?
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Electrical Engineering and Computer Science
6.003,Signals and Systems—Fall 2003
Problem Set 8 Solution
Exercise for home study
O&W 8.35
(a) From the system diagram,we see that
z(t)= x(t)cos (ω
c
t)
Using the multiplication property
1
Z(jω)=
2π
(X(jω)?FT{cos w
c
t})
FT of cos w
c
t is two impulses with area π at ±w
c
.
of X(jω) shifted to be centered at ω
c
and?ω
c
and scaled by
Therefore,Z(jw) is the spectrum
1
2
1
2π
×π =, The real and
imaginary parts of Z(jω) are shown below,
Re{ Z(j ω)}
1/2 1/2
ω
c
ω
ω
ω?ω
mc
ω +ω
mc
ω ω
c
+ω ω
cm c m
Im{ Z(jω)}
1/2
mc
mc mc
c
1/2
ω +ω
ω?ω?ω ω
+ω ω
ω
m
To?nd FT of p(t),let us?rst de?ne a signal,q(t),such that,
π
2,
2w
c
q(t)=
0,
|w|≤
π
w >
2w
c
| |
1
Therefore,
+∞
2π
p(t)= q(t)? δ(t? n
w
c
)? 1
n=?∞
Taking Fourier transform,
∞
4sin(
πω
)
P(jω)=
2ω
c
ω
c
δ(ω? kω
c
)? 2πδ(ω)
ω
k=?∞
Following is the plot of P(jw),Ifracturm{P(jw)} =0,
P(jω)
c
2ω?ω
c
c
3ωc?4ω
c
5ω
c
6ω cω
c
2ω
c
3ω
c
4ω
c
5ω c6ω
ω
0
4 4
Note that the impulse at the origin disappeared in the above graph because of the
subtraction with 2πδ(ω).
Now,using the multiplication property again,Y (jω)=
2
1
π
(Z(jω)? P(jw)).
The real and imaginary part of Y (jw) is shown below:
Re{ Y(jω)}
π
2ω 2ω
2
ω
2ω?ω
m
2ω
c
+ω
m
ω
m
ω
m 2ωc
ω
m
2ω
c
+ω
mc c c
2
π
ωY(j )Im{ }
ω
mc
2ω
+ω
mc
2ω
m
ω
ω
m
ω
mc
2ω
+ω
mc
2ω
ω
2
(b) To let x(t)= v(t),we must retain the frequency content between ±w
m
of Y (jω)and
scale it properly,Therefore,H(jω) is a low-pass?lter with cuto? frequency at ω
m
and
gain
π
2
,as sketched below,
H(jω)
π
2
ω
ω ω
m m
3
Problem 1 (O & W 7.34)
X(e
jω
) has characteristics as graphed below (where A is any real number),
j
X(e
ω
)
3π
14
3π
14
ω
A
π 2π?π
0
To let X(e
jω
) occupy the entire range between
3
,and therefore we need to change the sampling rate by a
π and π,we need to scale the spectrum
tofrom?
3π 3π
by a factor of
14
14 14
factor of
14
,Since we can only upsample and downsample by integer factors,we need to
3
upsample by a factor of 3,and then downsample by a factor of 14,
When we upsample by 3,we compress the spectrum of X(e
jω
) by a factor of 3,and then
pass this through a low-pass?lter with a gain of 3,The result is shown in the following
gure,
j ω
X (e )
p
14
π
14
π
ω
3A
π 2π?π
Next,when we downsample the above spectrum by 14,we expand the spectrum by a factor
of 14 and scale the height by
14
,as graphed below,
1
X
d
(e
j ω
)
3A
14
ω
π 2π?π
0
Therefore,L =3 and M = 14,
4
Problem 2
(a) x[n] is a real-valued DT signal whose DTFT for?π< ω< π is given by
X(e
jω
)
2
ω
0
=
3
π,ω
M
=
π
4
1
ω
0
ω
0 π
π
ω
0
ω
M
ω
0
ω
0
+ ω
M
ω
0
ω
M
ω
0
+ ω
M
Let
z
c
[n]= x[n]cos[w
o
n]
Using table 5.2 and taking Fourier transform of cos[w
o
n],
+∞
o
n]} = π {δ(w? w
o
2πl)+ δ(w + w
o
2πl)}DT FT {cos[w
l=?∞
[w
o
n]}DT FT {cos
π
π
0
ω
0
ω
0
ππ
ω
Using the multiplication property from table 5.1,Z
c
(e
jw
) is the periodic convolution
o
n]} over period 2π and thenscaledby
1
,Wetakeone
2π
of X(e
jw
)and DT FT {cos[w
period,from?π to π,of DT FT {cos[w
o
n]} and do regular convolution with X(e
jw
),
1
Centered at w = 0,weget thesuperposition of two X(e
jw
)scaled by
2
,Z
c
(e
jw
)is
shown below for the interval?π to π,
Z
c
(e
jw
)
1
=
1
π
2π 2
×
5π 11π5π11π
π ω12
12
ω
M 0
ω
M 12 12
π
5
Z
c
(e
jw
) is then passed through a low-pass?lter with cut-o? frequency w
M
and gain of
1,DTFT of x
c
[n] is shown below,
X
c
(e
jw
)
1
2
2π
(2π? ω
M
)
π?ω
M 0
ω
M π
(2π? ω
M
)
2π ω
Let
z
s
[n]= x[n]sin[w
o
n]
Using table 5.2 and taking Fourier transform of sin[w
o
n],
+∞
π
[w
o
n]} =
j
{δ(w? w
o
2πl)? δ(w + w
o
2πl)}DT FT {sin
l=?∞
We?nd Z
s
(e
jw
) using the periodic convolution as before,The superposition terms
centered at w =0 from X(e
jw
) (in dashed lines) are shown below,Adding the super-
position terms,resulting Z
s
(e
jw
)is shown for interval?π to π,
Z
s
(e
jw
)
5π
12
11π
12
5π
12
11π
12 π
π?ω
M
ω
M
1
2j
1
2j
ω
Z
s
(e
jw
) goes through the low-pass?lter with cut-o? frequency w
M
and gain of 1,we
nd DTFT of x
s
[n] as shown below,
6
X
s
(e
jw
)
1
2π
2π
π
π
(2π? ω
M
)
(2π? ω
M
)
ω
M
ω
M
2j
1
2j
ω
(b) Maximum possible downsampling is achieved once the non-zero portion of one period
of the discrete-time spectrum has expanded to?ll the entire band from?π to π,
Therefore,
π π
m = = =4
w
π
M
4
(c) Following is the system diagram to recover x[n],
2cos(ω
0
n)
y
c
[n] ↑ m ×
x
co
[n]
y
s
[n] ↑ m ×
x
so
[n]
+ x[n]
2sin(ω
0
n)
After upsampling by m,weget back x
c
[n]and x
s
[n]from y
c
[n]and y
s
[n] respectively,
Note that upsampling by m has zero-insertion block (up-arrow m) and a low-pass
lter for time-domain interpolation,DTFT of x
c
[n]and x
s
[n] are derived in part a,
According to the system diagram,
x
co
[n]= x
c
[n] × 2cos[w
o
n]
Using the multiplication property and doing periodic convolution,we get X
co
(e
jw
)as
shown below,
7
X
co
(e
jw
)
=
π
× 2π ×
1
2
1
2
1
2
π
w
o 0
w
o π ω
w
o
w
M
w
o
w
M
X
so
(e
jw
)
ππ
w
o
w
o
w
o
w
M
w
o
w
M
1
2
1
2
ω
Similarly,x
so
[n]= x
s
[n] × 2sin[w
o
n],and we get X
so
(e
jw
) as shown in the?gure,
Adding X
co
(e
jw
)and X
so
(e
jw
),we get back the spectrum of X(e
jw
),Thus,we recover
x[n],
8
Problem 3
(a)
2t
x(t)= e
t
u(? t)+2e u(t)
Using Laplace transforms of elementary functions (table 9.2),we?nd,
t
e u(? t)
1
,R
1
= R e{ s} <? 1←→
s +1
1
2t
e u(t) R
2
= R e{ s} >? 2←→
s +2
,
Using the linearity property,
2
X(s)=
1
+,ROC containing R
1
R
2
s +1 s +2
s
=
(s +1)(s +2)
ROC =? 2 < R e{ s} <? 1
The pole-zero plot is shown below,
×
2
×
1
Rfractur e
Ifractur m
X(s)p-z map of
9
(b)
x(t)=(e
t
cos t)u(? t)+ u(? t)
t
= e (
1
e
jt
+
1
e
jt
) u(? t)+ u(? t)
2 2
1 1
(1+j)t
u(? t)+ e
(1?j)t
= e u(? t)+ u(? t)
2 2
Using Laplace transforms of elementary functions (table 9.2),we?nd,
(1+j)t
1
R
1
= R e{ s} < 1e u(? t) ←→
s? (1+ j)
,
(1?j)t
1
R
2
= R e{ s} < 1e u(? t) ←→
s? (1? j)
,
u(? t)
1
R
3
= R e{ s} < 0←→
s
Using the linearity property,
1/2
X(s)=
1/2
+ +
1
,ROC containing R
1
R
2
R
3
s? (1+ j) s? (1? j) s
(1/2)s
2
+(1/2)s(1? j)? (1/2)s
2
+(1/2)s(1+ j)? s
2
+2s? 2
ROC= R e{ s} < 0=
s(s
2
2s +2)
1(2s
2
3s +2)
ROC= R e{ s} < 0
=
s(s
2
2s +2)
Solving the denominator,we see that poles are located at,
s =1 ± j
s =0
Solving the numerator,we see that zeros are located at,
3
√
7
s =
4
± j
4
The pole-zero plot is shown below:
10
Ifracturm
p-z map of X(s)
K =?1
1 ×
√
+
Rfracture×
1
√
1
×
3
4
7
7
4
4
2
Problem 4
(a) We are given,
s? 25
X(s)=
s? 25
=
s? s? 12 (s? 4)(s +3)
3 < Re{s} < 4
Using partial fraction expansion,
s? 25 A B
= +
(s? 4)(s +3) s? 4 s +3
multiply both sides by (s? 4) and plug-in s =4,
s? 25
= A
s +3
A =?3
multiply both sides by (s + 3) and plug-in s =?3,
s? 25
= B
s? 4
B =4
11
Therefore,
4
X(s)=
3
+? 3 < Re{s} < 4
s +3s? 4
Using the table of Laplace transforms for elementary functions (table 9.2) and given
ROC,we?nd
3t
x(t)=3e
4t
u(?t)+4e u(t)
(b) We are given,
2s
2
+7s +9 2s
2
+7s +9
X(s)= = Re{s} >?2
(s +2)
2
s
2
+4s +4
As the degree of the numerator is equal to that of the denominator,we need to use
long division to divide the numerator by the denominator before?nding the partial
fraction expansion,
X(s)= 2?
s? 1
(s +2)
2
Now we can?nd the partial fraction expansion of the second term on the right hand
side of the equality,
A Bs? 1
= +
(s +2)
2
(s +2)
2
s +2
multiply both sides by (s +2)
2
and plug-in s =?2,
s? 1= A + B(s +2)
A =?3
multiply both sides by (s +2)
2
,plug-in A and s =1,
s? 1= A + B(s +2)
B =1
Therefore,
1
X(s)=2?
3
+ Re{s} >?2
(s +2)
2
s +2
3 1
=2 +
(s +2)
2
s +2
Re{s} >?2
Using the table of Laplace transforms for elementary functions (table 9.2) and given
ROC,we?nd
2t
x(t)=2δ(t)+3te
2t
u(t)? e u(t)
12
Problem 5 (O & W 9.24 (f))
We are given X(s),|X(jw) for any w can be calculated as K
bc
where K is any real number
ad
|
and a,b,c,d are vector magnitudes as shown below in the pole-zero diagram of X(s),
×
2?1
×
11
2
a
b
c
d
∠a ∠b ∠c ∠d
w
Rfracture
Ifracturm
X(s)
X
1
(s |X
1
(jw)| = |X(jw)| = K
bc
p-z map of
We need to?nd ) such that,
ad
,and there are no poles and
zeros in the right-half plane,
1
2
jwRe?ecting the pole (at 1 on real axis) and zero (at on real axis) along -axis or imaginary
axis,we can conserve the magnitude d and c from the pole and zero respectively,The
resulting pole-zero diagram will be as follows,
×
2?1
×
1
2
a
b
c
d
w
Rfracture
Ifracturm
X
1
(s)p-z map of
From the plot above,the pole and zero at?1 will cancel each other,Therefore,
K(s +
1
2
)
X
1
(s)=
(s +2)
It is important to note that,from p-z map of X(s),as b = d,
bc c
X(jw)| = K = K = X
1
(jw)||
ad a
|
13
Now,we need to?nd X
2
(jw) such that ∠X
2
(jw)= ∠X(jw) and there are no poles or zeros
in the right-half plane of p-z map of X
2
(s),From the p-z map of X(s) shown on the previous
page,we can write the phase,∠X(jw),as
∠X(jw)= ∠b + ∠c? ∠a? ∠d
If we re?ect the pole at s = 1 along jw-axis,the contribution to overall phase from the
re?ected pole becomes?(π? ∠d)=?π + ∠d,Notice that the sign of contributed angle
(∠d) has?ipped,Now,lets convert that pole to a zero,The contribution to overall phase
from the resulting zero is +(π? ∠d)= π? ∠d,
The above two operations,re?ecting a pole along the jw-axis and changing it to zero,just
add +π to overall phase,In order to keep the phase unchanged,we can multiply the resulting
Laplace transform by?1as?1= e
jπ
will subtract π from the overall phase,
1
Similarly,if we re?ect the zero at s =
2
along jw-axis,change that zero to a pole,and
multiply resulting the Laplace transform with?1,the phase will remain unchanged,
Therefore,
X
2
(s)= K(?1)(?1)
(s +1)(s +1)
(s +2)(s +
1
2
)
K(s +1)
2
=
(s +2)(s +
1
2
)
14
Problem 6 (O & W 9.26 )
We need to?nd the Laplace transform of y(t) using the properties of Laplace transform,The
two properties,time shifting (table 9.1 of text book) and time scaling by? 1 (page 686? 687
of text book),that we will need are,
x(t? t
0
) e
st
0
X(s)←→
x(? t) ←→ X(? s)
The ROC is unchanged for the time shifting property,For the time scaling by? 1
property,ROC is re?ected about the jω or imaginary axis in s-plane,
y(t) = x
1
(t? 2)? x
2
(? t +3)
where x
1
(t) = e
2t
u(t)
and x
2
(t) = e
3t
u(t)
taking the Laplace transform
1
x
1
(t) ←→
s +2
,R e{ s} >? 2
1
x
1
(t? 2) e
2s
,R e{ s} >? 2←→
s +2
1
x
2
(t) ←→
s +3
,R e{ s} >? 3
1
x
2
(? t) ←→
s +3
,R e{ s} < 3
1
x
2
(? (t? 3)) e
3s
s +3
,R e{ s} < 3←→
Using the convolution property of Laplace transform (table 9.1),
Y (s)= X
1
(s)X
2
(s),at least R
1
∩ R
2
2s?3s
e e
=
s +2
×
s +3
5s
Y (s)=
2
e
,? 2 < R e{ s} < 3
s? s? 6
15
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Electrical Engineering and Computer Science
6.003,Signals and Systems—Fall 2003
Problem Set 8 Solution
Exercise for home study
O&W 8.35
(a) From the system diagram,we see that
z(t)= x(t)cos (ω
c
t)
Using the multiplication property
1
Z(jω)=
2π
(X(jω)?FT{cos w
c
t})
FT of cos w
c
t is two impulses with area π at ±w
c
.
of X(jω) shifted to be centered at ω
c
and?ω
c
and scaled by
Therefore,Z(jw) is the spectrum
1
2
1
2π
×π =, The real and
imaginary parts of Z(jω) are shown below,
Re{ Z(j ω)}
1/2 1/2
ω
c
ω
ω
ω?ω
mc
ω +ω
mc
ω ω
c
+ω ω
cm c m
Im{ Z(jω)}
1/2
mc
mc mc
c
1/2
ω +ω
ω?ω?ω ω
+ω ω
ω
m
To?nd FT of p(t),let us?rst de?ne a signal,q(t),such that,
π
2,
2w
c
q(t)=
0,
|w|≤
π
w >
2w
c
| |
1
Therefore,
+∞
2π
p(t)= q(t)? δ(t? n
w
c
)? 1
n=?∞
Taking Fourier transform,
∞
4sin(
πω
)
P(jω)=
2ω
c
ω
c
δ(ω? kω
c
)? 2πδ(ω)
ω
k=?∞
Following is the plot of P(jw),Ifracturm{P(jw)} =0,
P(jω)
c
2ω?ω
c
c
3ωc?4ω
c
5ω
c
6ω cω
c
2ω
c
3ω
c
4ω
c
5ω c6ω
ω
0
4 4
Note that the impulse at the origin disappeared in the above graph because of the
subtraction with 2πδ(ω).
Now,using the multiplication property again,Y (jω)=
2
1
π
(Z(jω)? P(jw)).
The real and imaginary part of Y (jw) is shown below:
Re{ Y(jω)}
π
2ω 2ω
2
ω
2ω?ω
m
2ω
c
+ω
m
ω
m
ω
m 2ωc
ω
m
2ω
c
+ω
mc c c
2
π
ωY(j )Im{ }
ω
mc
2ω
+ω
mc
2ω
m
ω
ω
m
ω
mc
2ω
+ω
mc
2ω
ω
2
(b) To let x(t)= v(t),we must retain the frequency content between ±w
m
of Y (jω)and
scale it properly,Therefore,H(jω) is a low-pass?lter with cuto? frequency at ω
m
and
gain
π
2
,as sketched below,
H(jω)
π
2
ω
ω ω
m m
3
Problem 1 (O & W 7.34)
X(e
jω
) has characteristics as graphed below (where A is any real number),
j
X(e
ω
)
3π
14
3π
14
ω
A
π 2π?π
0
To let X(e
jω
) occupy the entire range between
3
,and therefore we need to change the sampling rate by a
π and π,we need to scale the spectrum
tofrom?
3π 3π
by a factor of
14
14 14
factor of
14
,Since we can only upsample and downsample by integer factors,we need to
3
upsample by a factor of 3,and then downsample by a factor of 14,
When we upsample by 3,we compress the spectrum of X(e
jω
) by a factor of 3,and then
pass this through a low-pass?lter with a gain of 3,The result is shown in the following
gure,
j ω
X (e )
p
14
π
14
π
ω
3A
π 2π?π
Next,when we downsample the above spectrum by 14,we expand the spectrum by a factor
of 14 and scale the height by
14
,as graphed below,
1
X
d
(e
j ω
)
3A
14
ω
π 2π?π
0
Therefore,L =3 and M = 14,
4
Problem 2
(a) x[n] is a real-valued DT signal whose DTFT for?π< ω< π is given by
X(e
jω
)
2
ω
0
=
3
π,ω
M
=
π
4
1
ω
0
ω
0 π
π
ω
0
ω
M
ω
0
ω
0
+ ω
M
ω
0
ω
M
ω
0
+ ω
M
Let
z
c
[n]= x[n]cos[w
o
n]
Using table 5.2 and taking Fourier transform of cos[w
o
n],
+∞
o
n]} = π {δ(w? w
o
2πl)+ δ(w + w
o
2πl)}DT FT {cos[w
l=?∞
[w
o
n]}DT FT {cos
π
π
0
ω
0
ω
0
ππ
ω
Using the multiplication property from table 5.1,Z
c
(e
jw
) is the periodic convolution
o
n]} over period 2π and thenscaledby
1
,Wetakeone
2π
of X(e
jw
)and DT FT {cos[w
period,from?π to π,of DT FT {cos[w
o
n]} and do regular convolution with X(e
jw
),
1
Centered at w = 0,weget thesuperposition of two X(e
jw
)scaled by
2
,Z
c
(e
jw
)is
shown below for the interval?π to π,
Z
c
(e
jw
)
1
=
1
π
2π 2
×
5π 11π5π11π
π ω12
12
ω
M 0
ω
M 12 12
π
5
Z
c
(e
jw
) is then passed through a low-pass?lter with cut-o? frequency w
M
and gain of
1,DTFT of x
c
[n] is shown below,
X
c
(e
jw
)
1
2
2π
(2π? ω
M
)
π?ω
M 0
ω
M π
(2π? ω
M
)
2π ω
Let
z
s
[n]= x[n]sin[w
o
n]
Using table 5.2 and taking Fourier transform of sin[w
o
n],
+∞
π
[w
o
n]} =
j
{δ(w? w
o
2πl)? δ(w + w
o
2πl)}DT FT {sin
l=?∞
We?nd Z
s
(e
jw
) using the periodic convolution as before,The superposition terms
centered at w =0 from X(e
jw
) (in dashed lines) are shown below,Adding the super-
position terms,resulting Z
s
(e
jw
)is shown for interval?π to π,
Z
s
(e
jw
)
5π
12
11π
12
5π
12
11π
12 π
π?ω
M
ω
M
1
2j
1
2j
ω
Z
s
(e
jw
) goes through the low-pass?lter with cut-o? frequency w
M
and gain of 1,we
nd DTFT of x
s
[n] as shown below,
6
X
s
(e
jw
)
1
2π
2π
π
π
(2π? ω
M
)
(2π? ω
M
)
ω
M
ω
M
2j
1
2j
ω
(b) Maximum possible downsampling is achieved once the non-zero portion of one period
of the discrete-time spectrum has expanded to?ll the entire band from?π to π,
Therefore,
π π
m = = =4
w
π
M
4
(c) Following is the system diagram to recover x[n],
2cos(ω
0
n)
y
c
[n] ↑ m ×
x
co
[n]
y
s
[n] ↑ m ×
x
so
[n]
+ x[n]
2sin(ω
0
n)
After upsampling by m,weget back x
c
[n]and x
s
[n]from y
c
[n]and y
s
[n] respectively,
Note that upsampling by m has zero-insertion block (up-arrow m) and a low-pass
lter for time-domain interpolation,DTFT of x
c
[n]and x
s
[n] are derived in part a,
According to the system diagram,
x
co
[n]= x
c
[n] × 2cos[w
o
n]
Using the multiplication property and doing periodic convolution,we get X
co
(e
jw
)as
shown below,
7
X
co
(e
jw
)
=
π
× 2π ×
1
2
1
2
1
2
π
w
o 0
w
o π ω
w
o
w
M
w
o
w
M
X
so
(e
jw
)
ππ
w
o
w
o
w
o
w
M
w
o
w
M
1
2
1
2
ω
Similarly,x
so
[n]= x
s
[n] × 2sin[w
o
n],and we get X
so
(e
jw
) as shown in the?gure,
Adding X
co
(e
jw
)and X
so
(e
jw
),we get back the spectrum of X(e
jw
),Thus,we recover
x[n],
8
Problem 3
(a)
2t
x(t)= e
t
u(? t)+2e u(t)
Using Laplace transforms of elementary functions (table 9.2),we?nd,
t
e u(? t)
1
,R
1
= R e{ s} <? 1←→
s +1
1
2t
e u(t) R
2
= R e{ s} >? 2←→
s +2
,
Using the linearity property,
2
X(s)=
1
+,ROC containing R
1
R
2
s +1 s +2
s
=
(s +1)(s +2)
ROC =? 2 < R e{ s} <? 1
The pole-zero plot is shown below,
×
2
×
1
Rfractur e
Ifractur m
X(s)p-z map of
9
(b)
x(t)=(e
t
cos t)u(? t)+ u(? t)
t
= e (
1
e
jt
+
1
e
jt
) u(? t)+ u(? t)
2 2
1 1
(1+j)t
u(? t)+ e
(1?j)t
= e u(? t)+ u(? t)
2 2
Using Laplace transforms of elementary functions (table 9.2),we?nd,
(1+j)t
1
R
1
= R e{ s} < 1e u(? t) ←→
s? (1+ j)
,
(1?j)t
1
R
2
= R e{ s} < 1e u(? t) ←→
s? (1? j)
,
u(? t)
1
R
3
= R e{ s} < 0←→
s
Using the linearity property,
1/2
X(s)=
1/2
+ +
1
,ROC containing R
1
R
2
R
3
s? (1+ j) s? (1? j) s
(1/2)s
2
+(1/2)s(1? j)? (1/2)s
2
+(1/2)s(1+ j)? s
2
+2s? 2
ROC= R e{ s} < 0=
s(s
2
2s +2)
1(2s
2
3s +2)
ROC= R e{ s} < 0
=
s(s
2
2s +2)
Solving the denominator,we see that poles are located at,
s =1 ± j
s =0
Solving the numerator,we see that zeros are located at,
3
√
7
s =
4
± j
4
The pole-zero plot is shown below:
10
Ifracturm
p-z map of X(s)
K =?1
1 ×
√
+
Rfracture×
1
√
1
×
3
4
7
7
4
4
2
Problem 4
(a) We are given,
s? 25
X(s)=
s? 25
=
s? s? 12 (s? 4)(s +3)
3 < Re{s} < 4
Using partial fraction expansion,
s? 25 A B
= +
(s? 4)(s +3) s? 4 s +3
multiply both sides by (s? 4) and plug-in s =4,
s? 25
= A
s +3
A =?3
multiply both sides by (s + 3) and plug-in s =?3,
s? 25
= B
s? 4
B =4
11
Therefore,
4
X(s)=
3
+? 3 < Re{s} < 4
s +3s? 4
Using the table of Laplace transforms for elementary functions (table 9.2) and given
ROC,we?nd
3t
x(t)=3e
4t
u(?t)+4e u(t)
(b) We are given,
2s
2
+7s +9 2s
2
+7s +9
X(s)= = Re{s} >?2
(s +2)
2
s
2
+4s +4
As the degree of the numerator is equal to that of the denominator,we need to use
long division to divide the numerator by the denominator before?nding the partial
fraction expansion,
X(s)= 2?
s? 1
(s +2)
2
Now we can?nd the partial fraction expansion of the second term on the right hand
side of the equality,
A Bs? 1
= +
(s +2)
2
(s +2)
2
s +2
multiply both sides by (s +2)
2
and plug-in s =?2,
s? 1= A + B(s +2)
A =?3
multiply both sides by (s +2)
2
,plug-in A and s =1,
s? 1= A + B(s +2)
B =1
Therefore,
1
X(s)=2?
3
+ Re{s} >?2
(s +2)
2
s +2
3 1
=2 +
(s +2)
2
s +2
Re{s} >?2
Using the table of Laplace transforms for elementary functions (table 9.2) and given
ROC,we?nd
2t
x(t)=2δ(t)+3te
2t
u(t)? e u(t)
12
Problem 5 (O & W 9.24 (f))
We are given X(s),|X(jw) for any w can be calculated as K
bc
where K is any real number
ad
|
and a,b,c,d are vector magnitudes as shown below in the pole-zero diagram of X(s),
×
2?1
×
11
2
a
b
c
d
∠a ∠b ∠c ∠d
w
Rfracture
Ifracturm
X(s)
X
1
(s |X
1
(jw)| = |X(jw)| = K
bc
p-z map of
We need to?nd ) such that,
ad
,and there are no poles and
zeros in the right-half plane,
1
2
jwRe?ecting the pole (at 1 on real axis) and zero (at on real axis) along -axis or imaginary
axis,we can conserve the magnitude d and c from the pole and zero respectively,The
resulting pole-zero diagram will be as follows,
×
2?1
×
1
2
a
b
c
d
w
Rfracture
Ifracturm
X
1
(s)p-z map of
From the plot above,the pole and zero at?1 will cancel each other,Therefore,
K(s +
1
2
)
X
1
(s)=
(s +2)
It is important to note that,from p-z map of X(s),as b = d,
bc c
X(jw)| = K = K = X
1
(jw)||
ad a
|
13
Now,we need to?nd X
2
(jw) such that ∠X
2
(jw)= ∠X(jw) and there are no poles or zeros
in the right-half plane of p-z map of X
2
(s),From the p-z map of X(s) shown on the previous
page,we can write the phase,∠X(jw),as
∠X(jw)= ∠b + ∠c? ∠a? ∠d
If we re?ect the pole at s = 1 along jw-axis,the contribution to overall phase from the
re?ected pole becomes?(π? ∠d)=?π + ∠d,Notice that the sign of contributed angle
(∠d) has?ipped,Now,lets convert that pole to a zero,The contribution to overall phase
from the resulting zero is +(π? ∠d)= π? ∠d,
The above two operations,re?ecting a pole along the jw-axis and changing it to zero,just
add +π to overall phase,In order to keep the phase unchanged,we can multiply the resulting
Laplace transform by?1as?1= e
jπ
will subtract π from the overall phase,
1
Similarly,if we re?ect the zero at s =
2
along jw-axis,change that zero to a pole,and
multiply resulting the Laplace transform with?1,the phase will remain unchanged,
Therefore,
X
2
(s)= K(?1)(?1)
(s +1)(s +1)
(s +2)(s +
1
2
)
K(s +1)
2
=
(s +2)(s +
1
2
)
14
Problem 6 (O & W 9.26 )
We need to?nd the Laplace transform of y(t) using the properties of Laplace transform,The
two properties,time shifting (table 9.1 of text book) and time scaling by? 1 (page 686? 687
of text book),that we will need are,
x(t? t
0
) e
st
0
X(s)←→
x(? t) ←→ X(? s)
The ROC is unchanged for the time shifting property,For the time scaling by? 1
property,ROC is re?ected about the jω or imaginary axis in s-plane,
y(t) = x
1
(t? 2)? x
2
(? t +3)
where x
1
(t) = e
2t
u(t)
and x
2
(t) = e
3t
u(t)
taking the Laplace transform
1
x
1
(t) ←→
s +2
,R e{ s} >? 2
1
x
1
(t? 2) e
2s
,R e{ s} >? 2←→
s +2
1
x
2
(t) ←→
s +3
,R e{ s} >? 3
1
x
2
(? t) ←→
s +3
,R e{ s} < 3
1
x
2
(? (t? 3)) e
3s
s +3
,R e{ s} < 3←→
Using the convolution property of Laplace transform (table 9.1),
Y (s)= X
1
(s)X
2
(s),at least R
1
∩ R
2
2s?3s
e e
=
s +2
×
s +3
5s
Y (s)=
2
e
,? 2 < R e{ s} < 3
s? s? 6
15
