I'm following the derivation in this paper A Comb Filter Design Using Fractional-Sample Delay to obtain the objective function for the least-squares filter design.

N-order FIR filter:

$H(z) = \sum_{n=0}^N h(n)z^{-n}$

Frequency response:

$H(\omega) = \mathbf{h}^T\mathbf{e}(\omega) = \mathbf{e}^T(\omega)\mathbf{h}$

where $\mathbf{h}=[h(0)\quad h(1) \quad\dots \quad h(N)] $ and $\mathbf{e}=[1\quad e^{-j\omega} \quad\dots \quad e^{-jN\omega}] $

Least squares error:

$J(h)=\int_{\omega \in R^+\cup R^-} |H(\omega)-F_d(\omega)|^2 d\omega$

where $R^+=[0,\alpha\omega]$ and $R^-=[-\alpha\omega,0]$

Which is rewritten in the quadratic form: $J(h) =h^TQh - 2h^Tp + c$

$Q$, $p$, and $c$ are given in the paper as follows:


$F_d(\omega)= e^{-jD\omega}$ and $h(n)$ is real.

I've been having trouble with getting these expressions for the Q matrix and p vector. How can I obtain them?


1 Answer 1


$$ J(h) = \int_{R^+UR-}|e^Th-F_d|^2d\omega\tag{1}\\ = \int(e^Th-F_d)^H(e^Th-F_d)d\omega\\ = \int((e^Th)^H(e^Th) + F_d^HF_d -(e^Th)^HF_d - F_d^He^Th)d\omega\\ = h^H(\int(e^T)^He^Td\omega) h + \int |F_d|^2d\omega -\int (2 Re\{(e^Th)^HF_d\})d\omega $$ The second term in above integral is the integral of $L_2$ norm of $F_d$ in the region $R^+ U R^-$. Since $F_d(\omega)$ is symmetric about $\omega = 0$, this integral will be $2 \times \int_{R^+}|F_d(\omega)|^2d\omega$ which is $c$ in your questions.

The first term, the argument inside integral is the outer product of column vector $(e^H)^T$ with row vector $e^T$. Again, this will conjugate of each for the regions $R^+$ and $R^-$ (shown later in the appendix). So integral over both regions will result in the real value of integral $e^H e$. Also, since $H$ is real, $h^H=h^T$. So the first term can be rewritten as (for simplicity dropping $T$ notation as it is assumed to be a column vector) $$ h^T(\int_{R^+}2Re\{ ee^H\}d\omega )h\\ = h^TQh $$

The third term is the result of sum of 2 conjugate terms. Again this is symmetric about $\omega=0$. Also, since $H$ is real, $h^H=h^T$. Taking out $2h^T$ from the integral we have $$ 2h^T\int_{R+}2Re\{F_de^H \}d\omega = 2h^Tp $$ Therefore, summing and rearranging the 3 terms, $$ J(h) = h^TQh - 2h^Tp + c $$

Appendix: Showing that $ee^H$ is conjugate of each other in $R^+$ and $R^-$. $E_{mn}$, element of $ee^H$ is $e^{-jm\omega}e^{-jn\omega} = e^{-j(m+n)\omega}$. In $R^-$, this element is $e^{j(m+n)\omega}$ which is $E_{mn}^*$

  • $\begingroup$ So in the end, Q will be a Toeplitz matrix, will it not? $\endgroup$
    – Sndn
    Apr 13, 2020 at 3:33

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