# Calculating Data Rate using Bandwidth, Transmission Power, Noise Power Spectrum Density and Channel Gain

I'm a Software Engineering student researching on Load Balancing in IoT and I need to calculate data rate. However, I don't know how to use these values to attain data rate despite searching on the Internet a lot. Actually, I am not sure whether given values are sufficient or not.

I know that:

• Bandwidth (W) = 10 MHz
• Noise power density (N0) = -174 dBm/Hz
• Path loss model = 128.1 + 37.6log10d, where d refers to the distance in kilometers.
• Transmission power (P) = 23dbm

and the data rate:

$$r =Wlog_2(1 + \frac {P_{tr} H}{N_0W})$$ My questions are:

1. How should I calculate channel gain (H)? Is it related to path loss model?
2. Should channel gain change to the power 2?
3. Do parameters need any unit conversion?
• You don't have enough information to calculate the data rate. You're missing the required SNR at the receiver, and the constellation. Also, Shannon's theorem is irrelevant here since you don't seem to be using any coding. I have seen problems like this in some textbooks, though; is this a homework problem? – MBaz May 19 at 16:21
• @MBaz I've seen on dsp.stackexchange.com/questions/58420 and some papers that the name of this formula is Shannon, though it doesn't matter to me :). No, this is a part of my research in which I need to calculate the achievable transmit rate of Mobile Devices. – Benyamin T May 19 at 17:37
• The Shannon capacity is the maximum data rate you can transmit over a channel. What is the type of the channel you are considering. Is it wireless or wired? – BlackMath May 19 at 18:10
• @BlackMath I need to calculate a wireless channel data rate. – Benyamin T May 19 at 19:04
• @BenyaminT Note that a channel does not have a rate; it has a capacity. A communications system achieves a fraction of that capacity. As I said, you don't have enough data to calculate a rate. – MBaz May 19 at 19:13

Let me answer just in terms of the units (the other side-questions have been discussed in the comments already):

• Shannons formula is to be interpreted as $$W \log_2(1+{\rm SNR})$$, where SNR is the signal to noise ratio.
• A signal to noise ratio is a ratio of powers: received signal power divided by total noise power at the receiver.
• Your received signal power is the transmitted signal power $$P$$ reduced by what you lose on the way ("path loss").
• "Reduced by" means fraction in linear scale (1 Watt transmitted and a path loss of 1000 means you receive 1 Milliwatt).
• We often prefer to use logarithmic scales instead since division becomes subtraction in log scale, which is easier. To do this, we define logarithmic measures of power: The logarithmic equivalent of a power in W is $$10 \log_{10} \frac{P}{1W}$$ and measured in dBW (dB Watt). It means 1 W is the same as 0 dBW and every doubling adds 3 dBs (2 W = 3 dBW, 4 W = 6 dBW, 8 W = 9 dBW and so on). Alternatively, we can use Milliwatt as a reference and get $$10\log_{10} \frac{P}{1{\rm mW}}$$ in dBm (dB Milliwatt).
• You are transmitting 23 dBm and losing $$128.1+37.6 \log_{10} \frac{d}{1{\rm km}}$$ on the way. Hence, as an example, in 100m distance, you have lost 90.5dB. This means your received power is 23 dBm - 90.5 dB = -67.5 dBm. In linear scale this is $$10^{-67.5/10} = 0.177 µW$$.
• Your thermal noise power density is given as -174 dBm/Hz. This means $$10^{-17.4}$$W/Hz which at a bandwidth of 10 MHz becomes $$10^{-10.4} = 0.04 nW$$.
• Hence your SNR is approximately $$0.177/0.00004 = 4400$$. With perfect link adaptation this would give you $$\log_2(1+4400) = 12.1$$ bits per channel use (That's a crazy high number! I can tell you that you should not expect more than 2-4 bit per channel use in practice.)
• At a bandwidth of 10 MHz you could use symbol rates up to $$10^7$$ channel uses per second.

Just to be safe, a few words of warning when converting this to a data rate. This is risky since a lot of things are missing:

• Fading would lead to variations in the SNR which means you'd need to go lower in your modulation order to account for these.
• At 10 MHz you'd encounter frequency-selective fading, which requires suitable mechanisms to combat it, e.g., OFDM or SC-FDMA. For OFDM, you then have a frequency-dependent SNR and need to apply bit loading to make proper use of all the subcarriers. IOT devices would do something simpler like SC-FDMA but probably not at 10 MHz.
• Reliable communication will require coding, which comes with some overhead.
• You cannot fully use the 10 MHz, due to spectral leakage of whatever pulse shape you are using, you'd need to plan for some guard bands.
• The noise calculation is overly optimistic since no noise figure was included. Typically, you would need to increase your receiver noise power by the noise figure of your LNA. IOT devices may use simple LNAs with higher noise figures.
• In IOT you have multiple access problems to solve, i.e., the devices need to share the spectrum. This comes at some loss, since there will always be times where the spectrum is underutilized and you need to leave room for the initial access of new devices.

And that is still an incomplete list :)

• Thanks a lot for your thorough answer. You haven't mentioned anything about H in formula or channel gain. Is there any possibility that the high result of data rate relates to this? In one paper, which I'm referring to, the wireless channels between nodes are said to be i.i.d. frequency-flat block fading and small-scale fading channel power gain from the sender to the receiver at the 𝑡th time slot is ℎ𝑖 (𝑡). – Benyamin T May 20 at 17:22
• Subsequently, a formula represents the channel gain which is: 𝐻𝑖 (𝑡) = ℎ𝑖 (𝑡) 𝑔0 (𝑑0/𝑑𝑖)^𝜃 where h(t)= 1, path-loss constant 𝑔0=-40dB, the reference distance 𝑑0 = 1m, path-loss exponent 𝜃 = 4, and d the distance between sender and receiver. Is this formula accurate? – Benyamin T May 20 at 17:22
• A gain is the inverse of a loss, one can refer to a path loss or a channel gain but the terms are used quite interchangeably. If your channel gain is -90dB it means you lose 90 dB between TX and RX, so your path loss would be (+)90 dB. The formula for the channel gain you gave translates to a path loss of $40 {\rm dB} + 10\log_{10}([d/1m]^4) = 40 {\rm dB} + 40 \log_{10}(d/1m) = 160 {\rm dB} + 40 \log_{10}(d/1km)$. For the 100m example I was showing above this means 120 dB path loss (you'd receive less than a picowatt then). Quite a difference :) – Florian May 21 at 7:28
• Yes, the results are utterly different. However, as you mentioned and based on the context of my research, using channel gain formula, which by my calculation results in 23Mbps, is more realistic than 120Mbps of using that path loss model. But one question still remained for me, Isn't there any problem that you subtract dBm from dB? For example, you said my received power is 23 dBm - 90.5 dB = -67.5 dBm. Or using this gain formula I subtracted 23 dBm - 120 dB = -97 dBm – Benyamin T May 21 at 9:43