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Signals Problems March 9, 2014 1
Signal Problems
Given x n [ ] shown below, sketch the following a) x [1 − n ] b) x n [ −1] c) x [2 n +1] d) x [1 −2 ] n
3 0 1 2 3 4
0
1
2
n
x n [ ]
► Solution:
0 1 2 3 4 5
0
1 2
a)
3 2 1 0 1 2 3 4 5 2
1
0
1 2
b)
(^22 1 ) 1
0
1 2
c)
(^21 0 1 ) 1
0
1 2
d)
n
n n
n
For each of the following sequences, find and plot (^) xe [ ] n and (^) xo [ ] n.
0 1 2 3
0 1 2 3
0
2
a)
2
0
2
b)
4 3 2 1 0 1
2
0
2
c)
1
2
3
[ ]
[ ]
[ ]
x n
x n
x n
n
n
n
► Solution:
Signals Problems March 9, 2014 2
0 1 2 3
0
1
2 a)
3 2 1 0 1 2 3
1
0
1
2
3 2 1 0 1 2 3
2
1 0 1 2
b)
3 2 1 0 1 2 3
2
1 0 1 2
4 3 2 1 0 1 2 3 4
1
0
1
2
c)
4 3 2 1 0 1 2 3 4
1
0
1
2
n
n
n n
n
n
1 1
2 2
3 3
[ ] [ ]
[ ] [ ]
[ ] [ ]
e o
e o
e o
x n x n
x n x n
x n x n
Given x n [ ] = (1 − 2 ) [ j δ n + 2] − 2 [ ]δ n + (1 + j ) [ δ n − 1], find
a) Re{ x n [ ]}
b) Im { x n [ ]}
c) Even { x n [ ]}
d) Odd { x n [ ]}
e) Im Even{ { x n [ ]}}
f) Even Im{ { x n [ ]}}
► Solution:
a) Re{ x n [ ] } = δ[ n + 2] − 2 [ ]δ n + δ[ n −1]
b) Im { x n [ ] } = −2 [ δ n + 2] + δ[ n −1]
c) Even { x n [ ] } = ( 12 − j ) [ δ n + 2] + 12 (1 + j ) [δ n + 1] − 2 [ ]δ n + 12 (1 − j ) [ δ n − 1] + ( 12 − j ) [ δ n −2]
d) Odd { x n [ ] } = ( 12 − j ) [ δ n + 2] − 12 (1 + j ) [δ n +1] + 12 (1 + j ) [ δ n − 1] − ( 12 − j ) [ δ n −2]
e) Im Even{ { x n [ ] } } = − δ [ n + 2] − 12 δ[ n +1] + 12 δ[ n −1] + δ[ n −2]
f) Even Im{ { x n [ ] } } = − δ [ n + 2] + 12 δ[ n +1] + 12 δ[ n −1] − δ[ n −2]
For each of the following sequences, find (^) xe [ ] n and (^) xo [ ] n. a) (^) x n [ ] = 1 + ej^^4 π n b) (^) x n [ ] = (1 − j ) [ δ n + 1] + j δ[ ] n + ( j + 1) [δ n −1] c) (^) x n [ ] = δ[ n +1] − 2 [δ n −1] + 2 j δ[ n −2] d) x n [ ] = 2sin( 4 π n + 3 π)
► Solution: a)
( (^ )^ (^ ))
( (^ )^ (^ ))
4 4 4 4
(^12 ) (^12 )
[ ] 1 1 1 cos [ ] 1 1 sin
e^ j^ n^ j^ n j n j n o
x n e e n x n e e j n
π π
π π
π π
− −
b)
Signals Problems March 9, 2014 4
Similarly,
4 3 4 3 4 3 4 3 4 3 4 3 4 3 4 3 4 3 4
[ ] 2 cos( ) 2 cos( ) cos cos sin sin cos( ) cos sin( ) sin cos cos sin sin cos cos sin sin 2sin sin 3 sin
x o n n n n n n n n n n n n n
π π π π π π π π π π π π π π π π π π π π π π π
Determine which of the following sequences is periodic. If the sequence is periodic, compute the fundamental period, N. a) cos( 311 π n ) b) cos( ) n
c) cos( 52 π n +2)
d) cos( 73 π n ) +sin( 52 π n ) e) cos( 73 π^ n ) +sin(5 ) n f) cos( 73 π n ) sin( 52 π n )
► Solution:
a) 3 2 22 22 11 3 3
π N π k N π k k
= → = ^ =
, so^ k^ =^3 and^ N^ =^22.
b) Not periodic
c) 5 2 4 2 5
π N = π k → N = k
, so^ k^ =^5 and^ N^ =^4 .The phase term (2) is irrelevant to the issue of periodicity.
d) The first term is periodic with (^667) 7
N^ π^ k k
= ^ =
, which means that^ k^ =^7 and^ N^ =^6. The second term is periodic
with (^445) 5
N^ π^ k k
= ^ =
, which means that^ k^ =^5 and^ N^ =^4. Hence, the period is the greatest common multiplier or 4 and
6, namely N = 12. e) Not periodic.
f) cos( 73 π n ) sin( 52 π n ) = 12 ( sin( 52 π^ n + 73 π n ) + sin( 52 π n − 73 π n ) ) = 12 ( sin( 296 π n ) + sin( π 6 n )). The first term of the sum is periodic
with (^121029) 29
N^ π^ k k
= ^ =
,^ k^ =^29 and^ N^ =^12. The second term is also periodic with^ N^ =^12. Hence the entire sequence is
periodic with N = 12.
Given
, 3 3 ( ) 0, otherwise
n n x n
determine and sketch the following sequences: a) y n [ ] = 2 [ ] x n − 1 b) y n [ ] = x n [ −3] c) y n [ ] = (^13) ( x n [ + 1] + x n [ ] + x n [ −1])
Signals Problems March 9, 2014 5
d) [ ] [ ]
n k
y n x k =−∞
= (^) ∑
e) y n [ ] = x [2 − n ] f) y n [ ] = nx n [ ]
► Solution:
0 1 2 3
0
5
a)
0 1 2 3 4 5 6 0
2
4
b)
(^04 3 2 1 0 1 2 3 )
1
2
c)
(^03 2 1 0 1 2 )
10
20
d)
1 0 1 2 3 4 5 0
2
4
e)
(^103 2 1 0 1 2 )
0
10
f)
n
n
n
n
n
Determine whether each of the following systems is linear. For each, show whether additivity and scaling are satisfied. a) y n [ ] = nx n [ ] b) y n [ ] =cos (^) ( x n [ ]) c) y n [ ] = e j^^ ω x n [^ ] u n [ ] d) y n [ ] = x [2 ] n
► Solution: In the forgoing, let
1 1 2 2
[ ] { [ ]}
[ ] { [ ]}
y n T x n y n T x n
and conduct the “Experiments” A, B, C and D discussed in the Chapter.
a) y n [ ] = nx n [ ]is linear. Additivity : Experiment A: T (^) { x n 1 [ ] + x 2 (^) [ ] n (^) } = n x n ( 1 [ ] + x 2 (^) [ ] n (^) ) = nx n 1 [ ] + nx 2 [ ] n Experiment B: T (^) { x 1 (^) [ ] n (^) } + T (^) { x 2 (^) [ ] n (^) } = n x n ( 1 [ ] + x 2 (^) [ ] n (^) ) = nx n 1 [ ] + nx 2 [ ] n. The result of these two experiments is the same, so additivity is satisfied. Scaling : Experiment C: T (^) { kx n [ ] (^) } = n kx n ( [ ]) Experiment D: kT (^) { x n [ ] (^) } = k nx n ( [ ]) The result of these experiments is the same, so scaling is satisfied.
b) y n [ ] = cos (^) ( x n [ ])is non-linear. Additivity : Experiment A: T (^) { x 1 (^) [ ] n + x 2 (^) [ ] n (^) } = cos (^) ( x n 1 [ ] + x 2 [ ] n ) Experiment B: T (^) { x n 1 [ ] (^) } + T (^) { x 2 (^) [ ] n (^) } = cos( x n 1 [ ]) + cos( x 2 [ ]) n. The result of these two experiments is not the same, so additivity is not satisfied. Scaling :
