Solution of ECE 315 Test 8 F05
1.
A periodic DT signal is described over one fundamental period by
2 , 0 ≤ n < 2
. Its harmonic function, with N F = N 0 , is X [ k ] . Find the
x[n] = 
0 , 2 ≤ n < 5
numerical values of
(a)
X[0]
(b)
X [1]
X[ k ] =
1
NF
X [ 0 ] = 4/5
X [1] = X [1] = 0.524 − j 0.38 = 0.6472 e− j 0.628
∑ x[n]e
− j 2 π nk / N F
n = NF
1
(
= (1 / 5 ) ∑ 2 e− j 2 π nk / 5 = ( 2 / 5 ) 1 + e− j 2 π k / 5
0
)
X [ 0 ] = ( 2 / 5 ) (1 + 1) = 4 / 5
(
)
X [1] = ( 2 / 5 ) 1 + e− j 2 π / 5 = 0.524 − j 0.38 = 0.6472 e− j 0.628
2.
A periodic DT signal x [ n ] with fundamental period 12 has a harmonic function
x [ n / 3] , n / 3 an integer
Let y [ n ] = 
. Based on a
, otherwise
0
representation time N F = 3N 0 its harmonic function is Y[ k ] . If X [ 3] = 2 − j 5 ,
X [ k ] with N F = N 0 .
what are the numerical values of the harmonic function Y[ k ] at the following
points. (If it is impossible to determine a numerical value with the information
given just write “unknown”.)
Using the time-scaling property of the DTFS
x [ n / m ] , n / m an integer
, N F → mN F ⇒ Z [ k ] = (1 / m ) X [ k ]
z[n] = 
, otherwise
0
and X [ k ] = X [ k + qN F ] for any integer q and X [ k ] = X* [ − k ] ,
Since X [ k ] is periodic with period 12 and Y[ k ] = (1 / 3) X [ k ] , Y[ k ] also has a
period of 12, even though its representation time is N F = 3 × 12 = 36 .
1
2 − j5 2
5
(a)
Y[ 3] = X [ 3] =
= − j = 1.795 e− j1.19
3
3
3
3
2
5
(b)
Y[ 39 ] = Y  39 + ( −3) × 12  = Y[ 3] = − j = 1.795 e− j1.19
3
3
2
5
(c)
Y[ 33] = Y[ 33 − 3 × 12 ] = Y[ −3] = Y* [ 3] = + j = 1.795 e+ j1.19
3
3
(d)
Y[ −15 ]
Y[ −15 ] = Y[ −15 + 12 ] = Y[ −3] = Y* [ 3] =
(e)
Y[ −6 ]
Y[ −6 ] = unknown
2
5
+ j = 1.795 e+ j1.19
3
3
Solution of ECE 315 Test 8 F05
1.
A periodic DT signal is described over one fundamental period by
5 , 0 ≤ n < 2
. Its harmonic function, with N F = N 0 , is X [ k ] . Find the
x[n] = 
0 , 2 ≤ n < 5
numerical values of
(a)
X[0]
(b)
X [1]
X[ k ] =
1
NF
X[0] = 2
X [1] = X [1] = 1.309 − j 0.9511 = 1.618 e− j 0.628
∑ x[n]e
n = NF
− j 2 π nk / N F
1
= (1 / 5 ) ∑ 5 e− j 2 π nk / 5 = 1 + e− j 2 π k / 5
0
X [ 0 ] = 1 + 1 = 2 X [1] = 1 + e− j 2 π / 5 = 1.309 − j 0.9511 = 1.618 e− j 0.628
2.
A periodic DT signal x [ n ] with fundamental period 12 has a harmonic function
x [ n / 3] , n / 3 an integer
Let y [ n ] = 
. Based on a
, otherwise
0
representation time N F = 3N 0 its harmonic function is Y[ k ] . If X [ 3] = 1 + j 3 ,
X [ k ] with N F = N 0 .
what are the numerical values of the harmonic function Y[ k ] at the following
points. (If it is impossible to determine a numerical value with the information
given just write “unknown”.)
Using the time-scaling property of the DTFS
x [ n / m ] , n / m an integer
, N F → mN F ⇒ Z [ k ] = (1 / m ) X [ k ]
z[n] = 
, otherwise
0
and X [ k ] = X [ k + qN F ] for any integer q and X [ k ] = X* [ − k ] ,
Since X [ k ] is periodic with period 12 and Y[ k ] = (1 / 3) X [ k ] , Y[ k ] also has a
period of 12, even though its representation time is N F = 3 × 12 = 36 .
1
1 + j3 1
(a)
Y[ 3] = X [ 3] =
= + j = 1.0541e j1.249
3
3
3
1
(b)
Y[ 39 ] = Y  39 + ( −3) × 12  = Y[ 3] = + j = 1.0541e j1.249
3
1
(c)
Y[ 33] = Y[ 33 − 3 × 12 ] = Y[ −3] = Y* [ 3] = − j = 1.0541e− j11.249
3
(d)
Y[ −15 ]
1
Y[ −15 ] = Y[ −15 + 12 ] = Y[ −3] = Y* [ 3] = − j = 1.0541e− j11.249
3
(e)
Y[ −6 ]
Y[ −6 ] = unknown