I face an ambiguity when considering the carrier frequency, signal frequency, bandwidth and data rate.

When I consider the Shannon formula, I wonder if increasing the transmission power also means that the Data Rate is increased without modifying the bandwidth of the signal.

Or is modifying the power of the signal going to implicitly affect the bandwidth of that signal as well?

If you can give a thorough description through these terms and the relation between them, it would be very helpful


Increasing the signal increases signal to noise ratio therefore the data rate can rise: -

enter image description here

  • C is the channel capacity in bits per second;
  • B is the bandwidth of the channel in hertz (passband bandwidth in case of a modulated signal);
  • S is the average received signal power over the bandwidth (in case of a modulated signal, often denoted C, i.e. modulated carrier), measured in watts (or volts squared);
  • N is the average noise or interference power over the bandwidth, measured in watts (or volts squared); and
  • S/N is the signal-to-noise ratio (SNR) or the carrier-to-noise ratio (CNR) of the communication signal to the Gaussian noise interference expressed as a linear power ratio (not as logarithmic decibels).

The bandwidth remains unchanged although the extreme edges of the bandwidth may now spill over into closely localized channels due to overall higher power. In other words, in a practical world, tighter bandwidth filtering may be required. I'm thinking radio as a good example. But generally more signal means more bits per second.

  • \$\begingroup\$ awesome, so increasing the power of the signal results in higher data-rate... but why? When I try to visualize it, I think of a sine wave which now has farther edges (bigger amplitude). How does this bigger amplitude translate to higher data-rate. There must be something (probably modulation) happening in between taking advantage of this higher amplitude, right? Would you mind elaborating? Or is my assumption wrong... Then, data-rate gets better just because the formula says so? \$\endgroup\$ – Kristof Tak May 19 '15 at 20:38
  • 1
    \$\begingroup\$ Shannon's formula gives the maximum data rate for a transmission channel. What data rate your system actually uses depends on how you choose to modulate and encode your signal. \$\endgroup\$ – The Photon May 19 '15 at 22:06
  • \$\begingroup\$ @KristofTak With a bigger SNR you can use the analogue signal to "represent" more integers. A signal level of 1V with a noise of only (say) 100mVp-p (statistic abuse warning) is totally discernible from 1.5 volts, which in turn is totally distinguishable from 2.0V i.e. there is no cross-corruption that can make 1V look like 1.5 volts and ditto for the 2 volt signal. Thus, a given analogue amplitude might represent 2, 3 or more bits. Look up QAM64 - there are 64 different analogue values that represent a binary number. If the noise is low then range between 2 analog values is incorruptible. \$\endgroup\$ – Andy aka May 19 '15 at 22:18

Shannon's formula gives the maximum data rate for a transmission channel. It tells you nothing about the modulation scheme or encoding that you would need to use to achieve that maximum.

There are (at least) two ways you can change your system to change the data rate without changing the bandwidth:

  • Use multilevel modulation such as PAM or QAM to transmit more than one bit per symbol.

    For example, if you were using on-off coding to send data through a channel, and you increased the signal power, you might then be able to use 4-PAM or 16-QAM to send a higher bit rate through the same channel.

  • Use error correcting codes to reduce the data bits per symbol. This gives a lower bit rate but allows you to achieve essentially error-free transmission in a noisy channel.

    For example, if you were using a strong error-correcting code with high overhead to send data through a low-SNR channel, and then you increased the signal power, you might then be able to use a weaker code with less overhead to achieve a higher data rate.


The data rate gets better because the increase in signal to noise ratio allows more bits to be sent in a given time. This is because the larger signal to noise ratio makes it easier for the receiver to determine what bits were sent. For example, if one sends 2 bits in one time interval (4 possibilities), the receiver must determine which of 4 different levels were sent. If the signal level is increased, then the level shift between the 4 possibilities increases. This will decrease the error rate for the same bit rate. However, one could also send more bits in one time interval with the same difference in signal level between each possible combination. This is why increased signal level (power) allows an increase in bit rate. This explanation is simplified but the idea is valid.


I will throw in another answer into the mix.

Increasing the amplitude of the transmit signal (and keeping all other things constant) just increases the SNR of the transmission scheme and so it does not necessarily lead to an increased bit transmission rate but the higher SNR will improve the bit-error-rate (BER) of your scheme so although you will be receiving the same amount bits in a second, the number of errors will be reduced (so the number of error-free/information bits a second increases). Channel capacity measures the maximum number of error free bits that can be transmitted over a noisy AWGN channel for a particular SNR and thats why the channel capacity rises with an increase in SNR.

Now, if you were to keep the max transmit amplitude constant and try to achieve a higher bit-rate by switching to a bandwidth efficient scheme (such as switching to 16QAM from BPSK or something), you would get the a higher bit transmission rate but you would suffer from a higher bit error rate because the constellation symbols in a bandwidth efficient scheme like 16QAM are very close together and so the phase and amplitude immunity is very low.Increasing the amplitude however would allow you to increase you SNR and achieve the same BER as you achieved before but with a higher bit transmission rate this time.

Now, as another poster has already mentioned there are other ways to improve the data rate without changing the amplitude such as using source coding (i.e compressing the bits you want to send before you send them) or using channel coding (i.e increasing the bits you are sending by coding them in such a way that the transmission becomes resistant to noise).Doing this can increase the effective data transmission rate because the BER will be very low, this will obviously increase the complexity of the transmission scheme.


No increasing the transmit power does not mean the data rate increases. However, if you wish to transmit at higher speed then you do need to increase the transmit power.

Changing the amplitude does not increase the data rate, unless you are using an amplitude based modulation scheme to represent data bits (such as QAM - quadrature amplitude modulation) and you change the design and use more levels to enable more bits to be transmitted.

You need to understand what the modulation scheme employed is. Bandwidth can change (and often does) increase with increased data rate.

Once you understand what modulation scheme will be used, then you can lookup in published articles how to calculate the bandwidth for a given data rate. It is not necessarily trivial.

I'd advise: forget Shannon for the moment, that enables you to calculate the maximum information carrying capacity but it seems if you yet don't understand the terminology, definitions and relationships between bandwidth, power, data rate.


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.