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$$\begin{align} P_\text{out} &= 51.787\text{dBm}\\ P_\text{out} &= 10\log_{10}\frac{A_\text{out}^2}2\\ A_\text{out} &= \sqrt 2 \cdot 10^{\frac{P_\text{out}}{20}}\\ &= 549.3701 \end{align}$$

Is this calculation correct to yield the amplitude of my output signal when the output power in dBm is given? Or do I have to calculate power in dBW first? I assume an average output power of $P_{out} = A_{out}^2/2$

  • your second line, $P_\text{out} = 10\log_{10}{\frac{A^2}2}$ makes a physical assumption that doesn't generally hold. Ask yourself: What is the power of a voltage signal without specifying the impedance it drives or the current? – Marcus Müller Dec 18 '16 at 13:01
  • In other words: No, you cannot derive the physical amplitude of a signal from it's physical power if you don't specify the physical system you're observing/that the power flows into. For example, to generate 1W of heat over a 1MΩ resistor, you'd need a lot more amplitude than to generate 1W of heat over a 1 Ω resistor. – Marcus Müller Dec 18 '16 at 13:08
  • you're right, of course. I edited my question. –  Dec 18 '16 at 14:11
  • So my problem with your edited question is still: if you use physical units (dBm), amplitude square is still not a power, as discussed in my other comment. You're missing the units of your power-to-amplitude-square relation, and since that defines the answer to the question "should I be using dBm or dBW" (you probably should not use either, judging by the struggle you're having with the physical aspect of this), your question remains impossible to answer! – Marcus Müller Dec 18 '16 at 18:31

1 Answers1

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The Decibel Miliwatts Scale $dBm$ is the power ratio in Decibels, considering a reference of $P_0=1mW$.

$$P[dB]=10\text{log}_{10}(P/P_0)=10\text{log}_{10}(P[W]/0.001)$$

The general context is considering the signal amplitude $V$ in $volts$ and the power as an impedance load system $R$ in $\Omega$:

$$P=\frac{1}{R}V^2$$

Moreover, sinusoidal excitations are considered (for a given frequency also!), so we also can have: $$P=\frac{1}{R}V_{rms}^2=\frac{1}{2}\frac{1}{R}V_{pk}^2$$

Thus:

$$ P[dBm] = 10\text{log}_{10}\left(\frac{V[V]^2}{R[\Omega]}\frac{1}{P_0}\right) = 20\text{log}_{10}\left(\frac{1}{\sqrt{R[\Omega]}}\frac{V[V]}{\sqrt{P_0}}\right) $$

So, as reference:

  • $1W$ equals $30dBm$,
  • for a $50\Omega$ impedance system, this also equals $7.07V$, and
  • under the sinusoidal case, this also equals $10V_{pk}$ and $7.07V_{rms}$.
Alessandro Cuttin
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