RF Journey

01 / 12

Noise sets the weak-signal limit

A −88.5 dBm signal reaches a receiver input at 2.45 GHz. To decide whether the receiver can recover it, first compare the signal with the noise present over its bandwidth. A wider bandwidth admits more noise; the receiver’s components also add noise. Whether the resulting ratio is sufficient depends on the receiver’s detection method and the performance required. This signal-to-noise ratio, or SNR, compares wanted signal power with noise power in the same band.

Wanted signal at the input: −88.5 dBm at 2.45 GHz
Source noise temperature: 290 K
Noise bandwidth: 20 MHz
Required output SNR: 10 dB · hypothetical

First calculate source noise and compare it with the supplied signal. Then include noise generated inside the receiver. All signal and noise powers will refer to the same receiver input. Later sections add interference, strong unwanted signals, oscillator phase noise, and instrument settings, which can change how a real receiver performs.

The −88.5 dBm input is supplied for this example. To derive it from the transmitter EIRP in Fields & Waves, we would also need propagation loss, receiving-antenna properties, polarization, and feed losses.

State what successful reception requires

A receiver can sometimes recover a signal below the noise power in its band by combining samples using known signal timing or patterns. A stronger signal may still be 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

Measure the voltage at a receiver input while no wanted signal is present. A second recording will not repeat the first sample for sample. Nevertheless, recordings made under the same conditions can have similar average power. We describe that repeatable behavior using statistics. One finite recording is called a realization.

A zero average does not mean zero power

Take four equally spaced samples: −3, −1, +1 and +3 µV. Their average is zero. But current reverses with voltage in a resistor, so both positive and negative voltages deliver energy. Recall RMS from RF Power: square the values, average them, then take the square root.

\text{Mean}=\frac{-3-1+1+3}{4} =0\,\mathrm{\mu V}

\text{Mean square}=\frac{9+1+1+9}{4} =5\,(\mathrm{\mu V})^2

\text{RMS}=\sqrt5\,\mathrm{\mu V} \approx2.24\,\mathrm{\mu V}

These are the values for this four-sample recording.

To measure fluctuations around the average, first subtract that average from every sample. The average of the resulting squared values is the variance; its square root is the standard deviation. Add a fixed 2 µV offset to our four samples: the mean becomes 2 µV and RMS becomes 3 µV, but the standard deviation stays 2.24 µV. The fluctuations around the mean have not changed.

Run the short recording below. Watch current reverse while delivered energy keeps increasing, then compare the two offsets.

Current reverses. Energy keeps increasing.

A four-value recording applied to a resistor · slowed time

Hold each recorded voltage for 1 µs across 50 Ω. Voltage and current reverse together, so their product—delivered power—stays positive. A femtowatt (fW) is 10⁻¹⁵ W.

Delivered so far0 zJ

0.00 of 4 µs

Start a fresh record with

Elapsed portion only: mean — µV · RMS — µV. These settle to the complete-record values below at 4 µs.

Complete-record outcomes · same fluctuations about each mean
Quantity	No offset	+2 µV
Mean [µV]	0	2
RMS [µV]	√5 ≈ 2.24	3
Variance [(µV)²]	5	5
Energy [zJ]	400	720

The +/− signs fix the voltage reference; the copper arrow gives current direction. Resistor tint indicates power, not temperature. The dashed energy curve is the complete-record result; copper shows energy delivered so far. A zeptojoule (zJ) is 10⁻²¹ J. The display freezes at the last sample when the record ends.

When, how often, and at which frequencies?

The time trace shows when each voltage occurs. An amplitude histogramcounts samples in voltage ranges. A Gaussian distribution has a bell-shaped profile: values near the mean occur more often than large positive or negative excursions.

Power spectral density (PSD) describes how mean-square voltage is spread over frequency, in V²/Hz. Multiply each density by its frequency interval and add the results; then take one square root to obtain RMS voltage. For example, a flat 4 × 10⁻¹² V²/Hz across 100 Hz gives 4 × 10⁻¹⁰ V², or 20 µV RMS.

White noise has flat density over the stated band. Colored noise has more power at some frequencies than others; in the example below, neighboring samples tend to stay similar, emphasizing slow variations. Both models have Gaussian amplitude distributions. The impulsive model adds occasional large excursions. Compare the three views while changing the model.

Three views of the same noise model

One recording, three questions. Compare Gaussian white, colored Gaussian and impulsive noise on the same fixed scales.

Choose a noise model

When do the values occur?

Mean -0.01 µVRMS 0.96 µVVariance 0.92 µV²

Which values occur most often?

At which frequencies?

All three plots use the same record. Integrated PSD, using linear densities: 0.92 µV², equal to its RMS squared.

Try another recording or improve the estimate

Samples per recordSpectra to average

One sample every 1 µs. A longer record gives closer frequency bins; more averages make the spectrum less uneven. Neither changes the noise source. Each histogram bar covers 0.5 µV; its area is the fraction of samples in that interval.

Normalization and the finite-record model

$x_{\mathrm{RMS}}^2=\widehat{\sigma}^{\,2}+\widehat{\mu}^{\,2}$

Mean, RMS, histogram and variance use the first record; variance divides by N. The PSD keeps the DC bin and averages complete independent records. Its linear-density sum times the frequency spacing equals their average mean-square value.

Both Gaussian models have theoretical variance 1 µV²; colored samples are correlated. The impulsive model adds rare larger pulses to a smaller white background. The sampling rate is 1 MHz and the modeled band is 0–0.5 MHz. PSD is in µV²/MHz before converting to decibels; multiply its linear values by the bin width 1/N MHz to find mean-square voltage. This is an illustrative voltage scale, not a measured receiver record.

These are synthetic real discrete-time samples with a rectangular record window and no mean subtraction. Only interior positive-frequency bins are doubled; DC and Nyquist count once. A spectrum estimated from one record is called a periodogram. Longer records improve resolution; averaging reduces estimate variation. No infinite physical bandwidth is implied.

A spectrum estimated from a short recording has uneven peaks even when the underlying noise density is flat. At a fixed sampling rate, a longer record distinguishes more closely spaced frequencies. Averaging independent spectral estimates reduces their random variation. Neither operation changes the source noise itself.

Common misconceptionWhite noise means Gaussian noise.

White describes density across frequency. Gaussian describes how often amplitudes occur. A model can be one without the other; no physical source is white over infinite bandwidth.

Go deeperConnect the statistics to their mathematical definitions

The measured waveform is $x(t)=s(t)+n(t)$ : wanted signal plus noise. The operator $\mathrm E[\,\cdot\,]$ means an expected average over a statistical model, weighting possible values by how often they occur. It is not the electric field from the preceding module. A finite record estimates that average.

\mu_n=\mathrm E[n]

\sigma_n^2=\mathrm E[(n-\mu_n)^2]

n_{\mathrm{rms}}=\sqrt{\mathrm E[n^2]}

n_{\mathrm{rms}}^2=\sigma_n^2+\mu_n^2

For zero-mean noise, RMS equals standard deviation.

Amplitude spectral density (ASD) is the square root of PSD, in V/√Hz. Add PSD areas before taking the square root; do not add RMS contributions directly. In a sampled record, the highest uniquely represented frequency is half the sampling rate, the Nyquist frequency. The explorer excludes noise above that limit.

Go deeperWhen noise sources cannot be added independently

Uncorrelated fluctuations have zero average product after their means are removed. Their variances add. Two copies of the same noise are correlated: adding them doubles voltage and quadruples variance, rather than doubling variance.

\operatorname{var}(n_1+n_2) =\sigma_1^2+\sigma_2^2+2\operatorname{cov}(n_1,n_2)

\operatorname{cov}(n_1,n_2) =\mathrm E[(n_1-\mu_1)(n_2-\mu_2)]

For uncorrelated sources the covariance term is zero, leaving a sum of variances.

Autocorrelation applies the same idea to one waveform and a delayed copy, measuring how its samples relate across a time separation. In frequency-domain calculations, cross-spectral terms account for shared fluctuations between different sources.

03 / 12

From noise density to noise power

For a noise voltage measured across resistance R, divide its voltage PSD by R to obtain power density in W/Hz. This follows the power relation P = Vᵣₘₛ²/R from RF Power. Noise density states how much noise power falls in each hertz. If that density is flat, opening a wider band admits proportionally more noise. For example, 2 fW/Hz over a 10 kHz rectangular band gives 20 pW. If the density varies, divide the band into narrow bins, multiply each density by its bin width, and add those powers. This is the area under a density plot in linear units; areas on a dBm plot cannot be added directly.

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

Answer

Power doubles, increasing by about 3.01 dB. RMS voltage rises by √2 across the same resistance. Increasing bandwidth tenfold raises noise power by 10 dB.

Wider bandwidth. More noise power.

Keep the source unchanged. Widen the receiver’s passband and compare the shaded area with the output voltage.

290 K sourceIdeal filter50 Ω load

Receiver bandwidth

Passed noise 20.0 fWModel RMS 1.00 µV

Solid: 5 MHz outputDashed: 20 MHz reference

5 MHz admits 25% of the reference noise power. Its RMS voltage is 0.50 times the reference.

Both traces use the same input record. RMS describes the complete record; a short visible peak does not determine it.

About this illustration

The next section derives this source density from temperature. Here a matched source at 290 K supplies approximately 4.004 fW per MHz. The ideal unity-gain filter passes every frequency below its edge and rejects every frequency above it. Only source noise is counted; the load’s own noise is excluded.

The waveform combines 128 fixed-phase, equal-power frequency contributions over 0–20 MHz. Changing bandwidth retains the same input record. Its 12.8 µs period repeats in 48 seconds; the visible 0.8 µs history need not have the complete record’s RMS. This low-pass illustration is not the actual 2.45 GHz RF waveform. Filter delay and switching transients are omitted.

A narrower band admits less noise, but it must still pass the wanted signal. Removing part of that signal can make reception worse. The illustration isolates the noise contribution.

A real filter has sloping edges

A real filter passes some frequencies more strongly than others. Its response H(f) is output voltage divided by input voltage at frequency f. With equal input and output resistances, its power gain is $|H(f)|^2$ . A frequency interval passed at half the peak power gain contributes half as much white noise as an equally wide interval at the peak.

Equivalent noise bandwidth (ENBW) is the width of a rectangle at the filter’s peak power gain whose area equals the area under the full power response. A filter passing 1 MHz at unity power gain and another 1 MHz at half power has 1.5 MHz ENBW. The expressions below use one-sided density, meaning all noise power is accounted for over positive frequencies. The simple N = N₀B relation assumes unity peak power gain. For a different peak gain, divide output noise power by that linear gain to express an equivalent noise power at the input. This is an input-referred result.

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}

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)

Sₙ is input noise-power density in W/Hz; H is the filter response, not magnetic field. The integral adds the weighted frequency-bin powers as their widths become very small.

Compare the two shaded areas below: the filter shapes differ, but their total areas and passed white-noise powers are equal.

Different shapes. The same noise power.

Flat input noise · unity peak power gain. The shaded area sets how much noise passes.

Two-level filter

Power gain: 1, then ½

1½

012 MHz

1 MHz × 1 + 1 MHz × ½

Equivalent rectangle

Same peak gain: 1

1

012 MHz

1.5 MHz × 1

Equivalent noise bandwidth 1.5 MHzBoth areas are 1.5 MHz. With the same flat input density, both pass the same noise power. This is a power response, |H(f)|².

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.

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04 / 12

Thermal noise and the −174 dBm/Hz reference

Thermal motion of charge carriers in a resistor produces a fluctuating voltage, even without an applied signal. In the classical approximation used here, its available noise-power density is kT W/Hz, where T is source temperature in kelvins: this is the source contribution delivered to a matched load. The Boltzmann constant is exact: k = 1.380649 × 10⁻²³ J/K. At the standard noise temperature T₀ = 290 K, this is about −173.98 dBm/Hz, conventionally rounded to −174 dBm/Hz. At 300 K it is about −173.83 dBm/Hz. State the temperature when using either value.

A resistor also has an equivalent fluctuating noise voltage. For resistance $R$ , the open-circuit voltage PSD is $S_v=4kTR$ . Represent the resistor by that voltage source in series with its resistance, called a Thevenin model. An equal load forms a 2:1 voltage divider, so it sees half the open-circuit voltage and one quarter the mean-square voltage. That voltage across the load delivers the available noise power $kTB$ . Here $B$ is ENBW.

These voltage and power values describe the source resistor's contribution. A real load resistor also generates noise. At equal temperatures the two resistors exchange equal average noise powers, so there is no net heat transfer between them.

S_v =4kTR [\mathrm{V^2/Hz}]

e_n =\sqrt{4kTR} [\mathrm{V/\sqrt{Hz}}]

v_{n,\mathrm{oc}} =\sqrt{4kTRB}

v_{n,\mathrm{matched}} =\sqrt{kTRB}

R =50\,\Omega, T =290\,\mathrm K

B =20\,\mathrm{MHz}

P_n =kTB \approx80.08\,\mathrm{fW} \approx-100.96\,\mathrm{dBm}

Open-circuit density is about 0.895 nV/√Hz; matched-load density is about 0.447 nV/√Hz. The corresponding RMS voltages are about 4.00 µV and 2.00 µV.

In the two circuits below, compare the open output with an equal load. Change temperature, bandwidth or resistance and follow both voltages on their shared ruler.

One resistor, two voltage references

Temperature and noise bandwidth set available power. The circuit connection determines the voltage that delivers it.

Source temperature 290 KEquivalent noise bandwidth 20.0 MHz

Available density · kT: -173.98 dBm/Hz 4.004e-21 W/Hz
Available noise · kTB: -100.96 dBm 80.08 fW

Source and matched-load resistance 50 Ω

Compare RMS noise voltages · one fixed scale for both connections

Open circuit

4.002 µV RMS

894.9 pV/√Hz density · no load power

Matched load

2.001 µV RMS

447.4 pV/√Hz density · 80.08 fW source power delivered

Both drawings count the source resistor’s noise only. A real load also generates noise. Raising both resistances increases the voltages while available power stays at kTB.

Different filter shapes, equal noise area

Hold ENBW fixed; change the power response

Rectangular width 20.00 MHz · same 80.08 fW noise

Copper: filter response. Dashed forest: equal-total-area rectangle. For the one-pole response, about 16% of the total area lies beyond this window; the shaded part alone is not the complete noise integral. The frequency axis is normalized by ENBW, so a bandwidth change preserves the shape.

Voltage and filter relationships

$v_{n,\mathrm{oc}}=\sqrt{4kTRB},\quad v_{n,\mathrm{load}}=\frac12v_{n,\mathrm{oc}}$

$N=kTB_{\mathrm{ENBW}},\quad B_{\mathrm{ENBW}}=\frac\pi2f_{3\mathrm{dB}}$ for the unity-peak-gain one-pole low-pass. These are one-sided classical noise densities; all power calculations use the full response.

First answer for the opening receiver

The 290 K source supplies about −100.96 dBm of noise in 20 MHz. Our −88.5 dBm signal therefore has SNR = −88.5 − (−100.96) ≈ 12.46 dB: 2.46 dB above the hypothetical 10 dB requirement. This first answer assumes an ideal receiver that adds no noise. We will include the real receiver’s contribution after learning how to describe it.

Go deeperWhere the classical kT approximation stops being universal

Planck’s law replaces the classical approximation when the energy hf is no longer much smaller than kT. Here h is Planck’s constant, 6.62607015 × 10⁻³⁴ J·s, and f is frequency. 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

A warm resistor is one source of noise. The number of charges crossing a transistor junction in a short interval can fluctuate too, and an antenna collects unwanted radiation from sky, ground and nearby equipment. These mechanisms respond differently to temperature, current, frequency and surroundings. Use the table to identify a likely mechanism before choosing a calculation.

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	Independent charge arrivals fluctuate around the average current; ideal density increases with current
Flicker	Device and material processes	Stronger at low frequencies; often with density proportional to 1/f
Burst / popcorn	Discrete device-state changes	Random steps between device states; often not Gaussian
Atmosphere	Oxygen, water vapor, rain, and other propagation media	Noise level depends on frequency, elevation angle, weather, and propagation path
Galactic / celestial	Noise radiated by the sky and individual 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	Rounding recorded voltages to fixed levels	Can resemble white noise when the error is sufficiently uncorrelated with the signal
Spurs / interference	Deterministic leakage or other emitters	Lines or occupied signals—not automatically random noise

A repeating spectral line or another transmitter is interference, not automatically white thermal noise. Its frequency and timing help distinguish it from a broadband background; the appearance alone is not a complete diagnosis.

Go deeperWhy can a steady current fluctuate?

For the ideal shot-noise model, individual charges cross a barrier independently and randomly. The average current magnitude $I$ determines how many arrive per second, and $q$ is the charge of one electron in magnitude. The current PSD $S_i$ and its square-root density $i_n$ are:

S_i =2qI [\mathrm{A^2/Hz}]

i_n =\sqrt{2qI} [\mathrm{A/\sqrt{Hz}}]

q = 1.602176634 × 10⁻¹⁹ C exactly. The ideal model assumes independent arrivals, often called a Poisson process.

Real devices can have correlated arrivals or excess noise. A Fano factor multiplies the ideal shot-noise PSD to describe suppression below, or enhancement above, that independent-arrival value. Device bandwidth and operating point still matter.

Go deeperWhen converter error resembles noise

An analog-to-digital converter (ADC) represents sampled voltages with discrete levels, called codes. The rounding to these levels introduces quantization error; noise and distortion inside the converter are separate effects.

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

Here Δ is the converter’s voltage step between codes. The approximation treats errors as uniformly spread between −Δ/2 and +Δ/2 and uncorrelated with the signal. A periodic input sampled at the same phases on each repetition, or a very small input, can instead produce repeating errors and discrete spectral lines (spurs). Clipping and nonlinearity are different failure modes again.

06 / 12

Phase fluctuations around a carrier

An ideal oscillator repeats each cycle at exactly the expected time. If its phase fluctuates, the waveform reaches the same point in its cycle a little early or late. That is phase noise. Amplitude fluctuations instead make the waveform taller or shorter.

The comparison below isolates those two changes. Follow each phasor and its voltage trace, especially the zero crossings. These periodic variations demonstrate the mechanisms; they are not a random-noise recording.

Height changes. Timing changes.

Two disturbances of the same carrier · slow motion

Dashed forest: clean carrierSolid copper: disturbed carrier

Amplitude changes

Radius 1.18

Heights change. Zero crossings stay aligned.

Phase changes

Angle difference 21°

Zero crossings shift. Radius stays at 1.

Inspect time 0.00 T

A phasor is a rotating signal representation; we use its horizontal (cosine) component for voltage here. RF Signals used a vertical (sine) projection; either works with a consistent phase reference. Copper dots on the zero line mark the disturbed signal’s crossings. Voltage is relative to the clean carrier’s peak (1), on the same fixed scale in both rows; T is one carrier period.

These periodic variations isolate the mechanisms. Real amplitude and phase noise fluctuate unpredictably. The phase-noise density in the next example is supplied separately.

The signals behind the illustration

Write time as carrier cycles u. The clean carrier is cos(2πu); the slower variation repeats every six carrier periods.

$v_{\mathrm{AM}}(u)=\left[1+0.25\cos(2\pi u/6+\pi/4)\right]\cos(2\pi u)$

$v_{\mathrm{PM}}(u)=\cos\left[2\pi u+\frac\pi6\sin(2\pi u/6+\pi/4)\right]$

AM radius varies from 0.75 to 1.25 without changing the angle. PM radius is one, with up to ±30° phase difference. Changing phase does not give every crossing the same time delay.

Phase fluctuations spread some carrier power into nearby frequencies. We describe that spread as a noise density relative to the carrier.

Single-sideband phase-noise density $L(f_{\mathrm{offset}})$ compares noise in a 1 Hz bandwidth on one side of the carrier with the carrier power, in dBc/Hz. The offset is the frequency separation from the carrier. In the small-noise direct-spectrum approximation, add the carrier level in dBm to this relative density to get an absolute sideband density in dBm/Hz. State the carrier frequency and power, offset, sideband convention, and measurement conditions with the result.

From relative to absolute noise density

Separate example: a 2.45 GHz carrier has a specified phase-noise density of −100 dBc/Hz at 100 kHz offset. Keep that relative density fixed and change the carrier power.

Carrier power -20 dBm

−100 dBc/Hz + -20 dBm = -120 dBm/HzAbsolute single-sideband density at 100 kHz offset

This supplied density is attributed to phase fluctuations. An ordinary magnitude spectrum also includes amplitude noise and analyzer noise; it does not by itself isolate phase noise.

How a strong unwanted signal brings LO noise into the channel

Many receivers translate a radio signal to a lower frequency before processing it. A mixer combines the incoming signal with a receiver-generated tone from a local oscillator (LO), producing sum and difference frequencies. Filtering selects the wanted translated band. A strong unwanted incoming signal is a blocker, even when its carrier lies outside the wanted RF channel.

A real oscillator’s phase fluctuates, spreading some power into noise sidebands around its carrier. A blocker that reaches the mixer can mix with those LO sidebands, placing unwanted noise inside the translated channel. This is reciprocal mixing. Rejecting the blocker’s translated carrier afterward does not remove that in-channel noise; reducing the blocker before it reaches the mixer can help. Follow the three frequency paths below to see why the unwanted noise reaches the same output band as the wanted signal.

Three simultaneous paths through one mixer

For these difference-frequency paths, subtract LO frequency from RF frequency. The intermediate-frequency (IF) filter passes a narrow channel around 100 MHz.

RF input − LO component → IF output
RF input	LO component	IF result
Wanted signal 1000 MHz	Main LO 900 MHz	100 MHz Passed
Blocker 1010 MHz	Main LO 900 MHz	110 MHz Rejected
Same blocker 1010 MHz	LO noise near 910 MHz	Noise near 100 MHz Passed

The filter rejects the translated blocker carrier at 110 MHz, but noise from its mixing with LO sidebands shares the wanted 100 MHz channel. Rejecting the blocker before the mixer reduces both blocker paths.

Only relevant difference-frequency paths are shown. LO noise near 910 MHz is a small band of fluctuations, not a second pure tone. This is a frequency map, not a calibrated noise-power calculation.

Common misconceptionA low receiver noise figure guarantees good blocker performance.

Noise figure, developed below, describes small-signal noise performance. Strong signals can also cause compression (less gain than expected), intermodulation (new frequencies created by nonlinear response), or overload in an analog-to-digital converter. Those effects and reciprocal mixing can prevent reception even when additive noise is low.

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07 / 12

What a spectrum analyzer actually displays

Resolution bandwidth (RBW) sets the width of the analyzer’s frequency-selective filter. A narrower filter admits less broadband noise into each displayed bin, so the trace falls even though the source’s noise density is unchanged. The filter shape determines its ENBW.

Video bandwidth (VBW) smooths the detected display values; trace averaging combines repeated measurements. These reduce display variation rather than changing the source. Detector choice also matters: a peak detector reports different statistics from an RMS detector, and averaging logarithmic readings differs from averaging powers in watts.

Input attenuation reduces the signal before the analyzer electronics, protecting them from overload but worsening input-referred noise. A preamplifier boosts the signal before later electronics add their noise; it also adds noise of its own and can overload. In the experiment, first narrow RBW with the source unchanged, then increase the number of averages. Compare the change in mean power with the change in trace variation.

What makes the analyzer trace move?

The source stays at −160 dBm/Hz. Change the bandwidth to admit more or less power per bin. Then change how the display estimates that power.

All plotted powers are referred to the input connector, before the attenuator: we divide out the chain’s gain to compare source and instrument noise at the same point.

Resolution bandwidthResolution-filter shape

ENBW 10.60 kHz · source power per bin -119.75 dBm

Displayed estimate
Source
Analyzer-chain addition

In this model, Sample uses one power sample; RMS averages eight; positive peak selects the largest of eight. Try more trace averages to reduce display variation.

Video bandwidthTrace averagesDetector

Trace averaging combines repeated results in watts. VBW smooths display values. Neither changes source noise density or RF resolution.

Expected detected power -116.02 dBmmean in watts, then converted to dBm

Fixed −165 to −80 dBm axis; 0 displayed points clipped. Averaging a peak-selected reading does not remove its upward bias.

Noise added before the display

Input → attenuator at 290 K → analyzer

Input attenuation 10 dB

Preamp · 15 dB gain

Chain-added density, referred to input: -158.66 dBm/Hz
Chain-added power per bin: -118.41 dBm
Source + chain, before detection: -116.02 dBm
Illustrative acquisition time: 8.1 ms

Attenuation reduces overload risk but adds thermal noise and makes later noise larger when referred to the input. A preamp reduces the effect of later analyzer noise; it cannot remove the attenuator’s earlier contribution. Overload itself is not modeled.

Power accounting and instrument-model limits

$P_{\mathrm{bin}}[\mathrm{dBm}]=S_n[\mathrm{dBm/Hz}]+10\log_{10}\!\left(\frac{B_{\mathrm{ENBW}}}{1\,\mathrm{Hz}}\right)$

Source and chain powers add in watts. The matched linear chain has a 290 K attenuator, an optional 15 dB-gain preamplifier whose own input-referred added density is −176.30 dBm/Hz (NF 2 dB), and an analyzer whose own added density is −170 dBm/Hz. Each contribution is referred through the gains ahead of it. Chain-added noise alone is not the displayed average noise level (DANL), which includes the connected source or termination.

Rectangular ENBW equals RBW; the illustrative Gaussian-like filter uses 1.06 RBW. Independent exponential power samples represent Gaussian-noise detection. The displayed curve uses a repeatable realization so settings can be compared; a real trace also changes from acquisition to acquisition.

VBW applies a trailing average over approximately RBW/VBW display points. The time estimate is the larger of 81 × averages/RBW and 81/VBW seconds. These illustrate tradeoffs, not calibrated filter behavior or instrument sweep time. Actual DANL, detector correction and ENBW depend on the instrument and its settings.

Displayed average noise level is a settings-specific instrument quantity, not a universal physical floor. Report the front-end, filter and detector settings with a noise reading. 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 $S/N = 10^{1/10} - 1 = 0.2589$ . 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.

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08 / 12

SNR, C/N, C/N₀, and SINR

The signal-to-noise ratio (SNR) compares wanted signal power with noise power in the same measurement bandwidth. A −90 dBm signal and −100 dBm of noise have a 10 dB SNR: ten times as much signal power as noise power. When the signal is a clearly identified carrier, the same in-band comparison is called carrier-to-noise ratio, C/N. Use SNR for a general wanted waveform.

\mathrm{SNR} =\frac{P_s}{P_n}

\mathrm{SINR} =\frac{P_s}{P_n+P_i}

Pₛ is wanted signal power, Pₙ is noise power, and Pᵢ is interference power. These equations use watts and give linear ratios.

The signal-to-interference-plus-noise ratio (SINR) also includes unwanted signals in the denominator. Add independent noise and interference powers in watts before forming the ratio; a single dBm subtraction works only after they have been combined. Compare powers at the same connector, over the same frequency range and time interval.

Noise matters because a receiver must decide what was sent. The next simplified receiver makes two possible decisions from a value measured after filtering and processing.

When noise changes the decision

After receiver processing, a clean value of −1 represents bit 0 and +1 represents bit 1. The receiver chooses 0 below zero and 1 above zero.

Noise RMS · same eight values, scaled

○ Clean value → ● Received value · copper segment: added noise. A square marks a wrong decision.

Inspect a decision by selecting its received bit
Sent	1	1	0	1	0	0	1	0
Decided

Decision 2 · sent 1

+1.00 + (−1.36) = −0.36 → decide 0

Wrong side of zero: the noise changed this decision.

The same eight noise values are used at every setting. These selected examples illustrate decisions, not an error-rate measurement. The scale is normalized to clean decision values ±1; it is not the RF carrier voltage.

What this example assumes

Decision times are already known. Timing errors and other channel distortion are omitted. The eight constructed noise values have zero mean and the stated RMS. A short record does not establish a Gaussian distribution or a bit-error probability. No waveform between decision samples is implied.

Less noise makes a wrong decision less likely; it does not supply one universal SNR requirement. The required ratio depends on how the receiver detects the signal and the acceptable error rate.

Sometimes the bandwidth has not yet been chosen. C/N₀ compares carrier power with noise-power density, so its linear units are hertz and its logarithmic unit is dB-Hz. A −90 dBm carrier over a −170 dBm/Hz flat density has C/N₀ = 80 dB-Hz. Across 1 MHz, that becomes C/N = 20 dB: subtract $10\log_{10}(10^6)=60$ dB.

\frac{C}{N_0}\ [\mathrm{dB\!-\!Hz}] =C[\mathrm{dBm}] -N_0[\mathrm{dBm/Hz}]

\frac CN\ [\mathrm{dB}] =\frac C{N_0}\ [\mathrm{dB\!-\!Hz}] -10\log_{10}\left(\frac B{1\,\mathrm{Hz}}\right)

The conversion assumes flat noise density over B, the ENBW. The “Hz” in dB-Hz records the density normalization.

Go deeperEnergy per bit in a digital link

If a digital signal carries information at $R_b$ bits per second, its average signal power divided by that rate is the energy per information bit, $E_b$ . With consistent definitions, $E_b/N_0=(C/N_0)/R_b$ in linear units. This connects noise density to a bit-error requirement, but requires the bit rate, modulation, coding, and detector assumptions developed in Signals & Modulation (in preparation).

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 $10\log_{10}\!\left(B/(1\,\mathrm{Hz})\right)$ term.

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09 / 12

Noise factor, noise figure, and equivalent temperature

An amplifier makes an incoming signal larger. Does that also make it easier to distinguish from noise? Both amplifiers below receive the same wanted signal and the same noise. Change gain and compare their output powers and SNR.

Larger signal. Cleaner signal?

Dashed: wanted signal. Solid: signal with noise. Fixed voltage scales · slow motion.

Shared input

SNR 10.97 dB

Both amplifiers’ gain

Ideal amplifierReal amplifier+ its own noise

Ideal output · no added noise

SNR 10.97 dB

Signal: 39.8 fWNoise: 3.18 fW

Real output · adds noise

SNR 7.97 dB

Signal: 39.8 fWNoise: 6.35 fW

At “now”: ○ ideal output · ● real output. The copper bracket is noise added by the amplifier.

Gain increases both signal and noise. The ideal SNR stays unchanged; this real amplifier loses 3 dB of SNR at every gain setting.

Separate illustration: a 50 kHz test signal, approximately 200 kHz noise bandwidth, matched 50 Ω ports and a 290 K source. Powers are complete-record averages.

How the comparison is constructed

The signal has 1 µV peak amplitude and the source noise is 0.2 µV RMS. The real amplifier adds the equivalent of 0.1995 µV RMS at its input, giving a 3 dB noise figure at 290 K. Voltage grows by the square root of power gain; the equal resistances let us use that relation directly.

Two separate fixed 32-tone records represent uncorrelated source and amplifier noise powers. They repeat after 640 µs, shown in 48 seconds; they are not measured noise or an exact Gaussian process. All cross powers vanish over that complete record. The short visible history need not have those average powers. The copper bracket at “now” connects the ideal output to the output after amplifier noise is added. Switching transients, distortion and overload are excluded.

The ideal amplifier multiplies signal power and existing noise power by the same gain. Both become larger, so SNR stays the same. The real amplifier also produces noise of its own. That extra contribution makes total output noise larger and SNR lower.

To compare receivers consistently, noise factor F measures this SNR loss with a matched source at the reference temperature T₀ = 290 K. Divide input SNR by output SNR, using linear ratios. Noise figure NF expresses that factor in decibels.

Receiver-added noise can also be described at the input. Divide its output power by the receiver’s power gain: this is the input noise that would produce the same output contribution through an ideal amplifier. We call it input-referred noise. Its equivalent noise temperature Tₑ is the temperature a matched source would need to produce that amount of noise. It describes an electrical noise contribution, not the temperature of the receiver’s parts.

F =\frac{\mathrm{SNR}_{\mathrm{in}}}{\mathrm{SNR}_{\mathrm{out}}}

\mathrm{NF} =10\log_{10}F

T_e =T_0(F-1)

F =1+\frac{T_e}{T_0}

T_0 =290\,\mathrm K

N_{\mathrm{out}} = G k B(T_{s} + T_{e})

G is linear available power gain, Tₛ is source noise temperature, and B is ENBW. Every temperature is referred to the same input plane.

For F = 1.5, NF ≈ 1.76 dB and Tₑ = 145 K. The same receiver adds 145 K to a 290 K source or a 50 K source, giving totals of 435 K and 195 K. The colder source still gives less total noise. But the added 145 K is a larger fraction of its original 50 K, so its SNR loses a larger proportion. Compare the bars below.

Same receiver. Different source temperature.

The receiver adds 145 K in both cases: F = 1.5 and NF ≈ 1.76 dB, referenced to 290 K. Hold signal power, gain and bandwidth fixed.

Forest: source noiseCopper: receiver-added noise

290 K source → 435 K total

SNR degradation: 435/290 = 1.50 times · 1.76 dB

50 K source → 195 K total

SNR degradation: 195/50 = 3.90 times · 5.91 dB

0 KCommon scale →435 K

The cold source gives less total noise and a better output SNR. But adding the same 145 K is a larger proportional increase over its original noise. Only the 290 K case has actual SNR degradation equal to the specified NF.

For a source other than 290 K, calculate with Tₛ + Tₑ; the specified NF is not the actual SNR loss.

A cable adds noise too

A matched passive cable or filter attenuates the incoming signal and noise, while its resistive loss adds thermal noise. Define its linear loss $L$ as input power divided by output power, so $L\ge1$ and its gain is $G=1/L$ . At physical temperature $T_p$ :

T_e =(L-1)T_p

F =1+\frac{(L-1)T_p}{T_0}

Only when Tₚ = T₀ does F = L and noise figure equal loss in decibels.

Go deeperMeasure added noise using a hot and cold source

Measure receiver output noise with two calibrated input-source temperatures, hot and cold, using the same bandwidth and gain. Their linear output-power ratio is Y:

Y =\frac{P_{\mathrm{hot}}}{P_{\mathrm{cold}}} =\frac{T_{\mathrm{hot}}+T_e}{T_{\mathrm{cold}}+T_e}

T_e =\frac{T_{\mathrm{hot}}-YT_{\mathrm{cold}}}{Y-1}

Noise-source calibration often gives an excess noise ratio (ENR): the hot-to-cold temperature difference divided by 290 K, usually expressed in decibels. Real measurements must account for source calibration, 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.

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10 / 12

Why receiver stage order matters

A low-noise amplifier (LNA) adds relatively little noise while increasing the incoming signal. It cannot restore SNR already lost at its input. Its gain can, however, make the signal large compared with noise added by later stages. A lossy filter before the LNA weakens that signal and adds its own thermal noise.

To compare stages at the chain input, divide each later stage’s added-noise contribution by the power gain before it. The second stage is divided by G₁; the third by G₁G₂. Friis’ equation performs this accounting. Convert gain and noise figure from decibels to linear G and F before using it.

F_{\mathrm{total}} =F_1+\frac{F_2-1}{G_1} +\frac{F_3-1}{G_1G_2}+\cdots

T_{e,\mathrm{total}} =T_{e1}+\frac{T_{e2}}{G_1} +\frac{T_{e3}}{G_1G_2}+\cdots

Subscripts follow the signal path. These forms assume the stated matched, linear stages and independent added noise; use the gains ahead of each contribution.

Take an LNA with +15 dB gain and NF 1.5 dB, a −3 dB filter at 290 K with NF 3 dB, then a second amplifier with +30 dB gain and NF 5 dB. All three stages operate in the same band in this model. The gains multiply in linear units, so their decibel values add to 42 dB. The noise calculation is:

F_{\mathrm{total}} =1.41254 +\frac{1.99526-1}{31.6228} +\frac{3.16228-1}{31.6228\times0.501187}

F_{\mathrm{total}} \approx1.58044

\mathrm{NF} \approx1.99\,\mathrm{dB}

T_e \approx168.3\,\mathrm K

The chain has 42 dB total gain. Predict what happens to noise figure when the 3 dB filter moves ahead of the LNA, then swap the stages below. Compare their added-noise contributions while checking that total gain stays unchanged.

Same components. Different noise figure.

Move the filter before the LNA and compare noise figure. The total gain stays the same. Each bar shows how much that stage adds when referred back to the chain input.

Input cable loss · at 290 K 0.0 dB

Direct connection · no cable loss or added cable noise.

Total noise figure: 1.99 dB
Total power gain: 42.0 dB

1 → LNA
15 dB gain · 1.5 dB NF
LNA gain 15 dB
119.6 K added, referred to input
2 → Passive filter (290 K)
-3 dB gain · 3.0 dB NF
Matched passive filter at 290 K · fixed loss
9.1 K added, referred to input
3 → Amplifier
30 dB gain · 5.0 dB NF
Amplifier NF 5 dB
39.6 K added, referred to input

Equivalent input noise temperature: 168.3 K
Total noise factor: 1.5804

Each full bar width represents the current total added-noise temperature, including any cable. The numbers show absolute contributions. Moving a lossy filter before the LNA makes later noise more significant relative to the input signal.

Predict which change helps more

From the reset order and no cable, compare these separate changes:

Choose a prediction, then use the controls to test each case.

All stages are matched, linear and operate in the same band; their added noise is independent. Cable and filter are at 290 K. At a 290 K source temperature, the calculated NF is the SNR degradation in decibels.

Noise is only one ordering constraint. An input filter may be needed to reject strong blockers or protect the LNA. The order with the lowest calculated NF may still fail the receiver’s filtering or overload requirements.

Linear gain and Friis accounting

Sum the noise-factor contributions, then convert to decibels.
Stage	Gain G	Factor F	Contribution to F	Cumulative gain
LNA	31.623	1.4125	1.412538	15.0 dB
Passive filter (290 K)	0.501	1.9953	0.031473	12.0 dB
Amplifier	1000.000	3.1623	0.136430	42.0 dB

$F_{\mathrm{total}}=F_1+\frac{F_2-1}{G_1}+\frac{F_3-1}{G_1G_2}+\cdots$

The first term includes the reference 1; later terms use excess noise factors. Subtract that 1 before converting the first stage’s contribution into added kelvin. Total linear gain is 1.585e+4.

With the filter first, NF rises from about 1.99 to 4.71 dB. With a 290 K source, the receiver therefore needs about 2.72 dB more input signal for the same output SNR. Early filtering can still be necessary to keep a strong blocker from overloading the LNA.

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11 / 12

Antenna temperature and receiver sensitivity

An antenna receives unwanted radiation as well as the wanted signal. Its directional and polarization response determines how strongly each part of the surroundings contributes. A beam aimed mainly at warm ground can therefore deliver more noise than the same beam aimed at a colder part of the sky.

Express the collected noise as antenna temperature Tₐ: a matched thermal source at that temperature would deliver the same noise power per hertz, kTₐ, at the named antenna port. This describes received noise, not a thermometer reading of the antenna metal.

For a simplified lossless, single-polarization example, the sky’s brightness temperature is 20 K and the ground’s is 300 K. Brightness temperature expresses noise arriving from a direction using temperature units. Suppose 75% of the antenna’s receiving-response weight falls on sky and 25% on ground. Then Tₐ = 0.75 × 20 + 0.25 × 300 = 90 K. The weights describe the antenna’s response, not how much of the visible scene is sky. Change these weights in the experiment below.

An antenna receives noise from several directions

In this hypothetical two-region example, a lossless antenna gives a fraction w of its receiving-response weight to sky and the rest to ground. This weight is a response share, not a fraction of the visible sky.

Response weight on sky75% sky · 25% ground

Sky contribution

Sky brightness temperature20 K

0.75 × 20 K = 15.0 K

Ground contribution

Ground brightness temperature300 K

0.25 × 300 K = 75.0 K

Sum at the antenna port · Tₐ 90.0 KBar segments show the contributions: sky in forest, ground in copper. The full bar represents the current total.

Source temperature selected for the receiver experiment below: 290.0 K. Applying this temperature changes only the source; the receiver's own noise remains unchanged.

Brightness temperature describes the noise received from a direction. Actual antenna noise depends on its directional and polarization response, sky, atmosphere, ground and other objects; antenna loss also contributes. It is not generally the antenna metal’s temperature.

Go deeperFrom weighted directions to a continuous sky

Split the surroundings into many small patches, multiply each brightness temperature by the antenna response toward it, and divide the sum by the total response. In the continuous form, $T_b(\theta,\phi)$ is brightness temperature in the direction specified by two angles, $G_r(\theta,\phi)$ is the receiving pattern, and $\mathrm d\Omega$ is a small solid angle, measured in steradians:

T_a = \frac{\iint T_b(\theta,\phi)G_r(\theta,\phi)\,\mathrm d\Omega}{\iint G_r(\theta,\phi)\,\mathrm d\Omega}

This simplified expression uses one polarization and a lossless receiving antenna. The integration covers all directions.

A real estimate must account for polarization, frequency-dependent sky and ground brightness, atmosphere, antenna loss, and the antenna’s physical temperature. A lossy antenna both attenuates the collected noise and adds thermal noise from its material.

Combine the noise at the antenna port

When antenna and receiver meet directly at a matched input, add their equivalent noise temperatures there. If a cable lies between them, include it as the first receiver stage and use the cascade equation to refer its noise and the receiver’s noise back to the antenna port. Keep the wanted signal at that same reference plane.

T_{\mathrm{sys}} = T_{a} + T_{e}

N_{\mathrm{in}} =kB(T_a+T_e)

\mathrm{SNR}[\mathrm{dB}] =P_s[\mathrm{dBm}] -N_{\mathrm{in}}[\mathrm{dBm}]

Nᵢₙ is all source and receiver noise referred to the input, in the chosen ENBW. It is not just the noise physically supplied by the antenna.

Compare the result with a reception requirement

Sensitivity is the input signal level needed to meet a stated reception requirement. Here the requirement is an output SNR. Add the required SNR in dB to total input-referred noise in dBm to find the minimum signal level. Subtract that minimum from the supplied signal level to find the margin. A negative margin means this model falls short.

P_{s,\min}[\mathrm{dBm}] =N_{\mathrm{in}}[\mathrm{dBm}] +\mathrm{SNR}_{\mathrm{required}}[\mathrm{dB}]

\mathrm{margin} =P_s-P_{s,\min} =\mathrm{SNR}-\mathrm{SNR}_{\mathrm{required}}

The margin equation uses decibel levels and ratios. The criterion must also name the detector, error or detection target, and implementation assumptions.

The source-only SNR from Section 4 was about 12.46 dB. The LNA–filter–amplifier chain adds the equivalent of 168.3 K, making total noise temperature ≈ 458.3 K and input-referred noise ≈ −98.98 dBm in 20 MHz. The same −88.5 dBm signal now gives about 10.48 dB SNR: only 0.48 dB above the hypothetical 10 dB requirement. This assumes no interference, strong-signal distortion, fading or additional implementation loss.

When the source is exactly 290 K, the same calculation can use the familiar shorthand:

P_{s,\min}[\mathrm{dBm}] \approx-174+10\log_{10}\left(\frac B{1\,\mathrm{Hz}}\right) +\mathrm{NF}+\mathrm{SNR}_{\mathrm{required}}

NF and required SNR are in dB. For a different source temperature, return to kB(Tₐ + Tₑ).

Add input cable loss in the receiver experiment below. Watch the minimum required signal rise while the supplied signal at antenna port P0 stays fixed.

Is the signal strong enough?

Start with the supplied carrier and a hypothetical 10 dB output-SNR requirement. Then change the requirement, bandwidth or cable loss. All powers on this ruler are referred to the antenna port, P0.

Supplied carrier at P0-88.5 dBmNoise bandwidth20.0 MHzRequired output SNR10.0 dB

Signal margin +0.48 dBMeets the entered requirement. Calculated SNR: 10.48 dB.

Required signal = noise in dBm + required SNR in dB. The gap at the default setting is only 0.48 dB; use the numbers to judge it.

What sets the noise level?

Source temperature is set at P0. Receiver NF and Tₑ describe the same receiver at P1, after the cable; changing either updates the other.

Antenna noise temperature290.0 KCable loss · at 290 K0.0 dBReceiver NF · referenced to 290 K1.988 dBReceiver added temperature Tₑ168.3 K

Total noise at P0: -98.98 dBm126.6 fW in 20.0 MHz
Signal margin: +0.48 dB

Follow the noise, then refer it back to P0

Antenna P0 → direct connection → receiver input P1

Equivalent noise temperatures
Contribution	At P1	Referred to P0
Source	290 K	290 K
Cable	0 K	0 K
Receiver	168.3 K	168.3 K

The cable reduces source noise to Tₐ/L and adds 290(1 − 1/L) K at P1. The receiver adds Tₑ there. Multiplying these P1 contributions by L refers them back to P0, where signal and interference are specified.

Total referred to P0: 458.3 KTₐ + (L − 1) × 290 K + L × Tₑ. These describe noise power, not component operating temperatures.

Add in-channel interference

Interference at P0

Integrated interference powerexcluded

SNR: 10.48 dB · margin +0.48 dB

Independent interference and noise add in watts. Changing bandwidth changes noise; supplied integrated carrier and interference powers stay fixed in this experiment.

Supporting quantities, all referred to P0
Source density	-173.98 dBm/Hz
Cable + receiver NF / Tₑ	1.99 dB / 168.3 K
System temperature	458.3 K
Total noise density	-171.99 dBm/Hz
C/N₀ · excludes interference	83.49 dB-Hz
C/N · excludes interference	10.48 dB

This is a matched, linear receiver model with noise temperature and loss constant across the bandwidth. The requirement is hypothetical: modulation, coding, integration time, error rate, fading and implementation losses determine what a real receiver needs. Carrier frequency alone does not determine its noise; use component values for the operating frequency.

Noise and margin equations

$T_{e,\mathrm{network}}=(L-1)290\,\mathrm K+LT_e$

$N=kB(T_a+T_{e,\mathrm{network}})$

At a 290 K source this is also $kT_0 B F_{\mathrm{network}}$ . NF still uses the 290 K reference for a colder or warmer source. C is the supplied carrier power. Comparisons use full precision; displayed results are rounded.

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12 / 12

Apply it: receiver margin with a colder source

The antenna now supplies less noise than the 290 K source used earlier. First connect the receiver directly. Then add a warm cable ahead of it. Does the same wanted signal still meet the SNR requirement? Keep input comparisons at antenna port P0.

At the matched antenna port, P0, the available wanted signal power in the channel is −105 dBm at 1 GHz and antenna noise temperature is 100 K.
The receiver has 1 MHz ENBW, linear noise factor F = 2 referenced to 290 K, and 30 dB available power gain. The required output SNR is 6 dB.
First connect the receiver directly to P0. Then insert a matched cable between P0 and the receiver with linear loss L = 2, approximately 3.01 dB, at a physical temperature of 290 K. Keep the supplied signal and source temperature at P0 unchanged.
Use one-sided classical noise density and flat gain, loss, and noise factor across the band. Assume linear operation and no interference, reciprocal mixing, fading, or additional implementation loss.

Direct: antenna port P0 → receiver → output

With cable: antenna port P0 → L = 2 cable at 290 K → receiver → output

For the direct connection, find receiver added-noise temperature, total input-referred noise power, output SNR, sensitivity, and margin.
Use the 30 dB gain to find output signal and noise powers. Check that their ratio agrees with the input-referred calculation.
Refer the cable and receiver noise back to P0. Recalculate SNR and margin. Does SNR fall by exactly the cable’s 3.01 dB loss?
Write a short report stating the reference plane, bandwidth, temperatures, requirement, and decision. Would the given information establish performance near a strong off-channel transmitter?

Reveal the complete solution

1. Direct receiver connection

Convert noise factor to added-noise temperature: $T_e=290(2-1)=290\,\mathrm K$ . The source is 100 K, so $T_{\mathrm{sys}}=390\,\mathrm K$ . Using $k=1.380649\times10^{-23}\,\mathrm{J/K}$ :

N_{\mathrm{P0}} =k(10^6)(390) \approx5.3845\times10^{-15}\,\mathrm W \approx-112.69\,\mathrm{dBm}

\mathrm{SNR} \approx-105-(-112.69) =7.69\,\mathrm{dB}

P_{s,\min} \approx-112.69+6 =-106.69\,\mathrm{dBm}

\mathrm{margin} \approx-105-(-106.69) =+1.69\,\mathrm{dB}

2. Check at the output

Add 30 dB gain to both input-referred powers. The output signal is −75 dBm and total output noise is approximately −82.69 dBm. Their difference remains 7.69 dB. Receiver-added noise was already included when forming the input-referred noise; do not add it a second time at the output.

3. Add the cable while retaining P0

The cable contributes $(L-1)T_p=290\,\mathrm K$ and has gain 1/2. Refer the receiver’s 290 K contribution through that loss using the temperature cascade equation:

T_{e,\mathrm{network}} =290+\frac{290}{1/2} =870\,\mathrm K

T_{\mathrm{sys}} =100+870 =970\,\mathrm K

N_{\mathrm{P0}} \approx-108.73\,\mathrm{dBm}

\mathrm{SNR} \approx3.73\,\mathrm{dB}

\mathrm{margin} \approx3.73-6 =-2.27\,\mathrm{dB}

The cable causes about 3.96 dB SNR degradation, more than its 3.01 dB loss. It attenuates the cold source contribution while adding noise from its warmer material. Although the complete network NF increases by 3.01 dB at the 290 K reference temperature, that is not the actual SNR loss with this 100 K source.

4. Report the decision and its limit

“At antenna port P0, a −105 dBm, 1 GHz signal with 100 K antenna noise temperature has 7.69 dB output SNR through the direct F = 2 receiver in 1 MHz ENBW. It meets the specified 6 dB requirement with 1.69 dB margin. Adding the matched L = 2 cable at 290 K reduces SNR to 3.73 dB, missing the requirement by 2.27 dB under the stated linear, interference-free model.”

Strong-blocker performance is not established. The blocker’s power and offset, filtering before the mixer, LO phase noise, and receiver linearity are among the missing conditions. A real test report also needs the reception or error criterion behind the 6 dB requirement and the instrument settings and uncertainty.

Choose the calculation that answers the question

Density to total noise: add PSD area in linear units; use ENBW for flat input density.
Source and receiver noise: convert NF to Tₑ and refer every temperature to the signal’s reference plane.
A component before the receiver: include its loss, physical temperature, and position in the cascade.
A reception decision: compare SNR or SINR with an explicit criterion, then state the remaining interference and measurement limits.

Apply this with your own receiver: calculate cascaded gain and noise figure with explicit stage order, bandwidth, and temperature assumptions.

Final practice: analyse an RF link

Bring the five RF Fundamentals modules together by following a new signal from transmitter to receiver. This practice is free and ungraded.

A 1 GHz transmitter establishes a zero-offset sine of 0.5 V peak across the complete antenna input network, modeled at this frequency as ideal series 50 + j30 − j30 Ω.
The transmitting antenna’s linear gain toward the receiver is 2, including radiation efficiency. The receiving antenna’s gain toward the transmitter is 1. The antennas have aligned linear polarization and matched ports.
The separation is 1 km. Unobstructed free-space far-field conditions are established. Use c ≈ 3 × 10⁸ m/s and η₀ ≈ 376.7 Ω. For the phase comparison, a second hypothetical path is 75 mm longer in the same medium.
At the receiving antenna port P0, Tₐ = 290 K. The receiver has 5 MHz ENBW and F = 2, referenced to 290 K, and requires 15 dB output SNR.
After checking the direct connection, insert a matched 4 dB cable at 290 K between P0 and the receiver. Ignore interference, fading, and extra implementation loss; gain, loss, and noise factor are flat over the band.

The exercise signal path · givens only

Complete Tx antenna input0.5 V peak across
50 + j30 − j30 Ω1 GHz sine
Transmit antennaDirectional gain 2Efficiency already included
Free-space path1 kmFar-field use established
Receive antenna → P0Gain 1 · alignedMatched port · Tₐ = 290 K
Connection after P0Direct, then 4 dB cableCable at 290 K
ReceiverF = 2 · ENBW 5 MHzRequired output SNR 15 dB

Follow nodes 1–6 in order. Keep received signal and all input-referred noise at P0 when comparing the two connections. The optional 75 mm path difference is a separate phase comparison.

For this exercise, use the supplied receiving-antenna relation below. Effective aperture $A_e$ converts incident power density to available received power for the stated matched, polarization-aligned case; it need not equal the antenna’s physical area. Here $G_r$ is linear receiving gain.

A_e =\frac{G_r\lambda^2}{4\pi}

P_r =S_{\mathrm{avg}}A_e

Find period, wavelength, and the delay and phase lag caused by the additional 75 mm path.
Find the antenna input impedance, RMS voltage and current, accepted average power, and EIRP. Does cancellation of input reactance mean the inductor and capacitor store no energy?
At 1 km, find power density, RMS electric field, effective aperture, and available received signal power in dBm.
Decide whether the direct receiver meets the 15 dB requirement. Repeat with the 4 dB input cable, keeping P0 as the reference plane. Explain why subtracting the cable loss from SNR works in this case but did not in the preceding cold-source case.

Reveal the complete link solution

1. Time, wavelength, and phase

$T=1/f=1\,\mathrm{ns}$ and $\lambda=c/f\approx0.3\,\mathrm m$ . The 75 mm extra path takes 0.25 ns, one quarter of a period, so the arriving sine lags by 90° at the same observation time.

2. Voltage, impedance, and transmitted power

The reactive terms cancel at 1 GHz, leaving $Z=50\,\Omega$ . The 0.5 V peak sine has $0.5/\sqrt2\approx0.354\,\mathrm V$ RMS and 7.07 mA RMS current. Accepted power is $V_{\mathrm{rms}}^2/(50\,\Omega)=2.5\,\mathrm{mW}$ . The inductor and capacitor still store and return energy; their opposite reactances cancel in the total input impedance.

$\mathrm{EIRP}=2.5\,\mathrm{mW}\times2=5\,\mathrm{mW}$ , approximately +6.99 dBm. The stated gain already includes radiation efficiency.

3. Propagation and received power

S_{\mathrm{avg}} =\frac{0.005}{4\pi(1000)^2} \approx3.979\times10^{-10}\,\mathrm{W/m^2}

E_{\mathrm{rms}} =\sqrt{\eta_0S_{\mathrm{avg}}} \approx0.387\,\mathrm{mV/m}

A_e =\frac{(0.3)^2}{4\pi} \approx0.00716\,\mathrm{m^2}

P_r =S_{\mathrm{avg}}A_e \approx2.850\,\mathrm{pW} \approx-85.45\,\mathrm{dBm}

4. Noise margin and the cable decision

$F=2$ gives $T_e=290\,\mathrm K$ . With $T_a=290\,\mathrm K$ , the total is 580 K, so the 5 MHz input-referred noise is approximately −103.98 dBm. Subtracting that noise level from the received signal level gives about 18.52 dB SNR and 3.52 dB margin above the 15 dB requirement. These results use unrounded powers.

The 4 dB cable has $L=10^{4/10}\approx2.512$ . Referred to P0, the complete added-noise temperature becomes $(L-1)290+L\times290\approx1166.9\,\mathrm K$ . Including the source gives $T_{\mathrm{sys}}\approx1456.9\,\mathrm K$ and input-referred noise of −99.98 dBm. SNR is now 14.52 dB, about 0.48 dB below the requirement.

Here both the source and cable are at 290 K, so the total system noise temperature referred to P0 grows by exactly the linear cable loss. The SNR therefore falls by exactly 4 dB. The earlier 100 K source did not satisfy that temperature condition. Under this link model, the direct connection passes and the added cable does not.

Where to go after RF Fundamentals

Check your understanding across all five lessons with a free practice assessment. To earn an RF Fundamentals certificate, arrange a separate one-hour exam supervised by Ricardo Pereira.

Try the free assessment Schedule a supervised exam

The Example Exam requires a free verified account and does not award a certificate. Certification also requires a verified account, at least 18/20 and supervisor confirmation.

Choose what you want to understand next

Signals & Modulation (in preparation) — connect waveforms, spectra, and I/Q to how a radio carries information.
Transmission Lines & Matching (in preparation) — understand how cables and traces carry RF, why reflections occur, and how matching changes them.

Both courses are intended to be free when published. Start with the question most useful to your work.

Noise

Noise sets the weak-signal limit

Random signals have measurable structure

A zero average does not mean zero power

Current reverses. Energy keeps increasing.

When, how often, and at which frequencies?

Three views of the same noise model

When do the values occur?

Which values occur most often?

At which frequencies?

Try another recording or improve the estimate

From noise density to noise power

Wider bandwidth. More noise power.

A real filter has sloping edges

Different shapes. The same noise power.

Two-level filter

Equivalent rectangle

Thermal noise and the −174 dBm/Hz reference

One resistor, two voltage references

Open circuit

Matched load

Different filter shapes, equal noise area

First answer for the opening receiver

Noise sources leave different fingerprints

Phase fluctuations around a carrier

Height changes. Timing changes.

Amplitude changes

Phase changes

From relative to absolute noise density

How a strong unwanted signal brings LO noise into the channel

Three simultaneous paths through one mixer

What a spectrum analyzer actually displays

What makes the analyzer trace move?

Noise added before the display

SNR, C/N, C/N₀, and SINR

When noise changes the decision

Noise factor, noise figure, and equivalent temperature

Larger signal. Cleaner signal?

Shared input

Ideal output · no added noise

Real output · adds noise

Same receiver. Different source temperature.

290 K source → 435 K total

50 K source → 195 K total

A cable adds noise too

Why receiver stage order matters

Same components. Different noise figure.

1 → LNA

2 → Passive filter (290 K)

3 → Amplifier

Predict which change helps more

Antenna temperature and receiver sensitivity

An antenna receives noise from several directions

Sky contribution

Ground contribution

Combine the noise at the antenna port

Compare the result with a reception requirement

Is the signal strong enough?

What sets the noise level?

Add in-channel interference

Apply it: receiver margin with a colder source

1. Direct receiver connection

2. Check at the output

3. Add the cable while retaining P0

4. Report the decision and its limit

Check your understanding

Try 6 quick questions.

Final practice: analyse an RF link

1. Time, wavelength, and phase

2. Voltage, impedance, and transmitted power

3. Propagation and received power

4. Noise margin and the cable decision

Sources and further study

Constants, noise, and antennas

Receiver and measurement practice

Where to go after RF Fundamentals

Choose what you want to understand next