The analysis on monochromatic interferometry is nice and all but is not practical. In practice, due to the limitation of physics, hardware, and DSP algorithm, the observation is always made over a finite bandwidth \Delta\omega centered at \omega_c instead of a single frequency. \omega_c \pm \frac{\Delta\omega}{2}

Assume that the brightness magnitude over the narrow bandwidth (achieved by channelizations in DSP) is constant then the signal arrived at the station at each slice of the channel is: x(t) = M \int_{\omega_c-\Delta\omega/2}^{\omega_c+\Delta\omega/2} \cos(\omega t + \phi(\omega, t)) \, d\omega

So with that, let's redo the calculation for cosine interferometry output :

x_1(t) &= M \int_{\omega_c-\Delta\omega/2}^{\omega_c+\Delta\omega/2} \cos(\omega t + \phi(\omega, t)) \, d\omega \\ x_2(t) &= x_1(t-\tau_g)

Before we proceed, we insert a phase center (add an artificial delay \tau_0 for station b_1 ) like the figure below. We will explain why adding that \tau_0 is necessary.

v_{R_c}(t)|_T &= \langle x_2(t) , x_1(t-\tau_0) \rangle \\ &={\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}}{\color{blue} x_2(\psi)x_1(\psi-\tau_0)} \, \color{green}d\psi \\ \color{black} &={\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}}{\color{blue} x_1(\psi-\tau_g)x_1(\psi-\tau_0)} \, \color{green}d\psi \\ &={\color{green} \frac{M^2}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}}{\color{blue} \int_{\omega_c-\Delta\omega/2}^{\omega_c+\Delta\omega/2} \cos(\omega_2(\psi-\tau_g) + \phi(\omega_2, \psi-\tau_g)) \, d\omega_2 \int_{\omega_c-\Delta\omega/2}^{\omega_c+\Delta\omega/2} \cos(\omega_1 (\psi-\tau_0) + \phi(\omega_1, \psi-\tau_0)) \, d\omega_1 }\, \color{green}d\psi \\ \\ &={\color{green} \frac{M^2}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}}{\color{blue} \int_{\omega_c-\Delta\omega/2}^{\omega_c+\Delta\omega/2} \cos(\omega_2(\psi-\tau_g) + \phi(\omega_2, \psi-\tau_g)) \, d\omega_2 \int_{\omega_c-\Delta\omega/2}^{\omega_c+\Delta\omega/2} \cos(\omega_1 (\psi-\tau_0) + \phi(\omega_1, \psi-\tau_0)) \, d\omega_1 }\, \color{green}d\psi \\ &=M^2 {\color{blue} \int_{\Delta \omega} \int_{\Delta \omega} } {\color{green} \tfrac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2} } \cos\large(\small\omega_2(\psi-\tau_g) + \phi(\omega_2, \psi-\tau_g) \large)\small \cos\large(\small\omega_1 (\psi-\tau_0) + \phi(\omega_1, \psi-\tau_0) \large)\small {\color{green} d\psi} \color{blue} d\omega_2 d\omega_1 \\ &=M^2 {\color{blue} \int_{\Delta \omega} \int_{\Delta \omega} } {\color{green} \tfrac{1}{2T} \int_{\psi = t-T/2}^{\psi=t+T/2} } \large[ \small \large\cos(\small (\omega_2+\omega_1) \psi -\omega_2 \tau_g -\omega_1 \tau_0 +\phi(\omega_2,\psi-\tau_g) +\phi(\omega_1,\psi-\tau_0) \large)\small + \large\cos(\small (\omega_2-\omega_1) \psi -\omega_2 \tau_g +\omega_1 \tau_0 +\phi(\omega_2,\psi-\tau_g) - \phi(\omega_1,\psi-\tau_0) \large)\small \large]\small {\color{green} d\psi} \color{blue} d\omega_2 d\omega_1 \\ &=M^2 {\color{blue} \int_{\Delta \omega} \int_{\Delta \omega} } \left( \underbrace{ {\color{green} \tfrac{1}{2T} \int_{\psi = t-T/2}^{\psi=t+T/2} } \large\cos(\small (\omega_2+\omega_1) \psi -\omega_2 \tau_g -\omega_1 \tau_0 +\phi(\omega_2,\psi-\tau_g) +\phi(\omega_1,\psi-\tau_0) \large)\small {\color{green} d\psi} }_{=0} + {\color{green} \tfrac{1}{2T} \int_{\psi = t-T/2}^{\psi=t+T/2} } \large\cos(\small (\omega_2-\omega_1) \psi -\omega_2 \tau_g +\omega_1 \tau_0 +\phi(\omega_2,\psi-\tau_g) - \phi(\omega_1,\psi-\tau_0) \large)\small {\color{green} d\psi} \right) \color{blue} d\omega_2 d\omega_1

(the underlined integral term = 0 is from Time average of cosine for fast varying random phase :\text{1-0-2} )

=&M^2 {\color{blue} \int_{\Delta \omega} \int_{\Delta \omega} } \underbrace{ {\color{green} \tfrac{1}{2T} \int_{\psi = t-T/2}^{\psi=t+T/2} } \large\cos(\small (\omega_2-\omega_1) \psi -\omega_2 \tau_g +\omega_1 \tau_0 +\phi(\omega_2,\psi-\tau_g) - \phi(\omega_1,\psi-\tau_0) \large)\small {\color{green} d\psi} }_{A} \color{blue} d\omega_2 d\omega_1 \\ &\text{if } \omega_1 \neq \omega_2 \to A = 0 \because [\phi(\omega_2,\psi-\tau_g) - \phi(\omega_1,\psi-\tau_0)] \sim \mathcal{U}([0,2\pi]) \\ &\text{if } \omega_1 = \omega_2 , \; \tau_0 \approx \tau_g \to A = {\color{green} \frac{1}{2T} \int_{\psi = t-T/2}^{\psi=t+T/2} } \large\cos(\small -\omega_1 \tau_g +\omega_1 \tau_0 \large) {\color{green} d\psi} = \tfrac{1}{2} \cos( \omega_1 (\tau_g -\tau_0)) \\ &\text{since the only non-zero terms are when } \omega_1 = \omega_2, {\color{blue}\iint_{\Delta\omega}} \text{turn into a single integral}: {\color{blue}\int_{\omega = \omega_c -\Delta\omega/2} ^{ \omega_c +\Delta\omega/2}} \\ =& \frac{M^2}{2}\int_{\omega_c -\Delta\omega/2} ^{ \omega_c +\Delta\omega/2} \cos( \omega_1 (\tau_g -\tau_0)) d\omega \\ \\ &(\text{set } \tau = \tau_g-\tau_0 ) \\ \\ =& \frac{M^2}{2} \frac{1}{\tau} \large[ \sin(\omega\tau)\large]_{\omega_c -\Delta\omega/2} ^{ \omega_c +\Delta\omega/2}\\ =& \frac{M^2}{2} \frac{1}{\tau} \left[ \sin\left((\omega_c +\frac{\Delta\omega}{2})\tau\right) - \sin\left((\omega_c -\frac{\Delta\omega}{2})\tau\right) \right]\\ =& \frac{M^2}{2} \frac{1}{\tau} 2 \cos(\omega_c \tau) \sin( \frac{\Delta\omega \tau }{2})\\ =& \frac{M^2}{\tau} \cos(\omega_c \tau) \sin( \frac{\Delta\omega \tau }{2})\\ =& \frac{M^2 }{\tau} \frac{\frac{\Delta\omega}{2}}{\frac{\Delta\omega}{2}} \cos(\omega_c \tau) \sin( \frac{\Delta\omega \tau }{2})\\ =& \frac{M^2 \Delta\omega}{2 } \cos(\omega_c \tau ) \text{sinc}\left( \frac{\Delta\omega \tau }{2} \right)\\ &(\text{with }I = \frac{M^2}{2}) \\ =& I \Delta\omega \;\cos(\omega_c \tau ) \text{sinc}\left( \frac{\Delta\omega \tau }{2} \right) \\ =& I \Delta\omega \;\cos(\omega_c (\tau_g-\tau_0) ) \text{sinc}\left( \frac{\Delta\omega (\tau_g-\tau_0) }{2} \right) \\

The final quasi-monochromatic cosine interferometer visibility is:

v_{R_c}(t)|_T= &I \Delta\omega \cos(\omega_c (\tau_g-\tau_0) ) \text{sinc}\left( \frac{\Delta\omega (\tau_g-\tau_0) }{2} \right) \\ =& v_{R_c}(\tau_g,\tau_0)

then applying the geometric tricks we have shown you in 1.1 $\cos$ Interferometer - 1.1.4: Putting It All Together :

so:

v_{R_c}(\tau_g,\tau_0) &=I \Delta\omega \;\cos(\omega_c (\tau_g-\tau_0) ) \text{sinc}\left( \frac{\Delta\omega (\tau_g-\tau_0) }{2} \right)\\ \\ &=I \Delta\omega \;\cos(2\pi u (l_g-l_0) ) \text{sinc}\left( \frac{\Delta\omega}{2}\frac{ |\vec{b}|}{c} (l_g-l_0) \right) \\ &= v_{R_c}(l_g,l_0)

Compare to the visibility of monochromatic interferometer with phase center: v_{R_c(mono)}(\tau_g,\tau_0) &= I \;\cos(\omega_c (\tau_g-\tau_0) )\\ &= I \;\cos(2\pi u (l_g-l_0) )\\ &= v_{R_c(mono)}(l_g,l_0)

The visibility of quasi-monochromatic interferometer has this extra term of \Delta\omega \text{sinc}\left( \frac{\Delta\omega (\tau_g-\tau_0) }{2} \right)

Notice because of the \text{sinc} function, you should see why we need to insert a phase center, which is equivalent as adding the artificial delay \tau_0 which should be calculated to be very close to the geometric delay \tau_g such that you can squish the \text{sinc} function to \text{sinc}(0)=1 , or else \text{sinc} = 0 and you get 0 for the Visibility.

In real VLBI observation, you insert the \tau_0 by delaying one of the station's signal when calculating the correlation.

The interactive figure below shows that with the introduction of \text{sinc}\left( \frac{\Delta\omega (\tau_g-\tau_0) }{2} \right) , you can see that if the phase center \color{green}s_0 is not close to source \color{blue}s , the Visibility value v_{R_c} (the blue dot \color{blue}\cdot\text{ } ) is easily squashed to 0.

Now let's see the case for 2 sources s_x,s_y for station b_1,b_2 :

b_1 signal:

\color{brown} x_1(\psi) + y_1(\psi) \color{black}= \color{brown} \int_{\omega_c-\Delta\omega/2}^{\omega_c+\Delta\omega/2} M_x \cos(\omega \psi + \phi_x(\omega, \psi)) \, d\omega \ + \int_{\omega_c-\Delta\omega/2}^{\omega_c+\Delta\omega/2} M_y \cos(\omega \psi + \phi_y(\omega, \psi)) \, d\omega \

b_2 signal:

\color{red} x_2(\psi) + y_2(\psi) &= x_1(\psi-\tau_{gx}) + y_1(\psi-\tau_{gy})\\ &= \color{red} \int_{\omega_c-\Delta\omega/2}^{\omega_c+\Delta\omega/2} M_x \cos(\omega (\psi-\tau_{gx}) + \phi_x(\omega, \psi-\tau_{gx})) \, d\omega \ + \int_{\omega_c-\Delta\omega/2}^{\omega_c+\Delta\omega/2} M_y \cos(\omega (\psi-\tau_{gy}) + \phi_y(\omega, \psi-\tau_{gy})) \, d\omega \ \\

Now let's calculate the visibility with the phase center \tau_0 :

v_{R_c}(t)|_T =& \langle {\color{red} x_2(t)+y_2(t)} , {\color{brown} x_1(t-\tau_0)+ y_1(t-\tau_0)} \rangle \\ =&{\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}}{\color{red} [x_2(t)+y_2(t) ]}{\color{brown} [ x_1(t-\tau_0)+ y_1(t-\tau_0)]} \, \color{green}d\psi \\ =&{\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}} \\&{\color{red} \left( \underbrace{ \int_{\omega_c-\Delta\omega/2}^{\omega_c+\Delta\omega/2} \overbrace{ M_x \cos(\omega (\psi-\tau_{gx}) + \phi_x(\omega, \psi-\tau_{gx})) }^{x_{b2}} \, d\omega \ }_{X_{b2}} + \underbrace{ \int_{\omega_c-\Delta\omega/2}^{\omega_c+\Delta\omega/2} \overbrace{ M_y \cos(\omega (\psi-\tau_{gy}) + \phi_y(\omega, \psi-\tau_{gy})) }^{y_{b2}} \, d\omega \ }_{Y_{b2}} \right)} \\&{\color{brown} \left( \underbrace{ \int_{\omega_c-\Delta\omega/2}^{\omega_c+\Delta\omega/2} \overbrace{ M_x \cos(\omega (\psi-\tau_{0}) + \phi_x(\omega, \psi-\tau_{0})) }^{x_{b1}} \, d\omega \ }_{X_{b1}} + \underbrace{ \int_{\omega_c-\Delta\omega/2}^{\omega_c+\Delta\omega/2} \overbrace{ M_y \cos(\omega (\psi-\tau_{0}) + \phi_y(\omega, \psi-\tau_{0})) }^{y_{b1}} \, d\omega \ }_{Y_{b1}} \right) }\,\color{green}d\psi \\ =&{\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}} (X_{b2}+Y_{b2})(X_{b1}+Y_{b1}) \color{green}d\psi \\ =&{\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}} X_{b2}X_{b1} + X_{b2}Y_{b1} + Y_{b2}X_{b1} + Y_{b2}Y_{b1} \color{green}d\psi \\ =& {\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}} X_{b2}X_{b1} \color{green}d\psi \\ &+{\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}} X_{b2}Y_{b1} \color{green}d\psi \\ &+{\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}} Y_{b2}X_{b1} \color{green}d\psi \\ &+{\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}} Y_{b2}Y_{b1} \color{green}d\psi \\ =& {\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}} \int x_{b2} d\omega_2 \int x_{b1} d\omega_1 \color{green}d\psi \\ &+{\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}} \int x_{b2} d\omega_2 \int y_{b1} d\omega_1 \color{green}d\psi \\ &+{\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}} \int y_{b2} d\omega_2 \int x_{b1} d\omega_1 \color{green}d\psi \\ &+{\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}} \int y_{b2} d\omega_2 \int y_{b1} d\omega_1 \color{green}d\psi \\ =& \underbrace{ \int_{\Delta\omega}\int_{\Delta\omega}{\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}} x_{b2} x_{b1} {\color{green}d\psi} d\omega_2 d\omega_1 }_{XX}\\ &+ \underbrace{ \int_{\Delta\omega}\int_{\Delta\omega}{\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}} x_{b2} y_{b1} {\color{green}d\psi} d\omega_2 d\omega_1 }_{XY}\\ &+ \underbrace{ \int_{\Delta\omega}\int_{\Delta\omega}{\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}} y_{b2} x_{b1} {\color{green}d\psi} d\omega_2 d\omega_1 }_{YX}\\ &+ \underbrace{ \int_{\Delta\omega}\int_{\Delta\omega}{\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}} y_{b2} y_{b1} {\color{green}d\psi} d\omega_2 d\omega_1}_{YY} \\ =& XX+XY+YX+YY

The calculation for single source shown above can be applied directly to XX \& YY :

XX &= \frac{M_x^2}{2}\Delta\omega \cos(\omega_c (\tau_{gx}-\tau_0)) \text{sinc}\left( \frac{\Delta\omega (\tau_{gx}-\tau_0) }{2}\right)\\ &= I_x \Delta\omega\cos(\omega_c (\tau_{gx}-\tau_0)) \text{sinc}\left( \frac{\Delta\omega (\tau_{gx}-\tau_0) }{2}\right)\\ \\ YY &= \frac{M_y^2}{2} \cos(\omega_c (\tau_{gy}-\tau_0)) \text{sinc}\left( \frac{\Delta\omega (\tau_{gy}-\tau_0) }{2}\right) \\ &= I_y \Delta\omega\cos(\omega_c (\tau_{gy}-\tau_0)) \text{sinc}\left( \frac{\Delta\omega (\tau_{gy}-\tau_0) }{2}\right)

As for XY \& YX , they basically have the same form, so if we can prove XY=0 , then YX is also 0.

XY =& \int_{\Delta\omega}\int_{\Delta\omega}{\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}} x_{b2} y_{b1} {\color{green}d\psi} d\omega_2 d\omega_1 \\ =& \int_{\Delta\omega}\int_{\Delta\omega}{\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}} M_x \cos(\omega_2 (\psi-\tau_{gx}) + \phi_x(\omega_2, \psi-\tau_{gx})) M_y \cos(\omega_1 (\psi-\tau_{0}) + \phi_y(\omega_1, \psi-\tau_{gy})) {\color{green}d\psi} d\omega_2 d\omega_1 \\ =&M_x M_y \int_{\Delta\omega}\int_{\Delta\omega}{\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}} \frac{1}{2}[ \cos\left\{(\omega_2+\omega_1) \psi-\omega_2\tau_{gx}-\omega_1\tau_0 +\phi_x(\omega_2,\psi-\tau_{gx}) +\phi_y(\omega_1,\psi-\tau_{gy})\right\} \\ &+\cos\{(\omega_2-\omega_1) \psi-\omega_2\tau_{gx}+\omega_1\tau_0 +\phi_x(\omega_2,\psi-\tau_{gx}) -\phi_y(\omega_1,\psi-\tau_{gy})\} ] {\color{green}d\psi} d\omega_2 d\omega_1 \\ =&\frac{M_x M_y}{2} \int_{\Delta\omega}\int_{\Delta\omega} {\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}} \cos\left\{(\omega_2+\omega_1) \psi-\omega_2\tau_{gx}-\omega_1\tau_0 +\phi_x(\omega_2,\psi-\tau_{gx}) +\phi_y(\omega_1,\psi-\tau_{gy})\right\} {\color{green}d\psi} \\ &+ {\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}}\cos\{(\omega_2-\omega_1) \psi-\omega_2\tau_{gx}+\omega_1\tau_0 +\phi_x(\omega_2,\psi-\tau_{gx}) -\phi_y(\omega_1,\psi-\tau_{gy})\} {\color{green}d\psi} d\omega_2 d\omega_1 \\ \\ &\because \underbrace{ (\phi_x(\omega_2,\psi-\tau_{gx}) \pm \phi_y(\omega_1,\psi-\tau_{gy})) }_{=\Delta \phi}\mod 2\pi \sim \mathcal{U}([0,2\pi]) \\ \\ \\ =&\frac{M_x M_y}{2} \int_{\Delta\omega}\int_{\Delta\omega} \underbrace{ {\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}} \cos\left((\omega_2+\omega_1) \psi-\omega_2\tau_{gx}-\omega_1\tau_0 +\Delta\phi \right) {\color{green}d\psi} }_{=0} \\ &+ \underbrace{ {\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}}\cos((\omega_2-\omega_1) \psi-\omega_2\tau_{gx}+\omega_1\tau_0 +\Delta\phi ) {\color{green}d\psi} }_{=0} d\omega_2 d\omega_1 \\ =& 0\\ =&YX

(The two integrals = 0 are from Time average of cosine for fast varying random phase :\text{1-0-2} )

So finally:

V_{R_c}(t)|_T =& XX+YY\\ =& I_x\Delta\omega\cos(\omega_c (\tau_{gx}-\tau_0)) \text{sinc}\left( \frac{\Delta\omega (\tau_{gx}-\tau_0) }{2}\right)\\ &+ I_y \Delta\omega\cos(\omega_c (\tau_{gy}-\tau_0)) \text{sinc}\left( \frac{\Delta\omega (\tau_{gy}-\tau_0) }{2}\right) \\ =& V_{R_c}(\tau_{gx},\tau_{gy},\tau_0)

For N>2 sources, you can just simply add them up:

V_{R_c}(t)|_T =& \sum_{n=1}^{N} I_n\Delta\omega\cos(\omega_c (\tau_{gn}-\tau_0)) \text{sinc}\left( \frac{\Delta\omega (\tau_{gn}-\tau_0) }{2}\right)\\ \\ &(\text{with $l$ notation} )\\ \\ =& \sum_{n=1}^{N} I_n \Delta\omega \;\cos(2\pi u (l_{gn}-l_0) ) \text{sinc}\left( \frac{\Delta\omega}{2}\frac{ |\vec{b}|}{c} (l_{gn}-l_0) \right) \\ \\ &(\text{define $l_n$ as : } l_n = l_{gn}-l_0 )\\ \\ =& \sum_{n=1}^{N} I_n \Delta\omega \;\cos(2\pi u l_{n} ) \text{sinc}\left( \frac{\Delta\omega}{2}\frac{ |\vec{b}|}{c} l_n \right) \\

Now suppose we have N \to \infty sources in the sky with 2 stations, then then the summation over l_n above turn into integrals:

v_{R_c}(u) &=\Delta\omega \int_{l} I(l) \cos(2\pi u l) \text{sinc}\left( \frac{\Delta\omega }{2}\frac{|\vec{b}|}{c} l\right) dl\\ v_{R_s}(u) &= \Delta\omega \int_{l} I(l) \sin(2\pi u l) \text{sinc}\left( \frac{\Delta\omega }{2}\frac{|\vec{b}|}{c} l\right) dl\\

Assuming that:

• we are looking only at the line segment that is close to the phase center (l\approx 0 )
• bandwidth \Delta\omega is small

Then we can do the following approximation:

\text{sinc}\left( \frac{\Delta\omega }{2}\frac{|\vec{b}|}{c} l\right) \approx 1

so with the 2 assumptions above, we have: v_{R_c}(u) &\approx \Delta\omega \int_{l} I(l) \cos(2\pi u l) dl\\ v_{R_s}(u) &\approx \Delta\omega \int_{l} I(l) \sin(2\pi u l) dl\\

Now we put them together and create a new function V(u) , the Visibility Function:

\text{Visibility}: V(u) &= v_{R_c}(u) - j v_{R_s}(u) \\ &= \Delta\omega \int_l I(l) \cos( 2\pi u l) dl - j \Delta\omega \int_l I(l) \sin( 2\pi u l) dl \\ &= \Delta\omega \int_l I(l) [ \cos( 2\pi u l) -j \sin( 2\pi u l)] dl \\ &= \Delta\omega \int_l I(l) e^{-j 2\pi u l} dl \\

With

\boxed{ V(u) = \Delta\omega \int_l I(l) e^{-j 2\pi u l} dl}