분야
pdfG(s) = 1
s(s + 1) (1)
Solution of Problem 1 (20pt)
1. (5pt) Consider
G(jω) = 1
jω(jω + 1) (2)
2. (5pt) The magnitude and phase are obtained for ω ≥ 0 as follow:
|G(jω)| =
1
ω
√
1 + ω2
→ |G(jω)|dB = −20 log10 ω − 20 log10
p
1 + ω2 (3)
∠G(jω) = −90◦ − tan−1 ω (4)
3. (5pt) first at ω = 0
|G(jω)| = ∞ → |G(jω)|dB = ∞ (5)
∠G(jω) = −90◦
(6)
second at ω = 0.1
|G(jω)| ≈ 1
0.1
= 10 → |G(jω)|dB ≈ 20[dB] (7)
∠G(jω) ≈ −90◦
(8)
third at ω = 1
|G(jω)| =
1
√
2
→ |G(jω)|dB = −20 log10
√
2 = −3[dB] ≈ 0[dB] (9)
∠G(jω) = −135◦
(10)
third at ω = 10
|G(jω)| ≈ 1
10√
100
= 0.01 → |G(jω)|dB = −20 log10 ≈ −40[dB] (11)
∠G(jω) ≈ −180◦
(12)
