DC Constraint

The DTFT at frequency $ \omega$ is given by

$\displaystyle W\left(\omega \right)=\sum _{n=-L}^{L}w\left(n\right)e^{-j\omega n}.$ (4.69)

By zero-phase symmetry,
$\displaystyle W\left(\omega \right)$ $\displaystyle =$ $\displaystyle h\left(0\right)+2\sum _{n=1}^{L}h\left(n\right)\cos \left(n\omega \right)$  
  $\displaystyle =$ $\displaystyle \left[\begin{array}{cccc}
1 & 2\cos \left(\omega \right) & \cdots & 2\cos \left(L\omega \right)\end{array}\right]\left[\begin{array}{c}
h\left(0\right)\\
h\left(1\right)\\
\vdots \\
h\left(L\right)\end{array}\right]$  
  $\displaystyle =$ $\displaystyle d\left(\omega \right)^{T}h.$  

So $ W\left(0\right)=1$ can be expressed as

$\displaystyle \zbox {d\left(0\right)^{T}h=1.}$ (4.70)


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Positive Window-Sample Constraint