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barcode generator for ssrs + 0.2 cos 2w in Software
+ 0.2 cos 2w Recognizing Code 128B In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Code128 Creation In None Using Barcode creator for Software Control to generate, create USS Code 128 image in Software applications. Find h ( n ) and H ( ~ J " ) . Code 128A Recognizer In None Using Barcode recognizer for Software Control to read, scan read, scan image in Software applications. Code 128 Code Set A Printer In C#.NET Using Barcode generation for Visual Studio .NET Control to generate, create USS Code 128 image in Visual Studio .NET applications. We are given H,(eJw)and are asked to find
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Painting Data Matrix 2d Barcode In ObjectiveC Using Barcode encoder for iPad Control to generate, create DataMatrix image in iPad applications. Barcode Creation In .NET Using Barcode creator for VS .NET Control to generate, create bar code image in Visual Studio .NET applications. 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 discretetime 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 secondorder 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, ejN"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 halfinteger, N j 2 , and the symmetry condition isequivalent to Thus, the frequency response may be written as
Factoring out a linear phase term eJN"'/' and using the symmetry of h ( n ) . we have
Therefore, where which is the desired result.
N+1 b(k)=2h(Tk) 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 timereversal 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~(ejW) 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 ) = ejNWf/*(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 realvalued (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 realvalued 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 smallestorder 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

