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ECE114_2008FALL_EXAM1_PROFSOLN_[0]

ECE114_2008FALL_EXAM1_PROFSOLN_[0] - ECE 114 Fall 2008...

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Unformatted text preview: ECE 114 Fall 2008 Midterm Solutions 1. (13 points total) (a) (2 points): The system has a stable and causal realization and a stable and causal inverse. (b) (2 points): All zeros and poles of the system transfer flmction, HC (3), are in the left-half plane. ~ (c) (3 points): Not necessarily: although the poles from the left-half plane are mapped inside the unit circle (each pole pk of HC (3), which has negative real part, is mapped to eka), the zeros are not necessarily mapped inside the unit circle. This is because the method of impulse invariance does not directly map the s—plane to the z—plane. For completeness, take the following example. Let HC (3) be a minimum-phase analog system, given by: s + 10 HO = — (3) (s + 1) (s + 2) with a zero at s = —10 and poles at s = ~1, s = —2. Compute a partial fraction expansion: A B Hc = (s) (5+1)+ (3+2) A = lin_11 (s + 1) He (3) _ —1 + 10 — —1 + 2 = 9 B = lirn2 (s + 2) HC (3) _ —2 + 10 _ —2 + 1 = —8 Taking the inverse Laplace transform, we obtain: hc (t) = (QC—t — Se'Zt) u (t) Sampling the impulse response with period T = 1, we obtain: h [n] = he (71) - (9e‘n — 86’2") u [77.] Taking the z-transform, we obtain: 9 8 H (z) — 1_——nz—‘ ‘ 1—— 9 (1 — e’znz‘l) — 8 (1 — e‘"z‘1) (1 — e—nz‘l) (1 — e—an—l) 1 — 2‘1 (96'2” — 86‘") (1 — e—"z—l) (1 — e'znz‘l) 1 Notice that, as expected, the poles at s = —1 and s = —2 have been mapped to poles inside the unit circle at z = e‘" and z = (3—2”. However, the zero that was located at s = —10 has been shifted to a zero at z = 96—2" — 86—" = —1.725. Thus, the zero is located outside the unit circle. (d) (3 points): Yes: the bilinear transformation maps the entire left—half plane inside the unit circle; therefore, all zeros and poles of He (3) will be mapped inside the unit circle. (e) (3 points): Yes: Since H (z) = H1 (2) H2 (2), the poles and zeros of H (2) will be I the union of the poles and zeros of H1 (2) and H2 (2). . (10 points) H (z) = The paraconjugate is given by: The zeros of H" (2—1,) are: 21 = —2/j = 23', 22 = —5 The poles of H* (i) are: p1 = 5/4, 102 = 00 2t . (15 points) (22 + 5) (z — 2) H“) = (42 — 1) (2+2)2 To compensate for the magnitude distortions introduced by H, we should use a com- pensator HC (2) as: 1 Hmin (2) where Hmin (z) is the minimum-phase factor of the minimum-phase all—pass decompo- sition of H (z): Ha (z) = H (z) = Hmin (z) A (2) (Notice that the inverse of a minimum—phase system is also minimum—phase). Using the HC (2) given above, the overall system becomes an all-pass system: |H(w)Hc(W)l = HmmWMWEfiw) = |A(w)l = 0 since A (z) is an all—pass ftmction. Now compute the minimum-phase all—pass decomposition: H (2) has zeros at 21 = —5/2, 22 = 2 H (2) has poles at p1 = 1/4, p; = —2 (double pole). Thus, A (2) must include 21,22,122: (22 + 5) (2 — 2) (22 + 1)2 (52 + 2) (22 — 1) (2 + 2) and the minimum-phase component is given by: 2 (52 + 2) (22 — 1) A(2) = Hmin (Z) = 2 (42 — 1) (22 + 1) Therefore, _ (42 — 1) (22 + 1)2 H” (2) — 2 (52 + 2) (22 — 1) . (15 points) IH (“0'2 2 (17 — 8003(1))2 34 + 30 cos w First, by uSing the fact that cosw = % (ejw + e'j‘”), and 2 = 63"”: * 1 (17—4(z+z-1))2 C(z) = H(Z)H (7) = 34+ 15(z+z-l) Through some algebra, we obtain: 2-2 (172 — 422 — 4)2 2“1 (342 + 1522 + 15) (422 — 172 + 4)2 2 (1522 + 342 + 15) By factoring the numerator and denominator, we obtain: ((4z - 1) (z — 4))2 G = ——— (z) 2 (52 + 3) (32 + 5) Since we seek a minimum-phase H (2), we should take the poles and zeros that lie inside the unit circle: G (2) K (42 — 1)2 [1(2): 2(52+3) Then 11* (i) = K* (42-1 — 1)2 2* 2—1 (52‘1 + 3) and II ' * 1 (42 — 1)2 (42-1 — 1)2 H (z) H (E) (52 + 3) (52‘1 + 3) 2 2"2 (42 — 1)2 (4 — 2)2 2—1 (52 + 3) (5 + 32) (42 - 1)2 (2 — 4)2 2 (52 + 3) (32 + 5) |K| IKIZ Therefore, take K = 1: (42 — 1)2 “dim 5. (10 points total) (2z — 3)2 (z + 5) (z + 4) H“): (z—3)(4z+3)(22+1) (a) (2 points): zeros: 21 = 3/2, z; = 3/2, 23 = —5, 7.4 = —4 poles: p1 = 3, p2 = —3/4, 133 = —1/2, p4 = 00 (b) (4 points total): 0 Not possible (pole at 00) 0 3/4 < |z| < 3 0 3/4 < [21 < 3 o |z| < 1/2 (c) (2 points): The z’1 term will cancel the pole at z = 00 and insert a pole at z = 0. Therefore, the system 2—1H (2) can be causal but not anti-causal. (d) (2 points): The z term will add an additional pole at z = 00. Therefore, the system 2H (z) cannot be causal, but can be anti—causal. 6. (15 points total) 11(2) = 11(3) |S=g1_+z_-1r1 (l—z- ) (a) To analyze the frequency response of H (w), let 3 = jfl and z = 67"”: 1 + e‘j‘” 1 — 6‘1"“ e—jw/z (6101/2 + e—jw/z) e—jw/2 (aw/2 _ e—jw/z) = — j cot (w / 2) jQ ll Therefore, 9 = — cot (w/ 2). Examining the graph of cot, we see that w = 0 is mapped to Q = 00, and w = 7r is mapped to Q = 0. Additionally, the graph of cot (w / 2) is monotonic. This shows that, since H (9) is lowpass, the response of H (w) is highpass. (b) By introducing a second analog filter H2 (5) with HANG) * Mapping H2 (3) to H (z) by way of the bilinear transformation: H (z) = H2(8)|8=8;::1g 1 H (2) tea = H(s) |s_ _ 1—:— Notice that this strategy is the same way we designed a highpass filter in last week’s class. 7. (12 points) n n H H1: anti-imaging filter with cutoff: 7r/ D, gain: D H2: anti-aliasing filter with cutoff: 1r/C, gain: 1 H1 and H2 can be combined into one filter H with cutoff: 7r/M, gain: D, where M = max (C, D). ‘ 8. (10 points total) (a) (4 points): X(w)=%xc(3;) M s: X (w) is a triangular waveform (with maximum 2/T) for w E [—7r/2,1r/2] and zero for 7r/2 g |w| 5 7r. (b) (4 points): _ TX (9T) , |Ql SW/T Y” ((2) _ { 0 , otherwise Yc (Q) is a triangular waveform (with maximum 2) for Q 6 [—7r/ (2T) ,7r/ (2T)] and zero otherwise. (c) Notice that Yc (Q) = 2Xc (2(2). Therefore, _ i °° yea) — Mimi/cane] d9 % / Xc (2n) Wm Use substitution (2’ = 29: ya (15) i / X6 (9!) ejn'(t/2)dnl 27r _00 me (#2) ...
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