Electronic – What are the units used in Shannon formula calculation

cellularcommunicationgaintelecommunications

I'm a CompSci student with absolutely no background in wireless communication who recently needs to do something with the Shannon formula:
$$
R = W log_2\bigg(1+\frac{P_t H^2}{N_0 W}\bigg)
$$

I need to calculate the achievable data rate \$R\$ given:

  • Bandwidth \$W=2\$ MHz.
  • Transmit power \$P_t=0.2\$ Watt.
  • Path loss model: \$128.1 + 37.6log_{10}(d)\$, where d = 0.45 km.
  • Noise power density \$N_0 = -174\$ dBm/Hz.

This is how I'm calculating \$R\$:

  • Convert bandwidth \$W\$ to Hz, so now \$W=2\times10^6\$ Hz.
  • Leave the transmit power as is (Watt).
  • Given the path loss model, we have \$H=10^{-12.81}(450)^{-3.76}\approx 1.64\times10^{-23}\$.
  • Convert noise power density \$N_0\$ to W/Hz, so now \$N_0=10^{\frac{-174-30}{10}}=10^{-20.4}\$ W/Hz.
  • Finally, the data rate
    $$
    R = 2\times10^6\times log_2\bigg(1 + \frac{0.2\times(1.64\times10^{-23})^2}{10^{-20.4}\times 2\times10^6}\bigg) \approx 0 \text{ bps (?!?)}
    $$

The resulted data rate is extremely small – close to zero. I reckon because the channel gain is too small? Could someone please tell me what went wrong in my calculation above? I'm quite confused with all the different units of measurement in wireless communication.

I already went over this relevant question (Calculating Data Rate using Bandwidth, Transmission Power, Noise Power Spectrum Density and Channel Gain) but still couldn't figure out.


UPDATE 1:

I just found out that the path loss model: \$128.1 \text{dB} + 37.6log_{10}(d)\$ is equivalent to \$ \frac{P_r}{P_t}=\frac{10^{-12.81}}{d^{3.76}} \$. So that we can find the actual received power \$P_r\$ (transmit power minus the power lost by distance during the transmission). I guess the numerator in the formula (\$P_t H^2\$) should be replaced by this \$P_r\$. The data rate now is:
$$
R = 2\times10^6\times log_2\bigg(1 + \frac{0.2\times \frac{10^{-12.81}}{d^{3.76}}}{10^{-20.4}\times 2\times10^6}\bigg)\approx 0.5 \text{ Mbps}
$$

I'm not sure whether distance \$d\$ should be meter or kilometer. If meter, the data rate is 70.2Mbps, which is crazily high. If kilometer, we have spectral efficiency (the log part of the formula) \$=0.25\$ bps/Hz, and data rate \$R=0.5\$ Mbps, which makes more sense. Am I doing this right? Is this number realistic given all the settings above?

~~

Alternatively, I've seen the channel gain being calculated like this \$H=d^{-\alpha}\$, where distance \$d\$ is in meter, and path loss exponent is, say, 3.76. If I apply this to the original formula above. The data rate is now:
$$
R = 2\times10^6\times log_2\bigg(1 + \frac{0.2\times (450^{-3.76})^2}{10^{-20.4}\times 2\times10^6}\bigg) \approx 0.8 \text{ bps} \approx 8.1\times 10^{-7} \text{ Mbps}
$$

which seems quite small compared to what I got above (0.5 Mbps). I know the setting here (channel gain calculation) is different but I wasn't expecting such a large difference.

Is there anything wrong? Which calculation is more realistic in practice?


For those interested, I found these lecture notes quite useful. They contain some example calculations.

http://weber.itn.liu.se/~vanan11/TNK080/Lecture1.pdf

http://weber.itn.liu.se/~vanan11/TNK080/Lecture2.pdf

Best Answer

Your formula for data rate combines Shannon's equation and signal propagation loss into one row. Shannon's equation for an analog communication channel subject to additive white Gaussian noise (AWGN) of power N:

R = W∙log2(1+S/N)

Assuming your receiver amplifier's noise is negligible and the signal bandwidth is W, we can aspire to reach data rate as high as

R = W∙log2(1+P_R/(N_0∙W))

where P_R is a signal power at the receiver, N_0 is a noise power density.

With the freespace propagation loss model, the signal power at the receiver can be expressed as

P_R = P_T∙G_T∙G_R∙(lambda/(4πd))^2

where P_T is a full transmitted power, G_T and G_R are "directional gains" of transmitting and receiving antennas respectively, lambda is a wavelength, d is a distance between transmitting and receiving antennas.

Expressing wavelength lambda via frequency f in megahertz units (lambda = c/f, c = 299792km/sec), we arrive at the formula (d in kilometers)

P_R/P_T = G_T∙G_R∙0.57∙10^(-3)/(d∙f)^2

In engineering calculations, the ratio P_R/P_N is expressed in decibels:

(P_R/P_T)(dB) = (G_T)(dB) + (G_R)(dB) - (32.44 + 20∙lg(d) + 20∙lg(f))(dB) (lg is decimal logarithm)

The quantity 32.44 + 20∙lg(d) + 20∙lg(f) is termed "path loss in empty space", the losses, calculated under assumption that there are no signal reflection and no absorption of radiation in the medium.

You never mentioned the derivation of an H_0 quantity in your formulas, so I carry out the numerical calculation for half-wave dipole antennas at both transmitter and receiver ends. Therefore, the directional gains are

G_T = G_R ≈ 2.15 dB

Also, you do not specify the radio frequency f. I assume f = 2400 MHz.

With your data d = 0.45km and my assumed radio frequency f = 2400MHz,

(P_R/P_T)(dB) = 2∙2.15 - 32.44 - 20∙lg(0.45) - 20∙lg(2400) = -88.8

With your data P_T = 0.2W, N_0 = -174dBm/Hz = 10^(-20.4)W/Hz,

R = 2∙10^6∙log2(1+0.2∙10^(-8.88)/(10^(-20.4)∙2∙10^6)) ≈ 30 Mbps

With your data and under assumptions I specified above, we can achieve a data rate of approx. 30Mbps. That good value for a 2MHz bandwith is not unbelievable, because noise figure is quite low. Indeed, -174dBm/Hz is the thermal noise at 290K and therefore the low limit for conventional designs.