1. **Stating the problem:** We are given equations involving sums of $h(m)$ and $hb(m)$ multiplied by complex exponentials, specifically: $$H(-100) = \sum_{m=-2}^2 h(m) e^{-j(-100)m},\quad H(0) = \sum_{m=-2}^2 h(m) e^{-j(0)m},\quad H(100) = \sum_{m=-2}^2 h(m) e^{-j(100)m}$$ and similarly for $hb(m)$ with sums from $m=-1$ to $1$. The goal is to find $h(m)$. 2. **Understanding the formula:** These sums represent discrete Fourier transform (DFT) components of the sequence $h(m)$ at frequencies $-100$, $0$, and $100$. 3. **Key idea:** To find $h(m)$, we use the inverse discrete Fourier transform (IDFT) formula: $$h(m) = \frac{1}{N} \sum_{k} H(k) e^{j k m}$$ where $N$ is the number of points and $k$ are the frequency indices. 4. **Setting up the system:** Given $H(-100)$, $H(0)$, and $H(100)$, and $m$ ranging from $-2$ to $2$, we have 5 unknowns $h(-2), h(-1), h(0), h(1), h(2)$. We can write the system as: $$\begin{bmatrix} H(-100) \\ H(0) \\ H(100) \end{bmatrix} = \begin{bmatrix} e^{-j(-100)(-2)} & e^{-j(-100)(-1)} & e^{-j(-100)(0)} & e^{-j(-100)(1)} & e^{-j(-100)(2)} \\ e^{-j(0)(-2)} & e^{-j(0)(-1)} & e^{-j(0)(0)} & e^{-j(0)(1)} & e^{-j(0)(2)} \\ e^{-j(100)(-2)} & e^{-j(100)(-1)} & e^{-j(100)(0)} & e^{-j(100)(1)} & e^{-j(100)(2)} \end{bmatrix} \begin{bmatrix}h(-2) \\ h(-1) \\ h(0) \\ h(1) \\ h(2)\end{bmatrix}$$ 5. **Solving for $h(m)$:** Since we have 3 equations and 5 unknowns, the system is underdetermined. Additional constraints or data points are needed to solve uniquely. 6. **For $hb(m)$ with $m=-1$ to $1$:** We have 3 unknowns and 3 equations: $$H(-100) = \sum_{m=-1}^1 hb(m) e^{-j(-100)m},\quad H(0) = \sum_{m=-1}^1 hb(m) e^{-j(0)m},\quad H(100) = \sum_{m=-1}^1 hb(m) e^{-j(100)m}$$ This forms a $3 \times 3$ linear system: $$\begin{bmatrix} e^{-j(-100)(-1)} & e^{-j(-100)(0)} & e^{-j(-100)(1)} \\ e^{-j(0)(-1)} & e^{-j(0)(0)} & e^{-j(0)(1)} \\ e^{-j(100)(-1)} & e^{-j(100)(0)} & e^{-j(100)(1)} \end{bmatrix} \begin{bmatrix}hb(-1) \\ hb(0) \\ hb(1)\end{bmatrix} = \begin{bmatrix}H(-100) \\ H(0) \\ H(100)\end{bmatrix}$$ 7. **Invert the matrix:** Calculate the inverse of the $3 \times 3$ matrix of exponentials and multiply by the vector $[H(-100), H(0), H(100)]^T$ to find $hb(m)$. 8. **Summary:** - For $h(m)$ with $m=-2$ to $2$, more data or constraints are needed. - For $hb(m)$ with $m=-1$ to $1$, solve the $3 \times 3$ linear system: $$\mathbf{A} \mathbf{hb} = \mathbf{H}$$ where $\mathbf{A}$ is the matrix of exponentials, $\mathbf{hb}$ is the vector of unknowns, and $\mathbf{H}$ is the known vector. This yields $hb(m)$, which can be used to approximate or relate to $h(m)$ if needed.