MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Electrical Engineering and Computer Science
6.003,Signals and Systems|Fall 2003
Problem Set 11 Solutions
Problem 1 (O&W 10.29 (d))
In this problem we are asked to sketch the magnitude of the Fourier transform associated
with the pole-zero diagram,Figure P10.29 (d),In order to do so,we need to make some
assumptions:
(a) The ROC includes the unit circle to ensure the existence of the Fourier transform.
(b) The gain factor is one so that the system we are dealing with has the following form:
H(z) = 1(z z
1)(z + z1);
where z1 is a positive real number whose magnitude is less than 1.
To obtain the frequency response of a DT system,we need to evaluate the magnitude
and phase of H(z) along the unit circle in z plane,i:e:; z = ej! for 0 ! < 2, First,we
look at the magnitude plot,In general,for a xed ! we can think ofjH(ej!)jas
jH(ej!)j=
# of zeros
i=1 ( length of a vector connecting i th zero to e
j!)
# of polesj=1 ( length of a vector connecting j th pole to ej!)
If # of zeros or poles is zero,then we de ne the product above to be 1,In our case,there
are two poles and no zeros,Thus,the above expression can be s implied to:
jH(ej!)j= 1jv
1jjv2j;
where v1 and v2 are vectors shown in the gure below:
1
<e
=m
1
1
1
1
! \~v1
~v1
\~v2
~v2
z1 z1
ej!
From the plot,we see that when ! = 2,the product of jv1j and jv2j becomes maximum.
Thus,we expect to have the minimum magnitude at the frequency,Also,when ! = 0 or
,the product of the lengths or the two vectors becomes minimum; thus the magnitude of
H(ej!) becomes maximum,The magnitude plot of H(ej!) for 0 ! < is shown below:
jH(ej!)j
! 2
Although you are not asked to sketch the phase,here is a brie y outline on how to sketch
the phase.
2
The phase \H(ej!) can be described as:
\H(ej!) =
# of zerosX
i=1
( angle of vector connecting i th zero to ej!)
# of polesX
j=1
( angle of vector connecting j th pole to ej!):
Again for our speci c case,using the vectors v1 and v2 de ned above we have
\H(ej!) = (\v1 + \v2);
The phase starts o at 0 when ! = 0 and decreases to when ! = 2,Because of the
symmetric poles,the phase keeps decreasing to 2 at ! =, The phase plot is shown
below:
\H(ej!)
! 2
2
Problem 2 (O&W 10.34)
(a)
y[n] = y[n 1] + y[n 2] + x[n 1]
Taking the z-transform of this equation:
Y (z) = z 1Y (z) + z 2Y (z) + z 1X(z)
H(z) = Y (z)X(z) = z
1
1 z 1 z 2 =
z
z2 z 1
= z
z 1 +
p5
2
!
z 1
p5
2
!
H(z) has a zero at z = 0 and poles at z1 = 1+
p5
2 and z2 =
1 p5
2,Since the system is causal,the ROC of H(z) will be outside the circle containing its
outermost pole,jzj>jz1j,The pzmap and ROC are depicted below:
3
0
1+p5
2
1 p5
2
<e
=m
(b)
H(z) = z
1
z 1 + 1 +
p5
2
!
z 1 + 1
p5
2
!
= A
z 1 + 1 +
p5
2
+ B
z 1 + 1
p5
2
A =
z 1 + 1 +
p5
2
!
H(z)jz 1= 1+p5
2
= 1 +
p5
2p5
B =
z 1 + 1
p5
2
!
H(z)jz 1= 1 p5
2
= 1
p5
2p5
H(z) =
1p
5
1 + 21+p5z 1 +
1p
5
1 + 21 p5z 1 ; jzj>
1 +p5
2
4
Taking the inverse z-transform:
h[n] = 1p5
21 +p5
n
u[n] + 1p5
21 p5
n
u[n]
(c) The system is unstable,as its ROC does not contain the unit circle,The instability is
also apparent in h[n],as the
21 p5
n
term will grow inde nitely as n!1.
In order to make the system stable,the ROC must contain the unit circle,For stability,
the ROC should be,2 1 p5 <jzj< 2 1+p5.
The inverse z-transform of H(z) with this ROC is:
h[n] = 1p5
21 +p5
n
u[n] 1p5
21 p5
n
u[ n 1]
Problem 3 (O&W 10.42)
Because we are dealing with a system which has initial conditions,we may want to use
the unilateral z-transform,From the properties of the unilateral z-transform,we get the
following relationships:
y[n] ! Y (z)
y[n 1] ! z 1Y (z) + y[ 1]
In each part of this problem,the rst step is to take the unilateral z-transform of both sides
of the di erence equation,To nd the zero-input response (ZIR),set the input to 0 and
solve for Y (z),To nd the zero-state response (ZSR),set the initial conditions to zero and
solve for Y (z).
(a)
y[n] + 3y[n 1] = x[n]; y[ 1] = 1; x[n] =
1
2
n
u[n]
Taking the unilateral z-transform of both sides of the di erence equation,
Y (z) + 3z 1Y (z) + 3y[ 1] = X(z):
Solving for Y (z) gives,
Y (z) = X(z)1 + 3z 1
| {z }
ZSR
+ 3y[ 1]1 + 3z 1
| {z }
ZIR
:
5
To nd the ZIR,set X(z) = 0 and use the fact that y[ 1] = 1,
YZIR(z) = 31 + 3z 1
yZIR[n] = 3( 3)nu[n] = ( 3)n+1u[n]:
To nd the ZSR,set y[ 1] = 0 and use x[n] = 12 n u[n] as given in the problem,
X(z) = 11 1
2z
1
YZSR(z) = 1(1 + 3z 1) 1 1
2z
1 =
6
7
1 + 3z 1 +
1
7
1 12z 1
yZSR[n] = 67( 3)nu[n] + 17
1
2
n
u[n]:
(b)
y[n] 12y[n 1] = x[n] 12x[n 1]; y[ 1] = 0; x[n] = u[n]
Taking the unilateral z-transform of both sides of the di erence equation,
Y (z) 12z 1Y (z) 12y[ 1] = X(z) 12z 1X(z) 12x[ 1]:
Solving for Y (z) gives,
Y (z) = (1
1
2z
1)X(z)
1 12z 1 +
12x[ 1] + 12y[ 1]
1 12z 1
= X(z)|{z}
ZSR
+
1
2x[ 1] +
1
2y[ 1]
1 12z 1| {z }
ZIR
:
To nd the ZIR,set X(z) = 0 (note that this means x[ 1] = 0),and use the fact that
y[ 1] = 0,
YZIR(z) = 0 =) yZIR[n] = 0:
To nd the ZSR,set y[ 1] = 0 and use x[n] = u[n] as given in the problem,
X(z) = 11 z 1
YZSR(z) = 11 z 1
yZSR[n] = u[n]:
6
(c)
y[n] 12y[n 1] = x[n] 12x[n 1]; y[ 1] = 1; x[n] = u[n]
Since this is the same di erence equation as given in part (b),we can use the equation
for Y (z) derived above,To nd the ZIR,set X(z) = 0 (note that this means x[ 1] = 0),
and use the fact that y[ 1] = 1,
YZIR(z) =
1
2
1 12z 1
yZIR[n] = 12
1
2
n
u[n] =
1
2
n+1
u[n]:
To nd the ZSR,set y[ 1] = 0 and use x[n] = u[n] as given in the problem,This is
exactly what we did in part (b) above,so
yZSR[n] = u[n]:
Problem 4 (O&W 10.47)
(a) Recall that complex exponentials are the eigenfunction of LTI systems,Thus,the re-
sponse of LTI systems to complex exponentials is of the form:
zn !H(z)zn:
Since the output of the system to an input x[n] = ( 2)n is 0,we conclude that the ROC
of H(z) contains z = 2 and H(z) has a zero at z = 2.
The second input-output relation gives us:
H(z) = Y (x)X(z) =
1 + a1 1
4z
1
1
1 12z 1; jzj> 12
= (1 + a
1
4z
1)(1 1
2z
1)
1 14z 1
= (1 + a)
(z 14(1+a))(z 12)
z(z 14) ;
assuming that 1 + a6= 0,H(z) has two poles; at z = 0 and z = 14,In order to include
z = 2 in its ROC,we know that the system is causal and its ROC is outside of the
circle of radius 14,Since H(z) has a zero at z = 2,
1 + a 14z 1jz= 2 = 0)a = 98:
7
(b) Now we know the expression for H(z):
H(z) = (
1
8
1
4z
1)(1 1
2z
1)
1 14z 1,
The input in this case is a complex exponential x[n] = 1n.
The output will therefore be of the form:
y[n] = H(1) 1n =
3
8
1
2
3
4
= 14:
Problem 5 (O&W 10.50)
(a) From the pole-zero pattern,the system function takes the following form:
H(z) = z
1
a
z a:
Showing thatjH(ej!)jis constant is equivalent to showing thatjH(ej!)j2 is constant.
jH(ej!)j2 = H(ej!)H (ej!) = e
j! 1
a
ej! a
e j! 1a
e j! a
= 1
2
a cos(!) +
1
a2
1 2a cos(!) + a2 =
1
a2
a2 2a cos(!) + 1
1 2a cos(!) + a2
= 1a2,
Thus,jH(ej!)j= 1jaj which is constant and is completely determined by the location of
the pole and zero.
(b) Using the law of cosines,
jv1j2 = j1j2 +jaj2 2j1jjajcos(!)
= 1 + a2 2a cos(!):
(c) Following the same procedure as in (b),
jv2j2 = j1j2 +
1
a
2
2j1j
1
a
cos(!)
= 1a2 a2 + 1 2a cos(!) = 1a2jv1j2:
)jv2j =
v1a
,
8
It clearly shows that the length of v2 is proportional in length to v1 independently of !.
Problem 6 (O&W 11.25 (a))
The system given is:
G(z)H(z) = z 1z2 1
4
:
First,we would like to nd the poles and zeros of the system,It is easy to see that
G(z)H(z) can be expressed as:
G(z)H(z) = z 1 z + 1
2
z 1
2
, (1)
The system has poles at 12 and a zero at 1,The pole-zero map as is as shown below.
Note that dashed circle in all the gures in this problem denotes a unit circle.
12
1
2
10 <e
=m
Recall that the derivation of the angle criteria does not concern if the system is CT or
DT,Thus,by evoking the angle criteria we have:
\G(z)H(z) = odd integer multiple of if K > 0,
\G(z)H(z) = even integer multiple of if K < 0.
9
for all z on the locus.
Let’s look at the case where K > 0,Using the angle criteria,we can identify that the
two real line segments belong to the locus,One is ( 1; 12) and the other (12; 1),In this
case,these two segments completely specify the locus,From this,we can see that a pole at
1
2 will approach to the zero at 1 as K !1,The other pole,one at
1
2 will move to 1.Thus,as K !1,one of the poles will become unstable (Note that in feedback systems,we
only consider causal systems),Thus,we would like to know up to which K,the closed loop
system remains stable,i.e.,we would like to nd the value of K when the closed loop system
has a pole at 1,From Eqn (1),
G(z)H(z)jz= 1 = 1K
z 1
z + 12 z 12
z= 1
= 1K
1 1
1 + 12 1 12 =
1
K
) K = 38:
12
1
2
10 <e
=m
K = 38
Thus,the root locus for K > 0 is shown above.
10
Now let’s look at the case where K < 0,If we apply the angle criteria only on the real
axis,then we identify that again there are two segments belonging to the locus,One is
( 12; 12) and the other is (1;1),As we know,all the poles in G(z)H(z) will go to the zeros
in G(z)H(z) as jKj!1,Thus,at some point on ( 12; 12),the locus branches o from the
real axis and go to a point on (1;1) to reach to the zero at 1 and at 1,Thus,rst we
would like to know where those branching,or usually referred to bifurcation,points are,At
bifurcation points,multiple poles exist at those points; in our case there will be two identical
poles at bifurcation points,Since there are only two poles for the system,we want to nd a
value of K such that
1 + KG(z)H(z) = 0 (2)
has double-root.
1 + K z 1z2 1
4
= 0
z2 14 + K(z 1) = 0
z + K2
2
K
2
2
K 14 = 0 by completing the square.
Thus,in order to have multiple roots K needs to satisfy the following:
K
2
2
+ K + 14 = 0
) K = 2 p3:
As expected,both K values are indeed negative,and at those values of K,bifurcation takes
place and the corresponding locations are at z = K2 = 1
p3
2,
Can we nd the value of K up to which the closed loop system remains stable as K
changes from 0 to 1? In this case,it is not as easy as in the case of K > 0 since we
do not know where the locus crosses the unit circle,However,we do know that on the unit
circle,z = ej!0 for some !0,Thus,from Eqn (2),we have:
G(z)H(z)jz=ej!0 = 1K
0
ej!0 1
e2j!0 14 =
1
K0 ; (3)
where K0 corresponds to !0.
One thing to note is that G(z)H(z) is a rational function in z and all the coe cients are
real,Thus,the locus is symmetric about the real axis; speci cally the points where the locus
11
crosses the unit circle are of complex conjugate,Thus,with the same K0 as in Eqn (3) the
following holds:
G(z)H(z)jz=e j!0 = 1K
0
e j!0 1
e 2j!0 14 =
1
K0 ; (4)
and combining Eqns (3) and (4),we get:
e j!0 1
e 2j!0 14 =
ej!0 1
e2j!0 14
(e j!0 1)
e2j! 14
= (ej!0 1)
e 2j!0 14
ej!0 e2j!0 14e j!0 + 14 = e j!0 e 2j!0 14ej!0 + 14
5
4(e
j!0 e j!0) = (e2j!0 e 2j!0)
sin !0
sin 2!0 =
sin !0
2 sin!0 cos !0 =
4
5
) cos !0 = 58:
This result along with Eqn(3) yields
K0 = 54:
Finally,we want to determine the expression of the locus between the bifurcation points.
Let z = + j! and nd the relation between and !,From Eqn (2):
1 + K z 1z2 1
4
= 0
z2 14 + K(z 1) = 0
( + j!)2 + K( + j!) 14 K = 0
For real part,2 !2 + K 14 K = 0
For imaginary part,2 ! + K! = 0!K = 2
2 !2 2 2 14 + 2 = 0
) ( 1)2 + !2 =
p
3
2
!2
:
12
The locus traces a circle whose center is at ( ; !) = (1; 0) in the complex plane and whose
radius is
p3
2,Thus,the complete root locus for K < 0 case is shown below:
12
1
2
10
(1
p3
2 ; 0)
1 +
p3
2
<e
=m
K = 54
Problem 7 (O&W 11.59)
(a) Use Black’s equation to compute the transfer function from x[n] to e[n] for the system:
Y (z)
X(z) =
H(z)
1 + H(z) =
1
(z 1)(z + 12) + 1 =
1
z2 12z + 12
E(z)
X(z) = 1
Y (z)
X(z) =
z2 12z 12
z2 12z + 12
The transfer function has two poles at z = 14 j
q
7
16,Note that the magnitude of each
pole is 1=p2 0:707,so the system is stable.
Since the system is stable and LTI,it has a frequency response,and we can use a familiar
trick,To nd the steady-state response to a step input,evaluate the frequency response
13
at ! = 0 (z = ej! = 1):
x[n] = u[n] =) lim
n!1
e[n] = z
2 1
2z
1
2
z2 12z + 12
z=ej0=1
= 0
(b) Note that our equation for the E=X transfer function can be rewritten as follows:
E(z)
X(z) = 1
Y (z)
X(z) = 1
H(z)
1 + H(z) =
1
1 + H(z)
If x[n] = u[n],then X(z) = 11 z 1,and we have the following:
E(z) = zz 1 11 + H(z)
We assume the system is stable,so all the roots of 1 + H(z) must lie within the unit
circle,In particular,given the bounded input x[n] = u[n],we know that the output
e[n] must be bounded; all the poles of E(z) must lie within the unit circle,Since the
necessary Fourier transforms exist,we can use the familiar trick:
x[n] = u[n] =) lim
n!1
e[n] = 11 + H(z)
z=ej0=1
= 0
The last step follows because we assume that H(z) has a pole at z = 1,Thus the
denominator becomes unbounded as z!1,and the system response goes to zero.
(c) Plug the new transfer function into our equations:
E(z)
X(z) =
1
1 + H(z) =
1 z 1
1 z 1 + z 1 = 1 z
1
This system has one pole at z = 0; it is stable,By inspection,the impulse response of
the error is e[n] = [n] [n 1],and the step response of the error is e[n] = [n]; the
system tracks a unit step input without error after one time-step.
(d) Plug the new transfer function into our equations:
E(z)
X(z) =
1
1 + H(z)
= (1 +
1
4z
1)(1 z 1)
(1 + 14z 1)(1 z 1) + 34z 1 + 14z 2
= 1
3
4z
1 1
4z
2
1 34z 1 14z 2 + 34z 1 + 14z 2 = 1
3
4z
1 1
4z
2
This system has two poles at z = 0; it is stable,By inspection,the impulse response
of the error is e[n] = [n] 34 [n 1] 14 [n 2],and the step response of the error
is e[n] = [n] + 14 [n 1]; the system tracks a unit step input without error after two
time-steps.
14
(e) We are given the desired step response of the error signal e[n]:
e[n] =
N 1X
k=0
ak [n k]
=)E(z) =
N 1X
k=0
akz k
From our work in part (b),we also have the following equation for the step response of
e[n]:
E(z) = 11 z 1 11 + H(z)
Set the two equations equal and solve for H(z):
N 1X
k=0
akz k = 11 z 1 11 + H(z)
1 + H(z) = 11 z 1 1PN 1
k=0 akz k
H(z) = 1 (1 z
1)PN 1
k=0 akz
k
(1 z 1)PN 1k=0 akz k
(f) Find the transform of x[n] using the tables:
x[n] = (n + 1)u[n] = nu[n] + u[n]
=)X(z) = z
1
(1 z 1)2 +
1
1 z 1 =
1
(1 z 1)2
Plug the new transfer function H(z) into our equations:
E(z) = X(z) 11 + H(z) = 1(1 z 1)2 (1 + z
1)(1 z 1)2
(1 + z 1)(1 z 1)2 + z 1 + z 2 z 3
= 1 + z
1
(1 + z 1)(1 2z 1 + z 2) + z 1 + z 2 z 3
= 1 + z 1
By inspection,e[n] = [n]+ [n 1],so the system tracks perfectly after two time-steps.
15
Department of Electrical Engineering and Computer Science
6.003,Signals and Systems|Fall 2003
Problem Set 11 Solutions
Problem 1 (O&W 10.29 (d))
In this problem we are asked to sketch the magnitude of the Fourier transform associated
with the pole-zero diagram,Figure P10.29 (d),In order to do so,we need to make some
assumptions:
(a) The ROC includes the unit circle to ensure the existence of the Fourier transform.
(b) The gain factor is one so that the system we are dealing with has the following form:
H(z) = 1(z z
1)(z + z1);
where z1 is a positive real number whose magnitude is less than 1.
To obtain the frequency response of a DT system,we need to evaluate the magnitude
and phase of H(z) along the unit circle in z plane,i:e:; z = ej! for 0 ! < 2, First,we
look at the magnitude plot,In general,for a xed ! we can think ofjH(ej!)jas
jH(ej!)j=
# of zeros
i=1 ( length of a vector connecting i th zero to e
j!)
# of polesj=1 ( length of a vector connecting j th pole to ej!)
If # of zeros or poles is zero,then we de ne the product above to be 1,In our case,there
are two poles and no zeros,Thus,the above expression can be s implied to:
jH(ej!)j= 1jv
1jjv2j;
where v1 and v2 are vectors shown in the gure below:
1
<e
=m
1
1
1
1
! \~v1
~v1
\~v2
~v2
z1 z1
ej!
From the plot,we see that when ! = 2,the product of jv1j and jv2j becomes maximum.
Thus,we expect to have the minimum magnitude at the frequency,Also,when ! = 0 or
,the product of the lengths or the two vectors becomes minimum; thus the magnitude of
H(ej!) becomes maximum,The magnitude plot of H(ej!) for 0 ! < is shown below:
jH(ej!)j
! 2
Although you are not asked to sketch the phase,here is a brie y outline on how to sketch
the phase.
2
The phase \H(ej!) can be described as:
\H(ej!) =
# of zerosX
i=1
( angle of vector connecting i th zero to ej!)
# of polesX
j=1
( angle of vector connecting j th pole to ej!):
Again for our speci c case,using the vectors v1 and v2 de ned above we have
\H(ej!) = (\v1 + \v2);
The phase starts o at 0 when ! = 0 and decreases to when ! = 2,Because of the
symmetric poles,the phase keeps decreasing to 2 at ! =, The phase plot is shown
below:
\H(ej!)
! 2
2
Problem 2 (O&W 10.34)
(a)
y[n] = y[n 1] + y[n 2] + x[n 1]
Taking the z-transform of this equation:
Y (z) = z 1Y (z) + z 2Y (z) + z 1X(z)
H(z) = Y (z)X(z) = z
1
1 z 1 z 2 =
z
z2 z 1
= z
z 1 +
p5
2
!
z 1
p5
2
!
H(z) has a zero at z = 0 and poles at z1 = 1+
p5
2 and z2 =
1 p5
2,Since the system is causal,the ROC of H(z) will be outside the circle containing its
outermost pole,jzj>jz1j,The pzmap and ROC are depicted below:
3
0
1+p5
2
1 p5
2
<e
=m
(b)
H(z) = z
1
z 1 + 1 +
p5
2
!
z 1 + 1
p5
2
!
= A
z 1 + 1 +
p5
2
+ B
z 1 + 1
p5
2
A =
z 1 + 1 +
p5
2
!
H(z)jz 1= 1+p5
2
= 1 +
p5
2p5
B =
z 1 + 1
p5
2
!
H(z)jz 1= 1 p5
2
= 1
p5
2p5
H(z) =
1p
5
1 + 21+p5z 1 +
1p
5
1 + 21 p5z 1 ; jzj>
1 +p5
2
4
Taking the inverse z-transform:
h[n] = 1p5
21 +p5
n
u[n] + 1p5
21 p5
n
u[n]
(c) The system is unstable,as its ROC does not contain the unit circle,The instability is
also apparent in h[n],as the
21 p5
n
term will grow inde nitely as n!1.
In order to make the system stable,the ROC must contain the unit circle,For stability,
the ROC should be,2 1 p5 <jzj< 2 1+p5.
The inverse z-transform of H(z) with this ROC is:
h[n] = 1p5
21 +p5
n
u[n] 1p5
21 p5
n
u[ n 1]
Problem 3 (O&W 10.42)
Because we are dealing with a system which has initial conditions,we may want to use
the unilateral z-transform,From the properties of the unilateral z-transform,we get the
following relationships:
y[n] ! Y (z)
y[n 1] ! z 1Y (z) + y[ 1]
In each part of this problem,the rst step is to take the unilateral z-transform of both sides
of the di erence equation,To nd the zero-input response (ZIR),set the input to 0 and
solve for Y (z),To nd the zero-state response (ZSR),set the initial conditions to zero and
solve for Y (z).
(a)
y[n] + 3y[n 1] = x[n]; y[ 1] = 1; x[n] =
1
2
n
u[n]
Taking the unilateral z-transform of both sides of the di erence equation,
Y (z) + 3z 1Y (z) + 3y[ 1] = X(z):
Solving for Y (z) gives,
Y (z) = X(z)1 + 3z 1
| {z }
ZSR
+ 3y[ 1]1 + 3z 1
| {z }
ZIR
:
5
To nd the ZIR,set X(z) = 0 and use the fact that y[ 1] = 1,
YZIR(z) = 31 + 3z 1
yZIR[n] = 3( 3)nu[n] = ( 3)n+1u[n]:
To nd the ZSR,set y[ 1] = 0 and use x[n] = 12 n u[n] as given in the problem,
X(z) = 11 1
2z
1
YZSR(z) = 1(1 + 3z 1) 1 1
2z
1 =
6
7
1 + 3z 1 +
1
7
1 12z 1
yZSR[n] = 67( 3)nu[n] + 17
1
2
n
u[n]:
(b)
y[n] 12y[n 1] = x[n] 12x[n 1]; y[ 1] = 0; x[n] = u[n]
Taking the unilateral z-transform of both sides of the di erence equation,
Y (z) 12z 1Y (z) 12y[ 1] = X(z) 12z 1X(z) 12x[ 1]:
Solving for Y (z) gives,
Y (z) = (1
1
2z
1)X(z)
1 12z 1 +
12x[ 1] + 12y[ 1]
1 12z 1
= X(z)|{z}
ZSR
+
1
2x[ 1] +
1
2y[ 1]
1 12z 1| {z }
ZIR
:
To nd the ZIR,set X(z) = 0 (note that this means x[ 1] = 0),and use the fact that
y[ 1] = 0,
YZIR(z) = 0 =) yZIR[n] = 0:
To nd the ZSR,set y[ 1] = 0 and use x[n] = u[n] as given in the problem,
X(z) = 11 z 1
YZSR(z) = 11 z 1
yZSR[n] = u[n]:
6
(c)
y[n] 12y[n 1] = x[n] 12x[n 1]; y[ 1] = 1; x[n] = u[n]
Since this is the same di erence equation as given in part (b),we can use the equation
for Y (z) derived above,To nd the ZIR,set X(z) = 0 (note that this means x[ 1] = 0),
and use the fact that y[ 1] = 1,
YZIR(z) =
1
2
1 12z 1
yZIR[n] = 12
1
2
n
u[n] =
1
2
n+1
u[n]:
To nd the ZSR,set y[ 1] = 0 and use x[n] = u[n] as given in the problem,This is
exactly what we did in part (b) above,so
yZSR[n] = u[n]:
Problem 4 (O&W 10.47)
(a) Recall that complex exponentials are the eigenfunction of LTI systems,Thus,the re-
sponse of LTI systems to complex exponentials is of the form:
zn !H(z)zn:
Since the output of the system to an input x[n] = ( 2)n is 0,we conclude that the ROC
of H(z) contains z = 2 and H(z) has a zero at z = 2.
The second input-output relation gives us:
H(z) = Y (x)X(z) =
1 + a1 1
4z
1
1
1 12z 1; jzj> 12
= (1 + a
1
4z
1)(1 1
2z
1)
1 14z 1
= (1 + a)
(z 14(1+a))(z 12)
z(z 14) ;
assuming that 1 + a6= 0,H(z) has two poles; at z = 0 and z = 14,In order to include
z = 2 in its ROC,we know that the system is causal and its ROC is outside of the
circle of radius 14,Since H(z) has a zero at z = 2,
1 + a 14z 1jz= 2 = 0)a = 98:
7
(b) Now we know the expression for H(z):
H(z) = (
1
8
1
4z
1)(1 1
2z
1)
1 14z 1,
The input in this case is a complex exponential x[n] = 1n.
The output will therefore be of the form:
y[n] = H(1) 1n =
3
8
1
2
3
4
= 14:
Problem 5 (O&W 10.50)
(a) From the pole-zero pattern,the system function takes the following form:
H(z) = z
1
a
z a:
Showing thatjH(ej!)jis constant is equivalent to showing thatjH(ej!)j2 is constant.
jH(ej!)j2 = H(ej!)H (ej!) = e
j! 1
a
ej! a
e j! 1a
e j! a
= 1
2
a cos(!) +
1
a2
1 2a cos(!) + a2 =
1
a2
a2 2a cos(!) + 1
1 2a cos(!) + a2
= 1a2,
Thus,jH(ej!)j= 1jaj which is constant and is completely determined by the location of
the pole and zero.
(b) Using the law of cosines,
jv1j2 = j1j2 +jaj2 2j1jjajcos(!)
= 1 + a2 2a cos(!):
(c) Following the same procedure as in (b),
jv2j2 = j1j2 +
1
a
2
2j1j
1
a
cos(!)
= 1a2 a2 + 1 2a cos(!) = 1a2jv1j2:
)jv2j =
v1a
,
8
It clearly shows that the length of v2 is proportional in length to v1 independently of !.
Problem 6 (O&W 11.25 (a))
The system given is:
G(z)H(z) = z 1z2 1
4
:
First,we would like to nd the poles and zeros of the system,It is easy to see that
G(z)H(z) can be expressed as:
G(z)H(z) = z 1 z + 1
2
z 1
2
, (1)
The system has poles at 12 and a zero at 1,The pole-zero map as is as shown below.
Note that dashed circle in all the gures in this problem denotes a unit circle.
12
1
2
10 <e
=m
Recall that the derivation of the angle criteria does not concern if the system is CT or
DT,Thus,by evoking the angle criteria we have:
\G(z)H(z) = odd integer multiple of if K > 0,
\G(z)H(z) = even integer multiple of if K < 0.
9
for all z on the locus.
Let’s look at the case where K > 0,Using the angle criteria,we can identify that the
two real line segments belong to the locus,One is ( 1; 12) and the other (12; 1),In this
case,these two segments completely specify the locus,From this,we can see that a pole at
1
2 will approach to the zero at 1 as K !1,The other pole,one at
1
2 will move to 1.Thus,as K !1,one of the poles will become unstable (Note that in feedback systems,we
only consider causal systems),Thus,we would like to know up to which K,the closed loop
system remains stable,i.e.,we would like to nd the value of K when the closed loop system
has a pole at 1,From Eqn (1),
G(z)H(z)jz= 1 = 1K
z 1
z + 12 z 12
z= 1
= 1K
1 1
1 + 12 1 12 =
1
K
) K = 38:
12
1
2
10 <e
=m
K = 38
Thus,the root locus for K > 0 is shown above.
10
Now let’s look at the case where K < 0,If we apply the angle criteria only on the real
axis,then we identify that again there are two segments belonging to the locus,One is
( 12; 12) and the other is (1;1),As we know,all the poles in G(z)H(z) will go to the zeros
in G(z)H(z) as jKj!1,Thus,at some point on ( 12; 12),the locus branches o from the
real axis and go to a point on (1;1) to reach to the zero at 1 and at 1,Thus,rst we
would like to know where those branching,or usually referred to bifurcation,points are,At
bifurcation points,multiple poles exist at those points; in our case there will be two identical
poles at bifurcation points,Since there are only two poles for the system,we want to nd a
value of K such that
1 + KG(z)H(z) = 0 (2)
has double-root.
1 + K z 1z2 1
4
= 0
z2 14 + K(z 1) = 0
z + K2
2
K
2
2
K 14 = 0 by completing the square.
Thus,in order to have multiple roots K needs to satisfy the following:
K
2
2
+ K + 14 = 0
) K = 2 p3:
As expected,both K values are indeed negative,and at those values of K,bifurcation takes
place and the corresponding locations are at z = K2 = 1
p3
2,
Can we nd the value of K up to which the closed loop system remains stable as K
changes from 0 to 1? In this case,it is not as easy as in the case of K > 0 since we
do not know where the locus crosses the unit circle,However,we do know that on the unit
circle,z = ej!0 for some !0,Thus,from Eqn (2),we have:
G(z)H(z)jz=ej!0 = 1K
0
ej!0 1
e2j!0 14 =
1
K0 ; (3)
where K0 corresponds to !0.
One thing to note is that G(z)H(z) is a rational function in z and all the coe cients are
real,Thus,the locus is symmetric about the real axis; speci cally the points where the locus
11
crosses the unit circle are of complex conjugate,Thus,with the same K0 as in Eqn (3) the
following holds:
G(z)H(z)jz=e j!0 = 1K
0
e j!0 1
e 2j!0 14 =
1
K0 ; (4)
and combining Eqns (3) and (4),we get:
e j!0 1
e 2j!0 14 =
ej!0 1
e2j!0 14
(e j!0 1)
e2j! 14
= (ej!0 1)
e 2j!0 14
ej!0 e2j!0 14e j!0 + 14 = e j!0 e 2j!0 14ej!0 + 14
5
4(e
j!0 e j!0) = (e2j!0 e 2j!0)
sin !0
sin 2!0 =
sin !0
2 sin!0 cos !0 =
4
5
) cos !0 = 58:
This result along with Eqn(3) yields
K0 = 54:
Finally,we want to determine the expression of the locus between the bifurcation points.
Let z = + j! and nd the relation between and !,From Eqn (2):
1 + K z 1z2 1
4
= 0
z2 14 + K(z 1) = 0
( + j!)2 + K( + j!) 14 K = 0
For real part,2 !2 + K 14 K = 0
For imaginary part,2 ! + K! = 0!K = 2
2 !2 2 2 14 + 2 = 0
) ( 1)2 + !2 =
p
3
2
!2
:
12
The locus traces a circle whose center is at ( ; !) = (1; 0) in the complex plane and whose
radius is
p3
2,Thus,the complete root locus for K < 0 case is shown below:
12
1
2
10
(1
p3
2 ; 0)
1 +
p3
2
<e
=m
K = 54
Problem 7 (O&W 11.59)
(a) Use Black’s equation to compute the transfer function from x[n] to e[n] for the system:
Y (z)
X(z) =
H(z)
1 + H(z) =
1
(z 1)(z + 12) + 1 =
1
z2 12z + 12
E(z)
X(z) = 1
Y (z)
X(z) =
z2 12z 12
z2 12z + 12
The transfer function has two poles at z = 14 j
q
7
16,Note that the magnitude of each
pole is 1=p2 0:707,so the system is stable.
Since the system is stable and LTI,it has a frequency response,and we can use a familiar
trick,To nd the steady-state response to a step input,evaluate the frequency response
13
at ! = 0 (z = ej! = 1):
x[n] = u[n] =) lim
n!1
e[n] = z
2 1
2z
1
2
z2 12z + 12
z=ej0=1
= 0
(b) Note that our equation for the E=X transfer function can be rewritten as follows:
E(z)
X(z) = 1
Y (z)
X(z) = 1
H(z)
1 + H(z) =
1
1 + H(z)
If x[n] = u[n],then X(z) = 11 z 1,and we have the following:
E(z) = zz 1 11 + H(z)
We assume the system is stable,so all the roots of 1 + H(z) must lie within the unit
circle,In particular,given the bounded input x[n] = u[n],we know that the output
e[n] must be bounded; all the poles of E(z) must lie within the unit circle,Since the
necessary Fourier transforms exist,we can use the familiar trick:
x[n] = u[n] =) lim
n!1
e[n] = 11 + H(z)
z=ej0=1
= 0
The last step follows because we assume that H(z) has a pole at z = 1,Thus the
denominator becomes unbounded as z!1,and the system response goes to zero.
(c) Plug the new transfer function into our equations:
E(z)
X(z) =
1
1 + H(z) =
1 z 1
1 z 1 + z 1 = 1 z
1
This system has one pole at z = 0; it is stable,By inspection,the impulse response of
the error is e[n] = [n] [n 1],and the step response of the error is e[n] = [n]; the
system tracks a unit step input without error after one time-step.
(d) Plug the new transfer function into our equations:
E(z)
X(z) =
1
1 + H(z)
= (1 +
1
4z
1)(1 z 1)
(1 + 14z 1)(1 z 1) + 34z 1 + 14z 2
= 1
3
4z
1 1
4z
2
1 34z 1 14z 2 + 34z 1 + 14z 2 = 1
3
4z
1 1
4z
2
This system has two poles at z = 0; it is stable,By inspection,the impulse response
of the error is e[n] = [n] 34 [n 1] 14 [n 2],and the step response of the error
is e[n] = [n] + 14 [n 1]; the system tracks a unit step input without error after two
time-steps.
14
(e) We are given the desired step response of the error signal e[n]:
e[n] =
N 1X
k=0
ak [n k]
=)E(z) =
N 1X
k=0
akz k
From our work in part (b),we also have the following equation for the step response of
e[n]:
E(z) = 11 z 1 11 + H(z)
Set the two equations equal and solve for H(z):
N 1X
k=0
akz k = 11 z 1 11 + H(z)
1 + H(z) = 11 z 1 1PN 1
k=0 akz k
H(z) = 1 (1 z
1)PN 1
k=0 akz
k
(1 z 1)PN 1k=0 akz k
(f) Find the transform of x[n] using the tables:
x[n] = (n + 1)u[n] = nu[n] + u[n]
=)X(z) = z
1
(1 z 1)2 +
1
1 z 1 =
1
(1 z 1)2
Plug the new transfer function H(z) into our equations:
E(z) = X(z) 11 + H(z) = 1(1 z 1)2 (1 + z
1)(1 z 1)2
(1 + z 1)(1 z 1)2 + z 1 + z 2 z 3
= 1 + z
1
(1 + z 1)(1 2z 1 + z 2) + z 1 + z 2 z 3
= 1 + z 1
By inspection,e[n] = [n]+ [n 1],so the system tracks perfectly after two time-steps.
15
