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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
lefthalf plane. ~ (c) (3 points): Not necessarily: although the poles from the lefthalf 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 minimumphase
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 ztransform, 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 minimumphase factor of the minimumphase 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 allpass system: H(w)Hc(W)l = HmmWMWEﬁw)
= A(w)l
= 0 since A (z) is an all—pass ftmction.
Now compute the minimumphase 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 minimumphase 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+z1))2 C(z) = H(Z)H (7) = 34+ 15(z+zl) Through some algebra, we obtain: 22 (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 minimumphase 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* (421 — 1)2
2* 2—1 (52‘1 + 3)
and II ' * 1 (42 — 1)2 (421 — 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 anticausal. (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: antiimaging ﬁlter with cutoff: 7r/ D, gain: D
H2: antialiasing 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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