Measurements in Decibels

1. Introduction

In electrical and communication engineering we constantly deal with quantities that span enormous ranges. A radio receiver may need to detect a signal of 10⁻¹² watts at its antenna and still function when the same antenna delivers 10⁻³ watts from a nearby transmitter. That is a dynamic range of nine orders of magnitude. Similarly, human hearing responds to sound pressures from about 20 μPa (threshold of hearing) to 20 Pa (threshold of pain) — a ratio of one million to one.

Writing, plotting, and mentally comparing such numbers is awkward. The solution is the logarithmic (decibel) scale, which:

Compresses huge ratios into small, manageable numbers (a power ratio of 1,000,000,000,000 becomes simply 120 dB).
Converts multiplication into addition — cascaded gains and losses become simple sums, which is ideal for system (link-budget) analysis.
Matches human perception — both hearing and vision respond approximately logarithmically, so decibel measures correspond closely to what we subjectively perceive.
Matches relative measurements — most engineering questions are about ratios (gain, loss, SNR), not absolute values.

Key idea: A decibel never expresses an absolute quantity by itself. It always expresses a ratio of two like quantities. When we want an absolute measure, we fix one side of the ratio to an agreed reference level — that is how dBm, dBW, dBV, dBμV, dBrn and similar units are born.

Learning Objectives

Explain the historical and physical reasons for the logarithmic (decibel) scale.
Convert power, voltage and current ratios to decibels and back.
Use absolute-reference units: dBm, dBW, dBV, dBμV.
Interpret telephone-system noise units: dBrn, dBrnc and dBrnc0.
Apply decibels to amplifiers, attenuators, filters, antennas and complete communication links.

2. Origin of the Decibel

2.1 The transmission-unit problem

In the early days of the telephone (1880s–1920s), telephone engineers needed a convenient way to express how much a telephone line attenuated (weakened) the voice signal. Bell Telephone Laboratories introduced a unit called the TU (Transmission Unit) in 1923. One TU was defined as ten times the base-10 logarithm of the ratio of measured power to a reference power:

1 TU = 10 log₁₀ (P_measured / P_reference)

2.2 The Bel — honouring Alexander Graham Bell

The TU was soon renamed the bel (B), in honour of Alexander Graham Bell, inventor of the telephone. One bel represents a power ratio of 10:1:

1 bel = log₁₀ (P₁/P₂) ⇒ P₁/P₂ = 10 when the level difference is 1 bel

The bel proved too coarse for everyday work: the useful range of most systems spans only a few bels, and engineers needed finer resolution. The natural solution was to divide the bel into ten parts — deci-bel, exactly as a decimetre is a tenth of a metre.

1 decibel (dB) = 1/10 bel = 10 log₁₀ (P₁/P₂)

2.3 Why 10 log and not just log?

The human ear perceives a doubling of loudness for roughly every 10-fold (one bel) increase in acoustic power. Working in tenths of a bel therefore gives a unit that is both fine-grained and perceptually meaningful. Audio experience shows that a change of about 1 dB is the smallest difference an average listener can reliably detect.

Historical note: The decibel was formally standardized by international telephony bodies and adopted by the International Electrotechnical Commission (IEC) and the Institute of Electrical and Electronics Engineers (IEEE). Today "dB" appears in virtually every branch of electronics, acoustics, radio, optics and control engineering.

2.4 A perceptual anchor: sound level in dB SPL

Acoustics uses dB referenced to the threshold of human hearing, 20 μPa (0 dB SPL). The table below connects decibels to everyday experience:

Sound environment	Approx. level	Power ratio vs. threshold
Threshold of hearing	0 dB SPL	1
Quiet library / whisper	30 dB SPL	1,000
Normal conversation	60 dB SPL	1,000,000
Busy street traffic	80 dB SPL	100,000,000
Rock concert / jet at 30 m	110–120 dB SPL	10¹¹–10¹²
Threshold of pain	~130 dB SPL	10¹³

3. The Decibel (dB) — Relative Power and Amplitude Ratios

3.1 Power ratios

By definition, the decibel expresses the ratio of two powers:

L (dB) = 10 log₁₀ (P_out / P_in)

The factor 10 multiplies the base-10 logarithm so that 10 dB = 1 bel. Notice the symmetry:

If P_out > P_in → gain (positive dB)
If P_out = P_in → 0 dB
If P_out < P_in → loss / attenuation (negative dB)

3.2 Why voltage and current ratios use 20 log

Voltmeters and oscilloscopes measure amplitudes (volts), not power directly. Since power is proportional to the square of voltage (P = V²/R) or current (P = I²R) across the same resistance:

L (dB) = 10 log₁₀ (V_out²/V_in²) = 20 log₁₀ (V_out/V_in)

L (dB) = 20 log₁₀ (I_out/I_in)

Crucial condition: The factor 20 is valid only when the two voltages (or currents) are measured across equal impedances. Comparing 1 V across 50 Ω with 1 V across 1 MΩ is not 0 dB of power — the powers differ enormously. When impedances differ, always convert to power first: P = V²/Z.

3.3 Anchor values worth memorizing

Power ratio P_out/P_in	dB (10 log)	Voltage ratio V_out/V_in (same R)	dB (20 log)
1/100	−20 dB	1/10	−20 dB
1/10	−10 dB	1/√10 ≈ 0.316	−10 dB
1/2	−3.01 dB	1/√2 ≈ 0.707	−3.01 dB
1	0 dB	1	0 dB
2	+3.01 dB	√2 ≈ 1.414	+3.01 dB
10	+10 dB	√10 ≈ 3.162	+10 dB
100	+20 dB	10	+20 dB
1000	+30 dB	31.62	+30 dB

Mental math trick: Because log scales add, dB values add. A power ratio of 4000 = 4 × 1000 = 6.02 dB + 30 dB = 36.02 dB. Any ratio can be decomposed into powers of 10 (add 10 dB each) and 2 (add 3 dB each).

3.4 Worked examples

Example 3.1 — Amplifier power gain. An RF power amplifier delivers 25 W to an antenna with an input drive of 100 mW.
G (dB) = 10 log₁₀(25 / 0.1) = 10 log₁₀(250) = 23.98 dB ≈ 24 dB.

Example 3.2 — Voltage gain across equal impedances. An audio preamplifier raises 2 mV to 500 mV across the same input/output resistance.
G (dB) = 20 log₁₀(500/2) = 20 log₁₀(250) = 47.96 dB ≈ 48 dB.

Example 3.3 — Attenuator. A 6 dB pad outputs 1/4 of its input power: L (dB) = 10 log₁₀(0.25) = −6.02 dB, i.e. a 6 dB attenuation (equivalently a voltage ratio of 0.5 across equal impedances).

⚙ Interactive Calculator 1 — dB from Ratios

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⚙ Interactive Calculator 2 — Ratio from dB (inverse operation)

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4. dBm — Power Referenced to 1 Milliwatt

4.1 Definition

When the reference power is fixed at 1 milliwatt (1 mW = 10⁻³ W), the decibel becomes an absolute power unit called dBm ("dB referenced to one milliwatt"):

P (dBm) = 10 log₁₀ (P in mW / 1 mW) = 10 log₁₀ (P_W / 10⁻³)

0 dBm = 1 mW; +10 dB × 10 in power; 30 dBm = 1 W (0 dBW); 60 dBm = 1 kW

4.2 Converting between dBm and watts

P (W) = 10^{(P_dBm − 30)/10}
P (mW) = 10^P_dBm/10

Power (mW)	Power (W)	Power (dBm)	Typical context
0.001 nW = 10⁻⁹ mW	10⁻¹² W	−90 dBm	Deep sensitivity floor of a good receiver
0.0316 mW	3.16 × 10⁻⁵ W	−15 dBm	Typical Bluetooth / WLAN transmit power
1 mW	0.001 W	0 dBm	Reference level
10 mW	0.01 W	10 dBm	WLAN (10 dBm class)
100 mW	0.1 W	20 dBm	Typical cell-phone uplink
1 W	1 W	30 dBm = 0 dBW	Cellular base-station output
100 W	100 W	50 dBm	FM broadcast transmitter (small)
50 kW	50,000 W	77 dBm	TV broadcast transmitter

4.3 dBm and voltage across a known impedance

With P = V²/Z, substituting into the dBm definition gives a handy relation between a measured RMS voltage and dBm:

V_rms = √(Z × 1 mW × 10^P_dBm/10)
P (dBm) = 10 log₁₀(V_rms² / (Z × 10⁻³)) = 20 log₁₀(V_rms) + 30 − 10 log₁₀(Z)

Impedance Z	V_rms at 0 dBm	Constant in P(dBm) = 20 log(V_rms) + C
50 Ω (RF, most common)	224 mV	C = 13 dB
600 Ω (telephone/audio)	775 mV	C = 2.2 dB
75 Ω (coax video/CATV)	274 mV	C = 11.3 dB

dBμV and dBV. In RF field-strength and EMC work, voltage is referenced to 1 μV (dBμV) or 1 V (dBV). Across 50 Ω: 0 dBm = +107 dBμV = −13 dBV. Broadcast field strength is given in dBμV/m.

4.4 Worked examples

Example 4.1. Express 40 W in dBm: P(dBm) = 10 log₁₀(40 × 10³ mW / 1 mW) = 10 log₁₀(4×10⁴) = 46.02 dBm.

Example 4.2. A spectrum analyzer reads −27 dBm. Power = 10^{(−27−30)/10} = 10^−5.7 W ≈ 2 μW (10^−2.7 mW ≈ 2 mW… check: 10^−2.7 mW = 0.002 mW = 2 μW ✓).

Example 4.3. A signal generator set to −20 dBm into 50 Ω produces V_rms = √(50 × 10⁻³ × 10⁻²) = √(5×10⁻⁴) = 22.4 mV.

⚙ Interactive Calculator 3 — dBm ↔ mW / W

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⚙ Interactive Calculator 4 — dBm ↔ RMS Voltage

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5. dBrn — Noise Levels Above Reference Noise

5.1 Reference noise

In telephony, noise performance is specified using a dedicated unit: the dBrn (decibels above reference noise). The reference noise power is defined as 1 picowatt (10⁻¹² W) measured with a specified bandwidth. On the dBm scale this reference sits at:

P_ref = 10⁻¹² W = 10⁻⁹ mW ⇒ 10 log₁₀(10⁻⁹) = −90 dBm

dBrn = dBm + 90 (equivalently dBm = dBrn − 90)

So 0 dBrn = −90 dBm = 1 pW of noise. The unit was introduced by telephone administrations (the North American practice) because absolute noise floors in analog carrier and multiplex systems hovered near this picowatt level.

Noise power	dBm	dBrn	Comment
1 pW (10⁻¹² W)	−90 dBm	0 dBrn	Reference noise
10 pW	−80 dBm	10 dBrn
100 pW	−70 dBm	20 dBrn	Quiet long-haul telephone channel
1 nW (10⁻⁹ W)	−60 dBm	30 dBrn
1 μW (10⁻⁶ W)	−30 dBm	60 dBrn	Very noisy channel — unacceptable

5.2 dBrnc — C-message weighting

Not all noise frequencies annoy the human ear equally. Telephone noise measurement therefore applies a standardized frequency weighting curve that mimics the response of a telephone subscriber's ear, called the C-message weighting. Noise measured through this filter is quoted in dBrnc. The relationship to unweighted dBm of noise (flat, 3 kHz bandwidth) is approximately:

dBrnc ≈ dBm_noise(flat) + 90 − ~2 dB (the weighting typically removes about 2 dB of measured noise relative to flat response)

5.3 dBrnc0 — noise at the zero transmission level point

Because telephone networks contain many amplifiers (repeaters) at different points, noise measured at one point depends on the local signal level. To compare channels fairly, noise is normalized to the zero transmission level point (0 TLP) — the hypothetical point where the signal level is 0 dBm (1 mW). The suffix "0" means "referred to the 0 TLP":

dBrnc0 = noise measured in dBrnc − (signal level in dBm at the measurement point)

Design rule: A quality long-distance telephone channel is engineered for ≤ 30–34 dBrnc0 of noise. The classic noise figure definition of a telephone channel: total channel noise objective of about 33 dBrnc0 for a 2500-mile connection set the historical benchmark for system design.

5.4 Related noise units

dBm0 — power measured in dBm, normalized to the 0 TLP.
dBA / dBC — acoustic noise levels with A- and C-frequency weightings (environmental acoustics).
dBFS — digital full-scale reference (0 dBFS = maximum code), used in ADCs and digital audio.
Np (neper) — the natural-logarithm cousin of the dB: 1 Np = 8.686 dB (amplitude ratio e:1).

Example 5.1. A noise meter reads −55 dBm in a 3 kHz channel. The unweighted level in dBrn is −55 + 90 = 35 dBrn. Applying C-message weighting (subtract ≈2 dB) gives about 33 dBrnc. If this is measured at a point where the test-tone level is −10 dBm, the level at the 0 TLP is 33 − (−10) = 43 dBrnc0 — a noticeably noisy channel.

⚙ Interactive Calculator 5 — dBm ↔ dBrn ↔ dBrnc0

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C-message weighted? (subtract 2 dB)

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6. Applications of Decibels in Electrical & Communication Engineering

6.1 Cascaded systems — the "gain budget" becomes a sum

The greatest practical advantage of the decibel: the overall gain of cascaded blocks is the algebraic sum of individual dB values (multiplication of linear ratios becomes addition of logarithms):

G_total (dB) = G₁ + G₂ + G₃ + … + G_n

6.2 The link budget — the central tool of communication engineering

A link budget accounts for every gain and loss between transmitter and receiver to ensure the received power exceeds the receiver sensitivity by a safety margin:

P_RX (dBm) = P_TX + G_{TX ant} − L_FS − L_misc + G_{RX ant}

where free-space path loss (Friis) in dB is:

L_FS (dB) = 20 log₁₀(4πd/λ) = 20 log₁₀(d_km) + 20 log₁₀(f_MHz) + 32.44

Example 6.1 — Simple link budget. A transmitter delivers 40 dBm (10 W) into an antenna with 15 dBi gain, over 20 km at 900 MHz (FS loss = 20 log d_km + 20 log f_MHz + 32.44 = 26.0 + 59.1 + 32.44 = 117.5 dB), with 3 dB of miscellanneous losses, and a receiving antenna of 10 dBi.
P_RX = 40 + 15 − 117.5 − 3 + 10 = −55.5 dBm. If receiver sensitivity is −90 dBm, the link margin is 34.5 dB — a robust link.

6.3 Amplifiers, noise figure and sensitivity

Gain of amplifiers and attenuation of filters/cables are universally quoted in dB.
Noise figure (NF) is defined in dB: NF = 10 log₁₀(SNR_in/SNR_out).
Receiver sensitivity combines thermal noise floor (−174 dBm/Hz at room temperature, from kTB), noise figure and required SNR: Sensitivity (dBm) = −174 + NF + 10 log B + SNR_req.

6.4 Filters and frequency response

Filter specifications are given in dB: passband ripple (e.g. ±0.5 dB), stopband attenuation (e.g. 60 dB), and cutoff at the −3 dB (half-power) points. Bode plots express magnitude response in dB versus log frequency — multiplication of poles/zeros becomes addition of slopes (±20 dB/decade per pole/zero).

6.5 Antennas and propagation

Antenna gain in dBi (referenced to an isotropic radiator) or dBd (referenced to a half-wave dipole; dBd = dBi − 2.15).
EIRP = P_TX (dBm) + G_ant (dBi), in dBm — the effective radiated power used in regulatory limits.
Propagation losses (free space, reflection, diffraction, rain fade) are all tabulated and summed in dB.

6.6 Audio engineering

Sound levels in dB SPL (reference 20 μPa), mixing-console faders in dB, audio power in dBm/dBW into 600 Ω historically, and digital audio in dBFS. Every +6 dB roughly doubles perceived loudness for many program materials; +10 dB is "twice as loud" on average.

6.7 Transmission lines and fiber optics

Coaxial cable attenuation is quoted in dB per 100 m at a given frequency. In fiber optics, power budgets, connector losses (~0.3 dB each) and splice losses (~0.1 dB) are summed in dB, and optical power is measured in dBm.

6.8 Telephone network planning

Transmission levels at every point are tracked relative to the 0 TLP; noise objectives (dBrnc0), crosstalk (in dB) and echo (talker/listener echo ratings in dB) all trace back to decibel arithmetic.

⚙ Interactive Calculator 6 — Complete Link Budget

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7. Summary of Key Formulas and Units

Relative units

Power ratio: dB = 10 log₁₀(P₁/P₂)
Voltage/current (same Z): dB = 20 log₁₀(V₁/V₂)
Antenna gain: dBi, dBd (dBd = dBi − 2.15)

Absolute power units

dBm = 10 log₁₀(P/1 mW); 0 dBm = 1 mW
dBW = 10 log₁₀(P/1 W); 0 dBW = 30 dBm
dBμV (EMC), dBV, dBFS (digital)

Noise units (telephony)

0 dBrn = 1 pW = −90 dBm
dBrn = dBm + 90
dBrnc: C-message weighted noise
dBrnc0: noise at the 0 TLP

System relations

Cascade: G_tot = Σ G_i (dB)
Friis: L_FS = 20 log d_km + 20 log f_MHz + 32.44
Sensitivity = −174 + NF + 10 log B + SNR_req

Unit	Reference quantity	Field of use
dB	(ratio only)	General gains/losses
dBm	1 mW	RF/microwave power, fiber optics
dBW	1 W	High-power transmitters
dBμV / dBV	1 μV / 1 V	EMC, field strength, broadcast
dBi / dBd	isotropic / dipole antenna	Antennas
dBFS	full-scale digital code	Digital audio & ADCs
dBrn / dBrnc / dBrnc0	1 pW noise	Telephone network noise planning
dB SPL	20 μPa	Acoustics

8. Practice Problems

An amplifier has a voltage gain of 400 across equal input/output impedances. Find the gain in dB. (Answer: 52.0 dB)
Convert −35 dBm to watts and to dBrn. (Answer: 0.316 μW; 55 dBrn)
A coaxial cable attenuates 4.5 dB per 100 m at 1 GHz. What fraction of input power remains after 300 m? (Answer: −13.5 dB ⇒ 4.47%)
Three cascaded stages have gains of −10 dB, +25 dB and +6 dB. Overall gain and output power for 2 mW input? (Answer: 21 dB; 251.2 mW)
A 50 Ω RF source is set to +7 dBm. Find the RMS output voltage. (Answer: 0.5 V)
Express 250 W in dBm and dBW. (Answer: 53.98 dBm; 23.98 dBW)
A telephone channel shows noise of −62 dBm measured at a point where the signal level is −16 dBm. Find dBrn, and estimate dBrnc0. (Answer: 28 dBrn; ≈44 dBrnc0)
Free-space loss at 5 km and 2.4 GHz? A 14 dBi link with 20 dBm Tx power and 5 dB losses: received power? (L_FS = 114.0 dB; P_RX = −85 dBm)
Two +3 dB and two −6 dB elements are cascaded. Net gain? (0 dB overall: +3+3−6−6 = −6… careful: net = 0? +3+3 = +6, −6−6 = −12 ⇒ −6 dB)
Threshold voltage across 600 Ω for 0 dBm is 0.775 V. Verify and find the corresponding peak voltage. (V_rms = √0.6 = 0.775 V; peak = 1.096 V)