Solution of ECE 503 Test #3 Su03 6/20/03
1.
(2 pts) Why is there no CTFS integration property for periodic signals with non-zero
average values?
Because, if the average value is non-zero, the integral is not periodic and cannot, therefore, be
represented for all time by a CTFS.
2.
(10 pts) Let a signal be defined by
x( t) = rect (2 t) ∗ comb( t) =
∞
∑ rect (2(t − m)) .
m =−∞
Using a representation time, TF = 2, find the numerical values of the harmonic function, X[ k ], for
k = 0,1, 2, 3, 4 .
Using the fundamental period of x( t) ( TF = 1) as the representation time, we get directly from the
table,
1
 k
F
rect (2 t) ∗ comb( t) ←
→ sinc  .
 2
2
Changing the period to TF = 2 changes the harmonic function to
k
1
 k
an integer
 sinc  ,
4
2
X[ k ] =  2
0
, otherwise

Therefore,
X[0] =
1
1
 1 1
, X[1] = 0 , X[2] = sinc  =
, X[ 3] = 0 , X[ 4 ] = 0
 2 π
2
2
Alternate Solution:
1
X[ k ] =
TF
X[ k ] =
∫ x(t)e
TF
− j 2π ( kf F ) t


1

dt = ∫ x{
(t) cos
(πkt
) − j sin(πkt
)dt
4 4
3 1424
3
2 2 even 12

 even
odd
1
1
1
1
x( t) cos(πkt) dt = ∫ x( t) cos(πkt) dt = ∫ x( t) cos(πkt) dt
∫
2 2
2 −1
0
For k = 0,
X[ k ] =
1
1
4
1
1
1
1
x( t) dt = ∫ x( t) dt = ∫ dt + ∫ dt =
∫
2 −1
2
3
0
0
4
For k ≠ 0,
 πk 
 3πk 
sin  − sin

 4
 4 
 sin(πkt) 
 sin(πkt) 
X[ k ] = ∫ cos(πkt) dt + ∫ cos(πkt) dt = 
 +
 =
πk
 πk  0  πk  3
3
0
1
4
1
4
1
4
4
For any even k (except k = 0), X[ k ] = 0.
For any odd k,
and
1
 πk 
 3πk 
sin  = sin

 4
 4 
 πk 
2 sin 
 4 1
 k
X[ k ] =
= sinc 
 4
πk
2
and these solutions agreee with the previous ones.
X[0] =
1
1
 1 1
, X[1] = 0 , X[2] = sinc  =
, X[ 3] = 0 , X[ 4 ] = 0
 2 π
2
2
 2πn 
 2πn 
(14 pts) Find the DTFS harmonic function of x[ n ] = 2 cos
 − 5 sin
 using the
 8 
 12 
fundamental period of x[ n ] as the representation time, N F .
3.
N F = 24
Using the tables and the change-of-period property,
X[ k ] = (comb 24 [ k − 3] + comb 24 [ k + 3]) − j
4.
5
(comb24 [k + 2] − comb24 [k − 2]) .
2

1 
1
 
 
A DT signal has a DTFT, X( F ) = rect  5 F −   + rect  5 F +    ∗ comb( F ) .
 
 
4 
4

(a)
(8 pts) Find the signal’s total signal energy, E x .
2

1 
1
 
 
E x = ∫ rect  5 F −   + rect  5 F +    ∗ comb( F ) dF
1
 
 
4 
4

1 1
− +
4 10
1 1
+
4 10
Ex =
∫
dF +
1 1
− −
4 10
1 1
−
4 10
(b)
∫
dF =
2
5
(7 pts) Find the signal, x[ n ] , and express it as a combination of real-valued DT
functions (no j’s in the expression).
 n F
→ wrect ( wF ) ∗ comb( F )
sinc  ←
 w
 n F
sinc  ←
→ 5rect (5 F ) ∗ comb( F )
 5
1
 n F
sinc  ←
→ rect (5 F ) ∗ comb( F )
 5
5
πn
1
1
 
 n + j 2 F
←→ rect  5 F −   ∗ comb( F )
sinc  e
 5
 
4
5
πn
1
1
 
 n − j
F
→ rect  5 F +   ∗ comb( F )
sinc  e 2 ←
 5
 
4
5
πn
−j  1
 + j π2n
 
 n F 
→ rect  5 F −
+ e 2  sinc  ←
e


 
5

5

2  πn 
 
 n F 
cos  sinc  ←
→ rect  5 F −
 5
 
5  2

1 
1
 
  + rect  5 F +    ∗ comb( F )
 
4 
4
1 
1
 
  + rect  5 F +    ∗ comb( F )
 
4 
4
5.
F
e −πt ←
→ e −πf ,
2
(8 pts) Given the transform pair,
x( t) = e
−π ( 3( t − 2))
2
find the CTFT,
. What is the numerical maximum value of X( f ) ?
e
e
−π ( 3 t ) 2
−π ( 3( t − 2))
2
 f
X( f ) ,
of
2
1 −π  
←→ e  3 
3
F
 f
2
1 −π  
←→ e  3  e − j 4 πf
3
F
Maximum value occurs where f = 0. Value is
6.
2
1
.
3
( 2 pts) What is the fundamental reason that ideal filters cannot be actually built?
Their impulse responses are all infinite in extent and cannot, therefore, be the impulse responses of
causal systems.