G(i
—
10,
000(j’ +j+io,ooo) ^______
> (1) — ( i ' + 1 0 t + I O O ) i ( i + 5 ) — (j^ + IOj+ 1 0 ,0 0 0 ) 100' ^ 5 ' M0,0 00' '500(i’ +i+10,000)
' (*’ +15*’ + 150s + 5 0 0 )(j* + 10s +10,000)G(s) =
500(s’ +s+10,000)
(s^ + 15s"+150s+500)(s"+ 10s + 10,000)
(2) Step 2 of 4 ^In second order system it is easy to find the damping ratio using the classical method. However, it is not easy to follow the behavior of a higher order system by the same classical method. So evaluate the problem using MATLAB code.
Step 3 of 4
Refer Equation (2) and write the MATLAB program.
num=500*[1 1 10000];den =conv([1 15 150 500],[1 10 10000]);step{num,den) Figure 1 shows the output for the MATLAB program.
Figure 1
Step 4 of 4
Consider the values from Figure 1.
Maximum overshoot ( M , ) % Rise time ( t , ) sec Settling time 0 0.348 0.915 Table 1
Thus, the percentage maximum overshoot is , Rise time is |Q.348| and Settling time is 10.9151-
Problem 3.45PP I 'M ^ . 2 « (j) - - ( i2 + 2 i + 2)’ I'M 2(» + 3) *M ' 2(lZ + 2 l+ 2 ) ' i;m _ _ _________6 m (i + 3)(j2 + 2j + 2)'
In each case, provide estimates of the rise time, settling time, and percent overshoot to a unit- step input in r.
Three closed-loop transfer functions are given below:
i ^ t a n - K w - e t a n e n l i i t i n n
Step-by-step solution
Consider the closed loop transfer function.
« ( s ) (s* + 2j + 2)
^ ' (» = + 2j + 2 ) ' ' i? W = l
Consider the characteristic equation.
(2) Here,
^ is the damping ratio,
is the undamped natural frequency. Compare equation (1) and (2).
2 s * » , = 2 ... (3) = 2
(4)
Consider general formula for rise time. 1.8
— ... (5)
Here, is the rise time.
Substitute >12 for in equation (5). , _ l -8
' - - l i =1.272sec
Thus, the rise time / is 11.2721.
Consider general formula for settling time.
... (6)
Substitute ■ n for in equation (3). 2 f> j2 ~ 2
1 ^ ° l 2 = 0.707
Thus, the damping ratio g is 0.707.
Substitute 0.707 for g and ■Ji for in equation (6).
0.707xV2 =4.6sec
Thus, the settling time t, is m i
Consider general formula for overshoot.
Substitute 0.707 for f . -gxO.707
= 0.0432
Thus, the percentage overshoot is m u
Consider the closed loop transfer function.
d/..\
' ' 2 ( j ’ + 2 i + 2 ) ^ ^ S ( s ) = l
... , , , ' ' 2 ( j’ +2s + 2)
From equation (7), write the second order term. » * + 2 t + 2 = 0
Consider the characteristic equation. o>.
■(B) 5’ + 24»»,f + ffl’ Here,
^ is the damping ratio,
is the undamped natural frequency.
Compare equation (7) and (8).
2««». = 2 ... (9)
e i = 2r - ... (10) a , = >n
Consider general formula for rise time. 1.8
...(11) Here,
/^ is the rise time.
Substitute >12 for in equation (11). , _ l -8
' - - l 2 =1.272sec
Thus, the rise time /^ is I1.2721.
Consider general formula for settling time. ... (12)
Substitute ■ n for in equation (9). 2 f> j2 ~ 2
1 ^ ° l 2 = 0.707
Thus, the damping ratio g is 0.707.
Substitute 0.707 for g and ■Ji for in equation (12).
0.707xV2 =4.6sec
Thus, the settling time t, is E U
step 8 of 11 ^
Consider general formula for overshoot.
3 / , = . ^ Substitute 0.707 for g .
-gxO.707 ^ J i W
= 0.0432
Thus, the percentage overshoot is E m
J o f l l ^
In second order system it is easy to find the damping ratio using the classical method. However, it is not easy to follow the behavior of a third order system by the same classical method. So evaluate the problem using MATLAB code.
Consider the closed loop transfer function.
- 6 ( s + 3 ) ( s * + 2 s + 2 ) 6 Y { s ) ^ (j+ 3 ) (s^ + 2j+ 2 )R (s)
(13)
+ 3 )^ + 2j+ 2 JRefer equation (13) and write the MATLAB program. num=[6]:
den=conv([1 3], [1 2 2]); step(num, den), grid T = tf{num, den) damp(T)
Step 10 of 11
Figure 1 shows the output for the MATLAB program.
Figure 1
step 11 of 11
Consider the values from Figure 1.
Maximum overshoot ( M , ) % Rise time ( / , ) sec Settling time ( '. ) sec 3.65 1.69 4.54
Thus, the percentage maximum overshoot M^ \s |3.6Sl ■ Rise time /^ is and Settling time /. is 14.541
ransfer function(s) will meet an overshoot specitication of IWp < 5%?
:ransfer ftjnction(s) will meet a rise time specification of tr< 0.5 sec?
ransfer function(s) will meet a settling time specification of fs < 2.5 sec? 40 2 + 4i + 40)’ 40 + I)(j2 + 41 + 40)’ 120 -i-^tr.f2^4 v4.4 m ’ + l)(P + 4 i + 40)’ 120 + 3)(s? + 4 l+ 4 0 )’ 20(1 + 2) + l)(s? + 4 l+ 4 0 )’ 3«040/40I<j2 + 1+401) 2 + 4 l + 40)(l2 + i + 901)’ Step-by-step solution step 1 of 19 Step 2 of 19
le transfer function of the system. 4 0 (1) y ^ + 4 j+ 4 0 ) Step 3 of 19 re characteristic equation _£S_ + 2 f a ^ + a l (2) amping ratio
undamped natural frequency Equation (1) and (2). ...(3)
(4)
eneral formula for rise time. ....(5 ) 3e time. 6.32 for in Equation (5).
sec
isetime is [0 .2 8 4 1 Step 4 of 19eneral formula for settling time.
.... (6) 6.32 for in Equation (3). - 4 2 6.32 • 0.316 lamping ratio g is 0.316.
0.316 for g and 6.32 for in Equation (6). L6___
■x6.32
ec
lettlingtime is |2.3Q|
Step 5 of 19
eneral formula for overshoot.
0.316 for g . >^j« ruii* ; i 4 lercentage overshoot 3 /^ is 13541 Step 6 of 19
ition (1) and write the MATLAB program.
I 40]; den), grid
den)
□ obtains the step response plot
lows the output for the MATtAB program.
Step 7 of 19
le values from Figure 1.
overshoot Rise time Settling time ( ' / )
0.213 1.77
Id values are tabulated in table 2.
overshoot Rise time Settling time ( ' / )
<0.5 <2.5
from table 1 and 2. the following obsen/ations are made for the transfer ft quation (1).
rved K ) % is 35.1 but it should be less than 5%. Hence the equation Maximum peak overshoot condition.
rved is 0.213 sec but it should be less than 0.5 sec. Hence the equr d rise time condition.
rved J is 1.77 but it should be less than 2.5 sec. Hence the equation s sttling time condition.
Step 8 of 19
irder system it is easy to find the damping ratio using the classical metho ry to follow the behavior of a third order system by the same classical me le problem using MATtAB code.
Step 9 of 19
Step 10 of 19
le transfer function of the system. 40
t - l ) ( l * + 4 j + 4 0 ) (7)
Step 11 of 19
rtion (7) and write the MATtAB program.
'([1 1],[1 4 401); den), grid
den)
Step 12 of 19
le values from Figure 2.
overshoot Rise time Settling time ( / , ) sec ( '. ) sec
2.1 3.99
Id values are tabulated in table 4.
overshoot Rise time Settling time ( / , ) sec ( '. ) sec
<0.5 <2.5
from table 3 and 4. the following obsen/ations are made for the transfer ft quation (7).
rved K ) % is 0 but it should be less than 5%. Hence the equation sat aximum peak overshoot condition.
rved is 2.1 sec but it should be less than 0.5 sec. Hence the equatic required rise time condition.
rved J is 3.99 but it should be less than 2.5 sec . Hence the equation required Settling time condition.
Step 13 of 19
le transfer function of the system. 120
(8) l + 3 ) ( l ^ + 4 i + 4 0 )
Ition (8) and write the MATtAB program.
'([1 3],[1 4 40]); den), grid
den)
Step 14 of 19
le values from Figure 3.
overshoot Rise time Settling time ( ' / )
0.431 1.53
Id value is from the given condition is tabulated in table 6.
overshoot Rise time Settling time ( ' / )
<0.5 <2.5
from table 5 and 6. the following obsen/ations are made for the transfer ft quation (8).
rved K ) % is 0.869 but it should be less than 5% . Hence the equatio d Maximum peak overshoot condition.
rved is 0.431 sec but it should be less than 0.5 sec. Hence the equr d rise time condition.
rved is 1.53 but it should be less than 2.5 sec . Hence the equation sttling time condition.
Step 15 of 19
le transfer function of the system. 20(i + 2)
(9) j + l ) ( l ' + 4 i + 4 0 )
Ition (9) and write the MATtAB program.
01;
'(]1 1],[1 4 40]); , grid
den)
lows the output for the MATtAB program
Step 16 of 19
le values from Figure 4.
overshoot Rise time Settling time ( / , ) sec ( '. ) sec
1.42 3.32
Id value is from the given condition is tabulated in table 6.
overshoot Rise time Settling time ( ' / )
<0.5 <2.5
from table 7 and 8. the following obsen/ations are made for the transfer ft quation (9).
rved K ) % is 0 but it should be less than 5% . Hence the equation sa aximum peak overshoot condition.
rved is 1.42 sec but it should be less than 0.5 sec. Hence the equat e required rise time condition.
rved J is 3.32 but it should be less than 2.5 sec . Hence the equation e required Settling time condition.
Step 17 of 19
le transfer function of the system. 3 6 ,0 4 0 4 0 l ( i " + j + 4 0 l ) j ‘ + 4 i + 4 0 ) ( i ^ + s + 9 0 l ) 89.87 s‘ + 4 s + 4 0 ) ( i ’ + i + 9 0 l ) ( i ’ + 1 + 4 0 I ) 89.87 s‘ + 4 s + 4 0 ) ( j ’ + i + 9 0 l ) ( i ’ +1+ 4 0 l ) 89.87 j ’ + 4 i + 4 0 ) (j* + 2 s ’ + l,303s^ + l ,3 0 2 i + 3 6 l,3 0 l) 89.87 1* + 4 i + 4 0 ) ( i ' + 2 i ’ + 1303s’ + 1 3 0 2 1 + 361301) Ition (10) and write the MATtAB program.
(10)
'(]1 4 40],[1 2 1303 1302 361301]); den), grid
den)
lows the output for the MATtAB program
Step 18 of 19
le values from Figure 5.
overshoot Rise time Settling time ( / , ) sec ( '. ) sec
0.113 4.72
Id value is from the given condition is tabulated in table 6.
overshoot Rise time Settling time ( ' / )
<0.5 <2.5
from table 9 and 10. the following obsen/ations are made for the transfer quation (10).
rved K ) % is 46.8 but it should be less than 5%. Hence the equation required Maximum peak overshoot condition.
Step 19 of 19
red is 0.113 sec but it should be less than 0.5 sec. Hence the equati d rise time condition.
rved is 4.72 but it should be less than 2.5 sec. Hence the equation ( required Settling time condition.
r.?ir.1s-
(a) Sketch the unit-step responses for G1(sJ and G2(s^, paying close attention to the transient part of the response.
(b) Explain the difference in the behavior of the two responses as it relates to the zero locations.
(c) Consider a stable, strictly proper system (that is, m zeros and n poles, where m < n). Let y(t) denote the step response of the system. The step response is said to have an undershoot if it initially starts off in the “wrong” direction. Prove that a stable, strictly proper system has an undershoot if and only if its transfer function has an odd number of real RHP zeros.
(c) Consider a stable, strictly proper system (that is, m zeros and n poles, where m < n). Let y(t) denote the step response of the system. The step response is said to have an undershoot if it initially starts off in the “wrong” direction. Prove that a stable, strictly proper system has an undershoot if and only if its transfer function has an odd number of real RHP zeros.
Step-by-step solution
s t e p 1 of 18
(a)
The two non-minimum phase system are.
G , W =
( j + l ) ( i + 2)
’ (s + l ) ( i + 2 ) ( j + 3) The unit step response for Gy ( j ) i'
2 ( j - l ) - .
—r — ^ r for (7, ( e l in the above ( j + I ) ( * + 2 ) equation.
i ; w = i ( i + l ) ( l + 2)
Step 2 of 18
Apply partial fractions to simplify the function Yy ( ^ ) • ! ; ( , ) = _ + — . + — _ (1)
S 5+1 5+2
Use residue method to calculate coefficient x ■
_ 2 ( . - l )
I
• L (s + 1 ) (j+ 2 ) L 2 ( 0 - 0 ( 0 + l ) ( 0 + 2 ) ' 2 Step 3 of 18Use residue method to calculate coefficient g .
0 ( ^ + 2 ) J L .
_ i ( £ z O | 5(542(-<-0
- ( - l ) ( - l + 2 ) -1 Step 4 of 18Use residue method to calculate coefficient C- C = ( s * 2 ) Y , { s l ^ = ( * + 2 ) f 2 ( ^ - 0
*(*+oL
2( - 2 - 0 { - 2 ) ( - 2 + 0 ' 2 Step 5 of 18 Recall equation (1). \ A B C i ; ( j ) = - + — - + — - 5 5+1 5+2Substitute ( fo r X ’ - 4 for ^ ,a n d 3 for ^ in the above equation.
' ' 5 5+1 5 + 2 1 4 ^ 3 5 5+1 5+2 It is known that. i - ' i . i ‘ ■ ' f c ) " '
Apply Inverse Laplace transform. 3 ',(/) = l- 4 e ''+ 3 e '* '
Step 6 of 18
MATLAB code to sketch the unit step response for Gy (5) » t=0:0.1:10;
» yl=l-4.*exp(-t)+3.*exp(-2.*t); » plot(t,y1)
» title{'Step response for G1 (t)’) » xlabel('time(s)')
» ylabel('y1 (t)')
Step 7 of 18
The sketch of unit response for Gy (5) is shown in Figure 1. S t ^ respoase f o r G l ( t )
tim e (s ) Figure 1
Therefore, the sketch of unit response for Gy (5) is shown in Figure 1.
Step 8 of 18
The unit step response for <72(5) is.
(5 + I)(5 + 2)(5 + 3)for (72(5) in the above equation. r ( A - 2 ( » - I ) ( . - 2 )
' s ( j + l ) ( i + 2 ) ( * + 3 )
Apply partial fractions to simplify the function (5) . V / \ A B C D ,r>\ y , ( i ) = - + — - + — - + — - (2)
5 5+1 5 + 2 5+3
Step 9 of 18
Use residue method to calculate coefficient x ■
f l ) ( i + 2 ) ( j + 3 ) 2 ( » - l ) ( ^ - 2 ) (5 + 1){5 + 2)(5 + 3) 3 ( 0 - l ) ( 0 - 2 ) ( 0 + l) ( 0 + 2 ) ( 0 + 3 ) 3 ( - l ) ( - 2 ) ( 1 )( 2 ) (3 ) - I Step 10 of 18
Use residue method to calculate coefficient g .
= ( j + l ) 3 ( ^ - l ) ( » - 2 ) 5(5 + 1){5 + 2)(5 _ 3 M K f z 2 ) L . 1 5 (5 + 2 ) (5 + 3) 3 ( - 1 - 1 ) ( - 1 - 2 ) - ( - l ) ( - l + 2 ) ( - l + 3 ) - i i ‘ - 2 —9 s t e p 11 of 18
Use residue method to calculate coefficient C-
- ( . i J 1 3 ( ^ - l ) ( ^ - 2 ) | » + ') ( ^ + 3 ) L , 5(54 3 ( - 2 - l ) ( - 2 - 2 ) ( _ 2 ) ( - 2 + 1 ) ( - 2 + 3 ) _36 ' 2 - 1 8 Step 12 of 18
Use residue method to calculate coefficient / ) . f = ( * + 3 ) j ; M L _ - ( X I 3 ) [ 1 3 ( » - l ) ( ^ - 2 ) i ( i + l ) ( j + 2 ) [ _ j 3 ( - 3 - 1 ) ( - 3 - 2 ) ( _ 3 ) ( - 3 + 1 ) ( - 3 + 2 ) " - 6 — 10 Step 13 of 18 Recall equation (2). A B C D K ( 5 ) * — + ---+ --- + --- 5 5+1 5 + 2 5+3
Substitute ( fo r x ^ - 9 for q, 18 for -1 0 for C ill the above equation.
5 5+1 5 + 2 5+3 1 9 ^ 18 10 5 5 + 1 5 + 2 5 + 3 It is known that. i - ' i . i ‘ ■ ' f c ) " '
Apply Inverse Laplace transform. 3'2(/) = l - 9 e - ' + I85‘ *'-1 0 e **'
Step 14 of 18
MATLAB code to sketch the unit step response for G^ {5) ^ » t=0:0.1:10;
»y2=1-9.*exp(-t)+18.*exp{-2.*t)-10.*exp(-3.*t); » plot(t,y2)
» title{'Step response for G2(t)') » xlabel('time(s)')
» ylabel('y2(t)')
Step 15 of 18
The sketch of unit response for ^2(5) is shown in Figure 2. S t ^ respoase fo r G 2(t)
Find the relationships for the impulse response and the step response corresponding to Eq. for the cases where
(a) the roots are repeated.
(b) the roots are both real. Express your answers in terms of hyperbolic functions (sinh, cosh) to best show the properties of the system response.
(c) the value of the damping coefficient, Is negative.
H W = H W =_________£5_________ (i+ { <b,)2 + <o5 ( 1 - ! : 2 ) ' Step-by-step solution step 1 of 5 w
Li this case we have ^ = 1 ■ Find H(s).
m - U [ . )
( s + a i )
Step 2 of 5 ^ For the impiilse response U(s) = 1, we get k{t) = aj'te-"-'.
We can then integrate the in^ulse response to obtain the step response. Alternatively, for a unit step input, U ( s ) = —
By applying inverse Laplace we get
Step 3 of 5 (b) Fin<)H(s). ' U W H [ s ) : H {s ) = --- ^ --- - where |^] > 1
For the impulse response U{s) = 1. F in d ^(0 .
* ( 0 -
h{t)=
- lj
step 4 of 5
We can then integrate the in^ulse response to obtain the unit step response. Alternatively, for a unit step ii^u t, ) = - and using partial &action expansion, &nd
m - _________ 1_________ __________1_________ r ( r ) ' I Simplify further. Y (s) = 1 + 1 1
____
1 j ' ( s ) = i + — !— . - f i. ' 2 4^ r ( s ) = i + — r.:--- . - f " ' ^ ‘ 2 4 i > - \ Find^Cf). Thus, we get H O = ! - « ■ * Step 5 of 5 ^Now we have the remaining case where C ^ negative and | ^ < 1. since we already dealt with the case o f | ^ > 1 in the previous p art The impulse response and the step responses are exactly the same, that is ^(<) - —= ^ ^ = e ~ ^ sin(<P^), except now C
k j l - C
Problem 3.49PP
Consider the following second-order system with an extra pole;
« W = ;
( * + P )( ^ + 2f Show that the unit-step response isy(0 = 1 +Ae~^ +Be-^* an(coat - e), where
A =
« 3 -2 < a i« p + p 2 ’ P
- { P - { « * (a) Which term dominates y(t) as p gets large?
(b) Give approximate values for A and B for small values of p.
(c) Which term dominates as p gets small? (Small with respect to what?)
(d) Using the preceding explicit expression for y(t) or the step command in Matlab, and assuming that ojnr = 1 and ^ = 0.7, plot the step response of the preceding system for several values of p ranging from very small to very large. At what point does the extra pole cease to have much effect on the system response?
Step-by-step solution step 1 of 5 H (s) = (s+p)(s“ +2?a^s+i!^’ ) Y (s) °>»P s(s+p)(s“ +25ti>i,s+ti>«’ ) Y ( s ) = ^ + ^ + Step 2 of 5 Solving, w e get K i = l K j= A V And 2 2 ^ ( K ’ )(p=-2?avP+aw’ ) < 9 - ^ y f t ) = l +Ae'*‘ +Be’* sin(cDat-0)
^ Q) (1) ''
Step 3 of 5
(a) I * Term domnates
Step 4 of 5
(b)
“ 2^co,p P
Step 5 of 5
Consider the second-order unity DC gain system with an extra zero. H (A = a»g<J + 2)
-I- 2i<on5 + ai§)
(a) Show that the unit-step response for the system is given by
y ( 0 =
where
'c(IE((iy( + A ),
f t =UD
(b) Derive an expression for the step response overshoot. Mp, of this system.
(c) For a given value of overshoot. Mp, how do we solve for ^and ojn7
Step-by-step solution
step 1 of 7
Consider the second order unity DC gain S3rstem with an extra zero.
z(e^ + <z?'l
w
We write the transfer fimction in partial f ic tio n form, as
^ 2 . ^ ^ ^ 2 . I *.2 ■ We kn o w th at^(^) = . z dt Step 2 of 7 Find.y(i). - ^ = = c o s ( ® r f i - ^ ) + - ^ = = s i n ( a y - ^ ^ ( 0 = 1 - Where — 1^-1+ ijc o s ( a > ,( - ^ ) + - ^ 1 — s in (® ,i - -1 _ L _ Simplify further. ^ ( 0 = 1 g3 ^ ~ - ( ^ + A ) ] Where I ca^ ,-z Step 3 of 7
We combine the last two terms in the argument o f the cosine term, and find s /? + .
P - tan"
\ Hence it is proved that
Step 4 of 7
(b) At peak time we know that — ^—- = 0.
dt a
^ [./z’ + a>; - 2 ;a i,) cos (a>,f - fi,) ■
^ + a j - 2gai, )sin (a>^ - /?)