# Calculating signal/noise ratio and the noise level [closed]

If the received signal strength for a 5.2 dBi dipole was measured at -46 dBm and the noise level was -96 dBm, what would the signal/noise ratio be in dB, the received power in µW, and the noise level in nW?

• I don't completely understand what you're asking-- if the measurement was in dBm, the measurement is relative to 1mW. Are you trying to translate that into microWatt and nanoWatts? Are there multiple questions here? – Omegaman Jan 6 '16 at 23:25
• There are no multiple questions i cant figure out what it is asking me to do, i think it's asking me to find out what the signal/ noise ratio is and the received power is and the noise level from the given numbers. – user3395953 Jan 6 '16 at 23:32
• I think you're right about what's being asked of you. If you need help with a specific concept in the question, I suggest asking the question more specifically with a focus on the concept you're missing. – Omegaman Jan 7 '16 at 0:44
• These are straightforward calculations from logarithmic to linear scales. As Omegaman says, ask a more specific question if you need to. – MBaz Jan 7 '16 at 1:39

The decibels (dB) provides a logarithmic scale for the ratio of two power values. Mathematically it is defined as: $$R = 10 \log_{10} \left(\frac{P}{P_0}\right)$$ where $P$ and $P_0$ are power values in linear scale (such as Watts) and $R$ is the ratio expressed in logarithmic scale. Often a fixed reference power level (such as 1 Watt) is chosen for $P_0$. When that reference power level $P_0$ is 1 milliwatt (mW), the resulting unit for the $R$ quantity is called a dBm.
Given a known reference level, you can conversely obtain the power in linear scale from the power ratio value expressed in logarithmic scale by using the inverse formula: $$P = 10^{R/10}\cdot P_0$$
So for a measurement of -46dBm (remember that the dBm units means that the reference power level is 1mW), the resulting power is: \begin{align} P_s &= 10^{-46/10} \cdot 1 \mbox{mW} \\ &\approx 2.5 \times 10^{-5} \cdot 1 \mbox{mW} \\ &\approx 2.5 \times 10^{-8} \, \mbox{W} \\ &\approx 2.5 \times 10^{-2} \, \mbox{µW} \end{align} Similarly for a measurement of -96dBm, the resulting power is: \begin{align} P_n &= 10^{-96/10} \cdot 1 \mbox{mW} \\ &\approx 2.5 \times 10^{-10} \cdot 1 \mbox{mW} \\ &\approx 2.5 \times 10^{-13} \, \mbox{W} \\ &\approx 2.5 \times 10^{-4} \, \mbox{nW} \end{align}
Now for the signal-to-noise ratio (SNR), it is simply defined as the ratio of the signal power to the noise power. So in linear scale, that would be: \begin{align} \mbox{SNR} &= \frac{P_s}{P_n} \\ &\approx \frac{2.5 \times 10^{-8}}{2.5 \times 10^{-13}} \\ &\approx 10^{5} \end{align} which when expressed in decibels would give us: \begin{align} \mbox{SNR}_{\mbox{dB}} &= 10\log_{10} \left(\frac{P_s}{P_n}\right) \, \mbox{dB} \\ &= 10\log_{10} \left(10^{5}\right) \, \mbox{dB} \\ &= 50\mbox{dB} \end{align} We could of course also have taken a bit of a shortcut to compute that SNR directly from the power measurements in dBm by using logarithmic identities to get: \begin{align} \mbox{SNR}_{\mbox{dB}} &= 10\log_{10} \left(\frac{P_s}{P_n}\right) \, \mbox{dB} \\ &= 10\log_{10} \left(\frac{\frac{P_s}{P_0}}{\frac{P_n}{P_0}}\right) \, \mbox{dB} \\ &= 10\log_{10} \left(\frac{P_s}{P_0}\right) - 10\log_{10}\left(\frac{P_n}{P_0}\right) \, \mbox{dB} \\ &= \left(-46\,\mbox{dBm}\right) - \left(-96\,\mbox{dBm}\right) \\ &= 50\mbox{dB} \end{align}