Converting Modulation Profiles to Visibilities

In the absence of noise, the HESSI modulation profiles are sums of sinuoidal transforms like:

$$\Phi_j = \int\int cos(2\pi(u_j x + v_j y)+\alpha_j)\ f(x,y)\ dx dy, \eqno{(1)}$$

where the wave vector components u_j and v_j are functions of the orientation angle $\theta_j$ and the pitch p:

$$u_j = cos(\theta_j)/p,\quad v_j = sin(\theta_j)/p,$$

and $\alpha$ is the instantaneous collimator phase, to be obtained from the aspect system.

Our objective is to determine the visibilities, which are Fourier transforms very similar in form to (1), except having the sine and cosine terms in the real and imaginary parts:

$$\Phi_j =\int\int e^{2\pi i(u_j x + v_j y)}\ f(x,y)\ dx dy, \eqno{(2)}$$

The way to obtain the quantities (2) from the measured values (1) is, as shown below, to take a certain linear combination of successive values $j,j+1,j+2,\ldots$ and separate out the cosine and sine transforms. There is one complication, however, and that is that the wave vector (u_j,v_j) changes direction as j and $\theta_j$ change. To mitigate this effect, we shift the transforms to the ``phase center'' (x₀,y₀), which is the center of the map to be made later. Equation (1) then becomes:

$$\Phi_j = \int\int cos(2\pi (u_j(x-x_0) + v_j(y-y_0))+\gamma_j)\ f(x,y)\ dx dy, \eqno{(3)}$$

which looks very similar, except for the new phase shift $\gamma_j$.

$$\gamma_j = \alpha_j + 2\pi(u_j x_0 + v_j y_0)$$

From now on, we treat x-x₀ and y-y₀ as first-order quantities, and x₀ and y₀ as zero-order quantities in a series expansion. We do the same thing with the angle: $\theta_j=\bar{\theta}+ \Delta_j$, where $\bar{\theta}$ is zero-order and $\Delta_j$ is first-order.

So

$$cos(\theta_j)= cos(\bar{\theta}+\Delta_j)=cos(\bar{\theta}-\Delta_jsin(\theta_j)$$

and

$$sin(\theta_j)= sin(\bar{\theta}+\Delta_j)=sin(\bar{\theta})+\Delta_jcos(\theta_j)$$

Then

$$u_j=\bar{u} - \Delta_j sin(\theta_j)/p$$

and

$$v_j=\bar{v} + \Delta_j cos(\theta_j)/p.$$

Inspection of equation(3) shows that the first place where u_j and v_j appear has the first-order factors x-x₀ and y-y₀, so we can neglect the second-order products $(x-x_0)\Delta_j$ and $(y-y_0)\Delta_j$ in that part of the equation. Our translation to the new phase center has caused the first-order effects of rotation to appear only in the phase angle $\gamma_j$:

$$\gamma_j = \alpha_j + \bar{h} x_0 +\bar{k} y_0 - 2\pi \Delta_j (x_0 sin(\theta_j)- y_0 cos(\theta_j))/p$$

We clean up the mess by writing $\gamma_j=\bar{\gamma} + \delta_j$,where $\bar{\gamma}$ is the zero-order quantity $\alpha_j + \bar{h} x_0 +\bar{k} y_0$, while $\delta_j$, being proportional to $\Delta_j$, is first-order. Then the modulation harmonic term simplifies to:

$$\Phi_j = \int\int cos(2\pi(\bar{u}(x-x_0) + \bar{v}(y-y_0))+\delta_j)\ f(x,y)\ dx dy, \eqno{(4)}$$

This is very similar to (1) except that the new phase angle $\delta_j$ is a strong function of the rotation, and assuming $\alpha$ changes slowly during a modulation cycle, $\delta$ will change by about $360^\circ$.We can therefore pick two or more time-angle bins within a cycle, and have two or more independent sinusoid transforms $\Phi_j$. Using the cosine sum formula, we can then extract the cosine and sine transforms $\Phi_c$ and $\Phi_s$ :

$$\Phi_j = cos(\delta_j) \Phi_c - sin(\delta_j) \Phi_s \eqno{(5)}$$

where the cosine transform is:

$$\Phi_c = \int\int cos(2\pi[\bar{u}(x-x_0) + \bar{v}(y-y_0)])\ f(x,y)\ dx dy,$$

and the sine transform is:

$$\Phi_s = \int\int sin(2\pi[\bar{u}(x-x_0) + \bar{v}(y-y_0)])\ f(x,y)\ dx dy.$$

The solution to the HESSI visibility-extraction problem, then, is to do a regression model of the form (5) to the modulation profile values during a single cycle of modulation, and obtain the visibility as:

$$V(\bar{u},\bar{v})=\Phi_c + i \Phi_s.$$

A bare-bones IDL program to do this (modprof2visibility.pro), which determines the fit to (5) by a least-squares method, will soon be placed in the HESSI SSW tree.