Transient Response of System Without Disturbance for Ramp Input

8/21/2019 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)


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)


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)


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

Transient Response of System Without Disturbance for Ramp Input

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