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Find h ( n ) and H ( ~ J " ) .
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We are given H,(eJw)and are asked to find
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H (el") = HR(el") j Hr(el")
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Although we could find H,(ejw) using the discrete Hilbert transform, an easier approach is as follows. Because h(n) is causal, the wen and odd parts of h ( n )are
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TRANSFORM ANALYSIS OF SYSTEMS Therefore, the relationship between h,(n) and h,,(n) is as follows: h,,(n) = sgn(n) . h,(n)
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[CHAP. 5
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sgn(n) =
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n>O n =0 n<O
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The inverse DTFT of HR(eJW), which is the even part of h(n), is h,(n) = 6(n) Thus,
+ 0. IS(n - 2) + 0.16(n + 2)
+ 2)
h,(n) = sgn(n)h,(n) = 0. IS(n - 2) - 0.16(n
and Hl(eJ"), the discrete-time Fourier transform of h,(n), is Hl(eJ") = -0.2sin(2o) Therefore, and ~ ) = H(eJ") = ~ ~ ( e jj Hl(eJW) I
+ 0.2cos2o + 0.26(n
j0.2sin(2w) = I
+ 0.2e-*'"
h(n) = 6(n)
A second-order system has two poles at z = 0.5 and a pair of complex zeros at z = e * ~ = / ~ . Geometrically find the gain, A, of the filter so that I H ( e J W ) is equal to unity at w = 0. J
Because the length of the vectors from the two zeros at z = e'J"12 to the point z = 1 on the unit circle is equal to A,and because the distance from the two poles at z = 0.5 to z = 1 is equal to 0.5, the magnitude of the frequency response at o = 0 is
Therefore, the desired gain is
A = !8
Systems with Linear Phase
Derive Eq. (5.9)for the frequency response of a type I linear phase filter.
A type I linear phase filter satisfies the symmetry condition
h(n) = h(N and N is even. The symmetry condition is equivalent to
Therefore, the frequency response may be written as follows:
Factoring out a linear phase term, e-jN"f2, from each sum, and using the symmetry of h(n), we have
CHAP. 51
TRANSFORM ANALYSIS OF SYSTEMS
Therefore, we may write the frequency response as follows:
where
which is the desired result.
Derive Eq. (5.10) for the frequency response of a type I1 linear phase filter.
For a type I1 linear phase filter, h(n)= h(N
where N isodd. Therefore, h ( n ) is symmetric about the half-integer, N j 2 , and the symmetry condition isequivalent to
Thus, the frequency response may be written as
Factoring out a linear phase term e-JN"'/' and using the symmetry of h ( n ) . we have
Therefore,
where which is the desired result.
N+1 b(k)=2h(T-k)
k = 1 . 2. . . . ,
How would the derivations in the previous two problems be modified to find the form of the frequency response for types 111 and IV linear phase filters
The only difference between a type I and a type I11 linear phase filter is that h(n)= h(N - n ) for a type I filter, and h(n) = -h(N - n ) for a type 111 filter. Therefore, h ( N / 2 ) = 0, and Eq. (5.20) is modified as follows:
TRANSFORM ANALYSIS OF SYSTEMS Thus, it follows that the frequency response may be written as
[CHAP. 5
where The only modification required in Prob. 5.15 to find the form of the frequency response for a type IV linear phase filter is to use the fact that h ( n ) is odd to rewrite Eq. (5.21) as follows:
Therefore, the frequency response is
where
Show that a system with a complex unit sample response has generalized linear phase if
h(n) = fh*(N - n )
If ~ ( e j " is the DTFT of h ( n ) , it follows from the delay property and the time-reversal property that the DTFT of ) h ( N - n ) is h(N - n ) Applying the conjugation property. we then have h*(N - n )
DTFT D7FT
e lNw~(e-jW)
e 'NW~*(e'w)
- n).
Now, let us consider the case in which h ( n ) is conjugate symmetric, h ( n ) = hl*(N H ( e j w ) = e-jNWf/*(el"')
Then
) and, expressing H (el"') and H * ( e J Win terns of their magnitude and phase, we have H(eJw) = I ~ ( ~ j w ) l ~ ~ @ l t ( w ) and Therefore, it follows that
el@h(w)
~ * ( ~ i= ()~ ( ~ l ~ ) l ~ - j @ 1 1 ( 0 ) w
- ~ N [ ~ e - ~ ~ ~ ( w ~
2+h( w ) = - N o
+2nk(o)
CHAP. 51
TRANSFORM ANALYSIS OF SYSTEMS
where k(w) is an integer for each w. Solving for the phase, we have
where A(eJW) a real-valued (in general bipolar) function of w. Thus, h(n) has linear phase. is For the case in which h(n) is conjugate antisymmetric, h(n) = -h*(N - n) Eq. (5.22) becomes
Therefore,
24,,(w) = -N w
+ rr + 2 ~ k ( w )
where again k(w) is an integer for each w. Solving for the phase, we have
where A(eJw)is a real-valued function of w. and h ( n ) has generalized linear phase.
The relationship between the input and the output of an FIR system is as follows:
Find the coefficients h(k) of the smallest-order filter that satisfies the following conditions:
2. 3.
The filter has (generalized) linear phase. It completely rejects a sinusoid of frequency wo = n/3. The magnitude of the frequency response is equal to 1 at w = 0 and w = n.
To reject a sinusoid of frequency @ = 7r/3, the system function must have a pair of zeros on the unit circle at z = e*jnl3. Therefore, H(z) must contain a (linear phase) factor of the form
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