transient response of system without disturbance for ramp input
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Transient Response of system without disturbance for ramp input (Analytical Method )
Closed Loop Transfer Function,C(s )R ( s )
= G1G2(s)
1+G1G
2(s)
SupposeG
1( s )=
280(2 )( s+10 )
= 560
(s+10) andG
2( s )= 1
(s+0.2)(s+4 ) ;
C(s)R (s )
=
560
(s+10)(s+0.2)(s+4 )
1+ 560
(s+10)( s+0.2)( s+4)
C(s)= 560R (s )
s3+14.2 s2+42.8 s+568
For nit Ramp !nput
Apply nit Ramp !nput, R(s) "
1
s2 #
C(s) =560
s2(s3+14.2 s2+42.8 s+568)
=560
s2 ( s+14.034 )( s2+0.166 s+40.47)
Let C(s) =
560
s2 ( s+14.034 )(s2+0.166 s+40.47 ) =
A
s2+B
s+
C
s+14.034+
Ds+E
s2+0.166s+40.47
560=A ( s+14.034 )( s2+0.166s+40.47)+B(s)( s+14.034)( s2+0.166s+40.47)+C( s2 ) ( s2+0.166 s+40.47)+(
560=A [s3+0.166 s2+40.47 s+14.034 s2+2.329 s+568]+B[s4+14.2s3+42.8 s2+568 s ]+C[s4+0.166 s3+40.
s
[3+14.034 s2
]560=A [s3+14.2 s2+42.8 s+568 ]+B [s4+14.2 s3+42.8 s2+568 s]+C[s4+0.166 s3+40.47 s2]+D[s4+14.034 s3
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560=s4 (B+C+D )+s3 (A+14.2B+0.166C+14.034D+E )+s2 (14.2A+42.8B+40.47C+14.034E )+s (42.8
s$# 0 = B + C + D % (a)
s 0 = A+14.2B+0.166C+14.034D+E % (b)
s'# 0 = 14.2A+42.8B+40.47C+14.034E % (c)
s# 0 = 42.8A+568 B % (d)
s# 560 = 568A % (e)
From (a),
A =560
568 =0.9859 - (f)
Substitute (f) into (d),
568B = - (42.8 0.9859)
B =42.196
568 = - 0.07428 - (g)
Substitute (*) into (a),
C + D = 0.07428
C = 0.07428 D % (h)
Substitute (f), (*), (h) into (c),
" 14.2(0.9859) + 42.8(-0.07428) +40.47(0.07428-D) + 14.034E
14.034E = -13.987+3.179-3.00611+40.47D
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E =13.814+40.47D
14.034 = -0.9843 +2.883D % (i)
Substitute (f), (*), (h), and (i) into (b),
0 = (0.9859) + 14.2(-0.07428) +0.166(0.07428-D) +14.034D + (-
0.9843 +2.883D)
0.9859-1.0547+0.0123-0.166D+14.034D-0.9843+2.883D= 0
16.751D =1.0408
D = 1.040816.751 = 0.0621 % (+)
Substitute (+) into (h),
C = 0.07428 0.0621 = 0.0121 % ()
Substitute (+) into (i),
E = -0.9843 +2.883(0.0621) = -0.8052 % (l)
Thus, substitute the -alue of A as in (f), . as in (*), C as in (), / as in (+) and 0 as in (i) into
C(s) e1uation,
C(s) =0.9859
s2 +
0.07428s
+ 0.0121
s+14.034+ 0.0621 s0.8052
s2+0.166s+40.47
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C(s) =
0.083
s+
2+6.362
0.083
s+
0.9859
s2 +
0.07428s
+ 0.0121
s+14.034+0.0621(s+0.083)
Apply in-erse laplace transform to *et time%domain e1uation,
-1C(s) =
-1!
0.083
s+
2+6.362
0.083
s+
0.9859
s2 +
0.07428s
+ 0.0121
s+14.034+0.0621
(s+0.083)
"
C(t) " 23453t % 26$'47 2''e%$2&$t7 28'e%24&tcos82&8t 9 2'6$ e%24&tsin82&8t
Transient Response of system without disturbance for ramp input (MATLA. Analysis)
Commands:
;;s " tf(
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;;sys" tf(num,den)
;;step (sys)
;;*rid
;;=r,p,>"residue(num,den)
MATLA. responds with
Transfer function:
s
num "
58
den "
2 $2' $'24 5842
Transfer function:
58
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
s?5 7 $2' s?$ 7 $'24 s?& 7 584 s?'
r "
2'
2& 7 28&6i
2& % 28&6i
%26$&
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23453
p "
%$2&$'
%24'3 7 82&8&i
%24'3 % 82&8&i
"
=>
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Transient Response of System with Step /isturbance for ramp input (Analytical Method)
As in pre-iously stated that,
C( s)=
G2(s )G
1(s )R(s)
1+G2(s )G1(s)
G2( s)D(s)
1+G2(s)G1(s )
SupposeG
1( s )=
560
(s+10) andG
2( s)=
1
( s+0.2 )(s+4) #
C( s)=
560R (s)( s+10) (s+0.2)(s+4)
1+ 560
(s+10) ( s+0.2)(s+4)
D(s)(s+0.2 )(s+4 )
1+ 560
( s+10) (s+0.2)(s+4)
C( s)=
560R(s)
s3+14.2 s2+42.8 s2+568
(s+10)D(s)
s3+14.2 s2+42.8 s2+568
For nit Ramp !nput
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Apply nit Ramp !nput,R (s )=
1
s2 and Step /isturbance
D (s )=1
s #
C( s )= 560
(
1
s2 )s3+14.2 s2+42.8 s+568
(s+10
)(
1
s )s3+14.2 s2+42.8 s+568
C( s )=
s
s2(3+14.2 s2+42.8 s+568)
560s210 s
"
560s210 ss2(s+14.034)(s2+0.166s+40.47)
LetC( s )=
A
s2+
B
s+
C
s+14.034+
Ds+E
( s2+14.2 s+40.47 )
560s210 s
s2(s+14.034)(s2+0.166s+40.47) =
A
s2+
B
s+
C
s+14.034+
Ds+E
s2+0.166s+40.47
560s210s=A (s+14.034 )( s2+0.166 s+40.47)+B(s)(s+14.034)( s2+0.166 s+40.47)+C(s2 ) (s2+0.166 s+
s3+14.034 s2
560s210s=A[ s3+0.166 s2+40.47 s+14.034 s2+2.329s+568]+B [s4+14.2 s3+42.8 s2+568 s]+C[s4+0.1
s
[ 3+14.034 s2
]560s210s=A[ s3+14.2 s2+42.8s+568]+B[s4+14.2s3+42.8 s2+568s ]+C[s4+0.166 s3+40.47 s2]+D [s4
560s210s=s4 ( B+C+D )+s3 (A+14.2B+0.166C+14.034D+E )+s2 (14.2A+42.8 B+40.47C+14.034E )
s$# 0 = B + C + D - (a)
s3; 0 = A+14.2B+0.166C+14.034D+E - (#)
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s2; -1 = 14.2A+42.8B+40.47C+14.034E - ($)
s1; -10 = 42.8A+568B - (d)
s0; 560 = 568A - (%)
From (a),
A"560
568 " 23453 % (f)
Substitute (f) into (d),
568B = -10 - (42.8 0.9859)
B =52.1965
568 = - 0.09189 - (g)
Substitute (*) into (a),
C + D = 0.09189
C = 0.09189 D % (h)
Substitute (f), (*), (h) into (c),
-1 = 14.2(0.9859) + 42.8(-0.09189) +40.47(0.09189-D) + 14.034E
14.034E = (-1-13.7856) +40.47D
E = 14.7856
+40.47
D14.034 = -1.05356 +2.883D % (i)
Substitute (f), (*), (h), and (i) into (b),
" (23453) 7 $2'(%2343) 7288(2343%/) +14.034D 7 (%25&58 7'244&/)
23453%2&$4725'5%288/7$2&$/%25&587'244&/"
8265/ "2&56'
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/ "1.3572
16.751 " 24 % (+)
Substitute (+) into (h),
C = 0.09189 0.081= 0.01089 % ()
Substitute (+) into (i),
E = -1.05356 +2.883(0.081) = -0.82 % (l)
Thus, substitute the -alue of A as in (f), . as in (*), C as in (), / as in (+) and 0 as in (l) into
C(s) e1uation,
C(s) =0.9859
s2 +
0.09189s
+ 0.01089
s+14.034+
0.081 s0.82
(s2+0.166s+40.47 )
C(s) =
0.083
s+
2
+6.36
2
0.083
s+
0.9859
s2 +
0.09189s
+ 0.1089
s+14.034+0.081(s+0.083)
Apply in-erse laplace transform to *et time%domain e1uation,
-1C(s) =
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-1!
0.083
s+
2+6.362
0.083
s+
0.9859
s2 +
0.09189s
+ 0.01089
s+14.034+0.081
(s+0.083)
"
C(t) " 23453t % 23437 243e%$2&$t7 24e%24&tcos82&8t 9 2&e%24&tsin82&8t
Transient Response for system without disturbance for the ramp input throu*h simulin
(MATLA. Analysis)
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Transient Response for system with disturbance for the ramp input throu*h simulin (MATLA.
Analysis)
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Transient Response for system with disturbance for the ramp input and ramp reference throu*h
simulin (MATLA. Analysis)
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2.4 Root Locus using both the asymptotic and MATLAB
2.4.1 Root Locus using asymptotic method
i2 @pen Loop Transfer Function (@LTF),
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G ( s )=G1( s) G
2(s )
G ( s )=280 2
s+10
1
( s+0.2 ) (s+4 )
G ( s )= 560
(s+10)(s+0.2)(s+4 )
From @pen Loop Transfer Function (@LTF),
umber of open loop poles, n " %, %2', %$
umber of open loop Beroes, m " #
ii2 Root locus on real ais is at between %2' and %$ and between % to %D2
iii2 To find Asymptotes,
A= Pi Zi
(Poles )(Zeroes )
(10 )+(0.2)+(4 )
30
(14.2
0
)3
4.733
A= (2k+1)
(Poles )(Zeroes) ,
= 60,180 ,300 ,,k =0,1,2,
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i-2 To find breaaway point,
.y applyin* the properties of root locus which need to satisfy the condition as followed2
1+G ( s )H(s )=0 (s+10 ) (s+0.2) ( s+4 )+K=0
s
(3+4 s2+0.2 s2+0.8 s+10 s2+40 s+2 s+8)K=
K=( s3+14.2s2+42.8 s+8)
dKds=(
3 s2+28.4 s+42.8)
LetdK
ds=0,
(3 s2+28.4 s+42.8 )=0 s=1.881,7.586
Thusbreaaway" %244
Kbreakaway=
28.92
-2 !ntersection with +E ais,
s3+14.2 s2+42.8 s+8+K=0 &%' s=*
(!)3+14.2(!)2+42.8!+8+K=0
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!3+42.8 (! )=0 ! (!242.8 )=0
!=" 6.542rad s1
Substitute E,
K#+814.2!2=0 K#=599.728
-i2 The root locus plot is shown as raph 2
2.4.2 Root Locus using MATLAB
Commands:
;;num " =58>
;;den " = $2' $'24 4>
;;sys " tf(num,den)
;;rlocus(sys)
;;*rid
;;damp(sys)
;;=Gn,Beta,H> " damp(sys)
MATLA. responds with
num "
58
den "
2 $2' $'24 42
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Transfer function:
58
%%%%%%%%%%%%%%%%%%%%%%%%%%%
s?& 7 $2' s?' 7 $'24 s 7 4
0i*en-alue /ampin* Fre12 (radIs)
%'2e% 2e7 '2e%
%$2e7 2e7 $2e7
%2e7 2e7 2e7
Gn "
2'
$2
2
Beta "
H "
%2'
%$2
%2
And the root locus plot -ia MATLA. is shown in Fi*ure 2
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Fi*ure : Root Locus Hlot
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280(s)
D(s)
C(s)
2.5 Frequencyresponse method!Bode p"ots using both asymptote and MATLAB# and
$icho"%s chart
2.5.1 Frequencyresponse method !Bode p"ots using asymptote method#
- -
+ - +
@LTF, G ( s )=
C(s)R(s) =
280(2)(s+10)(s+0.2)(s+4 )
=
560
8( s
10+1)(
s
0.2+1)(
s
4+1)
G (! )= 70
(!
10+1)(
!
0.2+1)(
!
4+1)
20 ,g 70 = 36.9dB 37dB
/%
agn'd% as%
C ,%:,
%&C ,%
:,%
C ,%:,
%
(1+ !
0.2)1
0.2-
20dB-20dB
0.02
-45< -45< 2 45< -90
;;sys " tf(num,den)
;;bode(sys)
;;*rid
;;sys " allmar*in(sys)
MATLA. responds with
num "
58
den "
2 $2' $'24 42
Transfer function:
58
%%%%%%%%%%%%%%%%%%%%%%%%%%%s?& 7 $2' s?' 7 $'24 s 7 4
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sys "
ainMar*in: 26
MFre1uency: 825$'5
HhaseMar*in: 24'$4
HMFre1uency: 82&'&
/elayMar*in: 25
/MFre1uency: 82&'&
Stable:
and .ode plot -ia MATLA. shown in Fi*ure '2
Fi*ure ': @pen Loop .ode /ia*ram
Closed Loop .ode /ia*ram
Commands:
;;num " =58>#
;;den " = $2' $'24 584>#
;;sysp " tf(num,den)
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;;sys " feedbac(sysp,)#
;; w " lo*space(%,')#
;; bode (sys,w)#
;; =ma*, phase, w> " bode (sys, w)#
;; =Mp, > " ma (ma*)#
;;bandwidth " bandwidth(sys)
MATLA. responds with
Transfer function:
8
%%%%%%%%%%%%%%%%%%%%%%%%%%%
s?& 7 $2' s?' 7 $'24 s 7 84
bandwidth "
'2&68
and closed loop bode dia*ram -ia MATLA. is shown as Fi*ure &2
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Fi*ure &: Closed Loop .ode /ia*ram
From Fi*ure &, it shows that the closed loop bode dia*ram is an unstable system with resonant
pea of 524'd. and resonant fre1uency of 623radIsec2
Hroblem%bases learnin* (H.L) is a student%centered education whereby students learn a
sub+ect due to the problem sol-in*2 Thus, e-ery student had trained thinin* strate*ies and
domain nowled*e2 The main ob+ecti-e of H.L is to de-elop fleible nowled*e, problem
sol-in* sill, self learnin* and self moti-ation2 Therefore, H.L can be considered acti-e learnin*
in education as well2
A MATLA. *roup assi*nment had been *i-en with applied H.L concept2 Therefore, afew method had been carried out in order to sol-e the problem2 Such as, transient response, error
analysis, root locus, Fre1uency%response method and ichols Chart2
!n *eneral, the transient response is the response of the system that able to chan*e from
e1uilibrium2 !tKs also used to calculate response of a structure to time dependent loads2 sually
the application are structures sub+ect to earth1uaes, wind, eplosions and as well as pothole2
Transient response is in-ol-ed rise time, o-ershoot, settlin* time, delay time, pea time and as
well steady state error.
0rror analysis is one of the alternati-e methods of H.L2 This type of analysis isconcerned about the output of the model as the parameters to the model especially in numerical
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simulation or modelin* of real system as well2 Moreo-er, error analysis can be separate into
forward error analysis and bacward error analysis in numerical analysis2
Root Locus is one of the common techni1ue for desi*n criteria, this method is in-ol-ed
iteratin* on a desi*n by manipulatin* to the compensator *ain, poles and Beros in root locus
dia*ram2 sually, this method is used to tune the *ain of sin*le input and sin*le output (S!S@)
control system with specifyin* a desi*ned set of closed%loop pole location2
Fre1uency response method can be eplain as the steady state response of the system to a
sinusoidal input si*nal which in-ol-ed a uni1ue input si*nal, resultin* output si*nal for linear
system and si*nal from the system2 The fre1uency%response function usually is used to describe
the amplitude chan*e and also the phase shift as a function of fre1uency2
Lastly, ichols chart is one of the *raphical methods for fre1uency response2 !ts display
the ma*nitude in term of d. plotted a*ainst the phase in term of de*rees of the system response2
Furthermore, it useful to analyBe open loop and closed loop system for S!S@2
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