When Initial Conditions Is Zero

When initial conditions are zero, y(-l) = 0

The desired output is S(K) = yK(n) I „ = N = yK( N)                      ...(14)

for K = 0, 1, ... , N- 1



    Flow graph of 1st order complex.JPG


Fig. Flow graph of 1st order complex recursive computation of S(K).


Since both the input s(ii) and are complex, the computation of each new value of y%(ii) requires 4 real multiplications and 4 real additions.


System function of first order filter* is


         HK (z) = 1/ 1-WN-KZ-1


and its difference equation is


         yK (n) = WN-KyK (n- 1) + sK(n)


The complex multiplications and additions in Eq. (6.13) can be avoided by combining the pairs of single pole filters as it possesses complex-conjugate poles. Complex conjugate poles are those poles whose imaginary parts have opposite sign. For example: One pole is given by p1 = a + jb then its complex conjugate pole will be given by p2 = p1[= a - jb.


We can get second order filter from first order filter by multiplying both the numerator and denominator of a system function of first order filter


 denominator of a system function.JPG


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