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HW5 solutions – ECE351
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REVIEWS
For hw#5, you should review the following concepts:
(1) Fully understand how to use the following table representing of 4 classes of the signals.
(2) The relationship between sin/cos and exponential terms as follows:
sincos je j
sincos je j
j
eeee jjjj
22
sin;cos
(3) Shifting property:
)()()( afdxaxxf
(4) dueed uu )(
(5) For a continuous-time periodic signal (FS), when you find X[k], you may integrate over one period of
time T (fundamental period). However, you don’t have to strictly integrate from 0 to T. For example,
T tkjT T
T
tkjtkjtdetx
Ttdetx
Ttdetx
TkX 000
111
0
2
2)()()(][
/
/
for any real
HW5 solutions – ECE351
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Similar for a discrete-time periodic signal (DTFS),
nkj
N
n
nkjN
n
enxN
enxN
kX 00
11
0
11
][][][ for any integer
and N = fundamental period
Hints: The problem is asking to find X[k] from a given input x[n]. If x[n] composes of either sin() or cos(),
then we may simplify them into the form of exponential terms (i.e., )(je ). After that, we can determine X[k] by
inspection using
Otherwise, we need to compute X[k] directly from
(a) 317
6 nnx cos][
Find fundamental period N from knNnn 2317
6317
6317
6 cos)(coscos where k is the
smallest positive integer that makes N=integer
That is, kN 2176
3
17kN k =3 N = 17
Thus, 17
220
N
From ,cos2
jj ee
then 3176
3176
2
1
317
6
njnjeennx cos][
Because x[n] can be simplified into the form of exponential terms (i.e., )(je ), therefore, we can
determine X[k] by inspection using
nkjN
k
ekXnx 0
1
0
][][
That is: njjnjjnjnj
eeeeeenx))(())((
][ 172
3172
33176
3176
3
2
13
2
1
2
1
For a period of N=17, let 8778 ,,...,,k
][3X ][ 3X
HW5 solutions – ECE351
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Therefore,
(b) 1219
1019
14 nnnx cossin][
Find fundamental period 1N for n
1914sin
kN 211914
7
191
kN k =7 191 N
Find fundamental period 2N for n
1910cos
kN 221910
5
192
kN k =5 192 N
Find fundamental period sumN from 1921 NqNrN sum
;where r and q are smallest positive
integers
Find 19
220
N
For a period of N=19, let 9889 ,,...,,k
We do similar process as in part (a). That is:
1219
1019
14 nnnx cossin][
njnjnjnjnj
njnjnjnj
eeeejej
eeeej
19
20
19
25
19
25
19
27
19
27
19
01
19
01
19
41
19
41
2
1
12
11
)()()()()(
By inspection
X[k] =
j ; k= -7
0.5 ; k= -5
1 ; k= 0
0.5 ; k= 5
-j ; k= 7
0 ; otherwise on k = {-9,-8,…,8,9}
HW5 solutions – ECE351
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(c)
From the above graph x[n], the fundamental period N = 12.
Find 612
220
N
Because we cannot simplify x[n] into the exponential form. Therefore, we can compute X[k] directly
from
nkjN
n
enxN
kX 0
1
0
1
][][
Let 6545 ,,...,,n
Therefore,
(d)
From the above graph x[n], the fundamental period N =8.
Find 48
220
N
HW5 solutions – ECE351
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Compute X[k] directly from nkjN
n
enxN
kX 0
1
0
1
][][
Let 4323 ,,...,,n
Therefore,
(e)
From the above graph x[n], the fundamental period N =10.
Find 510
220
N
Compute X[k] directly from nkj
N
n
enxN
kX 0
1
0
1
][][
Let 4345 ,,...,,n
Therefore,
HW5 solutions – ECE351
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Hints: The problem is asking to find x[n] from a given input X[k]. If X[k] composes of either sin() or cos(),
then we may simplify them into the form of exponential terms (i.e., )(je ). After that, we can determine x[n] by
inspection using
Otherwise, we need to compute x[n] directly from
(b) kjkkX19
1419
10 2 sincos][
Find fundamental period 1N for k19
14sin
mN 211914 where m = positive integer
7
191
mN m=7 191 N
Find fundamental period 2N for k19
10cos
mN 221910
5
192
mN m =5 192 N
Find fundamental period sumN from 1921 NqNrN sum
;where r and q are smallest positive
integers
Find 19
220
N
Use 2
jj ee
cos and j
ee jj
2
sin in kjkkX19
1419
10 2 sincos][
For a period of N=19, let 9889 ,,...,,n
By inspection with nkjN
n
enxN
kX 0
1
0
1
][][ , we have
HW5 solutions – ECE351
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(c)
From the above graph X[k], the fundamental period N = 12.
Find 612
220
N
Because we cannot simplify X[k] into the exponential form. Therefore, we can compute x[n] directly
from
nkjN
k
ekXnx 0
1
0
][][
Let 6545 ,,...,,k
Therefore,
nkj
k
ekXnx 6
6
5
][][
njnjnjnjnjnjnj
eeeeeee 66
64
63
62
62
63
64
3212
)()()()()()()(
nnnn )()cos()cos()cos( 131242
323
2
Note:
From sincos je j .
Then nnj nnjne )()cos()sin()cos( 1
HW5 solutions – ECE351
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(d)
From the above graph, the fundamental period N =14.
Find 714
220
N
Compute x[n] directly from nkj
N
k
ekXnx 0
1
0
][][
Let 7656 ,,...,,n
Note that X[k] can be represented as ]}[arg{
][][kXjekXkX where ]}[arg{ kX is the phase of the
signal
Therefore,
Hints: The problem is asking to find X[k] from a given input x(t). If x(t) composes of either sin() or cos(),
then we may simplify them into the form of exponential terms (i.e., )(je ). After that, we can determine X[k] by
inspection using
Otherwise, we need to compute X[k] directly from
HW5 solutions – ECE351
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(a)
Find fundamental period 1T from
32
1 T
Find fundamental period 2T from
2
12 T
Find fundamental period sumT from 221 TqTrT sum ;where r and q are some positive constants
sumT
20
From 2
jj ee
cos and j
ee jj
2
sin
then
Compare to , then we have
(c) )()( mtetxm
mj
27
2
By inspection mj
e 7
2
, we can tell that
27
2
jmj
ee (the phase of the signal repeats every 2) when
m= 7.
Let },,,,,,{ 3210123 m . Thus, by looking at )( mt 2 , we can tell that the period T is from 0 to
2(7)=14.
7
20
T
Compute X[k] directly from
T tkjtkj
tdetxtdetxT
kX0
7
7
7
14
110
)()(][
By using the shifting property:
)()()( afdxaxxf , we have
t
Tt
1
23
sinsin
t
Tt
2
24
coscos
tkj
k
ekXtx 0
][)(
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7
7
7
14
1tdetxkX
tkj
)(][
3
3
)1(7
2)2(
7
3
3
7
2
7
7
73
3
7
2
7
7
7
3
3
7
2
14
1
14
1
)2(14
1
)2(14
1
m
kmjmkj
m
mj
tkj
m
mj
tkj
m
mj
eee
tdemte
tdemte
X[k]
(d)
From above figure, we can tell that )(sin)( ttx
T = 1 and 20
Compute X[k] directly from
T tkj
tdetxT
kX0
01
)(][
That is:
HW5 solutions – ECE351
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HW5 solutions – ECE351
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HW5 solutions – ECE351
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22
)24(2
)()(
j
n
nj
n
njj
e
en
enxeX
2)(
2|)(|
j
j
eX
eX
HW5 solutions – ECE351
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Method 1:
By inspection, nj
n
jj enxeeX
][)(2
That is, when n = -2, x[n] = 1. Otherwise, x[n] = 0.
In other words, ]2[][ nnx
Method 2:
Compute x[n] from
)(sin)(
)(sin)(
))((sin)(
)(
)()(
nn
nn
nn
eenj
njnj
2
12
2
12
2
1
22
1 22
2je
HW5 solutions – ECE351
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That means x[n]=0 except when n = -2. Note that when n = -2, denominator is equal to zero.
Therefore, we need to take the derivative of numerator and denominator then take the limit
approaching to -2.
1
222
)(cos
lim
)(
)(sin
limn
ndn
d
ndn
d
nn
That is
][][ 2 nnx
(c) )()( tutetx t
2
0
2
00
0
0
)1(
1
|)1
1(
1
1|)(
1
1
)(1
1
)(
jw
eejw
dteejw
eetjw
eedtjw
dtetejwX
jwtt
jwttjwtt
jwtt
jwtt
mj
m
mtj
m
m eadtemta
00
0
)( Shifting property
)()()( afdtattf
jm
mj
eaea
1
1
0
r
rn
n
1
1
0
where 1r
