the transfer function  (s). (dB) |^v | the magnitude and phase Bode plots shown below. Find an expression for (6) 0 40 30 20 10 0 -10 -20 -30 -40 10 -1 315 270 225 180 135 90 45 0 -45 10 -1 ww டட…l 10⁰ 10⁰ www 10 ¹ 10 ¹ 1 wwwww 10 10 2 f (Hz) 2 f (Hz) www 10 !!!! 10 3 3 www 10 لند 10 4 4 www ******* ---ml 10 TTTTT 10 5 5 10 6 10 6

Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Step 1: step 1
Initial slope.
0 dB/decade
=
after 10 HZ slope will change
→ Step 2: step 2
છ તક
then
to 10000 Hz
-4ø d
decade
Solution
o dB
decade
again slope will
de
20 dB
decade
o de
decade
So slope will be.
slope will change after 1000Hz to
So stope will be
between
decade
20 Transfer function
T
80 TF will be
20
0 TF =
0
-20
G (s) =
100H to 1000 Hz
given that
at 104 HZ
ode
decade
-40dB
+
1+
dicade
(
.
10'
ولد
+ S
I + S
Wy
до
magnitude
10
at 10²¹ HZ
K
10
Oda
to ods
decade
20 de to 20 de
-40 dB
decade
J
no pole at origin
2
⇒2 pole at w
=
lo
is ode
20loglok
20 de
20 dB
dicale
decade
104
dope will be o de
after 104 112
decade
o de
20
-20 to 6 de up
decade
[+] ² [ + ]]
[**($)]*
[(4)
+
phase will be
2 Zeros at W₂
I zero at or wz
I pole at
Wy
[¹* ² [+
8x8.14x lo
up
K = 10'
K = 10
["(usa)] [11
2x3.14x100
to 100Hz
90⁰ and
+ S
decade
f
→ (Hz)
S
2x 8.14 x 1000
2x3.14x 10000
Transcribed Image Text:Step 1: step 1 Initial slope. 0 dB/decade = after 10 HZ slope will change → Step 2: step 2 છ તક then to 10000 Hz -4ø d decade Solution o dB decade again slope will de 20 dB decade o de decade So slope will be. slope will change after 1000Hz to So stope will be between decade 20 Transfer function T 80 TF will be 20 0 TF = 0 -20 G (s) = 100H to 1000 Hz given that at 104 HZ ode decade -40dB + 1+ dicade ( . 10' ولد + S I + S Wy до magnitude 10 at 10²¹ HZ K 10 Oda to ods decade 20 de to 20 de -40 dB decade J no pole at origin 2 ⇒2 pole at w = lo is ode 20loglok 20 de 20 dB dicale decade 104 dope will be o de after 104 112 decade o de 20 -20 to 6 de up decade [+] ² [ + ]] [**($)]* [(4) + phase will be 2 Zeros at W₂ I zero at or wz I pole at Wy [¹* ² [+ 8x8.14x lo up K = 10' K = 10 ["(usa)] [11 2x3.14x100 to 100Hz 90⁰ and + S decade f → (Hz) S 2x 8.14 x 1000 2x3.14x 10000
the transfer function Ã₁ (s).
(dB)
|^v |
the magnitude and phase Bode plots shown below. Find an expression for
(6) 0
40
30
20
10
0
-10
-20
-30
-40
10 -1
315
270
225
180
135
90
45
0
-45
10
-1
www
டட…l
10⁰
0
10 º
www
10 ¹
10 ¹
1
wwwww
10
10
2
f (Hz)
2
f (Hz)
www
10
!!!!
10
3
3
www
10
لند
10
4
4
www
*******
---ml
10
TTTTT
10
5
5
10
6
10 6
Transcribed Image Text:the transfer function Ã₁ (s). (dB) |^v | the magnitude and phase Bode plots shown below. Find an expression for (6) 0 40 30 20 10 0 -10 -20 -30 -40 10 -1 315 270 225 180 135 90 45 0 -45 10 -1 www டட…l 10⁰ 0 10 º www 10 ¹ 10 ¹ 1 wwwww 10 10 2 f (Hz) 2 f (Hz) www 10 !!!! 10 3 3 www 10 لند 10 4 4 www ******* ---ml 10 TTTTT 10 5 5 10 6 10 6
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