Acquisition pointing and tracking (APT) loss calculation

Acquisition, pointing and tracking loss

Acquisition, pointing and tracking (APT) loss is caused by individual telescopes being pointed away from the other party.

In the case of the transmitter telescope, this causes the receiver to experience a part of the beam profile which is not maximal. The expected loss due to APT is the expected value of this reduction in intensity from maximum which the receiver telescope experiences. In the case of the receiver telescope, when its pointing error of is outside of its FOV, light from the transmitter is not coupled into receiving optics (an outage). The expected loss from this component is the probability of such an outage occurring.

Error assumptions

We assume that pointing error has the following properties:

• normally distributed in two dimensions with zero mean
• both dimensions have equal variance $\sigma^2$
• pointing errors in different directions are uncorrelated

Under these conditions, pointing error

$$\underline{e} = \begin{pmatrix} e_x \\ e_y \end{pmatrix}$$

(in radians) can be described by a covariance matrix

$$S=\begin{pmatrix} \sigma^2 & 0\\ 0 & \sigma^2 \end{pmatrix}$$

Such that

$$ P(\underline{e}) = \frac{1}{2 \pi \sqrt{S}} \exp\left(-\frac{1}{2} \underline{e}^T S \underline{e}\right) = \frac{1}{2 \pi \sigma^2} \exp\left(-\frac{|\underline{e}|^2}{2\sigma^2}\right)$$

These assumptions are safe provided that:

• the control system is (at least locally) linear
• it is exposed only to additive, white gaussian process and measurement noise
• an optimal kalman filter and observer are used

Transmitter telescope pointing

Each pointing error $\underline{e}$ incurs some reduction in received intensity due to the gaussian beam normalised intensity distribution $f()$ (see geometric loss calculation.

$$APT \ loss (\underline{e}) = \frac{f(\underline{e} l)}{f(\underline{0})}$$

Where l is the link loss, which for paraxial (small) pointing errors $\underline{e}$ converts pointing error angle into distance. APT is therefore a random variable. For link modelling, this is not useful, we need the expected APT loss.

$$ Expected \ transmitter \ APT \ loss = E[APT \ loss_t] = \int\int \frac{f(\underline{e} l)}{f(\underline{0})} P(\underline{e}) dA $$

Where here the area integral is over the space of possible pointing errors $\underline{e}$, i.e. the domain of $P(\underline{e})$.

From here, we assume that the transmitting telescope projects a gaussian beam such that $f(\underline{r})= \frac{1}{2 \pi w^2} \exp\left(-\frac{|\underline{r}|^2}{2w^2}\right)$. The case of a flat-top beam (which either uniformly illuminates, or does not, the receiver with some probability) can be dealt with in the same way as the receiver telescope pointing.

We now have both f(\underline{e}l) and P(\underline{e}) being gaussian functions, which are axisymmetric, so can simplify this integral by transforming into polar coordinates:

$$ e = |\underline{e}|, \qquad \phi = \angle{e}$$

$$ dA = e \ de \ d\phi $$

$$E[APT \ loss_t] = \int_{e=0}^{\inf} \int_{\phi=0}^{2\pi} \frac{\frac{1}{2w^2} \exp\left(-\frac{e^2l^2}{2w^2}\right)}{\frac{1}{2w^2}} \ \frac{1}{2\pi\sigma^2} \exp\left(-\frac{e^2}{2\sigma^2}\right) e de d\phi$$

$$E[APT \ loss_t] = \frac{1}{2\pi\sigma^2} \int_{e=0}^{\infty} \int_{\phi=0}^{2\pi} \exp\left(-\frac{e^2}{2}\left[\frac{l^2}{w^2}+\frac{1}{\sigma^2}\right]\right) e de d\phi$$

This integral is separable, as the functions in it are axisymmetric ($\phi$ independent).

$$E[APT \ loss_t] = \frac{1}{\sigma^2} \int_{e=0}^{\infty} \exp\left(-\frac{e^2}{2}\left[\frac{l^2}{w^2}+\frac{1}{\sigma^2}\right]\right) e de$$

Using a further substitution

$$a = e \sqrt{\frac{1}{2}\left(\frac{l^2}{w^2} + \frac{1}{\sigma^2}\right)}$$

$$E[APT \ loss_t] = \frac{1}{2\sigma^2}\left(\frac{l^2}{w^2} + \frac{1}{\sigma^2}\right)\int_{a=0}^{\infty} \exp\left(-a^2\right) a da$$

$$\int_{a=0}^{\infty} \exp\left(-a^2\right) a da$$ is a known integral with value $\frac{1}{2}$.

$$E[APT \ loss_t] = \frac{1}{4\sigma^2}\left(\frac{l^2}{w^2} + \frac{1}{\sigma^2}\right)$$

Finally, note that in a long-distance satellite link $\frac{w}{l}$ is approximately the transmitter FOV $\frac{FOV_t}{2}$.

$$E[APT \ loss_t] = \frac{FOV^2}{FOV^2 + \sigma^2}$$

Receiver telescope pointing

The receiver telescope pointing jitter is generally simpler. The FOV of the receiver telescope either contains the transmitter, or does not. If it does, then no loss is incurred. If it does not, then no signal is received and loss is 100%. In other words, receiver pointing jitter is another random variable:

$$Receiver \ APT \ loss (\underline{e})= \begin{cases} 1 & |\underline{e}|\leq\frac{FOV_r}{2}\\ 0 & |\underline{e}|>\frac{FOV_r}{2}\\ \end{cases}$$

Again, we need the expected APT loss

$$E[APT \ loss_r] = \int \int Receiver \ APT \ loss (\underline{e}) P(\underline{e}) dA$$

$$E[APT \ loss_r] = 1 - \int \int_{|\underline{e}|>\frac{FOV_r}{2}} P(\underline{e})dA$$

Using the same polar coordinate transformation as above

$$E[APT \ loss_r] = 1 - \int_{e=\frac{FOV_r}{2}}^{\infty} \int_{\phi=0}^{\infty} \frac{1}{2\pi \sigma^2}\exp\left(-\frac{e^2}{2\sigma^2}\right) e de d\phi$$

$$E[APT \ loss_r] = 1 - \frac{1}{\sigma^2}\int_{e=\frac{FOV_r}{2}}^{\infty} \exp\left(-\frac{e^2}{2\sigma^2}\right) e de$$

Using a special case of the substitution above $$a = \frac{e}{\sqrt{2}\sigma}$$

$$E[APT \ loss_r] = 1 - \int_{a=\frac{FOV_r}{2\sqrt{2}\sigma}}^{\infty} a \exp(-a^2) da$$

This again is a known indefinite integral result, or can be arrived at by noting that $a \exp(-a^2) = \frac{-1}{2} \frac{d}{da}\left[\exp(-a^2)\right]$.

$$E[APT \ loss_r] = 1 - \exp\left(-\frac{FOV_r^2}{8\sigma^2}\right)$$