ECE301 Summer II, 2006
Final Exam
August 03, 2006
1
Score:
/100
Name: Solution, Green/White Version
You must show all of your work for full credit. All answers should be simplified as much as
reasonably possible. Calculators may NOT be used.
1. (20 points)
Find the inverse DTFT of the function.
γe−jΩ
X(Ω) = 2 cos(kΩ)
(e−jΩ − γ)2
!
First, let us write the given function as
X(Ω) = 2 cos(kΩ)X1 (Ω)
where
γejΩ
.
(ejΩ − γ)2
!
X1 (Ω) =
Then from Table 9.1, entry 4, we have that
x1 [n] = nγ n u[n].
If we let X2 (Ω) = X1 (−Ω) we then have that
x2 [n] = −nγ −n u[−n].
Finally, applying the result of Problem 9.3-6 on page 894, or simply
writing the sine in terms of exponentials and applying the time
shifting property as listed in Table 9.2, we have that
x[n] = x2 [n + k] + x2 [n − k]
= (−n + k)γ −n+k u[−n + k] + (−n − k)γ −n−k u[−n − k].
ECE301 Summer II, 2006
4. (20 points)
Final Exam
August 03, 2006
5
Find the Discrete Time Fourier Series (DTFS) for the function
x[n] = 2 sin(1.3πn) + 3 cos(5.5πn)
and indicate the fundamental frequency, the number of terms in the DTFS, and which
harmonics are present.
Subtracting multiples of 2π to obtain angles in the range [0, 2π) yields
x[n] = 2 sin(1.3πn) + 3 cos(1.5πn).
13 is prime so GCF(13,15) is 1 and the fundamental
frequency is Ω0 = 0.1 rad/s. The number of terms in the DTFS will be
N0 = 2πm/Ω0 where m is the smallest positive integer that yields an
integer N0 . We find that N0 = 20 with m = 1. Writing x[n] in terms of
exponentials yields
x[n] = 2 ej1.3πn − e−j1.3πn /(2j) + 3 ej1.5πn + e−j1.5πn /2.
Simplifying and adding 2π to the negative angles yields
3 j0.5πn
3 j1.5πn
e
+
e
x[n] = −je
+ je
+
2
2
3 j15Ω0 n
3 j5Ω0 n
= −jej13Ω0 n + jej7Ω0 n +
e
+
e
2
2
j1.3πn
j0.7πn
and we see that the 5th, 7th, 13th, and 15th harmonics are present in
the DTFS.