Root Locus
Real Axis (seconds'*)
Figure 1
Thus, the root locus is plotted for the given transfer function and it is shown in Figurel.
Consider the following given funtXion.
. ___ s+3___
'*'“s(j+10)(s’+2s+2) M ATLAB program to obtain root locus:
s=tf('s');
sysL=(s+3)/(s*(s+10)‘(s''2+(2's)+2));
riocus(sysL)
The root locus plot is shown in Figure 2.
Thus, the root locus is verified from M ATLAB output.
(j+3)
Substitute — s‘(s+10)(s’+6t+25) zc-- 77— ;--- ofor 7(«1 in Equation (1). ^ s*(s+10)(s“+6s+25)
. _________ s j j __________
s*(s+10)(s+3+4y)(s+3-4y)
■T=0...(6)step 9 of 37 XV Compare the equation (6) and equation (3).
To find zeros put numerator JV(s) — 0 Thus, the zero is -3.
To find poles put denominator D(s) = 0 s’(s+I0)(s+3+4/)(s+3-4y) = 0...(7)
The roots of the equation (7) are 0, 0, - 10, — 3— 4y and — 3+4y Thus, the poles are 0, 0, - 10, — 3— 4y and — 3+4y.
Consider the formula for the asymptotes.
a . Z i c l f L n - m
(sum of finite poles)-(sum of finite zeros) (ntimberof finitepoies )-(numberof finilezeros) -10-3+4y-3-4y-(-3)
-13 4
Thus, the asymptotes is l-3.25i 4
Substitute 1 for i, 5 for rr and 1 for m in equation (5).
Thus, theangle of asymptotes are , jl3^, |— 45“|and j_^y3^.
Consider the following equation.
Consider the departure angles in the transfer function.
Stun of angleof vector to 180“- the complex pole A
fiom otherpoles Jsiimof angleof verXoistothe) [complex pole A fiom zeros
j61„=180“-[61+6l,+6!,]+61
Substitute all ff values in equation.
61„=180“-[126.93“+126.93“+90“+29.74“]+90“
Thus, the departure angle in the transfer function is I0L5^ at s = — 3+4y angleof depature 1
fiom a complex A J
step 12 of 37 xv Procedure to draw root locus plot:
• Take real and imaginary lines on X axis and Y axis respectively.
• Mark the poles on the real axis.
• Locate the asymptotes on the real axis, and draw the asymptotes from centroid at an angle of
• Draw the root locus.
45-The root locus plot is shown in Figure 3.
Real Axis (seconds'*)
Figure 3
Hence, the root locus is plotted for the given transfer function and it is shown in Figures.
Consider the following given funtXion.
i(,). (44-^)
' ’ s’(s+10)(s’+6s+25) M ATLAB program to obtain root locus:
s=tf('s');
sysL=(s+3)/(s''2‘(s+10)'(s''2+(6*s)+25));
riocus(sysL)
The root locus plot is shown in Figure 4.
Thus, the root locus is verified from M ATLAB output.
(0)
Compare the equation (8) and equation (3).
To find zeros put numerator JV(s) = 0 Thus, the zero are -3 and .3.
To find poles put denominator Dis) = 0 s’(s+I0)(s+3+4/)(s+3-4y) = 0...(9)
The roots of the equation (9) are 0, 0, - 10, — 3— 4y and — 3+4y Thus, the poles are 0, 0, - 10, — 3— 4y and — 3+4y.
Consider the formula for the asymptotes.
n - m
(sum of finite poles)-(sum of finite zeros) (ntimberof finitepoies )-(mimberof finilezeros) -10-3+4y-3-4y-[-3-3]
-10 3
Thus, the asymptotes is j^^33j. 3
Substitute 1 for i, 5 for n and 2for rrr in equation (5).
, 180“+360”(1-1)
^' k_o
Substitute 2 for i, 5 for n and 2for rrr in equation (5). =60“
, 180“+360“(2-l) 540“ 5-2 -180“ 3
Substitute 3 for i, 5 for n and 2for rrr in equation (5).
, 180“ + 360“(3-l) rr_o
e angle of asymptotes are jgQ^, jlgg^and
Consider the following equation.
Consider the departure angles in the transfer function.
sum of angleof vector to
180“-
the complex pole
Afiom otherpoles Jsiimof angleof vectoistothe]
[complex pole A fiom zeros j
61 = 180“-[61+61+61+61]+61+61
Substitute all ff values in equation.
61 = 180“-[126.93“+126.93“+90“+29.74“]+90“+90“
Thus, the departure angle in the transfer function is 13^^ at x = — 3+4y angleof depature [
fiom a complex A J
step 19 of 37 XV Procedure to draw root locus plot:
• Take real and imaginary lines on X axis and Y axis respectively.
• Mark the poles on the real axis.
• Locate the asymptotes on the real axis, and draw the asymptotes from centroid at an angle of
• Draw the root locus.
60-The root locus plot is shown in Figure 5.
Root Locus
Real Axis (seconds *)
step 20 of 37 xv
Hence, the root locus is plotted for the given transfer function and it is shown in Figures.
Consider the following given funrXion.
(443)‘
' ' r ’ (j+ 1 0 )(j’ +61+25)
M ATLAB program to obtain root locus:
s=tf('s');
sysL=(s+3)“2/(s“2‘(s+10}*(s“2+(6‘s}+25));
riocus(sysL)
The root locus plot is shown in Figure 6.
Step 22 of 37 . Thus, the root locus is verified from M ATLAB output.
(x+3)(x* +4s +68)
Substitute ---- 1— j’(s+10)(s’+4i+85) ---L-for Ds) in Equation (1).
= (4+3)( x *+4 x +68) s'(i+10)(s“+4!+85)
(x + 3)(l + 2 + 8y)(x+2-8y) j’(x+10)(s+2+9y)(s+2-9y) ...' ' Compare the equation (10) and equation (3).
To find zeros put numerator JV(x) = 0 Thus, the zero are-3, — 2+8y and — 2— 8 y.
To find poles put denominator D(s) = 0 s’(s+IO)(j+2+9/)(i+2-9y)=0 ...(11)
The roots of the equation (11) are 0, 0, -10, -2-9J and -2+9J- Thus, the poles are 0, 0, - 10, -2-9J and — 2+9J.
Consider the formula for the asymptotes.
n - m
(sum of finite poles)-(stim of finite zeros) (ntimberof finitepoies )-(mimberof finite zeros) -IO-2+9J-2-9y-[-3-2+8y-2+8y1
3
“ 2
Thus, the asymptotes is m
Substitute 1 for), 5 for n and 3for rrr in equation (5).
, 180“+360”(1-1)
=90“ 5-3
Substitute 2 for), 5 for n and 3for rrr in equation (5).
, 180”+360“(2-l)
f t “
5-3
X X540“
= 270“
2Thus, the angle of asymptotes are |9Q^and —90“ .
Consider the following equation.
Consider the departure angles in the transfer function.
sum of angleof vector to - the complex pole A
fiom otherpoles Jsiimof angleof verXoistotheJ [complex pole A fiom zeros j 61 = 180“-[«,+61+«,+6l]+[«,+61+61]
Substitute all ff values in equation.
61 = 180“-[102.52“+102.52“+90“+48.36“]+[90“+90“+83.65“]
Thus, the departure angle in the transfer function is |^^I002^ at s = — 2+9y.
angleof depature [ fiom a complex A J
Consider the following equation.
61 =180“- t a n - '^ i j
Consider the arrival angle in the transfer function. -45“
sum of angleof vector to - the complex zero A
fiom other poles zeros Jsumof angleof vetXoistothe[
[complex zero A fiom poles J 61 = 180-J61+«,]+[61+61+61+61+«,]
Substitute all ff values in equation.
61=180“-[90“+82.86“]+[l04.03“+104.03“+90“+45“+(-90“)] Procedure to draw root locus plot:
• Take real and imaginary lines on X axis and Y axis respectively.
• Mark the poles on the real axis.
• Locate the asymptotes on the leal axis, and draw the asymptotes from centroid at an angle of
• Draw the root locus. 9V
The root locus plot is shown in Figure 7.
Root Locus
Real Axis (seconds *) Figure 7
Hence, the root locus is plotted for the given transfer function and it is shown in Figure7.
Consider the following given funrXion.
i( x ) . (j+3)(x4+4x+68)
r’(j+10)(j’+4j+85) M ATLAB program to obtain root locus:
s=tf('s');
sysL=((s+3}*(s“2+(4‘s)+68})/(s''2‘(s+10)‘(s''2+(4*s)+85));
riocus(sysL)
The root locus plot is shown in Figure 8.
Step
30 of 37 Thus, the root locus is verified from M ATLAB output.
[(j+ O ’ + il
Compare the equation (12) and equation (3).
To find zeros put numerator JV(x) = 0 Thus, the zero are — 1 + y and — 1— j.
To find poles put denominator Dis) = 0 Thus, the poles are0,0,-2and — 3.
Consider the formula for the asymptotes.
n - m
(sum of finite poles)-(stim of finite zeros) (mimberof finitepoies )-(mimberof finilezeros) -2-3-1-1+7-1-Jl
-5+2
2Thus, the asymptotes is EEI!
2Substitute 1 for), 4 for n and 2for rrr in equation (5).
, 180“+360“(1-1)
=90“ 4-2
Substitute 2 for), 4 for n and 2for rrr in equation (5).
, 180“+360“(2-l) 540“ 4-2
= 270“
2=-90“
Thus, the angle of asymptotes are jg^and j^go^.
Consider the following equation.
Consider the arrival angle in the transfer function.
angleof arrival [ fiom a complex A J
sum of angleof vector to - the complex zero A
fiom other poles zeros Jsumof angleof vectors to the[
[complex zero A fiom poles J
61 =180“ -[61]+[61+61+6,+«,]
Substitute all ff values in equation.
61 =180“-[90“] + [l35 + 135 +45+26.56]
-431.56“
Thus, the arrival angle in the transfer function is |^^7L5^ at J = — 1+y.
Step 34 of 37 XV Procedure to draw root locus plot:
• Take real and imaginary lines on X axis and Y axis respectively.
• Mark the poles on the real axis.
• Locate the asymptotes on the leal axis, and draw the asymptotes from centroid at an angle of
9<r
• Draw the root locus.
The root locus plot is shown in Figure 9.
Root Locus
Real Axis (seconds'*)
Step 35
of 37 XV
Hence, the root locus is plotted for the given transfer function and it is shown in Figures.
Consider the following given funrXion.
r(44*)’ 4 i]
s‘(s+2)(z+3) M ATLAB program to obtain root locus:
s=tf('s');
sysL=((s+1 }"2+1 }/(s“2'(s+2}'(s+3»;
riocus(sysL)
The root locus plot is shown in Figure 10.
Step 37 of 37
Thus, the root locus is verified from M ATLAB output.
listed choices for L(s). Be sure to give the asymptotes and the arrival and departure angles at any complex zero or pole. After completing each hand sketch, verify your results using Matlab.
Turn in your hand sketches and the Matlab results on the same scales.
(a) L(s) = model for a case of magnetic levitation with lead compensation.
(b) L(s) = magnetic levitation system with integral control and lead compensation.
(c) L(f) = i
(d) L {s) = —j— . What is the largest value that can be obtained for the damping ratio of the stable complex roots on this locus?
- ( i - l ) ( ( i + 2 F + 3 ]
There are 3 poles and one zero.
Calculate the number of asymptotes, P.
P = « - «
Here, n represents number of open loop poles.
m represents number of open loop zeros.
P = 3 - l s 2
Step 2 of 17 ^
Determine the angle of asymptotes.
, = ,,2 .3 ...-™
Therefore, the asymptotes are |g()® apart from each other.
Determine the intersection point of asymptotes on the real a)s
n -m
There are 4 poles and one zero.
Calculate the number of asymptotes, P.
P = 4 - l
» 3
Step 5 of 17 ^
Determine the angle of asymptotes.
, = , . 2 . 3 . . . - „
Therefore, the asymptotes are 120* apart from each other.
Determine the intersection point of asymptotes on the real axis.
n -m 0 - 1 0 + l - l - ( - 2 )
2
Write the MATLAB code to obtain root locus for L { s y
num=[1 2];
)-There are 2 poles and one zero.
Calculate the number of asymptotes, P.
P = 2 - I s i
Step 8 of 17 ^
Determine the angle of asymptotes.
, = 1,2.3...-™
n - m , 360®(I-I)
3
= 0®
Determine the intersection point of asymptotes on the real axis.
Step 9 of 17 ^
Write the MATLAB code to obtain root locus for
Z.(j) •
num=[1 -1]: There are 5 poles and two zeros.
Calculate the number of asymptotes, P.
P = 5 - 2 s 3
Step 11 of 17 ^
Determine the angle of asymptotes.
= , = 1,2,3...-™
Determine the intersection point of asymptotes on the real axis.
a . I S l I d L n -m
0 - 2 0 - 2 0 + \+ l- { - l- \) 3
= - I 2
Determine the angle of departure from the pole i+ y .
A = = i8 0 ” - 2 ; ^ , + ! ; » '.
= l80*-140.45*-h 53.13*
=92.68*
The angle of departure from the conjugate pole l —J.
^ ^ = -(9 2 .6 8 * ) There are 4 poles and one zero.
Calculate the number of asymptotes, P.
P = 4 - l - 3
Determine the angle of asymptotes.
* = 5 5 M , = 1 ,2 .3 ...-™
Therefore, the asymptotes are 120* apart from each other.
Write the MATLAB code for £ ( j ) • There are 3 poles and no zeros.
Calculate the number of asymptotes, P.
P = 3 - 0 - 3
Determine the angle of asymptotes.
* = 5 5 M , = 1 .2 .3 ...-™
Therefore, the asymptotes are 120* apart from each other.
Step 16 of 17 ^
Determine the angle of departure from the pole —2+1.73J ■
= 1 8 0 * - ^ t a n - '^ ^ j + 9 0 * j
= 1 8 0 *-I20*
= 60*
The angle of departure from the conjugate pole —2 —1.73y.
= -60*
Put the characteristic equation of the system shown in Fig. in root-locus fonn with respect to the parameter a, and identify the con'esponding L(s), a(s), and b(s). Sketch the root locus with respect to the parameter a, estimate the closed-loop pole locations, and sketch the corresponding step responses when a = 0, 0.5, and 2. Use Matlab to check the accuracy of your approximate step responses.
Figure Control system
s(s+2) -O K
S te p - b y - s te p s o lu tio n
step 1 of 23
Refer to control system in Figure 5.52 in the textbook.
/ f ( i ) = l + O i
The above equation is in the form.
1 + J C ^ = 0 ... (2) a {s)
Step 2 Of 23
Equate equation (1) and (2).
a ( s ) = s* + 2 s + 5 A (s) = 5s
The open loop transfer function is.
s* + 2 s + 5
Therefore, the values of £ ( s ) . a ( s ) and ^ ^ s) are,
a ( s ) » s * + 2 s + 5 A ( s ) > Ss
Step 3 of 23
MATLAB code to plot the root locus:
num=[5 0];
den=[1 2 5];
sys=tf(num,den):
rlocus(sys)
Step 4 of 23
The root locus plot is shown in Figure 1.
R e a l A x is (secoods*^) Figure 1: Root locus
Step 5 of 23 ^
Calculate the closed loop transfer function.
r W - s ( s + 2 ) Calculate the closed loop pole locations when a — 0-
s* + ( 2 -h5o)s-i-5
The closed loop transfer function is, n s ) _ s
MATLAB code to plot the step response:
t=0:0.01:5;
y=1-exp{-t).*cos(2.*t)-0.5.*exp(-t).*sin{2.*t);
plotfty)
title('step response when a=0');
Step 9 of 23 -A.
The step response is shown in Figure 2.
s te p resp o B se w faea a= 0
title('step response when a=0');
Step 11 of 23
The step response when ^ s 0 shown in Figure 3.
The step responses in Figure 2 and Figure 3 are same.
Step 12 of 23 ^
The closed loop transfer function is.
5
MATLAB code to plot the step response:
t=0:0.01:5;
y=1-5.*exp(-2.*t)+4.*exp(-2.5.*t):
plot(t,y)
title('step response when a=0.5');
Step 15 of 23
The step response is shown in Figure 4.
stq > r e s p o n s e w k e n a = 0 ,5
The step responses in Figure 4 and Figure 5 are same.
Step 18 of 23 ^
Therefore, the closed loop pole locations when a —2 |-Q.432>-11.5S67|.
Step 19 of 23
The closed loop transfer function is,
I ' M 5
MATLAB code to plot the step response:
t=0:0.01:5;
y=1.0015-1.0404.*exp(-0.432.*t)+0.039.*exp(-11.5567.*t):
Plotfty)
title('step response when a=2');
Step 21 of 23 ^
The step response is shown in Figure 6.
s t ^ r e s p o n s e w h e n a = 2
title('step response when a=2');
Step 23 of 23
The step response when a — 2 shown in Figure 7.
s te p r e s p o n s e w h e n a = 2
Figure 7
The step responses in Figure 6 and Figure 7 are not same.
Use the Matlab function ritool to study the behavior of the root locus of 1 + KL(s) for
(i + o)
U s ) =
j(j + 1)(j2 + 8j + 52)
as the parameter a is varied from 0 to 10, paying particular attention to the region between 2.5 and 3.5. Verily that a multiple root occurs at a complex value of s for some value of a in this range.
Step-by-step solution
step 1 of 4
Step 1 of 4
The loop transfer function is.
s+a w / \ ________ J T « ______
' * ^ " i ( j + l ) ( i ’ + 8 s+ 5 2 )
Determine the characteristic equation of the system.
I + X Z ,( i ) = 0
I + i f
-j ( i + l ) ( -j * + 8 -j + 5 2)
Write the MATLAB code to draw the root locus when the parameter a is varied from 0 to 10 using ritool function.
a=0:0.5:10;
num=[1 a];
den=conv([1 1 0],[1 8 52]);
sys=tf(num,den);
rltool(sys)
Step 2 of 4 ^
Execute the code at the MATLAB command window and study the behavior of the root locus.
Step 3 of 4
Go to Analysis Plots. Select Pole/Zero and Response as Closed Loop r to y. Click on Show Analysis Plot.
CenlanlsafPWs---nod Response
12 3 4 s 6AM B r: r r n r nOoMdloMrtev n r' n r r n n Ootid LoMrtOM Q r r. r r n n Ootid leee du to V c r n r r □ n Ootid Uid^iev nr n r r n n Ootid Imp n to ir nr: n r' r' n n Oodiloopl nr r r r n n ComptAtMotC nn n r r n n nr n r r n nRIMO r n r r r n n SmotH
Step 4 of 4
Get the MATLAB output for the pole-zero plot.
Observe from the pole/zero plot that there is a multiple complex root in the region of a between 2.5 and 3.5.
Problem 5.11 PP
Use Routh’s criterion to find the range of the gain K for which the systems in Fig. are stable, and use the root locus to confirm your calculations.
Figure Feedback systems
Step-by-step solution
step 1 of 9
Step 1 of 9
(a)
Refer to Figure 5.54 (a) in the text book for the block diagram of a system.
The open loop transfer function is.
The characteristic equation of the system is, 1 + G ( j ) f f ( * ) = 0
For the system to be stable, all elements in the first column of the Routh table should be greater than zero.
2K>0 K > 0...(1)
-13A:’ + %72Ar*-67860Ar + 1513SI20 (6 1 8 -A :)’
The range of satisfying all the four conditions. (1). (2). (3) and (4) is, 0 < ^ < 4 8 4 . ! 6
Thus, the range of AT for stability is |Q< JC<484.16l-(3)
Step 4 of 9
The characteristic equation is, , . „ s’ + f + 2
r = 0 s (s + 5 ) ( i + 6 ) ( i ’ + 2 j + 1 )
Draw the root locus using MATLAB to find the range of the gain JC for stability.
num=[1 1 2]:
den=conv([1 11 30 0],[1 2 1]);
sys=tf(num,den);
rlocus(sys)
Step 5 of 9
Get the MATLAB output for the root locus.
Observe from the root locus that the gain should be less than 482 for the system to be stable.
Thus, the root locus result agrees with the Routh’s result.
Step 6 of 9
(b)
Refer to Figure 5.54 (b) in the text book for the block diagram of a system.
The open loop transfer function is.
The characteristic equation of the system is, l + G ( s ) H ( j ) = 0
As the first column element is infinity, the system is unstable for all values of .
Step 8 of 9
The characteristic equation is,
i + i c — — ---, = 0 5 (j- 2 ) (j* + 2j + 10)
Draw the root locus using MATLAB to find the range of the gain for stability.
num=[1 2];
den=conv([1 -2 0],[1 2 10]);
sys=tf(num,den);
rlocus(sys)
Step 9 of 9
Get the MATLAB output for the root locus.
Observe from the root locus that the locus does not touch the zero-axis. Hence, the system is unstable for all values of
IC-Thus, the root locus result agrees with the Routh’s result.
Sketch the root locus for the characteristic equation of the system for which
and determine the value of the root-locus gain for which the complex conjugate poles have the maximum damping ratio. What is the approximate value of the damping?
Step-by-step solution
step 1 of 5
Consider the General form of characteristics eouation.
Step 1 of 5
-Consider the general form of characteristics equation.
I + AX(i) = 0 (1) ( s +2)
Substitute for I / j ) in Equation (1).
a’ ( ^ + 5 )
1 +a: (2)
a* ( a + 5 )
Consider the roots of the general form of an equation by the root locus method.
... (3) D is ) Where,
The roots of A f(» )= 0 are called the zeros of the problem.
The roots of D ( j) = 0 are the poles.
Consider the number of poles and zeros from the characteristics equation.
Compare the Equation (2) and the Equation (3).
To find zeros put numerator J V (j)= 0 Thus, the one zero is - 2.
To find poles put denominator D (s) = 0 •
»“ ( i + 5 ) = 0 ... (4)
The roots of the equation (4) are 0,0, and - 5.
Thus, the three poles are 0.0, and - 5.
Step 2 of 5
Consider the formula for the asymptotes.
n - m
(su m o f fin ite p o les) - (su m o f fin ite zero s) (n u m b e ro f finite p o les ) ~ (n u n d > e ro f finite zero s) ( ° - S ) - ( - 2 )
■ (3)-(l)
= -1 .5
Thus, the asymptotes is E O
-Consider the formula for the angle of asymptotes.
^ l8 0 °+ 3 6 (y * (/-l) n —m Where.
Number of poles is n Number of zeros is m
/ = l,2 ,..J i—m
Substitute 1 for /, 3 for n and 1 for m in equation (5).
180°+360»(1-1) 3 - 1
= 90“
Substitute 2 for /, 3 for n and 1 for m in equation (5).
180” + 3 6 0 ° ( 2 - l) 3 - 1
= 270»
Thus, the ahgle of asymptotes are |9(y>|ahd |27Q**|.
Step 3 of 5 ^
Procedure to draw root locus plot:
• Take real and imaginary lines on X axis and Y axis respectively.
• Mark the poles on the real axis.
• Locate the asymptotes on the real axis, and draw the asymptotes from centroid at an angle of gffan a 270"
• Draw the root locus.
Consider the following given function.
' ’ * ’ (s + 5)
Write the MATLAB program to obtain root locus.
s=tf('s');
sysL=(s+2V(s''2*(s+5)):
rlocus(sysL) [K]=rlocfind{sysL)
Step 4 of 5
The MATLAB output for the value K.
Select a point in the graphics window selecteu_point —
-0.663 + 2.32i K = 10.7
The root locus plot is shown in Figure 1.
Figure 1
Step 5 of 5
From Figure 1, the maximum damping ^ is 0.275 when the value of gain K is approximately 10.7.
Thus, the maximum damping ^ is |Q.275l when the value of gain K is [|Q,7| and the root locus is plotted for the given transfer function and it is shown in Figurel.
Problem 5.13PP
(a) Find the locus of closed-loop roots with respect to K.
For the system in Fig.,
(b) Is there a value of K that will cause all roots to have a damping ratio greater than 0.5?
(c) Find the values of K that yield closed-loop poles with the damping ratio ^ = 0.707.
(d) Use Matlab to plot the response of the resulting design to a reference step.
Figure Feedback system
Obtain the characteristic equation for feedback system shown in Figure 5.54 to find the locus of closed -loop roots with respect to x ■
Since system contains cascaded blocks,
Closed loop transfer function of the system is, r ( j ) = G(»)
l + C ? (i)//(j) j - f l
V
J*+ 81 U + 1 3 j [ i ' ( f ‘ + 100)\s + \3 )
The characteristic equation of the system is.
■(1) departure angles of particular importance.
Step 2 of 10
Procedure to draw root locus:
RULE 1 : There are five branches to the locus, three of which approach to finite zeros and two of which approach to Infinity.
RULE 2 : The real-axis segment defined by —13 ^ ^ ] is part of the locus.
Draw pole-zero plot for the transfer function.
RULE 3 : calculate the centre of asymptotes.
5 - 3 _ -12
2 s - 6
The angles of asymptotes are at .
RULE 4 : Compute the departure angle from the pole at ^ +ylO-The angle at this pole we will define to be ^ . The other angles are marked in Figure 2.