Calculating Eb/No for PSK and QPSK

[Tjvelcro]

Homework Statement

A sine wave is to be used for two different signaling schemes: (a) PSK; (b) QPSK. The
duration of a signal element is 10^-5 s. If the received signal is of the following form
s(t) = 0.005 sin(2pi (10^6)t + theta) volts and if the measured noise power at the receiver is 2.5 X 10^-8 ,watts, determine the Eb/No(in dB) for each case.

Homework Equations

s(t) = 0.005 sin(2pi (10^6)t + theta) volts
Eb/No = ?
signal element = 10^-5s

The Attempt at a Solution

I'm not sure where to start this problem, I assume the 10^-5s should be put into the main equation but I'm not sure what theta equals and what variable I will be solving for. Any help would be appeciated.

Tjvelcro

 

[KataKoniK]

, thank you for your post. This is an interesting problem that involves understanding the relationship between signal power, noise power, and signal-to-noise ratio (SNR). Let's break down the problem and see if we can find a solution.

First, let's define some variables:
- s(t) = 0.005 sin(2pi (10^6)t + theta) volts is the received signal
- Pn = 2.5 X 10^-8 watts is the measured noise power at the receiver
- T = 10^-5 seconds is the duration of a signal element
- Eb/No = ? is the signal-to-noise ratio, which we will solve for

Now, let's look at the relationship between signal power, noise power, and SNR:
- Signal power (Ps) is the average power of the signal over a given time period. In this case, since the signal is a sine wave with an amplitude of 0.005, the signal power can be calculated as follows:
Ps = (0.005)^2/2 = 1.25 x 10^-5 watts

- Noise power (Pn) is the average power of the noise over a given time period. In this case, we already know that Pn = 2.5 x 10^-8 watts.

- Signal-to-noise ratio (SNR) is the ratio of signal power to noise power. It is typically expressed in decibels (dB) using the following formula:
SNR (dB) = 10log(Ps/Pn)

Now, let's plug in our values and solve for SNR:
SNR (dB) = 10log(1.25 x 10^-5 / 2.5 x 10^-8) = 10log(500) = 27 dB

This means that the SNR for the received signal is 27 dB.

Finally, to find the Eb/No (in dB), we can use the following formula:
Eb/No (dB) = SNR (dB) - 10log(T)

Plugging in our values, we get:
Eb/No (dB) = 27 dB - 10log(10^-5) = 27 dB - 50 dB = -23 dB

Therefore, the Eb/No (in dB) for this case is -23 dB.

For the second case, QPSK, we