Note that if H(z) = A(l - z-' the magnitude of the frequency response at w = 0 is

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H(eJw)lw,o = A and the magnitude of the frequency response at w = K is H(ejw)l0,=, = 3 A

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TRANSFORM ANALYSIS OF SYSTEMS

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Thus, no value for A will allow us to simultaneously satisfy both unit magnitude constraints, and it is necessary to add another linear phase term to H (2). To minimize the order of the filter, we will pick a factor of the form

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In this case, the filter becomes

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H ( z ) = A(l

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Now, with two free parameters, A and B , we should be able to satisfy the magnitude constraints. With

and solving for A and B , we find Therefore, the filter is

~ ( e ~ " ' ) )= 3A(2 - B ) = I ~,,

Show that if h ( n ) is real, and 2a is an integer, the constraint

is sufficient, but not necessary, for h(n) to be the unit sample response of a system with generalized linear phase.

To show that this symmetry condition is sufficient for a system to have generalized linear phase, if we let H ( e J W ) be the Frequency response, the symmetry condition implies that

) Expressing H ( e j W in polar form, H ( e j w )= ~ ~ ( e j ~ ) l e j @ h ( ~ '

this becomes

I ~ ( ~ J w ) l ~ j @ h ( w )~ ( ~ - i m ) l ~ j @ h ( - ~ ) ~ - j 2 " 0

Because the magnitude is an even function, I H(eJw)l IH(e-JW)I, the phase is odd, Q ~ ~ ( W ) = and = -&,(-0).

,J@h

=e-~@h(w)-j~W

Therefore, the terms in the exponentials must be equal to within an integer multiple of 2n,

where k is an integer. Solving for Q5,,(w), have we

and it follows that the system has generalized linear phase. To show that this condition is not necessary, note that if

then H (el'")has linear phase. However, the unit sample response is

which is not symmetric about an integer index unless 2a = nd is an integer.

CHAP. 51

TRANSFORM ANALYSIS OF SYSTEMS

An FIR linear phase filter has a unit sample response that is real with h(n) = 0 for n < 0 and n r 7. If h(0) = 1 and the system function has a zero at z = 0 . 4 e J = / ~ and a zero at z = 3, what is H ( z )

Because h(n) = 0 for n < 0 and for n > 7, H (z) has seven zeros. With a complex zero at z = 0.4ej"l3, because h(n) is real, there must be another zero at the conjugate location, z = 0 . 4 ~ - j " / ~This conjugate pair of zeros produces . the second-order factor

The linear phase constraint requires that there be a pair of zeros at the reciprocal locations. Therefore, H (z) must also contain the factor H2(z) = 0.16 - 0.42-I ;:-2

The system function also contains a zero at z = 3. Again, the linear phase constraint requires that there also be a zero at z = f . Thus, H(z) also has the factor H3(z) = ( I - 3z-')(1 and we have H(z) = A(l

- fz-')

- 0 . 4 ~ - '+ 0. 16zF2). (0.16 - 0 . 4 ~ - '$. z-') . ( I

- 3z-')(l

;z-1)

Finally, because the coefficient of the zero-order term in this polynonlial is 0. 16A, A must be equal to

in order to make h(0) = 1.

Let x ( n ) be a finite-length sequence that has a z-transform

with no conjugate reciprocal zeros, i.e., a # 1 /a; for any k and 1. Show that x ( n ) is uniquely defined k to within a constant by the phase of its discrete-time Fourier transform.

Let xl (n) be a finite-length sequence that has a z-transform with no conjugate reciprocal zeros and a DTFT with the same phase as x(n). Then XI^) = G(z)X(z) where G(z) is the system function of a filter that has zero phase. Thus, G(z) must have zeros in conjugate reciprocal

and/or poles in conjugate reciprocal pairs

However, conjugate reciprocal zeros in G(z) are not allowed, because this would imply that X I(z) has conjugate reciprocal zeros. Similarly, because X(n) is finite in length, G(z) cannot have poles in conjugate reciprocal pairs because this would imply that xl(n) is infinite in length. Therefore, G(z) must be a constant, and the result follows.