How Can QAM's Phase and Amplitude Resolution Impact Bit-Rate Capacity?

[aznelmo098]

wave signal, bit-rate help!

Homework Statement

Suppose we use QAM, in which the phases can be resolved only down to difference of π/12, π = pie symbol. And amplitudes down to differences of 1/23 of the range between the minimum and maximum amplitudes. Assuming that 1/9 of the bits are not used for sending the user data, what is the bit-rate capacity can be supported, assuming a bandwidth of W Hz can support a baud rate of 1.5 W ?

Homework Equations

Bit rate capacity = Wlog2(1 + SNR)

The Attempt at a Solution

no idea

 

[KataKoniK]

, please help!

Hello there!

I can help you with your question about wave signals and bit-rate. First, let's define some terms. QAM stands for Quadrature Amplitude Modulation, which is a type of modulation technique used in communication systems. It allows for the transmission of multiple bits of data at once by varying both the amplitude and phase of the carrier wave.

In this case, we are limited by the resolution of our phase and amplitude measurements. We can only resolve differences in phase down to π/12 and differences in amplitude down to 1/23 of the range between the minimum and maximum amplitudes. This means that for every 12 different phase values, we can only have 1 bit of information. Similarly, for every 23 different amplitude values, we can have 1 bit of information.

Now, you mentioned that 1/9 of the bits are not used for sending user data. This means that out of every 9 bits, 1 bit is reserved for other purposes. Therefore, the effective bit-rate capacity will be 8/9 of the total bit-rate capacity.

To calculate the bit-rate capacity, we can use the formula: Bit rate capacity = Wlog2(1 + SNR). Here, W is the bandwidth and SNR is the signal-to-noise ratio.

Given that the bandwidth is W Hz and the baud rate is 1.5 W, we can say that the signal occupies 1.5 W Hz of the bandwidth. This means that the noise occupies the remaining bandwidth, which is 0.5 W Hz.

Now, to calculate the SNR, we can use the formula: SNR = (S/N)^2, where S is the signal power and N is the noise power. Since we know that the signal occupies 1.5 W Hz of the bandwidth, we can say that the signal power is 1.5 W. Similarly, the noise power will be 0.5 W.

Therefore, the SNR will be (1.5/0.5)^2 = 9.

Plugging this into the formula for bit-rate capacity, we get: Bit rate capacity = Wlog2(1 + 9) = Wlog2(10)

Since we have already established that for every 12 different phase values, we can only have 1 bit of information, and for every