$\rho_{1_W}(1)=1$, $\rho_{\text{sgn}}(1)=q^4$, $\rho_{\text{sgn}_a}
(1)=\rho_{\text{sgn}_b}(1)=\frac{1}{2}q(q^2+1)$, $\rho_{r}(1)=\frac{1}{2}
q(q+1)^2$ and $\rho_{x_0}(1)=\frac{1}{2}q(q-1)^2$. By the explicit description of the Fourier matrices in [@C2 p. 471], we find that

$R_{x_0}=\frac{1}{2}(\rho_{r}-\rho_{\text{sgn}_a}-\rho_{\text{sgn}_b}+
\rho_{x_0})$.

If $q$ is odd, then the identification of $R_{x_0}$ with a characteristic function of a cuspidal character sheaf is explained in the appendix of Srinivasan [@Sr94]. Now assume that $q=2^f$ where $f{\geqslant}1$. Then, by [@L2d 22.2], there is a unique cuspidal character sheaf $A_0$ on $G$, and it is contained in $\hat{G}^{\text{un}}$. By the explicit description in [@LuSp1 2.7], we have $A_0=A_{x_0}$ and $A_0=\mbox{IC}(\overline{C},
{{\mathcal{E}}})[\dim C]$ where $C$ is the class of regular unipotent elements and ${{\mathcal{E}}}\not\cong \overline{{{\mathbb{Q}}}}_\ell$; we have $\dim G=10$ and $\dim C=8$. Let us fix $$g_1=x_a(1)x_b(1) \in C^F.$$ One checks that $g_1$ has order $4$ and that $g_1$ is conjugate in $G^F$ to $g_1^{-1}=x_b(1)x_a(1)$. Furthermore, $A(g_1)\cong {{\mathbb{Z}}}/2{{\mathbb{Z}}}$ is abelian and $F$ acts trivially on $A(g_1)$. Let $\lambda$ be the non-trivial character of $A(g_1)$. Then, as in Example \[exp2\], we obtain: $$\chi_{g_1,\lambda}(g)=\left\{\begin{array}{cl} q &\quad\mbox{if $g=g_1$},\\
-q & \quad \mbox{if $g=g_1'$},\\ 0 & \quad \mbox{if $g\not\in C^F$},
\end{array}\right.$$ where $g_1'\in C^F$ corresponds to the non-trivial element of $A(g_1)$. We now have $R_{x_0}=\zeta_{x_0}\chi_{g_1,\lambda}$. In order to determine $\zeta_{x_0}$ it is sufficient, by Remark \[exp1\], to 