PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

592 Practical MATLAB® Applications for Engineers

H(ejW)^ = h( 0 )e−^2 jW 2 cos( 2 W) + h( 1 )e−jW^22 cos(W) + h( 2 )e−^2 jW

H(ejW)^ = e−^2 jW[h( 0 ) 2 cos( 2 W) + 2 h( 1 )cos(W) + h( 2 )]

H(ejW) =  2 h( 0 )cos( 2 W) + 2 h( 1 )cos(W) + h( 2 ) for 0 ≤ n ≤ 4

then

∠H(ejW) = − 2 W

Observe that the magnitude of H(e−jW) is a real function of W, whereas the phase is a

linear function of W over the range of −π ≤ W ≤ +π.

R.6.125 Now consider the sequence

h(n) = −0.09
(n) + 0.29
(n − 1 ) + 0.53
(n − 2 ) + 0.29
(n − 3 ) − 0.09
(n − 4 )

which represents the impulse response of a type-1 FIR fi lter with N = 5.

ANALYTICAL Solution

The magnitude and phase response is then given by (using R.6.124)

H(ejW) = 2 (−0.09)cos( 2 W) + (^2) * (0.29)cos(W) + (0.53)
for n = 0, 1, 2, 3, 4 and ∠H(ejW) = − 2 W
R.6.126 The example below illustrates the analysis of a transfer function of an FIR type-2
fi lter with N = 8.
ANALYTICAL Solution
H(z) = h( 0 ) + h( 1 )z−^1 + h( 2 )z−^2 + h( 3 )z−^3 + h( 4 )z−^4 + h( 5 )z−^5 + h( 6 )z−^6 + h( 7 )z−^7
Then applying symmetry conditions, the following relations hold:
h( 0 ) = h( 7 )
h( 1 ) = h( 6 )
h( 2 ) = h( 5 )^
(^) h( 3 ) = h( 4 )
Then, its spectrum is given by
H(ejW) = [ 2 h( 0 )cos( 7 W/ 2 ) + 2 h( 1 )cos( 5 W/ 2 ) + 2 h( 2 )cos( 3 W/ 2 )

• 2 h( 3 )cos(W/ 2 )]e−jW7/ 2^
  Clearly, the (phase) ∠H(ejW) = −(7/ 2)W and the quantity inside the brackets
  represents its magnitude spectrum.
  R.6.127 Let us analyze an FIR type-3 fi lter with N = 9 given by
  H(z) = h( 0 ) + h( 1 )z−^1 + h( 2 )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