We have a one 1 MB file that in 1 second moved from 1 MHz channel with 60 dB attenuation. If the density of power to noise is -174 dBm/Hz, the transmitter power is equal to 1 microwatt.

My question is how this value is calculated? Any idea?

Update 1:

I add my note messy solution. Please help me understand what is this step and how the numerical value is calculated.

enter image description here

  • 1
    \$\begingroup\$ Using standard techniques found in communications theory textbooks and taught in college classes.. \$\endgroup\$ – Brian Drummond Aug 4 '16 at 9:32
  • \$\begingroup\$ Would you please give me links? I study more but couldn't find this relation? \$\endgroup\$ – user355834 Aug 4 '16 at 11:36
  • \$\begingroup\$ Sure. en.wikipedia.org/wiki/Channel_capacity \$\endgroup\$ – Brian Drummond Aug 4 '16 at 11:43
  • \$\begingroup\$ @BrianDrummond this is not very useful for me, I read it for one and half hour. thanks anyway. \$\endgroup\$ – user355834 Aug 4 '16 at 14:20
  • \$\begingroup\$ Well unless you tell us which part of the calculation you succeeded with and what you are stuck on, or what answer you got (and intermediate results) when you tried, you're not going to get help. \$\endgroup\$ – Brian Drummond Aug 4 '16 at 14:33

Basically, this exercise tries to illustrate what is the meaning of the capacity channel by using the Shannon capacity channel as the simplest model.

The information we have is:

  • How fast we have transmitted a determined amount of information. 1 MB in 1 second
  • How noisy the channel is. 60 dB
  • How many losses the channel has. 174 dBm/Hz
  • How wide the channel is. 1 MHz

Now, we want to know the input power of the signal. With all data above it is possible to figure out.

First goal is compute output signal power.

Capacity channel is the maximum transmission data rate possible without error in the transmitted data. This parameter depends on the power signal, the amount of noise added by the channel and the bandwidth. \$ R = W \log_2{(1+ \frac{S}{N})}\$, where R is the capacity channel, W the bandwidth, S the power signal in linear units, N the floor noise in linear units. Assuming we are transmitting at the maximum rate the channel can handle, we can compute R as the amount of data transmitted (1 MB) divided by the time employed to transmit it: $$R = \frac{1~MB }{1~s} \cdot \frac{10^6~byte}{1~MB} \cdot \frac{8~bit}{1~byte} = 8~Mb/s $$

According to noise, we now the spectral density noise (\$ N_0 \$), i.e. how much noise per Hz the channel is adding. To know the total noise power in the communication, \$ N = N_0 \cdot B \$.

Now, we just need to compute \$ S_0 \$ in the capacity channel expression taking into account that logarithm is base 2.

$$R = W \log_2{(1+ \frac{S}{N})}$$ $$8\cdot 10^6 = 10^6 \log_2{\left(1+\frac{S_0}{4\cdot 10^{-15}}\right)}$$ $$\frac{8\cdot 10^6}{10^6} = \log_2{\left(1+\frac{S_0}{4\cdot 10^{-15}}\right)} $$ $$8 = \log_2{\left(1+\frac{S_0}{4\cdot 10^{-15}}\right)} $$ Now, using the logarithm property \$ a = \log_b{x} \Rightarrow b^a = x\$ $$2^8 = 1+\frac{S_0}{4\cdot 10^{-15}} $$ $$256 = 1+\frac{S_0}{4\cdot 10^{-15}} $$ $$255 = \frac{S_0}{4\cdot 10^{-15}} $$ $$255 \cdot 4\cdot 10^{-15} = S_0 $$ $$1.02\cdot 10^{-12} = S_0$$ As \$S_0\$ is power, the conversion to dBm is: $$S_0~(dBm) = 10\log_{10}{\frac{1.02\cdot 10^{-12}}{10^{-3}}} \approx -90~ dBm$$

Finally, as the output signal is 60 dB smaller at the output than at the input due to channel losses: $$S_0 = S_i - L$$ $$S_i = S_0 + L = -90 - \left( -60\right) = -90 + 60 = -30~dBm = 10^{\frac{-30-30}{10}} = 10^{-6}~W = 1~\mu W$$ (Remember that dBm to W is \$ 10^{\frac{Power_{dBm}-30}{10}} \$)

  • \$\begingroup\$ very useful, but my problem really is in last section "finding input power " \$\endgroup\$ – user355834 Aug 7 '16 at 20:20
  • \$\begingroup\$ I've completed my answer with step-by-step process. Hope it could help. \$\endgroup\$ – Rubén Sánchez Aug 8 '16 at 11:09
  • \$\begingroup\$ "How many losses the channel has"? \$\endgroup\$ – user2943160 Aug 8 '16 at 11:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.