Calculate Power Margin Required to Prevent Satellite Link Drop

[niko2000]

Hi,

I've been looking at some example solution and still haven't found the physical/mathematical explanation for the given solution.

Problem:

Due to slow satellite stabilization rotation there are big differences between maximums and minimums of received signals. Calculate the power margin required to ensure the probability of link drop lower or equal to 1%.

Solution:

P/Pmax = sin^2(w*t)
w*t = pi/2 * probability; probability = 1%

a = 10*log(Pmax/P) = 20*log(sin(wt)) = 20*log(sin(pi/2*probability)) = 36.08 dB

where
P = current power
Pmax = maximum power
w = angular frequency
t = time

Physical explanation (my interpretation):

The power is proportional to square of the electric field strength, thus:

P = k * (abs(E))^2, where E = Emax*sin(wt)

so P = Pmax*sin^2(wt)

But from here I don't understand why the solution is wt = pi/2*probability!

There is certainly minimum power required in order to maintain the link at given parameters the function sin^2(wt) should be integrated over one period (2pi) and then the horizontal line (power margin) should be set at such level to divide upper and lower surface at ratio 99/1.

I would really appreciate if anyone of you could theoretically explain the given solution (36.08 dB)

Thank you!

Niko

 

[Evalesco]

Hi Niko,

I'm not sure I understand your question completely, but it looks like you're trying to understand why the solution is wt = pi/2*probability. If that's the case, then it's because the power margin is given by the ratio of the maximum power and the current power, which is equal to sin^2(wt). The probability of link drop is equal to 1%, so the power margin should be equal to the value at which sin^2(wt) is 99% of its maximum value. To calculate this value, we set wt = pi/2 * probability, which gives us the equation P/Pmax = sin^2(wt). Solving this equation gives us the result of 36.08 dB.

 

[astrokreng]

Hi Niko,

Thank you for your question. The solution provided is correct and I will try to explain it in more detail for you.

Firstly, the given problem states that there are big differences between maximum and minimum received signals due to slow satellite stabilization rotation. This means that the received signals are varying over time, with a maximum value (Pmax) and a minimum value (Pmin). In order to maintain a stable link, we need to ensure that the received signal does not drop below a certain threshold, which in this case is equal to 1% of the maximum signal (Pmin = 0.01*Pmax).

Now, let's look at the equation P/Pmax = sin^2(w*t). This equation represents the ratio of the received signal (P) to the maximum signal (Pmax) as a function of time (t). The term sin^2(w*t) represents the variation of the received signal over time, with w being the angular frequency and t being the time.

Next, we need to consider the probability of link drop, which is given as 1%. This means that we need to ensure that the received signal does not drop below the threshold (Pmin) with a probability of 1%. In other words, the probability of the received signal being greater than or equal to Pmin is 99%.

Now, let's look at the solution provided. It states that a = 10*log(Pmax/P) = 20*log(sin(wt)) = 20*log(sin(pi/2*probability)) = 36.08 dB. This is the power margin required to ensure the probability of link drop is lower or equal to 1%.

To understand this, we need to consider the equation for a, which is the decibel (dB) value of the power margin. The dB value is given by 10*log(Pmax/P), where Pmax is the maximum signal and P is the received signal. This means that we need to find the power margin (P-Pmin) in dB, where P is the received signal and Pmin is the minimum threshold for maintaining the link.

Now, let's substitute the value of Pmin (0.01*Pmax) into the equation for a. This gives us a = 10*log(Pmax/(P-0.01*Pmax)). We can simplify this to a = 10*log(Pmax/P) -