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Problem 5.04PP
Real poles and zeros. Sketch the root locus with respect to K for the equation 1 + KL(s) = 0 and 
the 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, venty your results using Matlab. 
Turn in your hand sketches and the Matlab results on the same scales.
(a) Us) -
(b) Us) =
((a) = 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 N(s) = 0 
Thus, there is no zero in the transfer function.
To find poles put denominator D(s) = 0 .
j (« + 1)(4+5)(j + I0) = 0 ..... (4)
The roots of the equation (4) are 0, - 1, - 5 and - 10.
Thus, the four poles are 0. - 1. - 5 and - 10.
Step 2 of 26 ''V
Consider the formula for the asymptotes. 
n -m
(sum of finite poles)-(sum of finite zeros) 
(numberof finite poies )-(nuinber of finite zeros) 
(0-l-5-l0)-(0)
(4)-(0)
= -4
Thus, the asymptotes is E 3
Consider the formula for the angle of asymptotes.
 ̂ l80°-t-360»(/-l) 
n -m
Where.
Number of poles is n 
Number of zeros is m 
/ = i,2,..ji-m
Step 3 of 26
Substitute 1 for /, 4 for n and 0 for m in equation (5).
I 8 0 ° + 3 6 0 ° ( I - I )
4 - 0
= 45*
Substitute 2 for /, 4 for n and 0 for m in equation (5).
f8 0 ° -t-3 6 0 ° (2 - l)
4 - 0
= 135»
Substitute 3 for /, 4 for n and 0 for m in equation (5).
, i8 0 » + 3 6 0 » (3 - l)
4 - 0
« 2 2 5 «
= -45»
Substitute 4 for /, 4 for n and 0 for m in equation (5).
l8 0 * + 3 6 0 » (4 - l)
4 - 0
-3 1 5 «
= -l35»
Thus, the angle of asymptotes are |4S«»|, [135**!, |-4 5 ° |and | - 135**|. 
There is no complex zero or pole in the given function.
Hence, there are no arrival and departure angles in the transfer function.
Step 4 of 26 s\-
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
4 r
• Draw the root locus.
Step 5 of 26
The root locus plot is shown in Figure 1.
Figure 1
Hence, the root locus is plotted for the given transfer function and it is shown in Figuret.
Step 6 of 26
Consider the following given function. 
£ ( * ) = -z(z+l)(s+5)(s+10)
Wnte the MATLAB program to obtain root locus. 
s=tf('s');
sysL=2/{s*(s+1 )*(s+5)*(s+10)); 
rlocus(sysL)
The root locus plot is shown in Figure 2;
Step 7 of 26
Thus, the root locus is verified from MATLAB output.
Step 8 of 26
(b)
(4 ^ 3 )
i ( z + l)(x + 5 )(z + I0) 
1 + J f - : -------------------------- r = 0
for L ( j) ir i Equation (1). 
(6)z(z + l)(j+5)(z+I0)
Consider the number of poles and zeros from the characteristics equation. 
Compare the Equation (6) and the Equation (3).
To find zeros put numerator N(s) = 0 
Thus, the one zero is - 3.
The roots of the equation (4) are 0, - 1, - 5 and - 10.
Thus, the four poles are 0. - 1. - 5 and - 10.
Step 9 of 26
Consider the formula for the asymptotes. 
n -m
(sum of finite poles) -(sum of finite zeros) 
(numberof finite poies )-(numberof finite zeros) 
(0-l-5-l0)-(-3)
(4)-(l)
= -4.33
Thus, the asymptotes is E n a 
Substitute 1 for /, 4 for n and 1 for m in equation (5). 
f80°+360o(l-l)
4-1
= 60»
Step 10 of 26
Substitute 2 for /, 4 for n and 1 for m in equation (5).
180°+360°(2-l)
4-1
= 180»
Substitute 3 for /, 4 for n and 1 for m in equation (5).
, 180°+360°(3-l)
4-1
= 300“
= -60»
Thus, the angle of asymptotes are |6()v|. | |8Qe|and | - 60^1 ■
There is no complex zero or pole in the given function.
Hence, there are no arrival and departure angles in the transfer function.
Step 11 of 26 -rv
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
6cr
• Draw the root locus.
The root locus plot is shown in Figure 3.
30
IUMLocm
■*45
K-poles 
o - zeros
a - Ceam of the «yni|]Mes
.. . . A
yy
y /
V -
a — 4J3
V " * ’’
VV
Vv
■Ov
•30 -2S -20 -IS -to -5
XnlAiit(ftc4)
S to IS
Figures
Hence, the root locus is plotted for the given transfer function and it is shown in Figure3.
Step 12 of 26
Consider the following given function.
i ( j ) = -------------------------------' ' i(i+l)(s+5)(s+10)
Write the MATLAB program to obtain root locus. 
s=tf('s');
sysL=(s+3)/(s*(s+1 )*(s+5)*(s+10)); 
rlocus(sysL)
The root locus plot is shown in Figure 4.
Figure 4
Step 13 of 26
Thus, the root locus is verified from MATLAB output.
Step 14 of 26
(c)
(s+2)(j+4)
u a : = 0 .4(z + 1)(j+5)(j+10)
Consider the number of poles and zeros from the characteristics equation. 
Compare the Equation (6) and the Equation (3).
To find zeros put numerator JV(j) = 0
(7)
Step 15 of 26 A
Thus, the two zeros are - 2 and - 4.
The roots of the equation (4) are 0, - 1, - 5 and - 10. 
Thus, the four poles are 0. - 1. - 5 and - 10.
Consider the formula for the asymptotes.
n -m
(sum of finite poles)-(sum of finite zeros) 
(numberof finite poles )-(number of finite zeros) 
(0-l-S-10)-(-2-4)
(4)-{2)
= -5
Thus, the asymptotes is E H
step 16 of 26 A
Substitute 1 for /, 4 for n and 2 for m in equation (5).
180°+360«(1-1)
4-2
= W>
Substitute 2 for /, 4 for n and 2 for m in equation (5).
180»+360°(2-l)
4-2
-270“
— 90“
Thus, the angle of asymptotes are |90|and | - 9(y>|.
There is no complex zero or pole in the given function.
Hence, there are no arrival and departure angles in the transfer function.
Step 17 of 26 ^
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.
The root locus plot is shown in Figure 5.
Root Locos
Figures
Hence, the root locus is plotted for the given transfer function and it is shown in Figures.
Step 18 of 26
Consider the following given function.
i ( j ) = ------(4 + 2 ) ( 4 -i- 4 ) -------
' ' i ( i+ l ) ( s + 5 ) ( s + 1 0 )
Write the MATLAB program to obtain root locus. 
s=tf('s');
sysL=(s+2)*(s+4)/{s*(s+1)*(s+5)*(s+10));
rlocus(sysL)
Step 19 of 26
The root locus plot is shown in Figure 6.
Step 20 of 26
Thus, the root locus is verified from MATLAB output.
Step 21 of 26
for A (j) in Equation (1).
(7)
(b)
(s+2)(i+6)
Substitute , V r ,
i ( z + l) (s + 5 ) (z + 10)
ua: 0.
z(z + l ) ( j+ 5 ) ( j+ 1 0 )
Consider the number of poles and zeros from the characteristics equation.
Compare the Equation (6) and the Equation (3).
To find zeros put numerator JV(j) = 0
Thus, the two zeros are - 2 and - 6.
The roots of the equation (4) are 0, - 1, - 5 and - 10.
Thus, the four poles are 0. - 1. - 5 and - 10.
Consider the formula for the asymptotes.
n -m
(sum of finite poles)-(sum of finite zeros)
(numberof finite poles )-(number of finite zeros) 
(0-l-5-10)-(-2-6)
(4)-(2)
= -4
Thus, the asymptotes is E 3
step 22 of 26 ^
Substitute 1 for /, 4 for n and 2 for m in equation (5).
180“+360“(1-1)
4-2
= 90“
Substitute 2 for /, 4 for n and 2 for m in equation (5).
, 180“+360“(2-l)
4-2
= 270“
—90“
Thus, the angle of asymptotes are |9(y>|and | - 9(y*|.
There is no complex zero or pole in the given function.
Hence, there are no arrival and departure angles in the transfer function.Step 23 of 26 ^
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
9 tt
• Draw the root locus.
Step 24 of 26
The root locus plot is shown in Figure 7.
Figure?
Hence, the root locus is plotted for the given transfer function and it is shown in Figure7.
Step 25 of 26
Consider the following given function.
(s*2){s+6)
' ' i(i+l)(s+5)(s+10)
Write the MATLAB program to obtain root locus. 
s=tf('s');
sysL=(s+2)*(s+6)/{s*(s+1)*(s+5)*(s+10));
rlocus(sysL)
The root locus plot is shown in Figure 8.
Step 26 of 26
Thus, the root locus is verified from MATLAB output.