D = 20 log (K)+20 x (k2 - k₁) log (wa) = 20 log (0.8)-20 x 1 x log (0.1) = 18.06 dB draw bode diagram with very detiales pls slve fast Bode Diagram: Problem 3 (17+1-4) Step 1. H(jw) G (j) = 0.8 x -(+1) Step 2. Draw table as below Factor Type (Corner/Break) Slope frequency (jw)' pole -20 x 1 (Starting slope) zero √8=2.82 √10 = 3.16 (+1) (+1-pole +20 x 2 -20 x 2 Step 3. Starting frequency w = 0.1. Step 4. Starting magnitude = 18.06 dB, Step 5. Phase angle =-90 + tan-(+) w(rad/sec) (deg) 0.1 0.28 -tan-15 1-to 2.8 3.16 28.3 31.6 -89.8 -89.2 -70.5 -72-90 -90 Ju +20 3 ta

Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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Question
Draw bode diagram
D
= 20 log (K)+20 x (k2 - k₁) log (wa)
= 20 log (0.8)-20 x 1 x log (0.1) = 18.06 dB
draw bode
diagram with very
detiales
pls slve fast
Bode Diagram: Problem 3
(17+1-4)
Step 1. H(jw) G (j) = 0.8 x
-(+1) Step 2. Draw table as below
Factor
Type
(Corner/Break)
Slope
frequency
(jw)'
pole
-20 x 1 (Starting slope)
zero
√8=2.82
√10 = 3.16
(+1)
(+1-pole
+20 x 2
-20 x 2
Step 3. Starting frequency w = 0.1. Step 4. Starting magnitude = 18.06 dB, Step 5. Phase angle
=-90 + tan-(+)
w(rad/sec)
(deg)
0.1 0.28
-tan-15
1-to
2.8 3.16 28.3 31.6
-89.8 -89.2 -70.5 -72-90
-90
Ju
+20
3
ta
Transcribed Image Text:D = 20 log (K)+20 x (k2 - k₁) log (wa) = 20 log (0.8)-20 x 1 x log (0.1) = 18.06 dB draw bode diagram with very detiales pls slve fast Bode Diagram: Problem 3 (17+1-4) Step 1. H(jw) G (j) = 0.8 x -(+1) Step 2. Draw table as below Factor Type (Corner/Break) Slope frequency (jw)' pole -20 x 1 (Starting slope) zero √8=2.82 √10 = 3.16 (+1) (+1-pole +20 x 2 -20 x 2 Step 3. Starting frequency w = 0.1. Step 4. Starting magnitude = 18.06 dB, Step 5. Phase angle =-90 + tan-(+) w(rad/sec) (deg) 0.1 0.28 -tan-15 1-to 2.8 3.16 28.3 31.6 -89.8 -89.2 -70.5 -72-90 -90 Ju +20 3 ta
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