass?
Model answer
Yes, narrowly: C/N is 10.4771 dB and margin is +0.4771 dB. A matched 1 dB cable ahead of the LNA changes that margin to −0.5229 dB under the stated 290 K assumptions.

Sources and further study

NIST fixes the exact constants; MIT, ITU-R, and IEEE provide theory and terminology; Keysight and Rohde & Schwarz document receiver and instrument practice. Manufacturer-specific behavior should always be checked against the exact instrument manual and settings.

Constants, noise, and antennas

Receiver and measurement practice

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