# A Method to Obtain the Transducers Impulse Response (TIR) in Photoacoustic Imaging
## 3 Algorithms  
For solving the photoacoustic pressure vector \(\mathbf{x}\) in Equation (12), it is multiplied by the transpose matrix of \(\hat{\mathbf{Y}}(\mathbf{K})\), and we obtain [36]  
\[\hat{\mathbf{Y}}^{T}\hat{\mathbf{Y}}\hat{\mathbf{x}}=0. \tag{19}\]  
We define it as the matrix \(\hat{\boldsymbol{E}}=\hat{\mathbf{Y}}^{T}\hat{\mathbf{Y}}\), which is a square matrix. The PAPS vector \(\hat{\mathbf{x}}\) is the solution of Equation (19). It must be the eigenvector with zero-eigenvalue of the matrix \(\hat{\boldsymbol{E}}\). If the matrix \(\hat{\boldsymbol{E}}\) has only one zero-eigenvalue, then the eigenvector corresponding to this zero-eigenvalue is the PAPS vector. The matrix \(\hat{\mathbf{Y}}\) consists of some C-matrices, and each C-matrix consists of a cyclic arrangement of PAPS vectors, the same as Equation (9). This vector is the convolution result of the PAPS vector and the TIR vector. Such characteristic matrix \(\hat{\mathbf{Y}}\) make the matrix \(\hat{\boldsymbol{E}}\) have special properties. The matrix \(\hat{\boldsymbol{E}}\) has only one zero-eigenvalue. By working out the eigenvector corresponding to this zero-eigenvalue of the matrix \(\hat{\boldsymbol{E}}\), the PAPS vector can be obtained. The above mathematical derivation and results assume that the length of the OPAS data is equal to the sum of TIR and the PAPS lengths, and they are all known. However, this assumption cannot be guaranteed in actual computation. The size of the PAPS is usually equal to that of the OPAS, or the lengths of the PAPS and TIR data are not known. In this situation, are the above derivation and results valid? In actual measurements, the OPAS is generated by finite biological tissues; therefore, the length of the OPAS is finite. We can obtain the whole of the OPAS as long as the record length is enough. In this case, the tail of the OPAS is characterized by gradual attenuation, so the deviation of reality from the assumption that the length of the OPAS is equal to the sum of TIR and PAPS lengths is little. In other words, the data in the part where the tail goes to zero can be assumed to have little effect on the results. The structure of the matrix \(\hat{\mathbf{Y}}\) is dependent on not only the OPAS data but also the length of PAPS. It is needed to know the length of the PAPS before solving Equation (19) to define the matrix \(\hat{\mathbf{E}}\). It is hard to know the length of the PAPS according to actual photoacoustic signal observation, but we can fix it based on the property of the matrix \(\hat{\mathbf{E}}\). When the deviation from the actual size of the PAPS occurs, the number of zero-eigenvalue of the matrix \(\hat{\mathbf{E}}\) will change. The number of zero-eigenvalue of the matrix \(\hat{\mathbf{E}}\) is equal to the deviation from the actual size of the PAPS. We therefore can determine the size of the PAPS vector in the light of the number of zero-eigenvalue of the matrix \(\hat{\mathbf{E}}\). Based on the above analysis, The following steps are used to obtain the PAPS vector. (1) Firstly, assuming \(K=M\), calculate the \(m\) minimum eigenvalues of \(\hat{\mathbf{E}}(M)\). Where \(m=\min\) (\(M_{j}\)) (since the signal length of the PAPS cannot be longer than the shortest OPAS length, i.e., \(K\leq m\), this involves all potentialities); and identify the number of the zero-eigenvalues of \(\hat{\mathbf{E}}(K)\). In the case of only one zero-eigenvalue, the value of the PAPS length \(K\) can be found. For simplicity, the singular value decomposition (SVD) of the matrix \(\hat{\mathbf{Y}}\) can be calculated to determine the zero-eigenvalues of the matrix \(\hat{\mathbf{E}}(M)\). \[\hat{\mathbf{Y}}=\mathbf{U}\mathbf{S}\mathbf{V}^{\mathrm{T}}. \tag{20}\]  
The column of the matrix \(\mathbf{V}\) is the right singular vector of \(\hat{\mathbf{Y}}\) and is the eigenvector of \(\hat{\mathbf{E}}=\hat{\mathbf{Y}}^{T}\hat{\mathbf{Y}}\). \(\mathbf{S}\) is a diagonal matrix.