4.27. Using the z-transform, redo Prob. 2.30.

From Fig. 2-23 and definition (4.3 )

x[ n ] = {1, 1, 1, I} +-+ X ( z ) = 1 + z – • + z -22 -3

h[ n ] = {l, 1, 1} +-+ H ( z ) = 1 + z – 1 + z -2

Thus, by the convolution property ( 4.26 )

Y( z ) = X ( z ) H ( z ) = (1 + z – 1 + z – 2 + z -3 )(1 + z – 1 + z – 2 )

= 1 + 2 z – 1 + 3z -2 + 3z -3 + 2 z -4 + z -5

Hence,

h [ n ] = {1, 2, 3, 3, 2, 1}

which is the same result obtained in Prob. 2.30.