13.1 Decibels, Power Ratios, and Cascaded Stage Calculations

13.1 Decibels, Power Ratios, and Cascaded Stage Calculations

In radio frequency (RF) engineering, signals span an enormous dynamic range. A sensitive HF receiver can detect an incoming signal delivering less than $0.0000000000001\text{ Watt}$ ($0.1\text{ picowatt}$ or $10^{-13}\text{ W}$), while an amateur high-power linear amplifier can generate $400\text{ Watts}$ ($4 \times 10^2\text{ W}$) of RF output power. Expressing and calculating such quantities using linear arithmetic requires unwieldy scientific notation and tedious multi-digit multiplications. Furthermore, human sensory perception of sound and signal strength follows a logarithmic rather than linear response (governed by the Weber-Fechner law).

To compress this vast numerical span into manageable numbers and simplify cascaded circuit calculations into basic addition and subtraction, engineers rely on logarithmic units: the Bel and the Decibel ($\text{dB}$).

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The Physics and Mathematics of Logarithmic Units

Origin of the Bel and Decibel

The Bel (symbol $\text{B}$) was formulated in the 1920s by Bell Telephone Laboratories to quantify signal loss along standard telephone cables, named in honour of Alexander Graham Bell. Because the Bel represents a tenfold increase in power ($1\text{ Bel} = \log_{10}(P_1 / P_2) = 1$), it proved too coarse for day-to-day electrical engineering. Engineers adopted the decibel (symbol $\text{dB}$), which is one-tenth of a Bel ($1\text{ B} = 10\text{ dB}$).

The decibel is strictly a dimensionless relative ratio comparing two power levels, $P_1$ (the output or measured power) and $P_2$ (the input or reference power):

$$\text{Ratio (dB)} = 10 \log_{10}\left(\frac{P_1}{P_2}\right)$$

• If $P_1 > P_2$, the decibel value is positive, indicating power gain (amplification).
• If $P_1 < P_2$, the decibel value is negative, indicating power loss (attenuation).
• If $P_1 = P_2$, the ratio is unity, and $\text{dB} = 10 \log_{10}(1) = 0\text{ dB}$ (no change).

Voltage and Current Ratios across Equal Impedances

While the decibel is fundamentally defined in terms of power, engineers frequently measure voltage ($V$) or current ($I$) rather than power directly. According to Joule's and Ohm's laws, the average power dissipated in a purely resistive load $R$ is:

$$P = \frac{V^2}{R} = I^2 R$$

Substituting these power expressions into the decibel formula yields:

$$\text{dB} = 10 \log_{10}\left(\frac{P_1}{P_2}\right) = 10 \log_{10}\left(\frac{V_1^2 / R_1}{V_2^2 / R_2}\right)$$

When the input and output measurements are taken across identical impedances ($R_1 = R_2 = R$, such as within a standard $50\ \Omega$ transmission system):

$$\text{dB} = 10 \log_{10}\left(\frac{V_1^2 / R}{V_2^2 / R}\right) = 10 \log_{10}\left(\frac{V_1^2}{V_2^2}\right) = 10 \log_{10}\left[\left(\frac{V_1}{V_2}\right)^2\right]$$

Applying the logarithmic power identity $\log_{10}(x^k) = k \log_{10}(x)$, the exponent 2 moves outside the logarithm as a multiplier:

$$\text{dB} = 10 \times 2 \log_{10}\left(\frac{V_1}{V_2}\right) = 20 \log_{10}\left(\frac{V_1}{V_2}\right)$$

Similarly, for current flowing through identical resistances ($R_1 = R_2$):

$$\text{dB} = 10 \log_{10}\left(\frac{I_1^2 R}{I_2^2 R}\right) = 20 \log_{10}\left(\frac{I_1}{I_2}\right)$$

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Absolute Power References: $\text{dBm}$ and $\text{dBW}$

Because the plain decibel ($\text{dB}$) is merely a relative ratio, it cannot express an absolute power level without a stated baseline. To quantify absolute power, engineers define reference standards by fixing the denominator $P_2$ to a known physical value.

                      Absolute Power Reference Scales

     Watts (W):    0.001 W       0.1 W        1 W          10 W         100 W        400 W
                      |            |          |             |             |            |
     dBm:          0 dBm       +20 dBm     +30 dBm       +40 dBm       +50 dBm      +56 dBm
                      |            |          |             |             |            |
     dBW:        -30 dBW       -10 dBW      0 dBW        +10 dBW       +20 dBW      +26 dBW
                      ^                       ^
                      |                       |
              1 milliwatt reference        1 Watt reference

1. The Decibel Relative to 1 Milliwatt ($\text{dBm}$)

The $\text{dBm}$ expresses power relative to $1\text{ milliwatt}$ ($1\text{ mW} = 10^{-3}\text{ W}$):

$$\text{Power (dBm)} = 10 \log_{10}\left(\frac{P}{1\text{ mW}}\right) = 10 \log_{10}\left(\frac{P}{10^{-3}\text{ W}}\right)$$

In standard $50\ \Omega$ RF systems, $0\text{ dBm}$ ($1\text{ mW}$) corresponds to specific RMS voltage and current values:

$$V_{\text{rms}} = \sqrt{P \times R} = \sqrt{0.001\text{ W} \times 50\ \Omega} = \sqrt{0.05} \approx 0.2236\text{ V}_{\text{rms}} = 223.6\text{ mV}_{\text{rms}}$$ $$V_{\text{pk}} = V_{\text{rms}} \times \sqrt{2} \approx 0.2236 \times 1.4142 \approx 0.3162\text{ V}_{\text{pk}}$$ $$V_{\text{p-p}} = 2 \times V_{\text{pk}} \approx 0.6325\text{ V}_{\text{p-p}}$$ $$I_{\text{rms}} = \frac{V_{\text{rms}}}{R} = \frac{0.2236\text{ V}}{50\ \Omega} \approx 4.472\text{ mA}_{\text{rms}} \quad (6.325\text{ mA}_{\text{pk}})$$

2. The Decibel Relative to 1 Watt ($\text{dBW}$)

The $\text{dBW}$ expresses power relative to $1\text{ Watt}$ ($1\text{ W}$):

$$\text{Power (dBW)} = 10 \log_{10}\left(\frac{P}{1\text{ W}}\right)$$

Because $1\text{ Watt} = 1000\text{ milliwatts}$, and $10 \log_{10}(1000) = 30\text{ dB}$, the relationship between $\text{dBm}$ and $\text{dBW}$ is a constant offset of $30\text{ dB}$:

$$\text{Power (dBm)} = \text{Power (dBW)} + 30\text{ dB}$$ $$\text{Power (dBW)} = \text{Power (dBm)} - 30\text{ dB}$$

Standard Conversion Benchmarks

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Receiver Microvolt Levels and Calibrated S-Meters

Receiver signal strength meters (S-meters) are calibrated in logarithmic units standardized by the International Amateur Radio Union (IARU) Region 1 technical recommendation:

1. High Frequency (HF, below $30\text{ MHz}$):
   • An $S9$ indication corresponds to an input terminal voltage of $50\ \mu\text{V}_{\text{rms}}$ into a matched $50\ \Omega$ load.
   • Computing the equivalent power: $P = \frac{V^2}{R} = \frac{(50 \times 10^{-6}\text{ V})^2}{50\ \Omega} = \frac{2.5 \times 10^{-9}}{50} = 5.0 \times 10^{-11}\text{ W} = 5.0 \times 10^{-8}\text{ mW}$ $\text{Power (dBm)} = 10 \log_{10}(5.0 \times 10^{-8}) = -73.01\text{ dBm} \approx -73\text{ dBm}$
2. Very High Frequency / Ultra High Frequency (VHF/UHF, above $30\text{ MHz}$):
   • Galactic and atmospheric noise levels are substantially lower at VHF/UHF than at HF. The IARU standard specifies $S9$ as $5\ \mu\text{V}_{\text{rms}}$ into $50\ \Omega$.
   • Because voltage is scaled by $1/10$, power decreases by a factor of 100 ($20\text{ dB}$): $\text{Power (dBm)} = -73\text{ dBm} - 20\text{ dB} = -93\text{ dBm}$
3. The S-Unit Progression:
   • Each single S-unit increment (from $S8$ to $S9$, or $S5$ to $S6$) represents a $6\text{ dB}$ change in signal level.
   • A $6\text{ dB}$ change represents a $2:1$ ratio in voltage ($20 \log_{10}(2) \approx 6.02\text{ dB}$) and a $4:1$ ratio in power ($10 \log_{10}(4) \approx 6.02\text{ dB}$).
   • Above $S9$, the scale is calibrated directly in decibels: $S9+10\text{ dB}, S9+20\text{ dB}, S9+40\text{ dB}$.

                    IARU HF S-Meter Scale (f < 30 MHz, 50 Ohm)

     S-Unit:     S1     S3     S5     S7     S8     S9     S9+10dB   S9+20dB
                 |      |      |      |      |      |         |         |
     Power:    -121   -109    -97    -85    -79    -73       -63       -53    (dBm)
     Voltage:   0.2    0.8    3.1   12.5     25     50       158       500    (uV rms)
                 <------------------ 6 dB per S-unit ------------------>

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Rules of Thumb and Mental Math Shortcuts

In field operations and technical examinations, rapid decibel estimation without an electronic calculator is an indispensable skill. All power and voltage ratios can be broken down using the Rule of 3s and 10s:

Fundamental Mental Math Building Blocks

• $+3\text{ dB} \approx 2\times$ power ($1.414\times$ voltage); $-3\text{ dB} \approx 0.5\times$ power ($50\%$ power, $0.707\times$ voltage, the half-power point).
• $+6\text{ dB} \approx 4\times$ power ($2\times$ voltage); $-6\text{ dB} \approx 0.25\times$ power ($25\%$ power, $0.5\times$ voltage).
• $+10\text{ dB} = 10\times$ power ($3.162\times$ voltage); $-10\text{ dB} = 0.1\times$ power ($0.316\times$ voltage).
• $+20\text{ dB} = 100\times$ power ($10\times$ voltage); $-20\text{ dB} = 0.01\times$ power ($1\%$ power, $0.1\times$ voltage).
• $+30\text{ dB} = 1000\times$ power ($31.62\times$ voltage); $-30\text{ dB} = 0.001\times$ power ($0.1\%$ power).

The Additive Decomposition Technique

Because decibels are logarithms, adding decibels corresponds to multiplying linear power factors, while subtracting decibels corresponds to dividing:

$$\text{Total Ratio} = 10^{\frac{\text{dB}_1 + \text{dB}_2 + \dots}{10}} = \text{Ratio}_1 \times \text{Ratio}_2 \times \dots$$

Any arbitrary decibel value can be factored into sums and differences of $10\text{ dB}$ and $3\text{ dB}$ blocks:

• Example A: Determine the power gain of $+16\text{ dB}$: $+16\text{ dB} = +10\text{ dB} + 3\text{ dB} + 3\text{ dB}$ $\text{Power Multiplier} = 10 \times 2 \times 2 = 40\times$
• Example B: Determine the power gain of $+27\text{ dB}$: $+27\text{ dB} = +10\text{ dB} + 10\text{ dB} + 10\text{ dB} - 3\text{ dB}$ $\text{Power Multiplier} = 10 \times 10 \times 10 \div 2 = 1000 \div 2 = 500\times$
• Example C: Estimate the power corresponding to $+56\text{ dBm}$: $+56\text{ dBm} = +30\text{ dBm} (1\text{ W}) + 20\text{ dB} + 3\text{ dB} + 3\text{ dB}$ $\text{Power} = 1\text{ W} \times 100 \times 2 \times 2 = 400\text{ Watts}$

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Cascaded RF Stage Gain and Loss Analysis

An RF system consists of multiple stages connected in series: signal generators, filters, preamplifiers, power amplifiers, attenuators, transmission lines, and antennas.

                        Cascaded RF System Architecture

      +-------------+    +------------+    +------------+    +-----------+
      | Transceiver |    | Coax Cable |    | Power Amp  |    | Feedline  |    |--+ Antenna
      | Output P_in |--->| Attenuation|--->|   Gain     |--->|   Loss    |--->|  | Gain
      |  (+10 dBm)  |    |  (-1.5 dB) |    |  (+20 dB)  |    | (-2.5 dB) |    |--+ (+11 dBi)
      +-------------+    +------------+    +------------+    +-----------+

In linear units, first calculate the conducted power reaching the antenna, then apply its isotropic gain to obtain EIRP:

$$P_{\text{ant}} = \frac{P_{\text{in}}g_{\text{amp}}}{l_{\text{coax}}l_{\text{feed}}}, \qquad \text{EIRP} = P_{\text{ant}}g_{\text{ant}}$$

Here $g$ and $l$ are linear gain and loss factors. In decibels, the diagram gives:

$$P_{\text{ant}}\text{ (dBm)} = 10 - 1.5 + 20 - 2.5 = 26\text{ dBm}, \qquad \text{EIRP}\text{ (dBm)} = 26 + 11 = 37\text{ dBm}$$

The antenna therefore receives about $0.40\text{ W}$; its EIRP in the specified direction is about $5.0\text{ W}$.

Complete Link Budget Analysis

A link budget accounts for all gains and losses from the transmitter power amplifier through the propagation medium to the receiver demodulator detector. The received power ($P_{\text{rx}}$) is formulated as:

$$P_{\text{rx}} = P_{\text{tx}} - L_{\text{tx\_line}} + G_{\text{tx\_ant}} - L_{\text{path}} + G_{\text{rx\_ant}} - L_{\text{rx\_line}}$$

where:

• $P_{\text{tx}}$ is transmitter output power in $\text{dBm}$ (or $\text{dBW}$),
• $L_{\text{tx\_line}}$ is the transmit feedline and connector attenuation in $\text{dB}$,
• $G_{\text{tx\_ant}}$ is the transmit antenna forward gain in $\text{dBi}$ (gain relative to an isotropic radiator),
• $\text{EIRP} = P_{\text{tx}} - L_{\text{tx\_line}} + G_{\text{tx\_ant}}$ is the Effective Isotropic Radiated Power,
• $L_{\text{path}}$ is the path loss through space in $\text{dB}$,
• $G_{\text{rx\_ant}}$ is the receiving antenna gain in $\text{dBi}$,
• $L_{\text{rx\_line}}$ is the receive feedline attenuation in $\text{dB}$.

The operating fade margin (safety buffer against fading, rain, or multipath) is:

$$\text{Fade Margin (dB)} = P_{\text{rx}} - P_{\text{receiver\_sensitivity}}$$

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Step-by-Step Worked Example: VHF Point-to-Point Link Budget

An amateur radio club designs a $144.20\text{ MHz}$ (2-meter) point-to-point link across a distance of $45\text{ km}$. The system parameters are:

1. Transmitter output power: $P_{\text{tx}} = 50\text{ Watts}$.
2. Transmit station feedline: $30\text{ metres}$ of RG-213 coaxial cable, having an attenuation of $2.8\text{ dB}$ per $30\text{ m}$.
3. Transmit antenna: 9-element Yagi with forward gain $G_{\text{tx}} = +13.1\text{ dBi}$.
4. Free-space path loss ($L_{\text{path}}$) over $45\text{ km}$ at $144.2\text{ MHz}$ is calculated as $108.7\text{ dB}$.
5. Receiving antenna: 5-element Yagi with forward gain $G_{\text{rx}} = +9.0\text{ dBi}$.
6. Receive station feedline: $15\text{ metres}$ of low-loss foam coax with $L_{\text{rx}} = 1.1\text{ dB}$.
7. Receiver sensitivity threshold for $12\text{ dB}$ SINAD: $-124\text{ dBm}$ ($0.14\ \mu\text{V}$). Calculate:
   • Transmitter power in $\text{dBm}$.
   • Effective Isotropic Radiated Power ($\text{EIRP}$).
   • Total received power ($P_{\text{rx}}$) in $\text{dBm}$.
   • The link fade margin in $\text{dB}$.
   • The equivalent S-meter reading on the VHF receiver.

Step 1: Convert transmitter power to $\text{dBm}$:

$$P_{\text{tx}}\text{ (dBm)} = 10 \log_{10}\left(\frac{50\text{ W}}{0.001\text{ W}}\right) = 10 \log_{10}(50000) = 10 \times 4.6990 \approx +47.0\text{ dBm}$$

Step 2: Calculate Effective Isotropic Radiated Power ($\text{EIRP}$):

$$\text{EIRP} = P_{\text{tx}} - L_{\text{tx}} + G_{\text{tx}} = +47.0\text{ dBm} - 2.8\text{ dB} + 13.1\text{ dBi} = +57.3\text{ dBm}$$

($+57.3\text{ dBm}$ is equivalent to $537\text{ Watts}$ of omnidirectional isotropic radiation).

Step 3: Calculate received signal power ($P_{\text{rx}}$):

$$P_{\text{rx}} = \text{EIRP} - L_{\text{path}} + G_{\text{rx}} - L_{\text{rx}}$$ $$P_{\text{rx}} = +57.3\text{ dBm} - 108.7\text{ dB} + 9.0\text{ dBi} - 1.1\text{ dB}$$ $$P_{\text{rx}} = -51.4\text{ dBm} + 7.9\text{ dB} = -43.5\text{ dBm}$$

Step 4: Calculate the link fade margin:

$$\text{Margin} = P_{\text{rx}} - P_{\text{sens}} = -43.5\text{ dBm} - (-124.0\text{ dBm}) = +80.5\text{ dB}$$

An $80.5\text{ dB}$ margin ensures exceptionally robust, uninterrupted communication.

Step 5: Determine the equivalent S-meter reading:

• On VHF ($> 30\text{ MHz}$), $S9 = -93\text{ dBm}$.
• Signal level above $S9$: $\Delta P = P_{\text{rx}} - (-93\text{ dBm}) = -43.5\text{ dBm} - (-93.0\text{ dBm}) = +49.5\text{ dB}$
• The receiving station will register an overwhelming signal of $S9+50\text{ dB}$.