?
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Electrical Engineering and Computer Science
6.003,Signals and Systems—Fall 2003
Problem Set 10 Solutions
Home study exercise (E1) O&W 11.32
(a)
Y (s) H(s)
= (via Black’s equation)
X(s) 1 + KG(s)H(s)
N
1
(s)D
2
(s)
= (multiply through by D
1
(s)D
2
(s))
D
1
(s)D
2
(s) + KN
1
(s)N
2
(s)
The zeros of this closed-loop system are the roots of N
1
(the zeros of H) and the roots of D
2
(the poles of G),
(b) If K = 0,then the system is operating without feedback of any sort,Naturally the poles and
zeros of the system without feedback are the poles and zeros of H(s),
More formally,we can take limits of the above equation,
N
1
(s)D
2
(s) N
1
(s)D
2
(s) N
1
(s)
lim = = = H(s)
K?0 D
1
(s)D
2
(s) + KN
1
(s)N
2
(s) D
1
(s)D
2
(s) D
1
(s)
(c) Check by simple substitution,
p(s) H(s)
Q(s) =
q(s)
·
1 + K
H(s)G(s)
N
1
(s)/p(s)
p(s)
D
1
(s)/q(s)
=
q(s)
·
1 + K
N
2
(s)/q(s) N
1
(s)/p(s)
D
2
(s)/p(s) D
1
(s)/q(s)
· ·
H(s)
=
1 + KG(s)H(s)
(d) We are given the following,
s + 1 s + 2
H(s) =,G(s) =
(s + 4)(s + 2) s + 1
In this simple example,we can?nd p and q by inspection,
1
p(s) = s + 1,q(s) = s + 2,H(s) =,G(s) = 1
s + 4
G(s) = Root locus equation,H(s)
1
=
1
=? s =?(K + 4)
s + 4
K
1
The closed-loop system has one?nite zero (s =?1),one,zero at in?nity”,one pole which is
not a?ected by K (s =?2),and one pole which is a?ected by K (s =?(K + 4)),
Note that the zeros in the feedback system can obscure certain complex frequencies in the
output; these frequencies will never show up in the error signal,This is another reason to avoid
pole-zero cancellation as a design technique,since it can cause unstable complex frequencies to
become unobservable in the error signal,
Problem 1
Let the output of G(s) be y(t),Then,we can obtain the transfer function from x(t) to y(t) using
Black’s formula,
Y (s) KG(s)
=,
X(s) 1 + KG(s)
Thus,the closed loop poles are s satisfying 1 + KG(s) = 0,To plot a root locus,you are sketching
the location of s as a function of K such that 1 + KG(s) = 0,
Having said that,all the points on the locus satisfy the following criteria usually referred to as
the angle criteria,
1 + KG(s
l
) = 0
1
G(s
l
) =,(1)?
K
where s
l
is on the locus,If K > 0,then
G(s
l
) be an odd integer multiple of?,while for K < 0,
then
G(s
l
) be an even integer multiple of?,
Also,from Eqn(1),it is easy to see that s
l
approaches to the poles of G(s) if K ∩ 0 and that | |
s
l
approaches to the zeros of G(s) if K ∩?,One thing to note here,however,is that by closing | |
the loop in the feedback con?guration shown in the problem,the number of the closed loop poles
is either the same as that of G(s) or less if pole-zero cancellation takes place,Thus,to make the
argument above applicable to any rational transfer functions,we need to have the same number of
poles and zeros for G(s),This leads us to de?ne poles or zeros at,in?nity” as mentioned in the
Home study exercise above,With all these concepts in your mind,let’s go through the questions,
(a) Given,
1
G(s) =,
s + 1
G(s) has a pole at?1 and no?nite zero.
For K > 0,In this case,with the angle criteria,all the points,s
l
on the locus should
satisfy
G(s
l
) = odd integer multiple of?.
So,if you take any point on the real line segment (,?1),
G(s
l
) = 0 ±? = ±?,
phase due to zeros
phase due to the pole
Also,this is the only segment that belongs to the locus,
2
×
For K < 0,The angle criteria states that:
G(s
l
) = even integer multiple of?.
So,if you take any point on the real line segment (?1,?),
G(s
l
) = 0 0 = 0.
phase due to zeros phase due to the pole
Combining the two cases,the root loci for both cases are plotted as shown below,The solid
line corresponds to the locus for K > 0 case and the dashed line for K < 0 case,
≤m
↓e?1
(b) Given,
1
G(s) =,
(s? 5)(s + 3)
G(s) has two poles at 5 and?3,and no?nite zero.
For K > 0, Evoking the angle criteria,we can?rst see that the real line segment (?3,5)
belongs to the locus,Now we would like to know at which point the locus branches out or
bifurcates from the real line,At the branching point,the closed loop poles are of double
pole,To?nd such a point,we can?rst?nd a value of K corresponding to the double pole,
1 + KG(s) = 0
K
1 + = 0
(s? 5)(s + 3)
(s? 5)(s + 3) + K
s
2
2s? 15 + K
(s? 1)
2
1? 15 + K
=
=
=
0
0
0 completing the square,
So,when?1? 15 + K = 0 or K = 16,the closed loop system has a double pole at 1,
Because of the symmetry of the poles about s = 1,the locus branches out from s = 1
along ↓e{s} = 1 parallel to the imaginary axis,This can be seen in the?gure below,
3
1
× ×
1
≤m
↓e
×
3
×
5
For K < 0, In this case,using the angle criteria,we can see that the real line segment
(5,?) belongs to the locus since both poles contributes 0 or even integer multiple of?,
Also,the real line segment (,?3) belongs to the locus since both poles contribute?
or odd integer multiple of?; the net phase of G(s
l
) is even integer multiple of?,
Thus,the root loci containing both K > 0 (solid line) and K < 0 (dashed line) cases are shown
below,
≤m
↓e?3 5
(c) Given,
s + 1
G(s) =
2
,
s
G(s) has a double-pole at 0 and one?nite zero at?1.
For K > 0, With the angle criteria,we can see that the real line segment (,?1)
belongs to the locus,Since both poles of G(s) do not belong to the segment,there will be
a double-pole point in the segment,To?nd where it is,use the same technique as in (b),
4
i.e.,complete the square and?nd the corresponding gain K and the bifurcation point.
1 + KG(s) = 0
s + 1
1 + K
s
2
= 0
s
2
+ Ks + K = 0
K
2
K
2
s +
2
+ K?
2
= 0
K
∩ K 1?
4
= 0
K = 0,4,
This result tells us that at K = 0 and K = 4 we have double-pole and the corresponding
locations are at s = 0 and s =
4
=?2.
2
How can we know the shape of the path that locus takes after branching out of s = 0 and
merging at s =?2? In the case in our hand,it is not hard to determine,Since the closed
loop poles for the K between 0 and 4 are a complex conjugate pair,we can assume that
s = ξ + j? for a given K,Then,using one of the equations above;
(ξ + j?)
2
+ K(ξ + j?) + K = 0
(ξ
2
+ 2jξ
2
) + Kξ + jK? + K = 0
For real part,ξ
2
2
+ Kξ + K = 0
For imaginary part,2ξ? + K? = 0 ∩ K =?2ξ
ξ
2
2
2ξ
2
2ξ = 0 (combining the two equations above)
ξ
2
+?
2
+ 2ξ = 0
(ξ + 1)
2
+?
2
= 1,
As you can see,the last equation is nothing but an equation of a circle whose radius is
1 and whose center is located at (ξ,?) = (?1,0),Thus,the locus is of the circle for
0 < K < 4,Then,at s =?2 the locus bifurcates to the zero at s =?1 and s = on
the real axis.
One thing to note here is that,in general,it is extremely di?cult to obtain expressions of
the locus for higher order systems.
For K < 0, In this case,we can see that the real line segments (?1,0) and (0,?) belong
to the locus,On the locus,one of the poles will move to the zero at s =?1 as K decreases
or?K increases,while the other pole moves out to? along the positive real axis,
Thus,the loci containing both K > 0 (solid) and K < 0 (dashed) cases are shown below,
5
×
≤m
2
1?2 e↓
Problem 2
1
(a) First let output of the plant G(s) =
s(s+10)
be y(t),Then,by Black’s formula,the closed
transfer function from the input x(t) to y(t) can be found as,
Y (s)
=
G(s)K(s)
X(s) 1 + G(s)K(s)
K
=
s(s+10)
1 +
K
s(s+10)
Y (s)
X(s)
=
K
s(s + 10) + K
,(2)
The error signal e(t) is the di?erence between x(t) and y(t),Thus from the linearity of Laplace
transforms,
e(t) = x(t)? y(t)
E(s) = X(s)? Y (s),(3)
So,with Eqns (2) and (3),we can obtain the transfer function from the input x(t) to the error
e(t),
6
E(s)
X(s)
= 1?
Y (s)
X(s)
Eqn (3)
K
= 1?
s(s + 1)
Eqn (2)
+ K
s(s + 10)
=
s(s + 10) + K
E(s) s(s + 10)
=, (4)
X(s) s
2
+ 10s + K
Thus,using Eqn (4),E(s) for the unit step input x(t) = u(t) is,
s(s + 10)
E(s) = X(s)
s
2
+ 10s + K
s(s + 10) 1
=
s
2
+ 10s + K s
s + 10
=
s
2
+ 10s + K
,
so if K → 0,we can see from Routh-Hurwitz criteria that E(s) has at least one of the poles in
the right half plane,i.e,the real part of the pole is positive,Hence,in this case,the steady
state tracking error diverges to?,
Now,consider the case when K > 0,Then,both poles of E(s) are in the left half plane and
the order of the numerator is strictly less than that of the denominator (the latter condition
is usually referred to as being strictly proper), Thus we can apply the?nal value theorem to
compute the steady state tracking error,
e(?) = lim e(t)
t
= lim sE(s)
s?0
s + 10
= lim s
s?0 s
2
+ 10s + K
= 0,
Thus,regardless of the value of K,as far as it is positive,e(?) = 0,
(b) To compute the error signal to the ramp input,we can use Eqn(4) with now X(s) =
s
1
2
,
s(s + 10) 1
E(s) =
2
s
2
+ 10s + K s
s + 10
=
s(s
2
+ 10s + K)
10 10
s + 1?
100
K
+
K K
=,
s s
2
+ 10s + K
E
1
(s)
E
2
(s)
7
= ,
e
1
(t),the corresponding time signal to E
1
(s) is a scaled step,E
2
(s) has two poles and they
remain in the left-half plane as long as K > 0,In this case e
2
(t),the corresponding time signal
to E
2
(s) will decay to 0 as t ∩?,Thus,e(?) =
10
K
,i.e.,the steady state error will not be 0,
On the other hand,if K →0,Then,E
2
(s) now has at least one of the poles in the right-half
plane,Thus,e
2
(?) ∩?as t ∩?; therefore e(?) ∩?,
(c) For this part,?rst we would like to get the transfer function from x(t) to y(t),Let’s denote
the output of K
f
,x
f
(t),then X
f
(s) = K
f
X(s),
1
Y (s) =
s(s + 10)
(X
f
(s) + K
s
X(s)?Y (s))
1
1
1 +
s(s + 10)
Y (s) =
s(s + 10)
(K
f
+ K
s
)X(s)
1
s(s+10)
(K
f
+ K
s
)
Y (s)
=
X(s) 1 +
1
s(s+10)
Y (s) K
f
+ K
s
X(s) s(s + 10) + 1
Since e(t) = x(t)?y(t),we can compute the Laplace transform of e(t),E(s) as follows,
E(s) = X(s)?Y (s)
Y (s)
E(s) = 1?
X(s)
X(s),
The input x(t) to the system is a ramp,so X(s) =
s
1
2
,Thus,E(s) is,
K
f
+ K
s
1
2
E(s) = 1?
s(s + 10) + 1 s
s
2
+ 10s + 1?(K
f
+ K
s
)
=
s
2
(s
2
+ 10s + 1)
∩sE(s) =
s
2
+ 10s + 1?(K
f
+ K
s
)
,
s(s
2
+ 10s + 1)
To be able to apply the?nal value theorem,we want to choose K
f
(s) and K
s
(s) such that the
poles of sE(s) are in the left-half plane,In addition,we need to choose K
f
(s) and K
s
(s) such
that
lim sE(s) = 0
s?0
to ensure that the stead state error to the ramp input is zero,Combining these two conditions,
we see that we need to make the numerator of sE(s) equal to s
2
,From this conclusion alone,
we only know that K
f
(s) + K
s
(s) = 10s + 1,Assuming that neither is 0,then we can choose
8
K
f
(s) = 10s and K
s
= 1,Then,
s
2
+ 10s + 1? (10s + 1)
sE(s) =
s(s
2
+ 10s + 1)
2
s
=
s(s
2
+ 10s + 1)
s
sE(s) =,
s
2
+ 10s + 1
Now,we can safely apply the?nal value theorem to obtain the steady state error and we indeed
obtain 0 as expected,
s
lim sE(s) = lim = 0,
s?0 s?0 s
2
+ 10s + 1
Problem 3 (O&W 11.27)
Given,
s + 2
H(s) =
s
2
+ 2s + 4
,G(s) = K,
First,let’s identify the poles and zeros of H(s),Since the system H(s) is a second order system,
is at?,The poles are the solutions of s
is as shown below,
2
s? ±
there are two poles and two zeros,It is clear that one of the zeros is of?nite at 2 and the other
+ 2 + 4 = 0,so 1 + 3,Thus,the pole-zero diagram
≤m
↓e
×
×
3
3
1?2
Since H(s) is causal,the ROC for H(s) is ↓e{s} >?1,
The closed loop poles are the solution to the following equation,
s + 2
1 + KH(s) = 1 + K = 0,
s
2
+ 2s + 4
9
or
2
s
2
+ 2s + 4 + K(s + 2) = s + (K + 2)s + 2K + 4 = 0,(5)
(a) For K > 0,for all s
l
on the locus
H(s
l
) is an odd integer multiple of?,Thus,we can?rst
see that the real line segment (,?2) belongs to the locus,Then,we would like to know at
which point on that segment there is a double-pole,On that point,Eqn (5) has a double root,
i.e.,
s
2
+ (K + 2)s + 2K + 4 = 0 completing the square
K + 2
2
K + 2
2
s +
2
2
+ 2K + 4 = 0
2
K + 2
∩?
2
+ 2K + 4 = 0
= 6.∩ K?2,
This tells us that there are two points where double pole occurs; one is when K = 6 > 0 and
the other is when K =?2 < 0,Here,we need to look at only the positive case,The negative
one will be used in part (b),For K = 6,the double-root is at?
K+2
=?4.
2
(b) For K < 0,using the angle criteria,we can see that the real line segment (?2,?) belongs to
the locus and the double-pole point is at?
K+2
= 0 where K =?2 as computed in (a),
2
(c) Since the closed-loop system is still just a second order system,the condition that the closed-
loop impulse response does not exhibit any oscillatory behavior simply means that the closed-
loop system is critically damped or overdamped,Since H(s) is underdamped i.e.,its poles are
not on the real axis,at the smallest K to be found,the denominator takes the following form,
2
s
2
+ 2
n
s +?
n
= 0
2
s
2
+ 2?
n
s +?
n
= 0,? = 1 critically damped
(s +?
n
)
2
= 0,
where?
n
is the undamped natural frequency of the closed loop system,Thus,the required
condition is nothing but the double-pole condition we found in (a),Since K is constrained to
be positive,the smallest K we are looking for is K = 6,
10
Problem 4
Note that z-transform is linear,
(a) Given,
x[n] = 2ω[n + 3]? ω[n? 2],
In this part of the problem,we would like to use one of the most fundamental z-transform
pairs,
ω[n? m]
z
m
,ROC,All z except for 0 if m > 0 or? if m < 0,(6)?∩ z
By applying this pair for both terms in the given x[n],we get:
2
X(z) = 2z
3
z
2z
5
1
=
2
,
z
with its ROC is all z except for 0 and?, Since the unit circle is included in the ROC,the
Fourier transform of the sequence exists,From the last expression,it can be seen that there are
two poles both of which are at 0,There are?ve zeros and they are the solutions of 2z
5
1 = 0,
1
2?
i.e,2
5
e
j?
z
where?
z
is integer multiples of
5
,
≤ m
↓ e
×
2
To see this,remember that any complex number z can be written in polar form as z = re
j?
where r is the length,or radius of the vector z and? is its phase,In our case using this idea,
we can?nd the zeros as follows,
11
2z
5
1 = 0
1
5
z =
2
1
5 j5?
z
r e =
2
√
1
1
5
1
r = = 2
5
2
j5?
z
= integer multiple of 2?,
The pole-zero diagram is depicted above,The 5 zeros are on the inner circle whose radius is
2
1/5
and the outer circle is a unit circle,2 at the origin denotes that there is a double pole at
the origin,The ROC is everywhere except for the origin and?,
(b) Given,
x[n] = 2
n
u[n? 1] + 4
n
u[? n],
x
1
[n] x
2
[n]
In this part,we would like to use another crucial z-transform pair,
a
n
u[n],z > a
z
1|
z
|
<
|
a
|
∩
1? az
1
,(7)
n
u[? n? 1],? a | | | |
Note that
x
1
[n] = 2
n
u[n? 1] = 2 · 2
n?1
u[n? 1],
and
x
2
[n] = 4
n
u[? n] = 4
n
u[? n? 1] + ω[n].
Let X
1
(z) and X
2
(z) be z-transforms of x
1
[n] and x
2
[n] respectively,Then,by applying Eqns
(6) and (7) to x
1
[n] and x
2
[n],we have,
2
X
1
(z) = 2 ·
1?
1
2z
1
· z
1
=
z? 2
,| z| > 2
1
X
2
(z) =? ·
1? 4z
1
+ 1 =
4
,| z| < 4,
z? 4
Thus,
X(z) = X
1
(z) + X
2
(z)
2 4
=
z? 2
z? 4
2z
X(z) =?
(z? 2)(z? 4)
12
with its ROC being an intersection of|z|> 2 and|z|< 4,i.e.,2 <|z|< 4,The ROC does not
contain the unit circle; thus the Fourier transform of the sequence does not exist,There are
two poles at 2 and 4 and is one zero at 0,Thus,the pole-zero plot with the ROC is depicted
below:
≤m
↓e0
×
2
×
4?1
One thing to note from Problem 4 is that it would be better to change the expression of z-
transforms into rational functions in z to?nd poles and zeros,while it would be better to
change the expressions into rational functions of z
1
to?nd the corresponding time sequences
since in many cases z-transform tables show almost all the transforms in terms of z
1
such as
the one on p.776 in O&W.
13
Problem 5
(a) Recall the z-transform pair,
m
z
z ].?∩ ω[n? m
Thus,we need to simply apply this to each term in the given expression,The answer therefore
is,
x[n] = 12ω[n? 4]? ω[n? 1] + 6ω[n] + 9ω[n + 2]? 8ω[n + 5],
(b) Given,
5 1 1
X(z) =
1 +
1
6
z
1
1
6
z
2
,< z <,
3
| |
2
First,we would like to perform partial fraction to reduce X(z) consisting of easily recognizable
terms,
5
X(z) =
1
6
z
1
5
1
6
z
2
1 +
=
1
3
z
1
1
2
z
1
1? 1 +
2 3
(8)
1
3
z
1
+=
1 +
1
2
z
1
,
1?
1
3
1
2
with ROC of < z <| |
Recall another very important z-transform pair,
,
1
z
a
n
u[n],z > a| |
(9)
n
1? az
1
∩
a u[? n? 1],z <
|
a
|
.| | | |
In Eqn (8),the?rst term corresponds to the right-sided signal pair above and the second pair
to the left-sided signal,Since z-transform is linear,we have,
1
n
1
n
x[n] = 2 u[? n? 1].u[n]? 3
3
2
14
Problem 6
Given,
y[n] = x
1
[? n? 2] ← x
2
[n + 4]
1
x
1
[n] =
2
u[n]?
1
x
2
[n] =
4
u[n],
In this problem,we would like to use Eqn(9) and two more basic properties of z-transform,First,
we can?nd X
1
(z) and X
2
(z),z-transforms of x
1
[n] and x
2
[n] respectively,
1
X
1
(z) =
1
1 + z
1
,z >| |
2
=
1 1
2 2
1
X
2
(z) =
1
z
1
,z >,| |
4
1
1?
4
x
2
[n + 4] is nothing but x
2
[n] ← ω[n + 4],Convolution in time domain corresponds to multiplication
in z-transform domain; thus
4
1
4
z
x
2
[n + 4]
z
( )X?∩ z z
2
=
1
z
1
,z >,| |
41?
4
Similarly,we want to?rst compute z-transform of x
1
[n? 2],
2
1
2
z
x
1
[n? 2]
z
( )X?∩ z z
1
=
1
1 + z
1
,z >,| |
2
2
Note that x
1
[? n? 2] is the time reversed version of x
1
[n? 2],Thus,using the time reversal property
of z-transform,we get,
>
1
1
2
2
1 1z
x
1
[? n? 2]
x
1
[? n? 2]
z
∩
z
∩
X
1
=
1
1 +
,
z
2
2z z z
z
z
1
+
1
,| z| < 2,
2
Combining together,
4
z z 1
Y (z) =
1 1
z
1
,< z
z
1
+
·
4
| |?| z| < 2
2
1?
4
5
z 1
Y (z) = 2 < z < 2.| |
1
z
1
,
(1 + 2z
1
) 41?
4
15
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Electrical Engineering and Computer Science
6.003,Signals and Systems—Fall 2003
Problem Set 10 Solutions
Home study exercise (E1) O&W 11.32
(a)
Y (s) H(s)
= (via Black’s equation)
X(s) 1 + KG(s)H(s)
N
1
(s)D
2
(s)
= (multiply through by D
1
(s)D
2
(s))
D
1
(s)D
2
(s) + KN
1
(s)N
2
(s)
The zeros of this closed-loop system are the roots of N
1
(the zeros of H) and the roots of D
2
(the poles of G),
(b) If K = 0,then the system is operating without feedback of any sort,Naturally the poles and
zeros of the system without feedback are the poles and zeros of H(s),
More formally,we can take limits of the above equation,
N
1
(s)D
2
(s) N
1
(s)D
2
(s) N
1
(s)
lim = = = H(s)
K?0 D
1
(s)D
2
(s) + KN
1
(s)N
2
(s) D
1
(s)D
2
(s) D
1
(s)
(c) Check by simple substitution,
p(s) H(s)
Q(s) =
q(s)
·
1 + K
H(s)G(s)
N
1
(s)/p(s)
p(s)
D
1
(s)/q(s)
=
q(s)
·
1 + K
N
2
(s)/q(s) N
1
(s)/p(s)
D
2
(s)/p(s) D
1
(s)/q(s)
· ·
H(s)
=
1 + KG(s)H(s)
(d) We are given the following,
s + 1 s + 2
H(s) =,G(s) =
(s + 4)(s + 2) s + 1
In this simple example,we can?nd p and q by inspection,
1
p(s) = s + 1,q(s) = s + 2,H(s) =,G(s) = 1
s + 4
G(s) = Root locus equation,H(s)
1
=
1
=? s =?(K + 4)
s + 4
K
1
The closed-loop system has one?nite zero (s =?1),one,zero at in?nity”,one pole which is
not a?ected by K (s =?2),and one pole which is a?ected by K (s =?(K + 4)),
Note that the zeros in the feedback system can obscure certain complex frequencies in the
output; these frequencies will never show up in the error signal,This is another reason to avoid
pole-zero cancellation as a design technique,since it can cause unstable complex frequencies to
become unobservable in the error signal,
Problem 1
Let the output of G(s) be y(t),Then,we can obtain the transfer function from x(t) to y(t) using
Black’s formula,
Y (s) KG(s)
=,
X(s) 1 + KG(s)
Thus,the closed loop poles are s satisfying 1 + KG(s) = 0,To plot a root locus,you are sketching
the location of s as a function of K such that 1 + KG(s) = 0,
Having said that,all the points on the locus satisfy the following criteria usually referred to as
the angle criteria,
1 + KG(s
l
) = 0
1
G(s
l
) =,(1)?
K
where s
l
is on the locus,If K > 0,then
G(s
l
) be an odd integer multiple of?,while for K < 0,
then
G(s
l
) be an even integer multiple of?,
Also,from Eqn(1),it is easy to see that s
l
approaches to the poles of G(s) if K ∩ 0 and that | |
s
l
approaches to the zeros of G(s) if K ∩?,One thing to note here,however,is that by closing | |
the loop in the feedback con?guration shown in the problem,the number of the closed loop poles
is either the same as that of G(s) or less if pole-zero cancellation takes place,Thus,to make the
argument above applicable to any rational transfer functions,we need to have the same number of
poles and zeros for G(s),This leads us to de?ne poles or zeros at,in?nity” as mentioned in the
Home study exercise above,With all these concepts in your mind,let’s go through the questions,
(a) Given,
1
G(s) =,
s + 1
G(s) has a pole at?1 and no?nite zero.
For K > 0,In this case,with the angle criteria,all the points,s
l
on the locus should
satisfy
G(s
l
) = odd integer multiple of?.
So,if you take any point on the real line segment (,?1),
G(s
l
) = 0 ±? = ±?,
phase due to zeros
phase due to the pole
Also,this is the only segment that belongs to the locus,
2
×
For K < 0,The angle criteria states that:
G(s
l
) = even integer multiple of?.
So,if you take any point on the real line segment (?1,?),
G(s
l
) = 0 0 = 0.
phase due to zeros phase due to the pole
Combining the two cases,the root loci for both cases are plotted as shown below,The solid
line corresponds to the locus for K > 0 case and the dashed line for K < 0 case,
≤m
↓e?1
(b) Given,
1
G(s) =,
(s? 5)(s + 3)
G(s) has two poles at 5 and?3,and no?nite zero.
For K > 0, Evoking the angle criteria,we can?rst see that the real line segment (?3,5)
belongs to the locus,Now we would like to know at which point the locus branches out or
bifurcates from the real line,At the branching point,the closed loop poles are of double
pole,To?nd such a point,we can?rst?nd a value of K corresponding to the double pole,
1 + KG(s) = 0
K
1 + = 0
(s? 5)(s + 3)
(s? 5)(s + 3) + K
s
2
2s? 15 + K
(s? 1)
2
1? 15 + K
=
=
=
0
0
0 completing the square,
So,when?1? 15 + K = 0 or K = 16,the closed loop system has a double pole at 1,
Because of the symmetry of the poles about s = 1,the locus branches out from s = 1
along ↓e{s} = 1 parallel to the imaginary axis,This can be seen in the?gure below,
3
1
× ×
1
≤m
↓e
×
3
×
5
For K < 0, In this case,using the angle criteria,we can see that the real line segment
(5,?) belongs to the locus since both poles contributes 0 or even integer multiple of?,
Also,the real line segment (,?3) belongs to the locus since both poles contribute?
or odd integer multiple of?; the net phase of G(s
l
) is even integer multiple of?,
Thus,the root loci containing both K > 0 (solid line) and K < 0 (dashed line) cases are shown
below,
≤m
↓e?3 5
(c) Given,
s + 1
G(s) =
2
,
s
G(s) has a double-pole at 0 and one?nite zero at?1.
For K > 0, With the angle criteria,we can see that the real line segment (,?1)
belongs to the locus,Since both poles of G(s) do not belong to the segment,there will be
a double-pole point in the segment,To?nd where it is,use the same technique as in (b),
4
i.e.,complete the square and?nd the corresponding gain K and the bifurcation point.
1 + KG(s) = 0
s + 1
1 + K
s
2
= 0
s
2
+ Ks + K = 0
K
2
K
2
s +
2
+ K?
2
= 0
K
∩ K 1?
4
= 0
K = 0,4,
This result tells us that at K = 0 and K = 4 we have double-pole and the corresponding
locations are at s = 0 and s =
4
=?2.
2
How can we know the shape of the path that locus takes after branching out of s = 0 and
merging at s =?2? In the case in our hand,it is not hard to determine,Since the closed
loop poles for the K between 0 and 4 are a complex conjugate pair,we can assume that
s = ξ + j? for a given K,Then,using one of the equations above;
(ξ + j?)
2
+ K(ξ + j?) + K = 0
(ξ
2
+ 2jξ
2
) + Kξ + jK? + K = 0
For real part,ξ
2
2
+ Kξ + K = 0
For imaginary part,2ξ? + K? = 0 ∩ K =?2ξ
ξ
2
2
2ξ
2
2ξ = 0 (combining the two equations above)
ξ
2
+?
2
+ 2ξ = 0
(ξ + 1)
2
+?
2
= 1,
As you can see,the last equation is nothing but an equation of a circle whose radius is
1 and whose center is located at (ξ,?) = (?1,0),Thus,the locus is of the circle for
0 < K < 4,Then,at s =?2 the locus bifurcates to the zero at s =?1 and s = on
the real axis.
One thing to note here is that,in general,it is extremely di?cult to obtain expressions of
the locus for higher order systems.
For K < 0, In this case,we can see that the real line segments (?1,0) and (0,?) belong
to the locus,On the locus,one of the poles will move to the zero at s =?1 as K decreases
or?K increases,while the other pole moves out to? along the positive real axis,
Thus,the loci containing both K > 0 (solid) and K < 0 (dashed) cases are shown below,
5
×
≤m
2
1?2 e↓
Problem 2
1
(a) First let output of the plant G(s) =
s(s+10)
be y(t),Then,by Black’s formula,the closed
transfer function from the input x(t) to y(t) can be found as,
Y (s)
=
G(s)K(s)
X(s) 1 + G(s)K(s)
K
=
s(s+10)
1 +
K
s(s+10)
Y (s)
X(s)
=
K
s(s + 10) + K
,(2)
The error signal e(t) is the di?erence between x(t) and y(t),Thus from the linearity of Laplace
transforms,
e(t) = x(t)? y(t)
E(s) = X(s)? Y (s),(3)
So,with Eqns (2) and (3),we can obtain the transfer function from the input x(t) to the error
e(t),
6
E(s)
X(s)
= 1?
Y (s)
X(s)
Eqn (3)
K
= 1?
s(s + 1)
Eqn (2)
+ K
s(s + 10)
=
s(s + 10) + K
E(s) s(s + 10)
=, (4)
X(s) s
2
+ 10s + K
Thus,using Eqn (4),E(s) for the unit step input x(t) = u(t) is,
s(s + 10)
E(s) = X(s)
s
2
+ 10s + K
s(s + 10) 1
=
s
2
+ 10s + K s
s + 10
=
s
2
+ 10s + K
,
so if K → 0,we can see from Routh-Hurwitz criteria that E(s) has at least one of the poles in
the right half plane,i.e,the real part of the pole is positive,Hence,in this case,the steady
state tracking error diverges to?,
Now,consider the case when K > 0,Then,both poles of E(s) are in the left half plane and
the order of the numerator is strictly less than that of the denominator (the latter condition
is usually referred to as being strictly proper), Thus we can apply the?nal value theorem to
compute the steady state tracking error,
e(?) = lim e(t)
t
= lim sE(s)
s?0
s + 10
= lim s
s?0 s
2
+ 10s + K
= 0,
Thus,regardless of the value of K,as far as it is positive,e(?) = 0,
(b) To compute the error signal to the ramp input,we can use Eqn(4) with now X(s) =
s
1
2
,
s(s + 10) 1
E(s) =
2
s
2
+ 10s + K s
s + 10
=
s(s
2
+ 10s + K)
10 10
s + 1?
100
K
+
K K
=,
s s
2
+ 10s + K
E
1
(s)
E
2
(s)
7
= ,
e
1
(t),the corresponding time signal to E
1
(s) is a scaled step,E
2
(s) has two poles and they
remain in the left-half plane as long as K > 0,In this case e
2
(t),the corresponding time signal
to E
2
(s) will decay to 0 as t ∩?,Thus,e(?) =
10
K
,i.e.,the steady state error will not be 0,
On the other hand,if K →0,Then,E
2
(s) now has at least one of the poles in the right-half
plane,Thus,e
2
(?) ∩?as t ∩?; therefore e(?) ∩?,
(c) For this part,?rst we would like to get the transfer function from x(t) to y(t),Let’s denote
the output of K
f
,x
f
(t),then X
f
(s) = K
f
X(s),
1
Y (s) =
s(s + 10)
(X
f
(s) + K
s
X(s)?Y (s))
1
1
1 +
s(s + 10)
Y (s) =
s(s + 10)
(K
f
+ K
s
)X(s)
1
s(s+10)
(K
f
+ K
s
)
Y (s)
=
X(s) 1 +
1
s(s+10)
Y (s) K
f
+ K
s
X(s) s(s + 10) + 1
Since e(t) = x(t)?y(t),we can compute the Laplace transform of e(t),E(s) as follows,
E(s) = X(s)?Y (s)
Y (s)
E(s) = 1?
X(s)
X(s),
The input x(t) to the system is a ramp,so X(s) =
s
1
2
,Thus,E(s) is,
K
f
+ K
s
1
2
E(s) = 1?
s(s + 10) + 1 s
s
2
+ 10s + 1?(K
f
+ K
s
)
=
s
2
(s
2
+ 10s + 1)
∩sE(s) =
s
2
+ 10s + 1?(K
f
+ K
s
)
,
s(s
2
+ 10s + 1)
To be able to apply the?nal value theorem,we want to choose K
f
(s) and K
s
(s) such that the
poles of sE(s) are in the left-half plane,In addition,we need to choose K
f
(s) and K
s
(s) such
that
lim sE(s) = 0
s?0
to ensure that the stead state error to the ramp input is zero,Combining these two conditions,
we see that we need to make the numerator of sE(s) equal to s
2
,From this conclusion alone,
we only know that K
f
(s) + K
s
(s) = 10s + 1,Assuming that neither is 0,then we can choose
8
K
f
(s) = 10s and K
s
= 1,Then,
s
2
+ 10s + 1? (10s + 1)
sE(s) =
s(s
2
+ 10s + 1)
2
s
=
s(s
2
+ 10s + 1)
s
sE(s) =,
s
2
+ 10s + 1
Now,we can safely apply the?nal value theorem to obtain the steady state error and we indeed
obtain 0 as expected,
s
lim sE(s) = lim = 0,
s?0 s?0 s
2
+ 10s + 1
Problem 3 (O&W 11.27)
Given,
s + 2
H(s) =
s
2
+ 2s + 4
,G(s) = K,
First,let’s identify the poles and zeros of H(s),Since the system H(s) is a second order system,
is at?,The poles are the solutions of s
is as shown below,
2
s? ±
there are two poles and two zeros,It is clear that one of the zeros is of?nite at 2 and the other
+ 2 + 4 = 0,so 1 + 3,Thus,the pole-zero diagram
≤m
↓e
×
×
3
3
1?2
Since H(s) is causal,the ROC for H(s) is ↓e{s} >?1,
The closed loop poles are the solution to the following equation,
s + 2
1 + KH(s) = 1 + K = 0,
s
2
+ 2s + 4
9
or
2
s
2
+ 2s + 4 + K(s + 2) = s + (K + 2)s + 2K + 4 = 0,(5)
(a) For K > 0,for all s
l
on the locus
H(s
l
) is an odd integer multiple of?,Thus,we can?rst
see that the real line segment (,?2) belongs to the locus,Then,we would like to know at
which point on that segment there is a double-pole,On that point,Eqn (5) has a double root,
i.e.,
s
2
+ (K + 2)s + 2K + 4 = 0 completing the square
K + 2
2
K + 2
2
s +
2
2
+ 2K + 4 = 0
2
K + 2
∩?
2
+ 2K + 4 = 0
= 6.∩ K?2,
This tells us that there are two points where double pole occurs; one is when K = 6 > 0 and
the other is when K =?2 < 0,Here,we need to look at only the positive case,The negative
one will be used in part (b),For K = 6,the double-root is at?
K+2
=?4.
2
(b) For K < 0,using the angle criteria,we can see that the real line segment (?2,?) belongs to
the locus and the double-pole point is at?
K+2
= 0 where K =?2 as computed in (a),
2
(c) Since the closed-loop system is still just a second order system,the condition that the closed-
loop impulse response does not exhibit any oscillatory behavior simply means that the closed-
loop system is critically damped or overdamped,Since H(s) is underdamped i.e.,its poles are
not on the real axis,at the smallest K to be found,the denominator takes the following form,
2
s
2
+ 2
n
s +?
n
= 0
2
s
2
+ 2?
n
s +?
n
= 0,? = 1 critically damped
(s +?
n
)
2
= 0,
where?
n
is the undamped natural frequency of the closed loop system,Thus,the required
condition is nothing but the double-pole condition we found in (a),Since K is constrained to
be positive,the smallest K we are looking for is K = 6,
10
Problem 4
Note that z-transform is linear,
(a) Given,
x[n] = 2ω[n + 3]? ω[n? 2],
In this part of the problem,we would like to use one of the most fundamental z-transform
pairs,
ω[n? m]
z
m
,ROC,All z except for 0 if m > 0 or? if m < 0,(6)?∩ z
By applying this pair for both terms in the given x[n],we get:
2
X(z) = 2z
3
z
2z
5
1
=
2
,
z
with its ROC is all z except for 0 and?, Since the unit circle is included in the ROC,the
Fourier transform of the sequence exists,From the last expression,it can be seen that there are
two poles both of which are at 0,There are?ve zeros and they are the solutions of 2z
5
1 = 0,
1
2?
i.e,2
5
e
j?
z
where?
z
is integer multiples of
5
,
≤ m
↓ e
×
2
To see this,remember that any complex number z can be written in polar form as z = re
j?
where r is the length,or radius of the vector z and? is its phase,In our case using this idea,
we can?nd the zeros as follows,
11
2z
5
1 = 0
1
5
z =
2
1
5 j5?
z
r e =
2
√
1
1
5
1
r = = 2
5
2
j5?
z
= integer multiple of 2?,
The pole-zero diagram is depicted above,The 5 zeros are on the inner circle whose radius is
2
1/5
and the outer circle is a unit circle,2 at the origin denotes that there is a double pole at
the origin,The ROC is everywhere except for the origin and?,
(b) Given,
x[n] = 2
n
u[n? 1] + 4
n
u[? n],
x
1
[n] x
2
[n]
In this part,we would like to use another crucial z-transform pair,
a
n
u[n],z > a
z
1|
z
|
<
|
a
|
∩
1? az
1
,(7)
n
u[? n? 1],? a | | | |
Note that
x
1
[n] = 2
n
u[n? 1] = 2 · 2
n?1
u[n? 1],
and
x
2
[n] = 4
n
u[? n] = 4
n
u[? n? 1] + ω[n].
Let X
1
(z) and X
2
(z) be z-transforms of x
1
[n] and x
2
[n] respectively,Then,by applying Eqns
(6) and (7) to x
1
[n] and x
2
[n],we have,
2
X
1
(z) = 2 ·
1?
1
2z
1
· z
1
=
z? 2
,| z| > 2
1
X
2
(z) =? ·
1? 4z
1
+ 1 =
4
,| z| < 4,
z? 4
Thus,
X(z) = X
1
(z) + X
2
(z)
2 4
=
z? 2
z? 4
2z
X(z) =?
(z? 2)(z? 4)
12
with its ROC being an intersection of|z|> 2 and|z|< 4,i.e.,2 <|z|< 4,The ROC does not
contain the unit circle; thus the Fourier transform of the sequence does not exist,There are
two poles at 2 and 4 and is one zero at 0,Thus,the pole-zero plot with the ROC is depicted
below:
≤m
↓e0
×
2
×
4?1
One thing to note from Problem 4 is that it would be better to change the expression of z-
transforms into rational functions in z to?nd poles and zeros,while it would be better to
change the expressions into rational functions of z
1
to?nd the corresponding time sequences
since in many cases z-transform tables show almost all the transforms in terms of z
1
such as
the one on p.776 in O&W.
13
Problem 5
(a) Recall the z-transform pair,
m
z
z ].?∩ ω[n? m
Thus,we need to simply apply this to each term in the given expression,The answer therefore
is,
x[n] = 12ω[n? 4]? ω[n? 1] + 6ω[n] + 9ω[n + 2]? 8ω[n + 5],
(b) Given,
5 1 1
X(z) =
1 +
1
6
z
1
1
6
z
2
,< z <,
3
| |
2
First,we would like to perform partial fraction to reduce X(z) consisting of easily recognizable
terms,
5
X(z) =
1
6
z
1
5
1
6
z
2
1 +
=
1
3
z
1
1
2
z
1
1? 1 +
2 3
(8)
1
3
z
1
+=
1 +
1
2
z
1
,
1?
1
3
1
2
with ROC of < z <| |
Recall another very important z-transform pair,
,
1
z
a
n
u[n],z > a| |
(9)
n
1? az
1
∩
a u[? n? 1],z <
|
a
|
.| | | |
In Eqn (8),the?rst term corresponds to the right-sided signal pair above and the second pair
to the left-sided signal,Since z-transform is linear,we have,
1
n
1
n
x[n] = 2 u[? n? 1].u[n]? 3
3
2
14
Problem 6
Given,
y[n] = x
1
[? n? 2] ← x
2
[n + 4]
1
x
1
[n] =
2
u[n]?
1
x
2
[n] =
4
u[n],
In this problem,we would like to use Eqn(9) and two more basic properties of z-transform,First,
we can?nd X
1
(z) and X
2
(z),z-transforms of x
1
[n] and x
2
[n] respectively,
1
X
1
(z) =
1
1 + z
1
,z >| |
2
=
1 1
2 2
1
X
2
(z) =
1
z
1
,z >,| |
4
1
1?
4
x
2
[n + 4] is nothing but x
2
[n] ← ω[n + 4],Convolution in time domain corresponds to multiplication
in z-transform domain; thus
4
1
4
z
x
2
[n + 4]
z
( )X?∩ z z
2
=
1
z
1
,z >,| |
41?
4
Similarly,we want to?rst compute z-transform of x
1
[n? 2],
2
1
2
z
x
1
[n? 2]
z
( )X?∩ z z
1
=
1
1 + z
1
,z >,| |
2
2
Note that x
1
[? n? 2] is the time reversed version of x
1
[n? 2],Thus,using the time reversal property
of z-transform,we get,
>
1
1
2
2
1 1z
x
1
[? n? 2]
x
1
[? n? 2]
z
∩
z
∩
X
1
=
1
1 +
,
z
2
2z z z
z
z
1
+
1
,| z| < 2,
2
Combining together,
4
z z 1
Y (z) =
1 1
z
1
,< z
z
1
+
·
4
| |?| z| < 2
2
1?
4
5
z 1
Y (z) = 2 < z < 2.| |
1
z
1
,
(1 + 2z
1
) 41?
4
15
