# 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? Jan 6, 2016 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. Jan 6, 2016 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. Jan 7, 2016 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, 2016 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}