1. No category

EE 179
Fourier Transform and Applications
Sample Midterm Examination Solutions
May 7, 2014
Handout #29
1. Filters.
a. The signal x(t) = cos4 2πt is the input to an ideal low-pass filter with cutoff frequency f0 =
3 Hz. Find the output y(t).
b. Find the impulse response of the nonideal bandpass filter whose transfer function is shown
below. The widths of the upper and lower bases of the trapezoids are B and 2B, respectively.
1
−fc
−fc + B
+fc
−fc + B/2
Solution
a. The bandwidth of the product of signals is the sum of their bandwidths. Therefore the
bandwidth of cos4 2πt is 4 Hz. By Euler’s formula, cos 2πt = 21 (ej2πt + e−j2πt ), and by the
binomial formula (a + b)4 = a4 + 4a3 b + 6a2 b2 + 4ab3 + b4 ,
4
1
cos4 2πt = 21 (ej2πt + e−j2πt ) = 16
ej8πt + 4ej4πt + 6 + 4e−j4πt + e−j8πt
(We could also use the convolution theorem to find the Fourier transform of cos4 2πt by
convolving the Fourier transform 12 δ(f − 1) + 12 δ(f − 1) with itself.) The low-pass filter
with cutoff frequency 3 Hz blocks the two complex exponential terms with frequency 4 Hz.
Therefore the output is
1
1
y(t) = 16
4ej4πt + 6 + 4e−j4πt = 16
(6 + 8 cos 4πt) = 38 + 12 cos 4πt .
The DC term arises because cos4 2πt is nonnegative.
b. The transfer function of this filter is the Fourier transform of a signal g(t) multiplied by
2 cos 2πfc t. The Fourier transform of g(t) is a trapezoid centered at f = 0:


−B/2 ≤ f ≤ B/2
1
G(f ) = 1 − 2B|f | B/2 < |f | < B


0
otherwise
The inverse Fourier transform can be found easily by decomposing the trapezoid into a sum
of triangles.
f
f
G(f ) = 2Λ
−Λ
B
B/2
Therefore the impulse response is
h(t) = 2B sinc2 (Bt) − 21 B sinc2 ( 12 BT ) cos 2πfc t .
2. Overmodulation. A signal m(t) = sin 2πt is transmitted using AM modulation:
ϕAM (t) = (1 + km(t)) cos 20πt .
a. Does the bandwidth of ϕAM (t) depend on the modulation index k?
b. Sketch the envelope of ϕAM (t) for k = 2.
c. Can the signal m(t) be recovered when k > 1? Explain briefly.
Solution
a. No. The spectrum of ϕAM (t) consists of impulses at ±10 Hz (carrier) and at ±9 Hz and
±11 Hz (sidebands). The modulation index affects the amplitude of the sidebands but not
the signal bandwidth.
b. The modulated signal and its envelope are shown below. Note the phase changes in the
modulated signal at (π + sin−1 (−1/2))/2π = 7/12 ≈ 0.5833 and (2π − sin−1 (−1/2))/2π =
11/12 ≈ 0.9167.
3
2
1
0
−1
−2
−3
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
c. Although an overmodulated signal cannot be recovered using envelope detection, a standard
coherent demodulator works. The local carrier can be kept in phase using the transmitted
carrier, although phase reversals in the received signal at the overmodulation points.
3. FM modulation. Consider the FM modulated signal
Z t
x(t) = cos 2πfc t + 2πkf
m(u) du .
0
The information signal is a square wave of period 2T0 :
m(t) = 5
∞
X
(−1)n Π((t − nT0 )/T0 ) .
n=−∞
Let fc = 10 kHz, kf = 2π · 200, and T0 = 0.001.
a. Use Carson’s rule to find the bandwidth of x(t). You will have to make a reasonable estimate of the bandwidth of m(t). Hint: the Fourier series for the square wave of period 1,
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EE 179, Spring 2014
P
n (−1)
n
Π(2(t − 12 n)), is
1
1
1
4
cos 2πt − cos 2π3t + cos 2π5t − cos 2π7t + · · ·
π
3
5
7
b. Sketch the instantaneous frequency fi (t) of x(t).
c. Sketch or describe x(t).
d. Suppose that this signal is demodulated with a zero-crossing detector. Find the minimum
time interval T of the zero-crossing detector such that there are at least 2 zero crossings in
every interval. For this T , will there ever be more than 2 zero crossings?
Solution
a. Carson’s rule states that the bandwidth of an WBFM signal is
2(∆f + B) ,
where ∆f is the maximum deviation of the modulated signal instantaneous frequency from
fc and B is the bandwidth of the modulating signal m(t).
The peak frequency deviation for FM is
kf mp
= 200 · 5 = 1000 .
2π
The fundamental frequency of the square wave is 500 Hz. Three terms of the Fourier series
provide a reasonable approximation to the square wave. So we estimate the bandwidth of
m(t) to be 5 · 500 = 2500 Hz. Then the bandwidth of this WBFM signal is
∆f =
2(1000 + 2500) = 7000 .
b. The instantaneous frequency alternates between 9 kHz and 11 kHz every millisecond.
12000
10000
8000
6000
0
1
2
3
4
5
−3
x 10
c. During each millisecond, x(t) oscillates alternately between 9 kHz (9 cycles/ms) and 11 kHz
(11 cycles/ms). The FM signal and its Fourier transform shown below.
1
0
−1
0
1
2
3
4
5
−3
x 10
Sample Midterm Examination Solutions
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−4
x 10
15
10
5
0
0
2000
4000
6000
8000
10000
12000
d. There are two zero crossings in any period of the FM signal. At 9 kHz the period is 1/9000 =
0.11. When the instantaneous frequency is 11 kHz, three zero crossings may occur, but usually
not.
4. Multipath channels. In a multipath channel, signal reflections give rise to multiple copies of the
signal arriving at the receiver at different times. Depending on the time difference between these
different arrivals, the signal may or may not be recoverable.
Consider the time-limited signal x(t) defined by
(
cos(2πt/T ) |t| < T
x(t) =
0
|t| > T
Suppose that the signal is passed through a channel with frequency response
H(f ) = 1 + cos(2πT0 f ) ,
resulting in channel output y(t) = h(t) ∗ x(t).
a. Find an expression for X(f ) and sketch X(f ). The plot sketches should indicate the general
shape of the function and all key values on the x- and y-axes.
b. Find h(t). How many received copies of the signal with different delays result from this
channel?
c. Sketch y(t) for T0 = 3T . How would you recover x(t) from y(t) ?
d. What is the minimum value for T0 such that x(t) can be recovered from y(t) with no distortion?
Solution
a. The signal x(t) is two cycles of cos(2πt/T ):
2πt t cos
x(t) = Π
2T
T
By the convolution theorem,
1 1 1 X(f ) = 21 δ f +
+ 2δ f −
∗ 2T sinc(2T f )
T
T
1 1 + T sinc 2T f −
= T sinc 2T f +
T
T
The sum of the two shifted sincs is shown below. The maximum value is T and zero crossings
are at multiples of 1/2T .
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EE 179, Spring 2014
0.1
0.08
0.06
0.04
0.02
0
−0.02
−0.04
−6
−4
−2
0
2
4
6
b. The inverse Fourier transform of 1 + cos(2πT0 f ) is
h(t) = δ(t) + 12 δ(f + T0 ) + 21 δ(f − T0 ) .
The three delta functions correspond to three different signal paths.
1
0.5
0
−0.5
−1
−0.4
−0.2
0
0.2
0.4
0.6
c. −0.6
We can recover x(t) by multiplying the signal by Π(t) in time. Note: this is not the same as
applying a low-pass filter or convolving with Π(t) in time.
d. If T0 < 2T the pulses will overlap and be unrecoverable. Therefore the minimum value of T0
is 2T .
5. Demodulation by sampling. In the communication system shown in the following figure, the
signal w(t) is obtained by sampling the channel output at intervals of T seconds with impulses
of area 1.
cos 2πfc t
Channel
z(t)
x(t)
×
v(t)
w(t)
H1 (f )
y(t)
H2 (f )
The information signal has Fourier transform
2

 3 |f | < B/2
X(f ) = 1 B/2 ≤ |f | < B


0 otherwise
The frequency response of the channel is


3
H1 (f ) = 1


0
|f | − fc < B/2
B/2 ≤ |f | − fc < B
otherwise
The carrier frequency fc is much larger than the bandwidth B of x(t).
Sample Midterm Examination Solutions
Page 5 of 7
a. Sketch Z(f ).
b. Sketch V (f ).
c. Find v(t).
d. The sampled signal w(t) and its Fourier transform are
∞
X
∞
1 X n
w(t) =
v(nT )δ(t − nT ) and W (f ) =
V f−
T n=−∞
T
n=−∞
Sketch W (f ) when T = 1/fc .
e. Find H2 (f ) so that y(t) = x(t) when T = 1/fc .
Solution
a. By the modulation theorem, Z(f ) = 12 X(f + fc ) + 12 X(f − fc ), which is sketched below.
1/2
1/3
−fc
+fc
b. Both Z(f ) and H1 (f ) are nonzero only in the frequency bands |f − fc | ≤ B. Therefore
 1
|f | − fc < B/2
=
1
3
·

 3
V (f ) = H1 (f )Z(f ) = 1 · 21 = 21 B/2 ≤ |f | − fc < B


0
otherwise
The Fourier transform V (f ) of the transmitted signal is sketched below.
1
1/2
−fc
+fc
c. The spectrum V (f ) can be written as
V (f ) = 12 V0 (f + fc ) + 21 V0 (f − fc )
where V0 (f ) is centered at f = 0.


2 |f − fc | < B/2
V0 (f ) = 1 B/2 ≤ |f − fc | < B


0 otherwise
The spectrum V0 (f ) can be decomposed into the sum of rectangles in several ways.
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EE 179, Spring 2014
−B
2
2
1
1
or
+B
−B
+B
The first decomposition is simpler:
f
f
+Π
⇒ v0 (t) = 2B sinc(2Bt) + B sinc(Bt)
V0 (f ) = Π
2B
B
In conclusion,
v(t) = B (2 sinc(2Bt) + sinc(Bt)) cos 2πfc t
d. The Fourier transform
is shown below for T = 1/fc .
∞
n
1 X V f−
W (f ) =
T n=−∞
T
2/T
...
...
1/T
−fc
+fc
e. We can reconstruct x(t) from y(t) using a filter that blocks the high-frequency components of
y(t) and equalizes the low-frequency components:
(
1/H1 (f ) |f | ≤ B
H2 (f ) =
0
otherwise
1 
 3 |f | − fc < B/2
= 1 B/2 ≤ |f | − fc < B


0 otherwise
This transfer function is sketched below.
1
1/3
−B
Sample Midterm Examination Solutions
+B
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