### 6.02 Practice Problems: Noise & Bit Errors

Problem .

Suppose the bit detection sample at the receiver is V + noise volts when the sample corresponds to a transmitted '1', and 0.0 + noise volts when the sample corresponds to a transmitted '0', where noise is a zero-mean Normal(Gaussian) random variable with standard deviation σNOISE.

1. If the transmitter is equally likely to send '0''s or '1''s, and V/2 volts is used as the threshold for deciding whether the received bit is a '0' or a '1', give an expression for the bit-error rate (BER) in terms of the erfc function and σNOISE.
Here's a plot of the PDF for the received signal where the red-shaded areas correspond to the probabilities of receiving a bit in error.

so the bit-error rate is given by 0.5*erfc(V/(sqrt(8)*sigma)). Note that sigma = sqrt(N0/2) using the definition of N0. This formula is related to the 0.5*sqrt(E_s/N0) from Chapter 5 and the lecture; E_s in our case is V*V/4.

2. Suppose the transmitter is equally likely to send zeros or ones and uses zero volt samples to represent a '0' and one volt samples to represent a '1'. If the receiver uses 0.5 volts as the threshold for deciding bit value, for what value of σNOISE is the probability of a bit error approximately equal to 1/5?
From part (A), 0.5*erfc(1/(sqrt(8)*σNOISE)) = 0.2, which gives us σNOISE = 0.594.

3. Will your answer for σNOISE in part (B) change if the threshold used by the receiver is shifted to 0.6 volts? Do not try to determine σNOISE, but justify your answer.
If move Vth higher to 0.6V, we'll be decreasing prob(rcv1|xmit0) and increasing prob(rcv0|xmit1). Considering the shape of the Gaussian PDF, the decrease will be noticeably smaller than the increase, so we'd expect BER to increase for a given σNOISE. Thus to keep BER = 1/5, we'd need to decrease our estimate for σNOISE. One can also work out the same result with some algebra; we saw in Chapter 6 how picking the mid-point threshold minimizes the bit error rate.

4. Will your answer for σNOISE in part (B) change if the transmitter is twice as likely to send ones as zeros, but the receiver still uses a threshold of 0.5 volts? Do not try to determine σNOISE, but justify your answer.
If we change the probabilities of transmission but keep the same digitization threshold, the various parts of the BER equation in (A) are weighted differently (to reflect the different transmission probabilities), but the total BER remains unchanged. This question is essentially the same as one on PSet 3.

Problem .

Messages are transmitted along a noisy channel using the following protocol: a "0" bit is transmitted as -0.5 Volt and a "1" bit as 0.5 Volt. The PDF of the total noise added by the channel, H, is shown below. It is not a Gaussian.

1. Compute H(0), the maximum value of H.
The area under the PDF is 1, so (0.5)*H(0)*(1+0.5) = 1 from which we get H(0) = 4/3.

2. It is known that a "0" bits 3 times as likely to be transmitted as a "1" bit. The PDF of the message signal, M, is shown below. Fill in the values P and Q.

We know that Q=3P and that P+Q=1, so Q=0.75 and P=.25.

3. If the digitization threshold voltage is 0V, what is the bit error rate?
The plot below shown the PDF of the received voltage in magenta. For a threshold voltage of 0, there is only one error possible: a transmitted "0" received as a "1". This error is equal to the area of the triangle formed by the dotted black line and the blue line = 0.5*0.5*0.5 = 0.125.

4. What digitization threshold voltage would minimize the bit error rate?
We'll minimize the bit error rate if the threshold voltage is chosen at the voltage where the red and blue lines intersect. By looking at the plot from the previous answer, let the ideal threshold be x and the value of the PDF at the intersection point be y. Then y/x=2/3 and y/(0.5-x)=1, so x = 0.3V.
Problem .

Ben Bitdiddle studies the bipolar signaling scheme from 6.02 and decides to extend it to a 4-level signaling scheme, which he calls Ben's Aggressive Signaling Scheme, or BASS. In BASS, the transmitter can send four possible signal levels, or voltages: $\left(-3A,-A,+A,+3A\right)$, where $A$ is some positive value. To transmit bits, the sender's mapper maps consecutive pairs of bits to a fixed voltage level that is held for some fixed interval of time, creating a symbol. For example, we might map bits “00” to $-3A$, “01” to $-A$, “10” to $+A$, and “11” to $+3A$. Each distinct pair of bits corresponds to a unique symbol. Call these symbols s_minus3, s_minus1, s_plus1, and s_plus3. Each symbol has the same prior probability of being transmitted.