Because the difference between \cos and \sin is the phase shift of \frac{\pi}{2} . , so Sine Interferometer is just a Cosine Interferometer with phase shift of \frac{\pi}{2} .
To achieve this phase shift, we add an artificial time delay \tau_{\sin} at station b_1 as shown in the figure below.
Figure
So with the artificial delay, now b_1 's signal at the correlator is:
x_1(t-\tau_{\sin}) = M\cos(\omega (t-\tau_{\sin}))1.2.1: Solving for
Now what value do we use for \tau_{\sin} ?
Since we we need a \frac{\pi}{2} phase shift, by looking at the figure below, phase = \frac{\pi}{2} corresponds to length = \frac{\lambda}{4} , from which we then can calculate the delay \tau_{\sin} :
Figure
So given the period of the wave is T , the delay \tau_{\sin} is calculated as:
\tau_{\sin} &=\frac{T}{4} \\ &(\text{given }T=\frac{1}{\nu})\\ &=\frac{1/\nu}{4} \\ &=\frac{1}{4\nu} \\1.2.2: Putting It All Together
Let's calculate the correlation v_{R_s} :
\large v_{R_s}(\bar t,\tau_g)|_T =&\large \frac{1}{2T} \int_{\bar t-T}^{\bar t +T} x_1(t-\tau_{\sin})x_2^*(t) dt\\ =&\large \frac{1}{2T} \int_{\bar t-T}^{\bar t +T} x_1(t-\tau_{\sin})x_1^*(t-\tau_g) dt\\ =&\large \frac{1}{2T} \mathcal{Re} \left\{ \int_{\bar t-T}^{\bar t +T} Me^{j\omega (t-\tau_{\sin}) } Me^{-j\omega (t-\tau_g)} dt \right\} \\ =&\large \frac{M^2}{2T} \mathcal{Re} \left\{ \int_{\bar t-T}^{\bar t +T} e^{-j\omega\tau_{\sin}}e^{j\omega \tau_g} dt \right\} \\ &\begin{matrix*}[l] \because & \tau_{\sin} = \frac{1}{4\nu} \\ & \omega = 2\pi \nu \\ \end{matrix*} \\ &\begin{matrix*}[l] \therefore & \omega \tau_{\sin} =2\pi \nu \frac{1}{4\nu} = \pi/2\\ \end{matrix*} \\ =&\large \frac{M^2}{2T} \mathcal{Re} \left\{ \int_{\bar t-T}^{\bar t +T} e^{-j \pi/2 }e^{j\omega \tau_g} dt \right\} \\ =&\large \frac{M^2}{2T} \mathcal{Re} \left\{-j e^{j\omega \tau_g} \int_{\bar t-T}^{\bar t +T} 1 dt \right\} \\ =&\large \frac{M^2}{2T} \mathcal{Re} \left\{ -j e^{j\omega \tau_g} 2T \right\} \\ =&\large M^2 \mathcal{Re} \left\{ -j e^{j\omega \tau_g} \right\} \\ =&\large M^2 \mathcal{Re} \left\{ -j (\cos(\omega \tau_g + j\sin(\omega \tau_g))) \right\} \\ =&\large M^2 \sin(\omega \tau_g) \\ =&\large I \sin(\omega \tau_g) \\
So the Sine Interferometer output v_{R_s} is:
v_{R_s}(\tau_g)= I \sin(\omega\tau_g ) = I \sin(2\pi ul ) =v_{R_s}(u,l)1.2.3: Fringe
Below figure you see the Fringe Pattern for Sine Interferometer. The difference in Fringe Pattern compare with Cosine Interferometer a \frac{\pi}{2} phase shift.
Feel free to drag around the blue triangle source \color{blue} s and the brown circle \color{brown} u , and observe that:
Figure :
Interactive \sin Fringe demonstration: draggable {\color{blue}\text{source } s}, {\color{brown}\text{baseline }u}