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barcode lib ssrs Multiplying the numerator and denominator by H*(eJw)= HR(eJW) jHI(eJW) yields in Software
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H, (eJ") = 1  a cos w
(a) For the first system, the frequency response is
+ ja sin w
Therefore, the phase is
(w) = tan' a sin w 1acosw
Because the group delay is r~(w= ) d dw
I a COEw
a sin w
Therefore, rl(w) =  (1  a c o s w ) a c o ~ w  (asinw)' (1  (Y cos w ) ~
+ ()2 FOURIER ANALYSIS which, after simplification, becomes q(w) = ( I  a c o s ~ ) a c o ~ w  ( a s i n w ) a 2  a c o s w ~ I +az2acosw (1  a cos w ) ~ ( a sin w)Z [CHAP. 2
Another way to solve this problem is to use the expression for the group delay derived in Rob. 2.18. With Hl(eJ") = 1  a c o s w + j a s i n w we see that HR(eJw) 1  a cos w = Because the unit sample response is h(n) = S(n)  aS(n then Hl(eJw)= a sin w  1) g(n) = nh(n) = a6(n  I ) and Therefore, the group delay is
G(ejw)= aejw = a cos w
+ ja sin w
which is the same as before. (b) Having found the group delay for Hl(ejW)= 1  aeJ", we may easily derive the group delay for H2(eJW), which is the inverse of Hl(eJw): Specifically, because 1 Hz(eJ") = Hz(eJ") &(w) = @I(w) and, therefore, ( c ) For the last system, H3(eJw) may be factored as follows: The group delay of H3(eJm) thus the sum of the group delays of these two factors. Furthermore, the group is delay of each factor may be found straightforwardly by differentiating the phase. However. the group delay of these terms may also be found from t2(w) in part (b) if we use the modulation property of the DTFT. Specifically, recall that if X(ejw) is the DTFT of x(n), the DTFT of ejn0x(n)is , Therefore, if the group delay of x ( n ) is ~ ( w )the group delay of ejdx(n) will be r(w  8). In part (b), we found that the group delay of H (ej") = 1/(I  cue'") is CHAP. 21
FOURIER ANALYSIS Thus, it follows from the modulation property that the group delay of H(ejw)= ]/(I  ae~(~O') is = and that the group delay of H(ejW) 1/(1  aej"u+e)) is
Therefore, the group delay of H3(eJW) the sum of these: is
Find the DTFT of each of the following sequences: (a) X I(n)= ($)"u(n
(b) xz(n) = CY" sin(nwo) u(n) ( c ) x3(n) = 1A;)" + 3) n = 0.2.4, . . . otherwise
(a) For the first sequence, the
DTFT may be evaluated directly as follows: (b) The best way to find the DTFTofxz(n)is toexpress the sinusoid as a sum of twocomplexexponentials as follows: Similarly, for the second term we have
Therefore, 1 I X2(eJW) = 2 j 1  aej(w"o) 1(c) Finally, for xj(n), we have
 (h wo)e j w + cr2eZjw cos
(a sin wo)ejw
Therefore, FOURIER ANALYSIS
[CHAP. 2
Because the DTFT of the output of a linear shiftinvariant filter with frequency response H ( e j w ) is where x ( e j w ) is the DTFT of the input, it follows that an LSI system cannot produce frequencies in the output that are not present in the input. Therefore, if a system introduces new frequencies. the system must be nonlinear and/or shiftvarying. For each of the following systems, find the frequencies that are present in the output when x(n) = cos(nwo): (a) With x(n) = cos(nmo), the output of the squarelaw device is y(n) = cos2(nw) Using the trigonometric identity cos2 A = it follows that y(n) = cos(2A) + f cos(2nm) wo. Therefore, although the only frequencies present in the input are w = f the frequencies in the output are w = 0, f2mn. Because this system is nonlinear, it creates frequencies in the output that are not in the input. (b) For the modulator, the output is Using the trigonometric identity 2cos A cos B = cos(A + B) + cos(A  B) it follows that y(n) = cos(nmo

