# Calculate power given PSD and bandwidth?

Given the spectral recommendation for an Asymmetric Digital Subscriber Line (ADSL) below, I'd like to calculate the power in Watts that a modem would have to emit for the upstream band. I want to do so for the purposes of determining the maximum signal voltage (based on a standard 100 $$\\Omega\$$ load) that would need to be transmitted in order to determine bit requirements.

I thought that to get power I should multiply the PSD by the upstream bandwidth as

$$\PSD \times bandwidth = 37\frac{dBm}{Hz} \times129349.125Hz = 4.786\times10^6 dBm \$$

And then convert $$\dBm\$$ to $$\dB\$$

$$\dBm = 10log(\frac{power}{1mW})\$$

$$\\frac{power}{1mW} = 10^{\frac{4.786 \times 10^6}{10}}\$$

I'm making a logical mistake somewhere because those numbers end up being way too large. It's probably very silly but what am I doing wrong here?

edit: mistake I made was to not convert from dBm to watts first. Calculation works out correct using $$\PSD \times bandwidth = 10^{\frac{-37}{10}} \times129349.125Hz = 20.65 dBm \$$

First: that's a maximally allowed spectral mask (more or less), so modems don't have to fully employ that spectrum. In fact, in an ADSL deployment, it's very probable that modems will be told to dial down their power to reduce crosstalk to other subscribers.

About your calculation: you're doing the classical mistake of ignoring the $$\-\$$ in front of $$\-37\,\text{dBm}\$$. 37 dBm is more than a watt, -37 less than a microwatt.

I'm doing a quick in-head calculation:

130 kHz is roughly 51 dBHz. -37 dBm/Hz over that bandwidth is (-37 + 51) dBm = 14 dBm. Since one milliwatt is a thousandst of a watt, 14 dBm = (14-30) dBW = -16 dBW. -10 dBW is a a tenth of a Watt, -6 is a factor of ¼, so we're at 1/40 W = 0.025 W output power.

I want to do so for the purposes of determining the maximum signal voltage (based on a standard 100 Ω load)

Won't perfectly work that way, because the impedance of a pair of telephone cabling isn't even remotely constant over such a big bandwidth. So, a single tone of 1mW of power measured somewhere on the cable at 100 kHz will need to have a different amplitude than a tone with the same power at 26 kHz.

that would need to be transmitted in order to determine bit requirements.

I don't understand what you mean with "bit requirements", but as communications engineers, we work with bit energies, not with bit amplitudes times duration. The reason for ADSL is very easy to grok:

Due to the large bandwidth, you don't use the ADSL uplink as single-carrier system, but as masssive multi-carrier system. You can't directly relate instantaneous voltage at the output of the transmitter to any single bit (not to mention that these bits are put into complex symbols first, which often even have constant amplitude/envelope), because the output signal is the sum of a lot of individual information-carrying signals at different frequencies.

• To produce 0.025 watts in 100 ohms, using P =V^2 / R, you need about 1.5 volt RMS or about 5 volts PP. Jun 1, 2019 at 9:39
• @analogsystemsrf that's only where you can link RMS to PP; see the bandwidth above: will a 2 km long twisted pair cable have constant impedance over three octaves? ADSL uses OFDM, and thus exposes a high PAPR; average power and peak amplitude hence aren't related as for sine signals. Jun 1, 2019 at 9:44