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𝑮(𝒔)
𝒖 𝒕
𝑼 𝒔
𝒚 𝒕
𝒀 𝒔
•
𝑮 𝒔 =
Transformada de la respuesta en estado nulo
Transformada de la señal de entrada
•
𝒂𝒏𝒚(𝒏)
𝒕 + ⋯ + 𝒂𝟏 ሶ
𝒚 𝒕 + 𝒂𝟎𝒚 𝒕 = 𝒃𝟎𝒖 𝒕 + 𝒃𝟏 ሶ
𝒖 𝒕 + ⋯ 𝒃𝒏𝒖(𝒎)
𝒕
𝒚 𝒕
𝒖 𝒕
𝒂𝒏 𝒃𝒏
•
• ℒ
𝒅𝒇(𝒕)
𝒅𝒕
= 𝒔ℒ 𝒇(𝒕) − 𝒇 𝟎
• ℒ
𝒅𝟐𝒇(𝒕)
𝒅𝒕𝟐 = 𝒔𝟐
ℒ 𝒇(𝒕) − 𝒔𝒇 𝟎 −
𝒅𝒇(𝟎)
𝒅𝒕
ℒ
𝒅𝒏𝒇(𝒕)
𝒅𝒕𝒏
= 𝒔𝒏
ℒ 𝒇(𝒕) − ෍
𝒌=𝟏
𝒏
𝒔𝒌−𝟏
𝒅𝒏−𝒌𝒇(𝟎)
𝒅𝒕𝒏−𝒌
•
𝒕 < 𝟎
𝒂𝒏𝒔𝒏𝒀(𝒔) + ⋯ + 𝒂𝟏 𝒔𝒀(𝒔) + 𝒂𝟎𝒀(𝒔) = 𝒃𝟎𝑼(𝒔) + 𝒃𝟏𝒔𝑼 𝒔 + ⋯ 𝒃𝒏𝒔𝒎𝑼 𝒔
𝒀 𝒔
𝑼 𝒔
𝒂𝒏 𝒃𝒏
• 𝒀(𝒔) 𝑼 𝒔
(𝒂𝒏𝒔𝒏 + ⋯ + 𝒂𝟏 𝒔 + 𝒂𝟎)𝒀(𝒔) = (𝒃𝟎+𝒃𝟏𝒔 + ⋯ 𝒃𝒏𝒔𝒎)𝑼 𝒔
𝑮 𝒔 =
𝒀(𝒔)
𝑼(𝑺)
=
𝒃𝟎 + 𝒃𝟏𝒔 + ⋯ 𝒃𝒏𝒔𝒎
𝒂𝒏𝒔𝒏 + ⋯ + 𝒂𝟏 𝒔 + 𝒂𝟎
𝑮 𝒔 =
𝒀(𝒔)
𝑼(𝑺)
=
𝒃𝟎 + 𝒃𝟏𝒔 + ⋯ 𝒃𝒏𝒔𝒎
𝒂𝒏𝒔𝒏 + ⋯ + 𝒂𝟏 𝒔 + 𝒂𝟎
•
𝒔
𝒔
• 𝑼 𝒔 ,
𝑷 𝒔 = 𝒂𝒏𝒔𝒏 + ⋯ + 𝒂𝟏 𝒔 + 𝒂𝟎
• 𝒏 𝑷 𝒔
• 𝒔 𝒀(𝒔)
𝒛𝟏, 𝒛𝟐, … , 𝒛𝒎 ,
𝒀 𝒔 = 𝒃𝟎 + 𝒃𝟏𝒔 + ⋯ 𝒃𝒏𝒔𝒎
= 𝟎
• 𝒔 𝑷(𝒔)
𝒑𝟏, 𝒑𝟐, … , 𝒑𝒏 ,
𝑷 𝒔 = 𝒂𝒏𝒔𝒏
+ ⋯ + 𝒂𝟏 𝒔 + 𝒂𝟎 = 𝟎
•
•
•
•
•
𝒔
•
•
•
•
𝒔
•
•
•
𝒔
• 𝒔
𝒔
𝒔
• 
 𝑠
•
𝒔
•
𝒔
𝒔
𝒔
ℝ𝑒
𝕀𝑚
𝒔
•
•
•
•
•
•
•
•
•
•
𝟕
𝒔 + 𝟓
𝟒
Time (sec.)
Amplitude
Step Response
0 0.2 0.4 0.6 0.8 1 1.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
From: U(1)
To:
Y(1)
Time (sec.)
Amplitude
Step Response
0 0.2 0.4 0.6 0.8 1 1.2
0
1
2
3
4
5
6
From: U(1)
To:
Y(1)
𝟏
𝒔 + 𝟐
Time (sec.)
Amplitude
Step Response
0 0.5 1 1.5 2 2.5 3
0
0.2
0.4
0.6
0.8
1
1.2
1.4
From: U(1)
To:
Y(1)
𝟕
𝒔 + 𝟓
𝟕
𝒔 + 𝟓
•
•
𝑲
𝒔(𝒔 + 𝟕)
+
−
𝒀(𝒔)
𝑹(𝒔)
𝒀(𝒔)
𝑼(𝒔)
•
𝑮 𝒔 =
𝒀(𝒔)
𝑼(𝒔)
=
𝑲
𝒔(𝒔 + 𝟕)
•
• 𝑠1 = 0
• 𝑠2 = −7
•
•
𝑮𝒍𝒄 𝒔 =
𝒀(𝒔)
𝑹(𝒔)
=
𝑲
𝒔 𝒔 + 𝟕 + 𝑲
•
𝒔𝟐 + 𝟕𝒔 + 𝑲 = 𝟎
•
𝒔𝟏,𝟐 = −𝟑. 𝟓 ± 𝟏𝟐. 𝟐𝟓 − 𝑲
• 𝑲
• 𝑲 = 𝟎. 𝟏
𝑮𝒍𝒄 𝒔 =
𝟎. 𝟏
𝒔𝟐 + 𝟕𝒔 + 𝟎. 𝟏
⇒ 𝒔𝟏 = −𝟔. 𝟗𝟖𝟓𝟕
𝒔𝟐 = −𝟎. 𝟎𝟏𝟒𝟑𝟏
• 𝑲 = 𝟏
𝑮𝒍𝒄 𝒔 =
𝟏
𝒔𝟐 + 𝟕𝒔 + 𝟏
⇒ 𝒔𝟏 = −𝟔. 𝟖𝟓𝟒𝟏
𝒔𝟐 = −𝟎. 𝟎𝟏𝟒𝟓𝟗
• 𝑲 = 𝟏𝟎
𝑮𝒍𝒄 𝒔 =
𝟏𝟎
𝒔𝟐 + 𝟕𝒔 + 𝟏𝟎
⇒ 𝒔𝟏 = −𝟓
𝒔𝟐 = −𝟐
• 𝑲 = 𝟏𝟐. 𝟐𝟓
𝑮𝒍𝒄 𝒔 =
𝟏𝟐. 𝟐𝟓
𝒔𝟐 + 𝟕𝒔 + 𝟏𝟐. 𝟐𝟓
⇒ 𝒔𝟏 = −𝟑. 𝟓
𝒔𝟐 = −𝟑. 𝟓
• 𝑲 = 𝟐𝟓
𝑮𝒍𝒄 𝒔 =
𝟐𝟓
𝒔𝟐 + 𝟕𝒔 + 𝟐𝟓
⇒ 𝒔𝟏 = −𝟑. 𝟓 + 𝒋𝟑. 𝟓𝟕𝟎𝟕
𝒔𝟐 = −𝟑. 𝟓 − 𝒋𝟑. 𝟓𝟕𝟎𝟕
• 𝑲 = 𝟏𝟏𝟐. 𝟐𝟓
𝑮𝒍𝒄 𝒔 =
𝟏𝟏𝟐. 𝟐𝟓
𝒔𝟐 + 𝟕𝒔 + 𝟏𝟏𝟐. 𝟐𝟓
⇒ 𝒔𝟏 = −𝟑. 𝟓 + 𝒋𝟏𝟎
𝒔𝟐 = −𝟑. 𝟓 − 𝒋𝟏𝟎
•
•
•
•
•
(𝟏 + 𝑮(𝒔)𝑯(𝒔)) 𝑲
𝒔
𝑮(𝒔)𝑯(𝒔)
•
𝟏 + 𝑮 𝒔 𝑯 𝒔 = 𝟎
⇒ 𝑮 𝒔 𝑯 𝒔 = −𝟏
• 𝒔
•
𝑮 𝒔 𝑯 𝒔 = 𝟏
•
∠𝑮 𝒔 𝑯 𝒔 = 𝟏𝟖𝟎° ± 𝟑𝟔𝟎°𝒌, 𝒌 = 𝟎, 𝟏, 𝟐, …
(𝟏 + 𝑮(𝒔)𝑯(𝒔) = 𝟎)
(𝑲 = 𝟎) 𝑮(𝒔)𝑯(𝒔)
(𝑲 = ∞) (𝑮(𝒔)𝑯(𝒔))
(𝒏) (𝒎)
(𝒏 − 𝒎)
𝜶𝒓 =
𝟐𝒓 + 𝟏 𝟏𝟖𝟎°
𝒏 − 𝒎
; 𝒓 = 𝟎, 𝟏, … , 𝒏 − 𝒎 − 𝟏
𝝈 =
σ𝒊
𝒏
𝑹𝒆 𝒑𝒊 − σ𝒋
𝒎
𝑹𝒆 𝒛𝒋
𝒏 − 𝒎
𝝀
±90º
𝑙𝜙𝑝 = 2𝑟 + 1 𝜋 + 𝜙; 𝑟 = 0,1,2, …
𝜙
𝜙 = σ 𝜃𝑧 − σ 𝜃𝑝
𝑙𝜙𝑧 = 2𝑟 + 1 𝜋 − 𝜙; 𝑟 = 0,1,2, …
𝜙
𝜙 = σ 𝜃𝑧 − σ 𝜃𝑝
𝒔 𝒋𝝎
𝝎 𝑲
𝒅𝑲
𝒅𝒔
= 𝟎
𝑲 = −
ς𝒊
𝒏
(𝒔 − 𝒑𝒊)
ς𝒋
𝒎
(𝒔 − 𝒛𝒋)
𝑲
𝑲 = −
ς𝒊
𝒏
𝒔 − 𝒑𝒊
ς𝒋
𝒎
𝒔 − 𝒛𝒋
𝑮(𝒔)𝑯(𝒔) =
𝒔 + 𝟑
𝒔𝟐 + 𝒔 + 𝟒
(𝑲 = 𝟎)
𝑮(𝒔)𝑯(𝒔)
𝒔
𝑠2
+ 𝑠 + 4 = 0
𝑮 𝒔 𝑯 𝒔
𝑝1 = −
1
2
+
1
2
𝑖 15
𝑝2 = −
1
2
−
1
2
𝑖 15
𝑛 = 2
(𝑲 = ∞)
𝑮(𝒔)𝑯(𝒔)
𝒔
𝑠 + 3 = 0
𝑧1 = −3
𝑚 = 1
𝑵 = 𝒏 − 𝒎
𝒏 = 𝑮𝑯 𝒔
𝒎 = 𝑮𝑯 𝒔
𝑵 =
𝒏 = 𝟐 𝒎 = 𝟏
𝑁 = 2 − 1
𝑁 = 1
𝜶𝒓 =
𝟐𝒓 + 𝟏 𝟏𝟖𝟎°
𝒏 − 𝒎
; 𝒓 = 𝟎, 𝟏, … , 𝒏 − 𝒎 − 𝟏
𝑁 = 1, 𝑟 = 0
𝛼0 =
2 0 + 1 180°
2 − 1
=
(1)180°
1
= 180°
𝝈 =
σ𝒊
𝒏
𝑹𝒆 𝒑𝒊 − σ𝒋
𝒎
𝑹𝒆 𝒛𝒋
𝒏 − 𝒎
𝑝1 = −
1
2
+
1
2
𝑖 15 𝑝2 = −
1
2
−
1
2
𝑖 15
𝑧1 = −3
𝜎 =
−
1
2
−
1
2
− (−3)
1
= 2
𝒔
𝑮(𝒔)𝑯(𝒔)
(−∞, −3)
ℝ𝑒
𝕀𝑚
𝑮𝑯(𝒔)
𝑺𝟏
∠𝐺𝐻(𝑠) = ෍ 𝜙𝑧 − ෍ 𝜙𝑝 = 180𝑜
(2𝑗 + 1)
ℝ𝑒
𝕀𝑚
𝝓𝒑𝟐
𝝓𝒛𝟏
𝝓𝒑𝟏
𝒙 = 𝟑 − 𝟎. 𝟓 = 𝟐. 𝟓
𝒚 = 𝕀𝒎 𝒑𝟏
𝒚 =
1
2
15
𝜙𝑝2
= 90°
𝜙𝑧1
= tan−1
𝑦
𝑥
𝜙𝑧1
= tan−1
1
2
15
2.5
= 37.761°
𝜙𝑧1
− 𝜙𝑝1
− 𝜙𝑝2
= 180°
𝜙𝑝1
= 𝜙𝑧1
− 𝜙𝑝2
− 180°
𝜙𝑝1
= 37.761° − 90° − 180°
𝜙𝑝1
= −232.24°
𝒔 𝒋𝝎
𝒔 → 𝒋𝝎 𝜔 𝑲
1 + 𝐾𝐺 𝑠 𝐻 𝑠 = 1 + 𝐾
𝑠 + 3
𝑠2 + 𝑠 + 4
= 0
𝑠2 + 𝑠 + 4 + 𝐾 𝑠 + 3 = 0
𝑠2 + (1 + 𝐾)𝑠 + 4 + 3𝐾 = 0
𝒔 = 𝒋𝝎
(𝑗𝜔)2+(1 + 𝐾)(𝑗𝜔) + 4 + 3𝐾 = 0
−𝜔2 + (1 + 𝐾)(𝑗𝜔) + 4 + 3𝐾 = 0
ℝ𝑒: −𝜔2 + 4 + 3𝐾 = 0
𝕀𝑚: 1 + 𝐾 𝜔 = 0
𝜔 = 0 𝜔
−(0)2+4 + 3𝐾 = 0
⟹ 𝐾 = −
4
3
𝑲 < 𝟎
𝒅𝑲
𝒅𝒔
= 𝟎
𝑲
1 + 𝐾
𝑠 + 3
𝑠2 + 𝑠 + 4
= 0
⟹ 𝐾 = −
𝑠2 + 𝑠 + 4
𝑠 + 3
𝒔
𝑑𝐾
𝑑𝑠
=
𝑑
𝑑𝑠
−
𝑠2 + 𝑠 + 4
𝑠 + 3
𝑑𝐾
𝑑𝑠
= −
2𝑠 + 1 𝑠 + 3 − (𝑠2 + 𝑠 + 4)(1)
(𝑠 + 3)2
𝑑𝐾
𝑑𝑠
= −
2𝑠2 + 7𝑠 + 3 − 𝑠2 − 𝑠 − 4
𝑠 + 3 2
𝑑𝐾
𝑑𝑠
= −
(2 − 1)𝑠2
+(7 − 1)𝑠 + 3 − 4
(𝑠 + 3)2
−
𝑠2 + 6𝑠 − 1
𝑠 + 3 2
= 0
𝑠2
+ 6𝑠 − 1 = 0
⟹ 𝑠1 = 0.16228
⟹ 𝑠2 = −6.1623
−∞, −3
𝒔𝟐 = −𝟔. 𝟏𝟔𝟐𝟑
𝑮(𝒔)𝑯(𝒔) =
𝑲
𝒔(𝒔 + 𝟒)(𝒔 + 𝟓)
(𝑲 = 𝟎)
𝒔
𝒑𝟏 = 𝟎, 𝒑𝟐 = −𝟒, 𝒑𝟑 = −𝟓
(𝑲 = ∞)
𝑵 = 𝒏 − 𝒎
𝒏 = 𝑮𝑯 𝒔
𝒎 = 𝑮𝑯 𝒔
𝑵 =
𝒏 = 𝟑 𝒎 = 𝟎
𝑵 = 𝟑 − 𝟎
𝑵 = 𝟑
𝜶𝒓 =
𝟐𝒓 + 𝟏 𝟏𝟖𝟎°
𝒏 − 𝒎
; 𝒓 = 𝟎, 𝟏, … , 𝒏 − 𝒎 − 𝟏
𝑵 = 𝟑, 𝒓 = 𝟎, 𝟏, 𝟐
• 𝛼0 =
2 0 +1 180°
3−0
=
(1)180°
3
=
180°
3
= 60°
• 𝛼1 =
2 1 +1 180°
3−0
=
(3)180°
3
= 180°
• 𝛼2 =
2 2 +1 180°
3−0
=
(5)180°
3
= 300° = −60°
𝝈 =
σ𝒊
𝒏
𝑹𝒆 𝒑𝒊 − σ𝒋
𝒎
𝑹𝒆 𝒛𝒋
𝒏 − 𝒎
• 𝒔𝟏 = 𝟎, 𝒔𝟐 = −𝟒, 𝒔𝟑 = −𝟓
•
𝝈 =
(𝟎 + −𝟒 + (−𝟓)) − (𝟎)
𝟑
= −𝟑
𝜶𝟎
𝝈 𝜶𝟐
𝜶𝟏
𝒔
𝑮𝑯(𝒔)
−∞, −5 ∪ [−4,0)
𝑮𝑯(𝒔)
𝑺𝟏
∠𝐺𝐻(𝑠) = ෍ 𝜑𝑝 − ෍ 𝜑𝑧 = 180𝑜(2𝑗 + 1)
𝜙𝑝1
+ 𝜙𝑝2
+ 𝜙𝑝3
= 180°
180° + 𝜙𝑝2
+ 0° = 180°
⟹ 𝜙𝑝2
= 0°
𝜙𝑝1
𝜙𝑝3
𝒔 𝒋𝝎
𝒔 → 𝒋𝝎 𝜔 𝑲
1 + 𝐺(𝑠)𝐻(𝑠) = 𝑠3
+ 9𝑠2
+ 20𝑠 + 𝐾 = 0
(𝑗𝜔)3
+ 9(𝑗𝜔)2
+ 20(𝑗𝜔) + 𝐾 = 0
−𝑗𝜔3 − 9𝜔2 + 𝑗20𝜔 + 𝐾 = 0
ℝ𝑒: −9𝜔2
+ 𝐾 = 0
𝕀𝑚: −𝜔3
+ 20𝜔 = 0
⇒ 𝜔 = 20
⟹ 𝐾 = 180
𝐾 = 180
𝑠3
+ 9𝑠2
+ 20𝑠 + 𝐾 = 0
𝑠3 + 9𝑠2 + 20𝑠 + 180 = 0
𝑠1 = −9, 𝑠2 = 𝑖 20, 𝑠3 = −𝑖 20
𝑠2 𝑠3
±𝑖 20
𝒅𝑲
𝒅𝒔
= 𝟎
𝑲
𝐾 = −𝑠3
− 9𝑠2
− 20𝑠
𝑑𝐾
𝑑𝑠
= −3𝑠2 − 18𝑠 − 20 = 0
−3𝑠2 − 18𝑠 − 20 = 0
𝑠1 = −4.5275 𝑠2 = −1.4724
(−∞, −5) ∪ (−4,0)
𝑠2 = −1.4724
Paso 1
Paso 2
hay
no
Paso 3
Paso 4
3
=
N

= 60
1
 
=180
2


−
= 60
3

Paso 5
3
−
=

Paso 6
3
−
3
−
Paso 7

=180
0
 
= 0
4


=180
5

Paso 8
20
=

180
=
K
𝑗 20
−𝑗 20
Paso 9
4724
.
1
−
=
s
Este es el lugar de las raíces
del sistema.
𝒛
• 𝒛
𝒛
𝒛
• 
 𝒛
•
𝒛
•
𝒛
𝒛
𝒛
ℝ𝑒
𝕀𝑚
1
(𝟏 + 𝑮(𝒛)𝑯(𝒛) = 𝟎)
(𝑲 = 𝟎) 𝑮(𝒛)𝑯(𝒛)
(𝑲 = ∞)
(𝑮 𝒛 𝑯(𝒛))
(𝒏) (𝒎)
(𝒏 − 𝒎)
𝜶𝒓 =
𝟐𝒓 + 𝟏 𝟏𝟖𝟎°
𝒏 − 𝒎
; 𝒓 = 𝟎, 𝟏, … , 𝒏 − 𝒎 − 𝟏
𝝈 =
σ𝒊
𝒏
𝑹𝒆 𝒑𝒊 − σ𝒋
𝒎
𝑹𝒆 𝒛𝒋
𝒏 − 𝒎
𝝀
±90º
𝑙𝜙𝑝 = 2𝑟 + 1 𝜋 + 𝜙; 𝑟 = 0,1,2, …
𝜙
𝜙 = σ 𝜃𝑧 − σ 𝜃𝑝
𝑙𝜙𝑧 = 2𝑟 + 1 𝜋 − 𝜙; 𝑟 = 0,1,2, …
𝜙
𝜙 = σ 𝜃𝑧 − σ 𝜃𝑝
𝒛 𝒋𝝎
𝝎 𝑲
𝒅𝑲
𝒅𝒛
= 𝟎
𝑲 = −
ς𝒊
𝒏
(𝒛 − 𝒑𝒊)
ς𝒋
𝒎
(𝒛 − 𝒛𝒋)
𝑲
𝑲 = −
ς𝒊
𝒏
𝒛 − 𝒑𝒊
ς𝒋
𝒎
𝒛 − 𝒛𝒋
𝑮(𝒛)𝑯(𝒛) =
𝒛 − 𝟏
𝒛𝟐 − 𝟑𝒛 + 𝟓
(𝑲 = 𝟎)
𝑮(𝒛)𝑯(𝒛)
𝒔
𝑧2
− 3𝑧 + 5 = 0
(𝑲 = 𝟎)
𝑮 𝒛 𝑯 𝒛
𝑝1 =
3
2
+
1
2
𝑖 11
𝑝2 =
3
2
−
1
2
𝑖 11
𝑛 = 2
(𝑲 = ∞)
𝑮(𝒛)𝑯(𝒛)
𝒔
𝑧 − 1 = 0
𝑧1 = 1
𝑚 = 1
𝑵 = 𝒏 − 𝒎
𝒏 = 𝑮(𝒛)𝑯 𝒛
𝒎 = 𝑮(𝒛)𝑯 𝒛
𝑵 =
𝒏 = 𝟐 𝒎 = 𝟏
𝑁 = 2 − 1
𝑁 = 1
𝜶𝒓 =
𝟐𝒓 + 𝟏 𝟏𝟖𝟎°
𝒏 − 𝒎
; 𝒓 = 𝟎, 𝟏, … , 𝒏 − 𝒎 − 𝟏
𝑁 = 1, 𝑟 = 0
𝛼0 =
2 0 + 1 180°
2 − 1
=
(1)180°
1
= 180°
𝝈 =
σ𝒊
𝒏
𝑹𝒆 𝒑𝒊 − σ𝒋
𝒎
𝑹𝒆 𝒛𝒋
𝒏 − 𝒎
𝑝1 =
3
2
+
1
2
𝑖 11 𝑝2 =
3
2
−
1
2
𝑖 11
𝑧1 = 1
𝜎 =
3
2
+
3
2
− (1)
1
= 2
𝒛
𝑮(𝒛)𝑯(𝒛)
(−∞, 1)
ℝ𝑒
𝕀𝑚
𝑮(𝒛)𝑯(𝒛)
𝒁𝟏
∠𝐺𝐻(𝑠) = ෍ 𝜙𝑧 − ෍ 𝜙𝑝 = 180𝑜
(2𝑗 + 1)
ℝ𝑒
𝕀𝑚
𝝓𝒑𝟐
𝝓𝒛𝟏
𝝓𝒑𝟏
𝒙 = 𝟏. 𝟓 − 𝟏 = 𝟎. 𝟓
𝒚 = 𝕀𝒎 𝒑𝟏
𝒚 =
1
2
11
𝜙𝑝2
= 90°
𝜙𝑧1
= tan−1
𝑦
𝑥
𝜙𝑧1
= tan−1
1
2
11
0.5
= 73.221°
𝜙𝑧1
− 𝜙𝑝1
− 𝜙𝑝2
= 180°
𝜙𝑝1
= 𝜙𝑧1
− 𝜙𝑝2
− 180°
𝜙𝑝1
= 73.221° − 90° − 180°
𝜙𝑝1
= −196.78°
𝒛 𝒋𝝎
𝒛 → 𝒋𝝎 𝝎 𝑲
1 + 𝐾𝐺 𝑧 𝐻 𝑧 = 1 + 𝐾
𝑧 − 1
𝑧2 − 3𝑧 + 5
= 0
𝑧2 − 3𝑧 + 5 + 𝐾 𝑧 − 1 = 0
𝑧2 + 𝐾 − 3 𝑧 + 5 − 𝐾 = 0
𝒛 = 𝒋𝝎
(𝑗𝜔)2+ 𝐾 − 3 𝑗𝜔 + 5 − 𝐾 = 0
−𝜔2 + 𝐾 − 3 𝑗𝜔 + 5 − 𝐾 = 0
ℝ𝑒: −𝜔2 + 5 − 𝐾 = 0
𝕀𝑚: 𝐾 − 3 𝜔 = 0
𝐾 = 3 𝐾
−𝜔2
+ 5 − (3) = 0
⟹ 𝜔 = ± 2
𝑲
𝑧2 + 𝐾 − 3 𝑧 + 5 − 𝐾 = 0
𝑧2 + 3 − 3 𝑧 + 5 − (3) = 0
𝑧2
+ 2 = 0
⟹ 𝜔 = ±𝑗 2
±𝒋 𝟐
𝑲 = 𝟑.
𝒅𝑲
𝒅𝒛
= 𝟎
𝑲
1 + 𝐾
𝑧 − 1
𝑧2 − 3𝑧 + 5
= 0
⟹ 𝐾 = −
𝑧2 − 3𝑧 + 5
𝑧 − 1
𝒛
𝑑𝐾
𝑑𝑧
=
𝑑
𝑑𝑧
−
𝑧2 − 3𝑧 + 5
𝑧 − 1
𝑑𝐾
𝑑𝑧
= −
2𝑧 − 3 𝑧 − 1 − (𝑧2 − 3𝑧 + 5)(1)
(𝑧 − 1)2
𝑑𝐾
𝑑𝑧
= −
2𝑧2 − 5𝑧 + 3 − 𝑧2 + 3𝑧 − 5
𝑧 − 1 2
𝑑𝐾
𝑑𝑧
= −
(2 − 1)𝑠2
+ −5 + 3 𝑧 + 3 − 5
(𝑧 − 1)2
−
𝑧2 − 2𝑧 − 2
𝑧 − 1 2
= 0
𝑧2 − 2𝑧 − 2 = 0
⟹ 𝑧1 = 2.732
⟹ 𝑧2 = −0.732
−∞, 1
𝒛𝟐 = −𝟎. 𝟕𝟑𝟐
ℝ𝑒
𝕀𝑚
•
𝑝1 =
3
2
+
1
2
𝑖 11
𝑝2 =
3
2
−
1
2
𝑖 11
•
𝑧1 = 1
•
𝑁 = 1
•
𝛼0 = 180°
•
𝜎 = 2
𝜶𝟎
𝟐
𝒑𝟏
𝒑𝟐
𝒛𝟏
•
(−∞, 1)
•
𝜙𝑝1
= −196.78°
𝜙𝑝2
= 196.78°
•
𝐾 = 3
𝜔 = ±𝑗 2
•
𝑧2 = −0.732
ℝ𝑒
𝕀𝑚
2
𝒑𝟏
𝒑𝟐
𝒛𝟏
−0.732
𝝓𝒑𝟐
𝝓𝒑𝟏
+ 2
− 2
09.sdcd_lugar_geometrico_raices
09.sdcd_lugar_geometrico_raices

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