Solution to ECE Test #2 Su04
1.
For any even periodic function of time, its complex Fourier series harmonic
function using the fundamental period as the representation time is always purely real.
2.
For any real-valued signal, the CTFS harmonic function has a magnitude that is
an even function of harmonic number, k.
3.
The signal, x [ n ] = 1, has a fundamental period, N 0 = 1 .
(a)
Find its DTFS harmonic function using that fundamental period as the
representation time.
X [ k ] = comb1 [ k ] = 1
(b)
n
 n
an integer
x   ,
Now let z [ n ] =   4 
. Find the DTFS harmonic function for
4
0
, otherwise

z [ n ] using its fundamental period as the representation time.
Z[ k ] =
(c)
Verify that z [ 0 ] = 1 and that z [1] = 0 by using the DTFS representation of z,
z[n] =
z[0] =
z [1] =
4.
1
1
X[ k ] =
4
4
∑
k= 4
1
∑ 4e
k= 4
0
∑ Z[ k ]e
j 2 π ( kF0 ) n
.
k = NF
=
1 1 1 1
+ + + = 1 Check
4 4 4 4
π
3π
j
j
 1
1 j 2 π4 k 1  0
e
=  e + e 2 + e jπ + e 2  = (1 + j − 1 − j ) =0
4
4
 4
Check
A signal, x ( t ) = 4 cos ( 30π t ) − 3 can be represented for all time by a CTFS but its
t
integral,
∫ x ( λ ) dλ ,
−∞
cannot. Why? Because the integral is not periodic.
5.
Using Parseval’s theorem, find the numerical value of the signal power of the
signal, x ( t ) = 3 cos (10π t ) − 2 sin (10π t ) from its CTFS harmonic function.
3
3

3

δ [ k − 1] + δ [ k + 1]) − j (δ [ k + 1] − δ [ k − 1]) = δ [ k − 1]  + j  + δ [ k + 1]  − j 
(
2

2

2
X[ k ] =
Px =
∞
∑ X[ k ]
k = −∞
2
∞
3
= ∑ δ [ k − 1]  +
2
k = −∞
∞

3
j  + δ [ k + 1]  −
2


j

2
2
3
2
3

Px = ∑  δ [ k − 1] + δ [ k + 1] + (δ [ k − 1] − δ [ k + 1])


2
k = −∞ 2

 

9 2
2
2
2



Px = ∑
δ [ k − 1] + δ [ k + 1] + 2 δ [ k − 1]δ [ k + 1] + δ [ k − 1] + δ [ k + 1] − 2 δ [ k − 1]δ [ k + 1]
4

  = δ k −1

k = −∞
 = δ [ k −1]
=0
=0
− δ [ k + 1]
− δ [ k + 1]
[ ]
9 9
13
Px = + + 1 + 1 =
= 6.5
4 4
2
∞
6.
A DT signal, x [ n ] with fundamental period, N 0 = 6 , has a DTFS harmonic
function, X [ k ] , with the following numerical values at the given harmonic numbers, k.
k
−3 −2 −1 0 1
X [ k ] 1 − j 2 −1 4 −1
(a)
8 9
j2 1
Find the numerical value of x [ 0 ] .
x[n] =
(b)
2 3 4
5 6 7
j 2 1 − j 2 −1 4 −1
∑ X[ k ]e
j 2π
kn
N0
k= 6
⇒ x[0] =
5
∑ X[ k ]e = ∑ X[ k ] = 4 − 1 + j 2 + 1 − j 2 − 1 = 3
0
k= 6
k=0
Find the numerical average value of x [ n ] .
The average value of x [ n ] is X [ 0 ] which is 4.
(c)
Find the numerical value of the average signal power of x [ n ] .
From Parseval’s theorem,
Px =
∑
k = NF
5
X [ k ] = ∑ X [ k ] = 16 + 1 + 4 + 1 + 4 + 1 = 27
2
k=0
2