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Ptc.Controls.Worksheet.Printing.EngineeringPages+EngPagesPaginator 12) PV/R = 0.68/(1 + 1.55s) where R = 1/s for unit step input PV = 0.68/(1 +1.55s)(1/s) PV = 0.68/[s(1 +1.55s)] Taking the inv...

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Ptc.Controls.Worksheet.Printing.EngineeringPages+EngPagesPaginato
12)
PV/R = 0.68/(1 + 1.55s) where R = 1/s for unit step input
PV = 0.68/(1 +1.55s)(1/s)
PV = 0.68/[s(1 +1.55s)]
Taking the inv Laplace transform
pv(t) = XXXXXXXXXXe^-0.645t) (Time domain response)
Using the final value theorem:
Value = = 0.68 (Final value)
13)
Given transfer function: 3.16/[s(1+2s)]
From calculations and graphing (see next page) we get:
Phase margin = 22°
Gain margin = 24dB
Stable = Yes
Frequency (MHz)
S
(d
B
)
-80
-60
-40
-20
0
20
40
60
80
ang( S[2,1] ) (deg)
-180
-135
-90
-45
0
45
90
135
180
Frequency (MHz)
XXXXXXXXXX
Gain vs Phase
S ang( S[2,1] )
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D
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D
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D
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D
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D
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D
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1)20log 3.16
D
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D
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2) - 20dB/decade
D
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3)-20dB/decade
D
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D
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D
Line
D
Line
D
Line
D
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W0dB
D
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D
Line
D
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D
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D
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W-180°
D
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D
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D
Line
D
Line
D
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24dB
D
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D
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D
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D
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D
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D
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2)-90°
D
Line
D
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1) 0°
D
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-135°
D
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-180°
D
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CF = 0.5
s
D
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D
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D
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D
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GH = 3.16/ [(1 + 2s)s]
Gjw = 3.16/ [ jw(1 + 2jw)]
= 3.16/ [jw(1 + jw/0.5)]
Start Freq (SF) = 0.5/10 = 0.05
s ; Entry Point = 20 log 3.16/0.05 = 36dB
Entry slope -20dB/decade due to s term in denominator.
Corner Freq (CF) = 0.5
s; slope after CF -40dB/decade
Entry phase is -90° due to s term in denominator.
At CF, phase shift is -90° - 45° = -135°; At 10*CF, Phase shift = -180°

D
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Rad/s
D
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D
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(10dB)
D
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D
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D
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D
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Calculations:
D
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Frequency (MHz)
S
(d
B
)
-80
-60
-40
-20
0
20
40
60
80
ang( S[2,1] ) (deg)
-180
-135
-90
-45
0
45
90
135
180
Frequency (MHz)
XXXXXXXXXX
Gain vs Phase
S ang( S[2,1] )
D
Line
D
Text Box
SF=0.05
s
D
Line
D
Text Box
36dB
D
Line
D
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CF=0.5
s
D
Line
D
Text Box
16dB
D
Line
D
Line
D
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5
s
D
Line
D
Line
D
Line
D
Line
D
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D
Text Box
-24dB
D
Line
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w0 = 1.2
s
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Line
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-158°
D
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0dB
D
Line
D
Line
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24 dB
D
Line
D
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Phase margin = -180° XXXXXXXXXX°) = 22°
Gain Margin = 24dB
System is stable because at -180° phase shift, gain = -24dB
D
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Rad/s
D
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SP = 36dB
Answered Same Day Dec 21, 2021

Solution

Robert answered on Dec 21 2021
130 Votes
Solution 1:
Load or distu
ance is the un desired effect produced in the process which is measured at output and
then is fed back to the comparator. The SET POINT is the input to the process which acts as a reference.
Solution 8:
The system is stable. Hence the solution is inco
ect:
Since at phase shift of +180 which is equivalent to <-180 the gain is +10 dB given means the gain margin
at frequency 1 rad/sec
Secondly at gain 0 dB it has been given that...
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