Solution: 9.10.4.1 H = (1;T d ) 1 T c =  1; d d  t t  =  d ; d d  t t  = ( d ; d ) t t d Then the degree (or order) of the numerator is deg( d ; d )+(deg)( t );; and the degree of the denominator of H is deg( t )+(deg)( d );; Thus for H to haveapolezero excess of zero or greater wemust have deg( d ; d )+(deg)( t )  deg( t )+(deg)( d ): Similarly, G c = T c T d 1 G p = t d p d d p : Then the degree (or order) of the numerator of G c is deg( d )+deg( d )+deg( p );; and the degree of the denominator of G c is deg( d )+(deg)( p )+deg( d ): Thus for G c to haveapole zero excess of zero or greater wemust have deg( d )+deg( d )+deg( p )  deg( d )+(deg)( p )+deg( d ): 1