0
250views
Explain in detail polyphase decomposition.
1 Answer
1
1views

Solution:

The polyphase structures for the decimators and the interpolator require the same number of filter operations as the efficient direct form structures.

$\rightarrow$ Consider a system function $H[z]$ of length 9 $H[z]=h[0]+h[1] z^{-1}+h[z] z^{-2}+h[3] z^{-3}+h[4] z^{-4}$ $+h[5] z^{-5}+h[6] z^{-6}+h[7] z^{-7}+h[8] z^{-8}$

$\longrightarrow$ This can be expressed as a sum of two terms, with one term containing the even indexed coefficients and the other containing the odd indexed coefficients. as follows,

$ \begin{aligned} H[z]=& {\left[h[0]+h[2] z^{-2}+h[4] z^{-4}+h[6] z^{-6}+h[8] z^{-8}\right] } \\ &+\left[h[1] z^{-1}+h[3] z^{-3}+h[5] z^{-5}+h[7] z^{-1}\right] \\ =& {\left[h[0]+h[2] z^{-2}+h[4] z^{-4}+h[6] z^{-6}+h[8] z^{-8}\right] } \\ &+z^{-1}\left[h[1]+h[3] z^{-2}+h[5] z^{-4}+h[7] z^{-6}\right] \\ H[z]=& E_0\left[z^2\right]+z^{-1} E_1\left[z^2\right] \end{aligned} $

where $E_0[z]=h[0]+h[z] z^{-1}+h[4] z^{-2}+h[6] i^{-3}+h[8] d^4$

$ E_1[z]=h[1]+h[3] z^{-1}+h[5] z^{-2}+h[7] z^{-3} $

The decomposition of $\mathrm{H}[2]$ in the eqn. is known as polyphase decomposition. $\rightarrow$ A two branch polyphase structure of $H[z]$ is

shown in figure, where the two functions $E_0\left[z^2\right]$ and $E_1\left[z^2\right]$ are subfilters connected in parallel.

These subfilters in the polyphase realization of an FIR filter system function are also FIR filters and can be realized in either direct form or transposed direct form.

In general, an FIR filter system function of length $N$ can be decomposed in an $M$-branch poly phase form as follows. $ \begin{aligned} H[z]=\sum_{k=0}^{M-1} i^{-k} E_k\left[z^M\right] \ \text { where } E_k[z] &=\sum_{n=0}^M h\left[M_n+k\right] z^{-n} \ &=\sum_{n=0}^{N / k \mid} e_k[n] z^{-n} ; 0 \leq k \leq M-1 \end{aligned} $ where $e_k[n]=h[M n+k], 0 \leqslant k \leqslant M-1$ This is called Type I polyphase decomposition

enter image description here

Please log in to add an answer.