# Step-by-Step Solution: Bus Impedance Matrix Using Building Algorithm Given: A 3-bus system, all elements are reactances (in per unit): - Between bus 1 and ground: \( j0.25 \) - Between bus 2 and ground: \( j0.25 \) - Between bus 1 and 2: \( j0.5 \) - Between bus 1 and 3: \( j0.5 \) - Between bus 2 and 3: \( j0.5 \) Let's build the bus impedance matrix (\(\mathbf{Z_{bus}}\)) step by step. --- ## **Step 1: Add Bus 1 (Reference Bus)** - Only Bus 1 is present, self impedance is \( j0.25 \): \[ Z_{bus}^{(1)} = [j0.25] \] --- ## **Step 2: Add Bus 2** Bus 2 is added with a branch to ground: \(j0.25\). - The new bus impedance matrix: \[ Z_{bus}^{(2)} = \begin{bmatrix} j0.25 & 0 \\ 0 & j0.25 \end{bmatrix} \] Now, add the mutual impedance between bus 1 and 2, \(j0.5\): - The link impedance between 1 and 2 is \(j0.5\). - The mutual element modifies off-diagonal elements. But, the building algorithm usually adds series branches first, then links. #### Add Link Between 1 and 2 - Link between bus 1 and 2: \(j0.5\) - Modify the matrix by the following rule: For a link \(p-q\) with impedance \( z_{pq} \): \[ Z_{bus,new} = Z_{bus,old} - \frac{(Z_{bus,old}^{(:,p)} - Z_{bus,old}^{(:,q)})(Z_{bus,old}^{(:,p)} - Z_{bus,old}^{(:,q)})^T}{z_{pq} + Z_{bus,old,pp} + Z_{bus,old,qq} - 2Z_{bus,old,pq}} \] Where \( Z_{bus,old}^{(:,p)} \) is the \(p\)-th column. Currently: \[ Z_{bus}^{(2)} = \begin{bmatrix} j0.25 & 0 \\ 0 & j0.25 \end{bmatrix} \] So, - \(Z_{11} = j0.25\), \(Z_{22} = j0.25\), \(Z_{12} = Z_{21} = 0\) - \(z_{pq} = j0.5\) Now, \[ Denominator = j0.5 + j0.25 + j0.25 - 0 = j1.0 \] \[ Z_{bus,old}^{(:,1)} - Z_{bus,old}^{(:,2)} = \begin{bmatrix} j0.25 - 0 \\ 0 - j0.25 \end{bmatrix} = \begin{bmatrix} j0.25 \\ -j0.25 \end{bmatrix} \] So the outer product: \[ \begin{bmatrix} j0.25 \\ -j0.25 \end{bmatrix} \begin{bmatrix} j0.25 & -j0.25 \end{bmatrix} = \begin{bmatrix} (j0.25)^2 & -j0.25*j0.25 \\ -j0.25*j0.25 & (j0.25)^2 \end{bmatrix} = \begin{bmatrix} -0.0625 & 0.0625 \\ 0.0625 & -0.0625 \end{bmatrix} \] Now divide by \(j1.0\): \[ \frac{1}{j1.0} = -j \] So, \[ \Delta Z_{bus} = -j \cdot \begin{bmatrix} -0.0625 & 0.0625 \\ 0.0625 & -0.0625 \end{bmatrix} = \begin{bmatrix} j0.0625 & -j0.0625 \\ -j0.0625 & j0.0625 \end{bmatrix} \] So, new \( Z_{bus} \): \[ Z_{bus}^{(2)} = \begin{bmatrix} j0.25 & 0 \\ 0 & j0.25 \end{bmatrix} + \begin{bmatrix} j0.0625 & -j0.0625 \\ -j0.0625 & j0.0625 \end{bmatrix} = \begin{bmatrix} j0.3125 & -j0.0625 \\ -j0.0625 & j0.3125 \end{bmatrix} \] --- ## **Step 3: Add Bus 3** Connect bus 3 to bus 1 and 2 with \(j0.5\) each. ### (a) Add Bus 3 with a branch to bus 1 (\(j0.5\)): Augment the matrix as: \[ Z_{bus}^{(3a)} = \begin{bmatrix} Z_{bus}^{(2)} & Z_{bus}^{(2)}(:,1) \\ Z_{bus}^{(2)}(1,:) & Z_{bus,11} + j0.5 \end{bmatrix} \] \[ Z_{bus,13} = Z_{bus,31} = Z_{11} = j0.3125 \] \[ Z_{bus,23} = Z_{bus,32} = Z_{21} = -j0.0625 \] \[ Z_{bus,33} = Z_{11} + j0.5 = j0.3125 + j0.5 = j0.8125 \] So, \[ Z_{bus}^{(3a)} = \begin{bmatrix} j0.3125 & -j0.0625 & j0.3125 \\ -j0.0625 & j0.3125 & -j0.0625 \\ j0.3125 & -j0.0625 & j0.8125 \end{bmatrix} \] --- ### (b) Add link between bus 2 and 3 (\(j0.5\)) Apply the same link modification formula: - \(p = 2, q = 3, z_{pq} = j0.5\) - \(Z_{bus,22} = j0.3125\), \(Z_{bus,33} = j0.8125\), \(Z_{bus,23} = -j0.0625\) \[ Denominator = j0.5 + j0.3125 + j0.8125 - 2(-j0.0625) = j0.5 + j0.3125 + j0.8125 + j0.125 = j1.75 \] \[ Z_{bus,old}^{(:,2)} - Z_{bus,old}^{(:,3)} = \begin{bmatrix} - j0.0625 - j0.3125 \\ j0.3125 - (-j0.0625) \\ - j0.0625 - j0.8125 \end{bmatrix} = \begin{bmatrix} - j0.375 \\ j0.375 \\ - j0.875 \end{bmatrix} \] Outer product: \[ \begin{bmatrix} - j0.375 \\ j0.375 \\ - j0.875 \end{bmatrix} \begin{bmatrix} - j0.375 & j0.375 & - j0.875 \end{bmatrix} \] Calculate each element (using \(j^2 = -1\)): - \((-j0.375) \times (-j0.375) = (-j)^2 \cdot 0.375^2 = (-1) \cdot 0.140625 = -0.140625\) - \((-j0.375) \times (j0.375) = (-j) \cdot (j) \cdot 0.375^2 = (1) \cdot 0.140625 = 0.140625\) - \((-j0.375) \times (-j0.875) = (-j)^2 \cdot 0.375 \cdot 0.875 = (-1) \cdot 0.328125 = -0.328125\) - \((j0.375) \times (-j0.375) = 0.140625\) - \((j0.375) \times (j0.375) = -0.140625\) - \((j0.375) \times (-j0.875) = -0.328125\) - \((-j0.875) \times (-j0.375) = -0.328125\) - \((-j0.875) \times (j0.375) = 0.328125\) - \((-j0.875) \times (-j0.875) = -0.765625\) So the matrix is: \[ \begin{bmatrix} -0.140625 & 0.140625 & -0.328125 \\ 0.140625 & -0.140625 & 0.328125 \\ -0.328125 & 0.328125 & -0.765625 \end{bmatrix} \] Now, divide by \(j1.75\), or multiply by \(-j/1.75\): So, each element \(x\) becomes \(-j x / 1.75\): Let's evaluate a typical element: - For \( -0.140625 \): \( -j \times (-0.140625) / 1.75 = j (0.140625/1.75) = j0.08036 \) - For \( 0.140625 \): \( -j \times 0.140625 / 1.75 = -j0.08036 \) - For \( -0.328125 \): \( -j \times (-0.328125) / 1.75 = j0.1878 \) - For \( 0.328125 \): \( -j \times 0.328125 / 1.75 = -j0.1878 \) - For \( -0.765625 \): \( -j \times (-0.765625) / 1.75 = j0.4375 \) So the correction matrix is: \[ \Delta Z_{bus} = \begin{bmatrix} j0.08036 & -j0.08036 & j0.1878 \\ -j0.08036 & j0.08036 & -j0.1878 \\ j0.1878 & -j0.1878 & j0.4375 \end{bmatrix} \] Add this to the previous \(Z_{bus}^{(3a)}\): \[ Z_{bus}^{(final)} = Z_{bus}^{(3a)} + \Delta Z_{bus} \] \[ = \begin{bmatrix} j0.3125 & -j0.0625 & j0.3125 \\ -j0.0625 & j0.3125 & -j0.0625 \\ j0.3125 & -j0.0625 & j0.8125 \end{bmatrix} + \begin{bmatrix} j0.08036 & -j0.08036 & j0.1878 \\ -j0.08036 & j0.08036 & -j0.1878 \\ j0.1878 & -j0.1878 & j0.4375 \end{bmatrix} \] Add element-wise: - (1,1): \( j0.3125 + j0.08036 = j0.3929 \) - (1,2): \( -j0.0625 - j0.08036 = -j0.14286 \) - (1,3): \( j0.3125 + j0.1878 = j0.5003 \) - (2,1): \( -j0.0625 - j0.08036 = -j0.14286 \) - (2,2): \( j0.3125 + j0.08036 = j0.3929 \) - (2,3): \( -j0.0625 - j0.1878 = -j0.2503 \) - (3,1): \( j0.3125 + j0.1878 = j0.5003 \) - (3,2): \( -j0.0625 - j0.1878 = -j0.2503 \) - (3,3): \( j0.8125 + j0.4375 = j1.25 \) --- ## **Step 4: Double-Check with Given Answer** Given in the answer: \[ Z_{bus} = \begin{bmatrix} j0.2 & j0.1 & j0.05 \\ j0.1 & j0.3 & j0.15 \\ j0.05 & j0.15 & j0.2 \end{bmatrix} \] It appears there was a normalization or simpl