Solution of EE 503 Test #1 S04
1.
A CT signal is defined by x( t) = 10 + 3t − t 2 . Find the numerical values of
x(−2) = 0
(a)
x(−2) = 10 + 3(−2) − (−2) = 10 − 6 − 4 = 0
2
 t − 1
y( 3) = 11.25 where y( t) = x

 4 
(b)
1
3 1
 3 − 1
 1
 1  1
y( 3) = x
 = x  = 10 + 3  −   = 10 + − = 11 = 11.25
 2  2
 4 
 2
4
2 4
2
2.
A DT signal, x[ n ] , is periodic with period, 8. It is described over one period by
x[ n ] = n (u[ n ] − u[ n − 5]) , 0 ≤ n < 8
Find the numerical value of x[1251] = 3.
For any integer, n, x[1251] = x[1251 + 8 n ] . Let n be -156. Then
x[1251] = x[1251 + 8 × (−156)] = x[1251 − 1248] = x[ 3]
x[1251] = 3(u[ 3] − u[ 3 − 5]) = 3
3.
A CT signal is defined by x( t) = sin(2πt) rect (5 t)e − t . Find the numerical value of
1
2
∫ x(t)dt = 0.
−
1
2
The integrand is an odd function and the limits are equal in magnitude and opposite in sign.
4.
 n − 3
Find the numerical value of the signal energy of x[ n ] = −2 tri
.
 2 
Ex =
∞
∑ x[n]
n =−∞
2
2
5
 2  1 2 2  1 2

 n − 3
2  n − 3
=
4
0 +   + 1 +   + 02  = 6
=
4
= ∑ −2 tri
tri 



∑
 2 
 2
 2
 2 
n =−∞
n =1


∞
5.
A DT system is described by 2 y[ n ] − 5 y[ n − 1] = x[ n ] where x is the excitation and y is the
response. Is it stable?
Characteristic Equation:
2α − 5 = 0 ⇒ α =
5
>1
2
Unstable because the magnitude of at least one eigenvalue is greater than or equal to one.
6.
A CT system is described by y′′ ( t) − 2 y′ ( t) + y( t) = x( t) − x′ ( t) where x is the excitation
and y is the response. Is it stable?
λ2 − 2λ + 1 = 0 ⇒ (λ − 1)(λ − 1) ⇒ λ1,2 = 11
,
Unstable because the real part of at least one eigenvalue is greater than or equal to zero.
7.
If y[ n ] = x[ n ] ∗ h[ n ] and x[ n ] = u[ n ] − u[ n − 2] and h[ n ] = 3δ [ n ] + 2δ [ n + 1] what is the
numerical value of y[1] ?
y[ n ] = (u[ n ] − u[ n − 2]) ∗ ( 3δ [ n ] + 2δ [ n + 1])
y[ n ] = u[ n ] ∗ 3δ [ n ] − u[ n − 2] ∗ 3δ [ n ] + u[ n ] ∗ 2δ [ n + 1] − u[ n − 2] ∗ 2δ [ n + 1]
y[ n ] = 3 u[ n ] − 3 u[ n − 2] + 2 u[ n + 1] − 2 u[ n − 1]
y[1] = 3 u[1] − 3 u[1 − 2] + 2 u[1 + 1] − 2 u[1 − 1] = 3 u[1] − 3 u[−1] + 2 u[2] − 2 u[0] = 3 − 0 + 2 − 2 = 3
8.
Let x( t) = [δ ( t) − 4δ ( t − 2)] ∗ comb( t) . Find the numerical value of its CTFS harmonic
function, X[ k ], at k = 3 using its fundamental period as the representation time.
x( t) = δ ( t) ∗ comb( t) − 4δ ( t − 2) ∗ comb( t) = comb( t) − 4 comb( t − 2)
14243
= comb( t )
x( t) = comb( t) − 4 comb( t) = −3 comb( t) ⇒ X[ k ] = −3 comb1[ k ] = −3
Alternate Solution:
X[ 3] = −3
x( t) = δ ( t) ∗ comb( t) − 4δ ( t − 2) ∗ comb( t) = comb( t) − 4 comb( t − 2)
FS
comb( t) − 4 comb( t − 2) ←
→ comb1[ k ] − 4 comb1[ k ]e − j 2πk(1)( −2)
[
FS
comb( t) − 4 comb( t − 2) ←
→comb1[ k ] 1 − 4 e j 4 πk
1424
3
=1
j12π
= −3
X[ k ] = 1 − 4 e j 4 πk ⇒ X[ 3] = 1 − 4 e{
=1
]
 2π ( n − 3) 
Let x[ n ] = 5 cos
 . Let X[ k ] be its DTFS harmonic function


8
using its fundamental period as the representation time. X[−25] can be expressed as a single
complex number in the form, X[−25] = a + jb . Find the numerical values of the real numbers, a
and b.
9.
(11 pts)
 2πn  FS 1
cos
 ←→ comb N 0 [ k − 1] + comb N 0 [ k + 1]
2
 N0 
(
)
 2πn  FS 5
5 cos
 ←→ (comb 8 [ k − 1] + comb 8 [ k + 1])
 8 
2
2πk ( 3)
−j
 2π ( n − 3)  FS 5
8
5 cos
 ←→ (comb 8 [ k − 1] + comb 8 [ k + 1])e


8
2
3πk
−j
 2π ( n − 3)  FS 5
4
1
1
5 cos
comb
+
comb
+
←

→
k
−
k
e
[
]
[
]
(
)

8
8


8
2
3π ( −25)
−j
5
4
X[−25] = (comb 8 [−25 − 1] + comb 8 [−25 + 1])e
2
 j 75π
5
26
24
X[−25] =  comb 8 [− ] + comb 8 [− ] e 4
2  14243 14243
=0
 75π

+ 2 nπ 

4
5 j 754π 5 j 
X[−25] = e
= e
2
2
X[−25] = −
 75π

−18π 

4
5 j
= e 
2
=1
 75π 72π 
−

4
4 
5 j
= e 
2
5 j 34π 5  2
2
= e
= −
+j
2
2 2
2 
5 2
5 2
5 2
5 2
+j
⇒a=−
and b =
4
4
4
4