ECE 2025 Spring 2011, Problem Set #10 Solutions
10.1* The objective is to find the forward Fourier Transforms for the expressions below. The key to
solving these types of problems is understanding (1) the Fourier transform pairs that you can use
from Table 11-2 in your text, and (2) the Fourier transform properties that you can apply from
Table 11-3. For ease, the tables are appended to this homework solution.
(a)
x(t) = sin(2t) ∗[u(t − 7) − u(t − 9)]
Understanding that a convolution in continuous time (CT) become a multiplication in the
frequency domain (FD) we take the Fourier transform (FT) of both signals. Essentially we will
do the following:
€
X( jω ) = X1 ( jω ) • X 2 ( jω )
For the first signal we have:
sin(2t) ⇒ X1 ( jω ) = − jπδ (ω − 2) + jπδ (ω + 2)
€
For the next signal we have a delayed square wave. So we take the FT of a square pulse:
F
€
[u(t +1) − u(t −1)]↔ X 2 ( jω ) =
and multiply it by:
€
e − j 8ω
sin(ω )
ω /2
to yield
X 2 ( jω ) = 2e − j 8ω
To find the overall result we need multiply the two FTs:
sin(ω )
ω
X( jω )€= X1 ( jω ) • X 2 ( jω ) = [− jπδ (ω − 2) + jπδ (ω + 2)] • 2e − j 8ω
€
sin(ω )
ω
Given the Sampling property of the CT impulse we have:
⎛
⎞
⎛ sin(−8) ⎞
sin(8)
X( jω ) = j2π ⎜ e − j16
δ (ω + 2) − e j16 ⎜
⎟δ (ω − 2)⎟
⎝ −8 ⎠
8
⎝
⎠
sin(8) − j16
X( jω ) = j2π
e δ (ω + 2) − e j16δ (ω − 2))
(
8
€
X( jω ) = j0.777(e − j16δ (ω + 2) − e j16δ (ω − 2))
(b)
€
x(t) = 0.3e −0.1(t −0.2)u(t − 0.2)
Understanding that we have a delayed one-sided exponential, we take the FT of a one-sided
exponential:
€
F
0.3e −0.1t u(t)↔
€
0.3
0.1+ jω
and multiply it by:
(c)
e − j 0.2ω
sin(30πt)
x(t) = €
cos(80πt)
πt
to yield
X( jω ) = e − j 0.2ω
0.3
0.1+ jω
€
Examining the FT table we see that we the pair for a sinc function but not a sinc multiplied by a
cosine. We can apply the cosine Modulation property from the table to the FT of the sinc.
Beginning with the FT of the sinc we have:
€
X1 ( jω ) = [u(ω + 30π ) − u(ω − 30π )]
The modulation property will cause frequency shifts of 80pi and a multiplication by ½. Thus:
1
1
X( jω ) = [u(ω + 30π − 80π ) − u(ω − 30π − 80π )] + [u(ω + 30π + 80π ) − u(ω − 30π + 80π )]
2
2
1
1
X( jω ) = [u(ω − 50π ) − u(ω −110π )] + [u(ω +110π ) − u(ω + 50π )]
2
2
€
€
Now we can plot:
X( jω )
| X( jω ) |
z
1/2 €
(d)
€
1/2 x(t) = 8e −0.7t cos(πt)u(t)
Examination of the FT pairs indicates that we have a match for a one-sided exponential. The
hints leads up to using the inverse Euler’s function to expand the cosine. We have:
€
⎛ e jπt + e − jπt ⎞
(−0.7+ jπ )t
x(t) = 8e −0.7t ⎜
u(t) + 4e(−0.7− jπ )t u(t)
⎟u(t) = 4e
2
⎝
⎠
If we apply the FT to each expression in x(t) we have:
€
4
4
4
4
+
=
+
0.7 + jπ + jω 0.7 − jπ + jω 0.7 + j(π + ω ) 0.7 − j(π − ω )
⎡
⎤
1.4 + j2ω
X( jω ) = 4 ⎢
2
2 ⎥
⎣ 0.49 + j1.4ω + (π − ω ) ⎦
X( jω ) =
€
10.2* The objective is to find the inverse Fourier Transforms for the expressions below. Similar to 10.1,
the key to solving these types of problems is understanding (1) the Fourier transform pairs that you
can use from Table 11-2 in your text, and (2) the Fourier transform properties that you can apply
from Table 11-3.
(a)
X( jω ) = 5 jδ (ω −13) − j5δ (ω +13)
Using the FT pairs we can see that this is the Fourier transform of a sine function except we need
to scale by -5/pi.
€
5
5
π
x(t) = − sin(13t) = cos(13t + )
π
π
2
(b)
X( jω ) =
€
3 − j4ω
5 + j6ω
The hint directs us to express
X( jω ) = A +
€
We can first rewrite the expression:
B
C + jω
X( jω ) =
€
The next step is to perform
the long division:
X( jω ) = 5 /6 + jω ) 3/6€− jω 2 /3 = −2 /3 +
3/6 − jω 4 /6
5 /6 + jω
19 /18
5 /6 + jω
We do the inverse transform of each and use superposition:
€
2
19
x(t) = − δ (t) + e −5t / 6 u(t)
3
18
An alternative method is:
€
X( jω ) =
3 − j4ω
3
j4ω
=
−
5 + j6ω 5 + j6ω 5 + j6ω
We use the superposition property and find the inverse transform of each expression separately.
€
For the first expression:
X1 ( jω ) =
3
3/6
=
which has a time domain
5 + j6ω 5 /6 + jω
representation as a one-sided exponential.
x1 (t) = 0.5e −5t / 6 u(t)
For the second
€ expression we have a jω term. Looking at the table of properties we see that the
differentiation property applies. Thus we have:
€
X 2 ( jω ) = − jω
x 2 (t) =
€
4
4 /6
= − jω
5 + j6ω
5 /6 + jω
d
10
2
5
2
−2 /3e −5t / 6 u(t)) = e −5t / 6 u(t) − e −5t / 6δ (t) = e −5t / 6 u(t) − e −5t / 6δ (t)
(
dt
18
3
9
3
Using the sampling property of the impulse:
€
5
2
5
2
x 2 (t) = e −5t / 6 u(t) − e 0δ (t) = e −5t / 6 u(t) − δ (t)
9
3
9
3
Summing the results we have:
€
⎛ 1
⎞
5
2
19
2
x(t) = ⎜ e −5t / 6 + e −5t / 6 ⎟ u(t) − δ (t) = e −5t / 6 u(t) − δ (t)
⎝ 2
⎠
9
3
18
3
Taking to route with the hint is much easier!
€ (c)
X( jω ) = u(ω + 7) + u(ω + 2) − u(ω − 2) − u(ω − 7)
= u(ω + 2) − u(ω − 2) + u(ω + 7) − u(ω − 7)
Again, we can use superposition and note that we have the inverse transforms of the sum of sinc
functions. Thus:
€
x(t) =
€
The trig identity used is:
(d)
⎤ 2
sin(7t) sin(2t) 2 ⎡ 1
1
+
= ⎢ sin((4.5 + 2.5)t) + sin((4.5 − 2.5)t) ⎥ = sin(4.5t)cos(2.5t)
⎦ πt
πt
πt
πt ⎣ 2
2
1
1
sin(a)cos(b) = sin(a + b) + sin(a − b)
2
2
X( jω ) = −3 + j8sin(5ω )
€
If we view the expression as a sum of complex exponentials we can use superposition, the delay
€
property and the inverse transform of the complex expontial to obtain a sum of impulses.
⎛ e j 5ω − e − j 5ω ⎞
j0
jω
− j 5ω
X( jω ) = −3e + j8⎜
⎟ = −3e + 4e − 4e
2j
⎝
⎠
j0
x(t) = −3δ (t) + 4δ (t + 5) − 4δ (t − 5)
€
€ 10.3* For this problem we exercise our ability to design filters and examine the resulting frequency
response and its impact on two input functions.
(a) To plot the time domain signal for the second filter we have:
h2 (t) =
sin(8πt) 1 sin(8πt)
= •
10πt
10
πt
We have a sinc function with a maximum value of 8/10 at t=0 and zero values at integer
multiples of 1/8. Note that the amplitude diminishes since the time value is in the denominator.
€
(b) Now we move to the frequency domain and obtain the response of the cascaded filter by
multiplying the FT of each filter. To do so, we need to obtain the FT of h2(t).
X 2 ( jω ) =
€
€
1
[u(ω + 8π ) − u(ω − 8π )]
10
1
1
X 3 ( jω ) = X1 ( jω ) • X 2 ( jω ) = j ω • [ u(ω + 8π ) − u(ω − 8π )]
2
10
ω
1
X 3 ( jω ) = e jπ / 2 • [ u(ω + 8π ) − u(ω − 8π )] = H( jω )
2
10
To make a sketch of the magnitude and phase it may be helpful to visualize the constant slope of
the first filter multiplied by the scaled square pulse of the second filter. To better understand the
effect of the cascade filter we see that for frequencies beyond ±8π the signal is nulled.
Furthermore, the signal is also nulled at DC.
Magnitude
€Phase
(c) Our goal is to now apply an input to our cascaded filter and examine the response.
x(t) = cos(7πt) +
sin(15πt)
+ cos(20πt)
0.2πt
We can see that the signals that will pass through our filter is the cosine at a frequency of 7pi and
the sinc function. The cosine at 20pi gets nulled by the filter.
€
To determine the overall output we use superposition and apply each input directly and sum the
responses.
Now we evaluate the frequency response of with the input of:
For the magnitude the filter scales by :
ω 7π
=
20 20
cos(7πt)
and for the phase a shift of pi/2 results.
€
Thus the output due to the cosine is:
y1 (t) =
7π
π
7π
cos(7πt + ) = €
− sin(7πt)
20
2
20
Next we apply the sinc function. Note that we can multiply the FT of the sinc by the system
function to obtain the output.
€
Y2 ( jω ) = H( jω ) • 5[ u(ω +15π ) − u(ω −15π )] = 5H( jω )
As a result we obtain the impulse response scaled by a factor of 5.
To obtain the inverse FT of the impulse response we se that we can use the differentiation
property:
€
1
[ u(ω + 8π ) − u(ω − 8π )]
20
⎤
1 d sin(8πt) 1 ⎡ 8cos(8πt) π
h(t) =
= ⎢
− 2 sin(8πt) ⎥
⎦ 20 dt πt
20 ⎣
t
t
H( jω ) = jω
Therefore
5h(t) =
€
⎤
1 d sin(8πt) 1 ⎡ 8cos(8πt) π
= ⎢
− 2 sin(8πt) ⎥ ⎦
20 dt πt
4 ⎣
t
t
The total response is the sum: ⎤
7π
1 ⎡ 8cos(8πt) π
y (t) = − sin(7πt) + ⎢
− 2 sin(8πt) ⎥
⎦
20
4 ⎣
t
t
€
€
(d) Now we apply a different input to our cascaded filter and examine the response.
x(t) = cos(7πt) + 5δ (t)
€
Again we use the superposition property. Thinking about our filter we see that have the same
cosine as above as an input plus the scaled impulse. In part (c) we determined the response to
the cosine at 7pi.
y1 (t) =
7π
π
7π
cos(7πt + ) = − sin(7πt)
20
2
20
Now we have to think about the scaled impulse. Not that in part (c) we also figured out the
inpulse response scaled by 5.
€
5h(t) =
⎤
1 ⎡ 8cos(8πt) π
− 2 sin(8πt) ⎥ ⎢
⎦
4 ⎣
t
t
Thus we obtain the same results for parts (c) and (d).
€
€
y (t) = −
(e) Although the signals are different in parts (c) and (d), the response of the filter nulls frequencies
great than ±8π and truncates the FT of the sinc function. Effectively we have a scaled inputlse
response. Therefore, the respective responses are equivalent.
€
⎤
7π
1 ⎡ 8cos(8πt) π
sin(7πt) + ⎢
− 2 sin(8πt) ⎥
⎦
20
4 ⎣
t
t
10.4* For this problem we further exercise our ability to design filters and meet the necessary frequency
response specifications.
(a) We create a new filter by subtracting the second system from the first in the frequency domain:
H n ( jω ) = H1 ( jω ) − H 2 ( jω ) = {[ u(ω + 80π ) − u(ω − 80π )] − [ u(ω + 60π ) − u(ω − 60π )]}e − jϖ / 3
To plot the magnitude we see that we end up with two bands, a bandpass filter, that goes from
±60π to ±80π . With respect to the phase, it’s a linear function of frequency. Thus we have:
€
€
€
(b) Now we define another bandpass filter hb(t) where
hb (t) = α cos(ω b t) • h3 (t)
that gives us the same frequency response the first part of this problem,
€
H b ( jω ) = H n ( jω )
We can see that we need to use the modulation property to move the square pulse over to center
at frequencies of ±70π . Therefore ω b = 70π . We also can see that a scaling factor 2 is
€
necessary. Therefore α = 2 .
€
€
€
Table of Fourier Transform Pairs
Signal Name
Time-Domain: x.t /
Right-sided exponential
e
at u.t /
.a > 0/
Left-sided exponential
e bt u. t /
.b > 0/
Œu.t C T =2/
Square pulse
u.t
Frequency-Domain: X.j!/
1
a C j!
1
b j!
sin.!T =2/
!=2
T =2/
sin.!0 t /
t
“sinc” function
Œu.! C !0 /
ı.t /
Impulse
ı.t
Shifted impulse
t0 /
2ı.!
Ae j ı.!
Cosine
cos.!0 t /
ı.!
Sine
sin.!0 t /
jı.!
1
X
General periodic signal
Impulse Train
kD 1
1
X
ak e
j!t0
e
A cos.!0 t C /
General cosine
!0 /
1
e j!0 t
Complex exponential
u.!
ı.t
!0 / C Ae
!0 / C jı.! C !0 /
2ak ı.!
kD 1
1
X
2
T
nT /
nD 1
j ı.! C ! /
0
!0 / C ı.! C !0 /
1
X
j k!0 t
!0 /
ı.!
k!0 /
2k=T /
kD 1
Table of Fourier Transform Properties
Time-Domain x.t /
Frequency-Domain X.j!/
ax1 .t / C bx2 .t /
aX1 .j!/ C bX2 .j!/
Conjugation
x .t /
X . j!/
Time-Reversal
x. t /
X. j!/
Scaling
x.at /
1
X.j.!=a//
jaj
Property Name
Linearity
Delay
x.t
Modulation
x.t /e j!0 t
Modulation
x.t / cos.!0 t /
Differentiation
Convolution
Multiplication
td /
j!td
X.j!/
X.j.!
!0 //
e
1
2 X.j.!
!0 // C 21 X.j.! C !0 //
d k x.t /
dt k
.j!/k X.j!/
x.t / h.t /
X.j!/H.j!/
x.t /p.t /
1
2 X.j!/ P .j!/
Date: 1-May-2008