Thermal noise is present in all circuits possessing a positive temperature (above 0^oK). Years of research has established that this type of noise, N, is a normally distributed random variable with mean 0 and variance ² (the variance depends on the temperature). The Ace Circuitry Company has determined that the variance of the noise in all of its digital circuits is ² = 0.25. Their receivers accept a digital waveform, S, that is ideally either 3.5 volts. Once inside the receiver, the waveform is S + N volts. Aces decision circuitry will declare that a + voltage was sent if any voltage in the range +2.5 to +4.5 is measured and declares a bit was sent if a voltage in the range 2.5 to 4.5 is found. With any other measured voltage, the receiver declars Huh?_{. What is the probability of a correct decision? Whats the probability of a decision} error? What is the probability of a Huh? (sometimes called an erasure).

With respect to the last problem, suppose that a new digital waveform must be dealt with wherein the voltages transmitted are 0.5 volts. Ace changes its decision criteria to the following: If the measured voltage in the receiver was positive, then a voltage of +0.5 volts was sent. If the measured voltage is negative, then a voltage of 0.5 was sent. Note that this decision criteria eliminates erasures. What is the probability of error? [FYI: When the digital revolution began, most chips operated using 12 volts. This then gave way to 5 volts and then to 3.5 volts. Current chips now operate using internal logic voltages of 1.28 volts or less.

With respect to the previous problem, the transmitter communicates a 0 bit to the receiver by holding the voltage at 0.5 volts for 4 microseconds. Likewise, the transmitter communicates a 1 bit to the receiver by holding the voltage at +0.5 volts for 4 microseconds. The receiver (magically) can determine where each bit begins and ends and uses this information to take three independent samples of the transmitted waveform 1 microsecond after the start of the bit, 2 microseconds after the start, and 3 microseconds after the start. The Ace receiver uses the same decision criteria in the pervious problem to determine the transmitted polarity of each sample and then uses a majority rule to decide the polarity that was transmitted (if the receiver determines that 2 or more of the samples indicate a 0 bit was transmitted, the receiver proclaims that a 0 bit was transmitted, otherwise a 1 bit