# Vector Signals and Invariant Systems: Re-Tooling Linear Systems Theory
## 5 Frequency Domain Analysis for LRITI Systems
### Example 2 (Illustrating the Orthogonal Bivector Exponential via Rotations)  
Suppose we are given two bivector exponential signals, denoted by \(\mathbf{x}_{1}(t)\) and \(\mathbf{x}_{2}(t)\), each rotating in distinct planes with different angular frequencies \(\omega_{1}\) and \(\omega_{2}\). Specifically, \(\mathbf{x}_{1}(t)\) rotates in the \(\mathbf{e}_{1}\mathbf{e}_{2}\) plane, while \(\mathbf{x}_{2}(t)\) rotates in the \(\mathbf{e}_{3}\mathbf{e}_{4}\) plane:  
\[\mathbf{x}_{1}(t) = \mathbf{e}_{1}\mathbf{e}^{\mathrm{j}\,\omega_{1}t},\quad\mathrm{j }_{1}=\mathbf{e}_{1}\mathbf{e}_{2} \tag{51a}\] \[\mathbf{x}_{2}(t) = \mathbf{e}_{3}\mathbf{e}^{\mathrm{j}\,\omega_{2}t},\quad\mathrm{j }_{2}=\mathbf{e}_{3}\mathbf{e}_{4}. \tag{51b}\]Similar to example 1, we just need to examine whether the bivectors \(\mathbf{j}_{1}\) and \(\mathbf{j}_{2}\) commute:  
\[\mathbf{j}_{1}\mathbf{j}_{2} = (\mathbf{e}_{1}\mathbf{e}_{2})(\mathbf{e}_{3}\mathbf{e}_{4}) \tag{52a}\] \[= \mathbf{e}_{1}\mathbf{e}_{2}\mathbf{e}_{3}\mathbf{e}_{4}\] (52b) \[\mathbf{j}_{2}\mathbf{j}_{1} = (\mathbf{e}_{3}\mathbf{e}_{4})(\mathbf{e}_{1}\mathbf{e}_{2})\] (52c) \[= \mathbf{e}_{1}\mathbf{e}_{2}\mathbf{e}_{3}\mathbf{e}_{4}. \tag{52d}\]  
Clearly \(\mathbf{j}_{1}\mathbf{j}_{2}=\mathbf{j}_{2}\mathbf{j}_{1}\). Consequently, the bivector exponentials commute, i.e., \(\mathbf{e}^{\mathbf{j}\omega_{1}t}\mathbf{e}^{\mathbf{j}\omega_{2}t}=\mathbf{e }^{\mathbf{j}\omega_{2}t}\mathbf{e}^{\mathbf{j}\omega_{1}t}\). Therefore, the rotations generated by these bivector exponentials are orthogonal. Examples 1 and 2 demonstrate that the orthogonality of bivector exponentials depends not only on their distinct angular frequencies but also on the planes in which the rotations occur. Theorem 8 (Eigenfunctions of VV Convolution and LRITI Systems): _Bivector exponential signals are eigenfunctions for VV convolution and LRITI systems._  
Proof. : Suppose the input to an LRITI system is the bivector exponential signal given in (47e). Then, using the VV convolution definition (43) gives  
\[\mathbf{y}(t) = \mathbf{x}(t)*\mathbf{h}(t) \tag{53a}\] \[= \left[\alpha\mathbf{u}_{||\mathbf{j}_{0}}\mathbf{e}^{\mathbf{j} \omega_{0}t}+\beta\mathbf{u}_{\perp\mathbf{j}_{0}}\right]*\mathbf{h}(t). \tag{53b}\]  
Using the polar representation gives  
\[\mathbf{u}_{||\mathbf{j}_{0}}=\mathbf{e}_{1}\mathbf{e}^{\mathbf{j}\omega_{0} }\,\ \mathbf{j}_{\mathbf{u}}= \begin{cases}\frac{\mathbf{e}_{1}\wedge\mathbf{u}_{|| \mathbf{j}_{0}}}{\|\mathbf{e}_{1}\wedge\mathbf{u}_{||\mathbf{j}_{0}}\|},\ \text{when}\ \|\mathbf{e}_{1}\wedge\mathbf{u}_{|| \mathbf{j}_{0}}\|\
eq 0\\ 0,\ \text{otherwise}\end{cases}  \tag{54}\]  
where \(\phi\) is the angle between \(\mathbf{e}_{1}\) and \(\mathbf{u}_{||\mathbf{j}_{0}}\), and  
\[\mathbf{u}_{\perp\mathbf{j}_{0}}=\mathbf{e}_{1}\mathbf{e}^{\mathbf{j}\omega _{0}}\,\ \mathbf{j}_{\mathbf{v}}= \begin{cases}\frac{\mathbf{e}_{1}\wedge\mathbf{u}}{\| \mathbf{e}_{1}\wedge\mathbf{u}\|},\ \text{when}\ \|\mathbf{e}_{1}\wedge\mathbf{u}\|\
eq 0\\ 0,\ \text{otherwise}\end{cases}  \tag{55}\]  
where \(\theta\) is the angle between \(\mathbf{e}_{1}\) and \(\mathbf{u}_{\perp\mathbf{j}_{0}}\).