-
Chapter 1 Answers
1.1
Converting
from polar to Cartesian coordinates:
1
j
?
1
1
?
cos
?
?
?
e
2
2
2
1
2
e
?
j
?
j
?
1
1
?
cos(
?
?
)
?
?
2
2
e
e
j
?
2
?
cos(
)
?
j
sin(
)
?
j
?
?
2
2
5
?
j
2
< br>?
e
j
9
?
4
j
?
2
?
j
<
/p>
?
1
?
j
?
cos(
)
?<
/p>
j
sin(
)
?
?
j
2
2
?
j
?
?
4
2
e
?
2(cos(
)
?
j
sin(
))<
/p>
?
1
?
j
e
?
2
?
?
4
4
2
e
?
2
e
j
?
4
2
e
?
9
?
j<
/p>
4
?
2
e
?
?
j
4
?
1
?
j
2
e
j
?
?
4
?
1
?
j
?
2
1.2
converting from Cartesian to polar
coordinates:
5
?
5
e
j
0
,
?
2
?
2
e
j
?
,
?
3<
/p>
j
?
3
e
2
?
j
?
2
?
?
j
1
3
?
j
4
,
,
2
1
?
j
?
2
?
j
?
e
e
2
2
?
1
?
< br>j
?
?
2
e
1
?
j
3
?
j
?
?
j
(1
?
j
)
?
e
4
,
1
?
j
?
4
2
?
j
2
?
?
12
e
e
?
1
?
j
.
(a)
E
?
=
?
e
0
?
?
4
t
dt
?
?
1
,
P
?
=0, because
E
?
?
?
4
2
(b)
x
2
(
t
)
?
e
j
(2
t
?
4
< br>)
,
x
(
t
)
?
1
.Therefore,
E
?
=
P
?
=
lim
1
T
??
?
x
2
(
t
)
??
??
2
dt
=
?
2
??
??
dt
=
?
,
(
t
)
2
T
?
x
2
?
T
T
T
2
dt
?
lim
1
T
??
2
T
?<
/p>
T
?
T
dt
?
?
T
??
??
lim1
?
1<
/p>
2
??
(c)
x
(
t
)
=cos(t). Therefore,
E
2
2
T
T
??
?
T
< br>T
??
n
=
?
x
3
(
t
)
dt
=
?
??
??
cos(
t
)
dt
=
?
,
n
P
?
=
lim
1
?
cos(
t
)
dt
?
lim
1
?
2
T
2
T
,
n
2<
/p>
1
1
?
COS<
/p>
(2
t
)
1
dt
?
?
T
2
2
(d) <
/p>
?
1
?
?
1
?
x
[
n
]
?
?
?
u
[
n
]
x
1
[
n
]
?
?
?<
/p>
u
[
n
]
?
4
?
?
2
?
P
?
=0
，
because
E
?
<
?
.
?
n
?
.
Therefore,
E
?
=
??
??
?
x
1
[
n
]
2
?
?
?
n
?
0
?
?
1
?
?
?
4
?
?
4<
/p>
3
2
=
?
,
(e)
[
n
]
=
e
?
j
(
2
?
8
)
,
[
n
]
2
=1.
therefore,
E
?
=
??
x
2
x
2
?
x
2
[
n
]
??
P
?
=
lim
N
1
?
N
??
2
N
?
1
n
??
N
n<
/p>
x
2
[
n
]
2
N
1
?
lim
1
?
1
.
?
N
??
2
N
< br>?
1
n
??
N
2
2
2
(f)
x
[
n
]
=
?
?
?
. Therefore,
E
?
=
??
=
??
?
=
??
?
,
[
n
]
3
?
cos
?
x
cos(
n
)
?
3
cos(
n
)
?
?
??
?
4
?
??
4
??
4
?
?
?
?
.
(a) The
signal x[n] is shifted by 3 to the right. The
shifted signal will be zero for
n<1,
And n>7.
(b)
The signal x[n] is shifted by 4 to the left. The
shifted signal will be zero for
n<-6.
And n>0.
(c)
The signal x[n] is flipped signal will be zero for
n<-1 and n>2.
(d) The
signal x[n] is flipped and the flipped signal is
shifted by 2 to the right. The
new
Signal will be zero for n<-2 and n>4.
(e) The signal x[n] is flipped and
the flipped and the flipped signal is shifted by 2
to
the left.
This new signal will be zero for n<-6
and n>0.
. (a)
x(1-t)
is
obtained
by
flipping
x(t)
and
shifting
the
flipped
signal
by
1
to
the
right.
Therefore, x (1-t) will be zero for
t>-2.
P
?
=
lim
N
?
?
1
cos
?
?
N
??
2
N
?
1
n
?
?
N
4
n
1<
/p>
?
cos(
n
)
1
2
)
?
1
lim
(
?
N
??
2
N
?
1
n
?
2
2
??
N
N
?
(b) From (a), we know that x(1-t) is
zero for t>-2. Similarly, x(2-t) is zero for
t>-1,
Therefore, x (1-t) +x(2-t) will be zero
for t>-2.
(c)
x(3t) is obtained by linearly compression x(t) by
a factor of 3. Therefore, x(3t)
will
be
zero for t<1.
(d) x(t/3) is obtained by linearly
compression x(t) by a factor of 3. Therefore,
x(3t)
will be
zero for t<9.
(a)
x
1
(
t
) is not periodic
because it is zero for t<0.
(b)
x
2
[
n
]=1 for all n. Therefore,
it is periodic with a fundamental period of
1.
(c)
< br>x
3
[
n
] is as shown in the Figure .
1
1
1
…
x
3
[
…
-3
5
-4<
/p>
0
1
n
4
-1
-1
-1
Therefore, it is periodic with a
fundamental period of 4.
.
(a)
1
1
[
n
]
=
x<
/p>
[
n
]
?
x
[
?
n
]
?
(
u
[
n
]
?
u
[
n
?
4]
?
u
[
?
n
]
?
u
[
?
n
?
4])
x
?
v
1
1
2
1
2
Therefore,
x
[
n
]
is zero for
x
[
n
]
>3.
?
v
1
(b) Since
x
1
(
t
) is an odd
signal,
x
[
n
]
is zero for all values of
t.
?
?
?
?
?
?
?
1
?
v
?
2
?
(c)
1
[
n
]
?
?
?
x
3
?
v
2
n
?
n
?
< br>?
1
?
?
1
?
?
1
?
?
x
1
[
n
]
?
x
1
[
?
n
]
?
2
< br>?
?
?
u
[
n
?
3]
?
?
?
u
[<
/p>
?
n
?
3]
?
?
2
?
?
?
?
?
2
?
?
?
< br>
Therefore,
v
1
?
v
?
x
[
n
]
?
is
zero when
n
<3 and when
n
3
?
5
t
5
t
4
4
?
?
.
(d)
1
1
(
t
)
?
?
(
x
(
t<
/p>
)
?
x
(
?
t
))
?
?
e
?
x
?
2
2
?
< br>u
(
t
?
2)
?
e
u
(
?
t
?
2)
?
?
Therefore,
?
v
?
x
(
t
)
?
is zero
only when
t
4
?
?
.
. (a)
?
l
{
x
1
(
t
)}
?
?
2
?
2
e
cos(0
t
?
?
)
(b)
?
?
l
{
x
(
t
)
}
?
?
2
co
s(
?
)cos(3
t
?
2
?
)
?
cos(3
t
)
?
e
0
t
cos(3
t
?
0)
2
4
?
< br>?
0
t
(c)
?
?
l
{
x
(
t
)}
< br>?
?
e
?
t
sin(3
?
?
< br>t
)
?
e
?
t
sin(3
t
< br>?
?
)
3
2
(d)
?
?
l
{
x
(
t
)}
?<
/p>
?
?
e
?
2
t
sin(100
t
)
?
e
?<
/p>
t
sin(100
t
?
?
)
?
e
?
2
t
co
s(100
t
?
)
4
2
. (a)
(
t
)
is
a periodic complex exponential.
?
x
1
2
x
(
t
)
?
j
e
1
j
10
t
?
e
?
?
?
j
?
10
t
?
?
2
?
?
(b)
x
(
t
)
is a complex exponential
multiplied by a decaying exponential. Therefore,
is not
periodic.
（
c
）
x
3
[
n
]
is a
periodic signal.
x
[
n
]
is a complex exponential with a fundamental period
of
2
?
?
?
.
3
2<
/p>
x
3
[
n
]
=
e
j
7
?
n
=
e
j
?
n
.
(d)
x
[
n
]
is a
periodic signal. The fundamental period is given
by N=m(
3
2
?
)
4
3
?
/
5
=
m
(
10
)
.
By choosing m=3. We obtain the
fundamental period to be 10.
5
(e)
x
[
n
]
is
not
periodic.
x
[
n
]
is
a
complex
exponential
with
w
=3/5.
We
cannot
find
any
5
0
integer m such
that m(
2
?
) is
also an integer. Therefore,
w
.
x
[
n
]
5
is not
periodic.
0
x
(
t
)=2cos(10t
＋
1)-sin(4t-1)
Period of first term in the RHS =
2
?
?
?
< br>.
10
4
5
Period of first term in the RHS =
2
?
?
?
.
2
Therefore, the overall
signal is periodic with a period which the least
common
multiple of the
periods of the first and second terms. This is
equal to
?
.
x[n] = 1+
e
j
4
?
n
?
< br>e
j
5
?
n
2
.
7
Period of first
term in the RHS =1.
Period
of second term in the RHS =
?
2
?
?
=7
（
when
m=2
）
?
?
?
4
/
7<
/p>
?
Period of second term in the
RHS =
?
2
?
?
?
?
?
2
?
/
5
?<
/p>
=5 (when m=1)
Therefore,
the
overall
signal
x[n]
is
periodic
with
a
period
which
is
the
least
common
Multiple of the
periods of the three terms inn x[n].This is equal
to 35.
. The signal x[n] is
as shown in figure . x[n] can be obtained by
flipping u[n] and then
Shifting the flipped signal by 3 to the
right. Therefore, x[n]=u[-n+3]. This implies
that
M=-1 and
no=-3.
X[
n
]
-
-
-
0
1
2
3
n
3
2
Figure S
1
?
0
,
t
?
?
2
y
(t)=
x
(
?
)
dt
=
(
?
(
?<
/p>
?
2
)
?
?
(
?
?
2
))
dt
=
,
?
1
,
?
2
?
t
?
2
?
?
?
?
?
?<
/p>
?
0
,
t
?
2
?
t
?
t
Therefore
E
?
?
2
?
dt
?
4<
/p>
?
2
The signal
x
(t) and its derivative
g
(t) are shown in Figure
.
x
(t)
g
(t)
1
-1
1
-1
0
1
2
0
t
-3
-3
-2
Figure S
Therefore
g
(
t
)
?
3
?
?
(
t
?
2
k
)
?
3
< br>?
?
(
t
?
2
k
?
1
)
k
?
??
k
?
??
?
?
2
t
This implies
that A
1
=3,
t
1
=0,
A
2
=-3, and
t
2
=1.
(a) The signal x
2
[n], which
is the input to S
2
, is the
same as y
1
[n].Therefore
,
1
x
2
[n-3]
2
1
= y
1
[n-2]+
y
1
[n-3]
2
y
2
[n]=
x
2
[n-2]+
=2x
1
[n-2]
+4x
1
[n-3]
+
1
(
2x
1
[n-3]+
4x
1
[n-4])
2
=2x
1
[n-2]+
5x
1
[n-3] +
2x
1
[n-4]
The input-output relationship for S is
y[n]=2x[n-2]+ 5x [n-3] + 2x
[n-4]
(b)
The
input-output
relationship
does
not
change
if
the
order
in
which
S
1
and
S
2
are
connected
series reversed. .
We can easily prove this assuming that
S
1
follows
S
2
. In this case , the
signal x
1
[n],
which is the input to S
1
is
the same as
y
2
[n].
Therefore
y
1
[n]
=2x
1
[n]+
4x
1
[n-1]
= 2y
2
[n]+4
y
2
[n-1]
=2( x
< br>2
[n-2]+
1
1
x
2
[n-3]
)+4(x
2
[n-3]+
x
2
[n-4])
2
2
=2 x<
/p>
2
[n-2]+5x
2
< br>[n-3]+ 2 x
2
[n-4]
The input-output relationship for S is
once again
y[n]=2x[n-2]+
5x [n-3] + 2x [n-4]
(a)The
system is not memory less because y[n] depends on
past values of x[n].
(b)The
output of the system will be y[n]=
?
[
n
]
?
[
n
?
2
]
=0
(c)From
the
result
of
part
(b),
we
may
conclude
that
the
system
output
is
always
zero
for
inputs of the form
?
[
n
?
k
]
,
k
?
?
.
Therefore , the system is not invertible
.
(a)
The
system
is
not
causal
because
the
output
y(t)
at
some
time
may
depend
on
future
values
of
x(t). For instance ,
y(-
?
)=x(0).
(b) Consider two arbitrary inputs
x
1
(t)and
x
2
(t).
x
1
(t)
?
y
1
(t)=
x
1
(sin(t))
x
2
(t)
?
y
2
(t)=
x
2
(sin(t))
Let
x
3
(t)
be
a
linear
combination
of
x
1
(t)
and
x
2
(t).That
is
,
x
3
(t)=a
x
1
(t)+b
x
2
(t)
Where a and b are arbitrary scalars .If
x
3
(t) is the input to the
given system ,then the
corresponding
output y
3
(t) is
y
3
(t)=
x
3
( sin(t))
=a x
1
(sin(t))+
x
2
(sin(t))
=a y
1
(t)+
by
2
(t)
Therefore , the system is
linear.
.(a) Consider two
arbitrary inputs x
1
[n]and
x
2
[n].
x
1
[n]
?
y
1
[n]
=
x
2
[n ]
?
y
2
[n] =
n
?
n
0
k
?
n
?
n
0<
/p>
n
?
n
0
?
x
[
k
]
1
k
?
n
?
n
0
?
x
[
k
]
2
Let
x
3
[n] be a linear
combination of x
1
[n] and
x
2
[n]. That is :
x
3
[n]=
ax
1
[n]+b
x
2
[n]
where a and b are arbitrary scalars. If
x
3
[n] is the input to the
given system, then the
n
?
n
0
corresponding
output y
3
[n] is
y
3
[n]=
n
?
n
0
k
?
n
?
n
0<
/p>
?
x
[
k
]
3
n
?
n
0
n
?
n
0
1
2
k
?
n
?
n
0
k
?<
/p>
n
?
n
0
=
k
?
n
?
n
0
?
(
ax
[
k
]
?
bx
[
k
])
=a
?
x
[
k
]
+b
?
x
[
k
]
1
2
= ay
1
[n]+b
y
2
[n]
Therefore the system is
linear.
(b) Consider an
arbitrary input x
1
[n].Let
n
?
n
0
y
1
[n] =
k
?
n
?
n
0
?
x
[
k
]
1
be
the
corresponding
output
.Consider
a
second
input
x
2
[n]
obtained
by
shifting
x
1
[n]
in
time:
x
2
[n]= x
1
[n-n
1
]
< br>
The output corresponding to this
input is
n
?
n
0
y
2
[n]=
k
?
n
?
n
0
?
x
[
k
]
=
?
x
[
k
?
n
]
=
?
x<
/p>
[
k
]
2
1
1
k
?
n
?
n
0
1
k
?
n
?
n
1
?
n
0
n
?<
/p>
n
1
?
n
0
1
k
?
n
?
n
1
?
n
0
n
?
n
0
n
?
n
1
?
n<
/p>
0
Also note that
y
1
[n-
n
1
]=
?
x
[
k
]
.<
/p>
Therefore ,
y
2
[n]=
y
1
[n-
n
1
]
This implies that the system is time-
invariant.
(c) If
x
[
n
]
<
B, then y[n]
?
(2
n
0
+1)B.
Therefore ,C
?
(2
n
0
+1)B.
(a) (i) Consider two arbitrary inputs
x
1
(t) and
x
2
(t).
x
1
(t)
?
y
1
(t)=
t
x
1
(t-1)
x
2
(t)
?
y
2
(t)= t
x
< br>2
(t-1)
Let
x
3
(t)
be
a
linear
combination
of
x
1
(t)
and
x
2
(t).That
is x
3
(t)=a
x
1
(t)+b
x
2
(t)
where a and b are arbitrary scalars. If
x
3
(t) is the input to the
given system, then the
corresponding
output y
3
(t) is
y
3
(t)=
t
x
3
(t-1)
=
t
(ax
1
(t-1)+b
x
2
(t-1))
= ay
1
(t)+b
y
2
(t)
Therefore , the system is
linear.
(ii) Consider an
arbitrary inputs x
1
(t).Let
y
1
(t)= t
x
< br>1
(t-1)
be the
corresponding output .Consider a second input
x
2
(t) obtained by shifting
x
1
(t) in
time:
x
2
(t)= x
1
(t-t
0
)
< br>
The output corresponding to this
input is y
2
(t)=
t
x
2
(t-1)=
t
x
1
(t- 1-
t
0
)
Also note that
y
1
(t-t
0
)= (t-t
0
)
x
1
(t- 1-
t
0
)
?
y
2
(t)
Therefore the system is not time-
invariant.
(b) (i) Consider
two arbitrary inputs
x
1
[n]and
x
2
[n].
x
1
[n]
?
y
1
[n] = x
1
2
[n-2]
x
2
[n ]
?
y
2
[n] = x
2
2
[n-2].
Let
x
3
(t) be a linear
combination of x
1
[n]and
x
2
[n].That is
x
3
[n]=
ax
1
[n]+b
x
2
[n]
2
2
2
2
2<
/p>
2
2
2
wher
e a and b are arbitrary scalars. If
x
3
[n] is the input to the
given system, then the
corresponding
output y
3
[n] is
y
3
[n] = x
3
2
[n-2]
=(a
x
1
[n-2] +b x
2
[n-2])
2
=
a
2
x
1
2<
/p>
[n-2]+b
2
x
2
2
[n-2]+2ab
x
1
[n-2]
x
2
[n-2]
ay
1
[n]+b
y
2
[n]
Therefore the system is not
linear.
(ii) Consider an
arbitrary input x
1
[n]. Let
y
1
[n] =
x
1
2
[n-2]
?
be
the
corresponding
output
.Consider
a
second
input
x
2
[n]
obtained
by
shifting
x
1
[n]
in
time:
x
2
[n]=
x
1
[n-
n
0
]
The output corresponding to this input
is
y
2
[n] =
x
2
2
[n-2].=
x
1
2
[n-2-
n
0
]
Also note that
y
1
[n-
n
0
]=
x
1
2
[n-2-
n
0
]
Therefore ,
y
2
[n]=
y
1
[n-
n
0
]
This implies that the system is time-
invariant.
(c) (i) Consider
two arbitrary inputs
x
1
[n]and
x
2
[n].
x
1
[n]
?
y
1
[n]
= x
1
[n+1]-
x
1
[n-1]
x
2
[n ]
?
y
2
[n] =
x
2
[n+1 ]-
x
2
[n -1]
Let x
3
[n] be a
linear combination of x
1
[n]
and x
2
[n]. That is
:
x
3
[n]=
ax
1
[n]+b
x
2
[n]
where a and b are arbitrary scalars. If
x
3
[n] is the input to the
given system, then the
corresponding output
y
3
[n] is
y
3
[n]=
x
3
[n+1]-
x
3
[n-1]
=a x
1
[n+1]+b
x
2
[n +1]-a
x
1
[n-1]-b
x
2
[n -1]
=a(x
1
[n+1]- x
1
[n-1])+b(x
2
[n +1]-
x
2
[n -1])
= ay
1
[n]+b
y
2
[n]
Therefore the system is
linear.
(ii) Consider an
arbitrary input x
1
[n].Let
y
1
[n]=
x
1
[n+1]-
x
1
[n-1]
be the corresponding output .Consider a
second input x
2
[n] obtained
by shifting x
1
[n] in
time: x
2
[n]= x
1
[n-n
0
]
The output corresponding to
this input is
y
2
[n]=
x
2
[n +1]-
x
2
[n -1]=
x
1
[n+1-
n
0
]-
x
1
[n-1-
n
0
]
Also note that y
< br>1
[n-n
0
]=
x
1
[n+1-
n
0
]-
x
1
[n-1-
n
0
]
Therefore ,
y
2
[n]= y
1
< br>[n-n
0
]
This implies that the system is time-
invariant.
(d) (i) Consider
two arbitrary inputs x
1
(t)
and x
2
(t).
x
1
(t)
?
y
1
(t)=
?
d
?
x
1
(t)
?
x
2
(t)
?
y
2
(t)=
?
d
?
x
2
(t)
?
Let x
3
(t) be a
linear combination of x
1
(t)
and x
2
(t).That is
x
3
(t)=a
x
1
(t)+b
x
2
(t)
where a and b are arbitrary scalars. If
x
3
(t) is the input to the
given system, then the
corresponding
output y
3
(t) is
y
3
(t)=
?
d
?
x
3
(t)
?
=
?
d
?
ax
1
(t)
+
b
x
2
(t)
?
=a
?
d
?<
/p>
x
1
(t)<
/p>
?
+b
?
d
?
x
2
(t)
?
=
ay
1
(t)+b
y
2
(t)
Therefore the system is
linear.
(ii) Consider an
arbitrary inputs
x
1
(t).Let
y
1
(t)=
?
d
?
x
1
(t)
?
=
x
1
(t)
-
x
1
(
?
t
)
2
be the corresponding output
.Consider a second input
x
2
(t) obtained by shifting
x
1
(t) in
time:
x
2
(t)= x
1
(t-t
0
)
< br>
The output corresponding to this
input is
y
2
(t)=
?
d
?
x
2
(t)
?
=
x
2
(t)
-
x
2
(
?
t
)
2
=
x
1
(t
-
t
0
)
-
x
1
(
?
t
?
t
0
)
Also note that y
1
2
(t-t
0
)=
x
1
(t
-
t
0
)
-
x
1
(
?
t
?
t
0
)
2
?
y
2
(t)
Therefore the system is not time-
invariant.
(a)
Given
x
(
t
)
=
e
2
jt
y(t)=
e
x
(
t
)
=
e
?
2
jt
j
3
t
?
j
3
t
Since the
system liner
Therefore
(b) we know that
y(t)=
e
?
j
3
t
x
1
(
t
)
?
1
/
2
(
e
j
2
t
?
e
?<
/p>
2
jt
)
y
(
t
)
=
1/2(
e
j
3
t
+
e
1
)
x
(t)=
1
cos(2t)
y
(
t
)
=cos(3t)
1
x
2
(t)=cos(2(t-1/2))= (
e
?
j
e
2
jt
+
e
j
e
?
2
jt
)/2
Using
the linearity property, we may once again
write
x
1
(t)=
1
?
j
< br>j
(
e
e
2
jt
+
e
2
e
?
2
j
t
)
y
1
(
t
)
=(
e
?
j
e
3
jt
+
e
e<
/p>
?
3
jt
)=
cos(3t-1)
j
Therefore,
x
1
(t)=cos(2(t-1/2))
y
(
t
)
=cos(3t-1)
1
signals are sketched in
figure .
2
x(t-1)
x
(2-t
1
-1
t
0
1
2
3
-1
0
-1
x
(4-t/2)
4
6
2
1
8
10
12
1
t
2
1
1
2
2
3
4
t
x
(2t+1)
1
-1
2
1
0
t
[
x
(
t
)
?
x
(
?
t
)]
u
(
t
)
0
1
t
-3/2
3/2
t
Figure
The
signals are sketched in figure
The even and odd parts are sketched in
Figure
x[3-
n]
x[n-4
1
1/2
0
0
1
2
3
7
n
1
2
-1/2
3
1
1/2
7
n
(b)
1
2
x[n]u[n-3]=x[n]
x[3n+1]
1
x[3n]
1/2
2
n
n
1
-1
0
1
2
-1
0
n
(d)
-1/2
0
1
2
(c)
(f)
n
x[3- n]/2
+(-1)
x[n]/2
1
1
1/2
-4
0
1
2
n
-2
0
2
n
(h)
-1
(g)
Figure
x
0
(t)
x
0
(t)
1/2
1/
2
-2
-1
0
1
2
t
-2
0
2
t
(a)
-1/2
x
0
(t)
x
0
(t)
1
1/
1/
2
2
-1
-2
0
1
2
t
-2
-1
0
1
2
t
-1/2
-1/2
(b)
x
0
(t)
x
0
(t)
-3t/
3t/2
2
-t/2
(c)
0
t
0
t
Figure
x[n]
1
xo[n]
1
-7
7
-7
0
0
7
n
n
-1/
-1/