sol homework 1

7/28/2019 Sol Homework 1

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EE160 Fall 2005 San Jose State University

Solution Homework # 1

1. ISM bands

(a) The acronym ISM stands for Industrial, Scientific and Medical. An ISM band is alicense-free band that can be used for noncommercial purposes. Wireless local areanetworks (WLANs), such as those specified in the IEEE 802.11a/g standards operate intwo of these bands.

(b) From FCC ONLINE TABLE OF FREQUENCY ALLOCATIONS(http://www.fcc.gov/oet/spectrum/table/fcctable.pdf),

The following bands:

13553-13567 kHz (centre frequency 13560 kHz),26957-27283 kHz (centre frequency 27120 kHz),40.66-40.70 MHz (centre frequency 40.68 MHz),902-928 MHz in Region 2 (centre frequency 915 MHz),2400-2500 MHz (centre frequency 2450 MHz),5725-5875 MHz (centre frequency 5800 MHz), and24-24.25 GHz (centre frequency 24.125 GHz)

are also designated for industrial, scientific and medical (ISM) applications. Radiocom-munication services operating within these bands must accept harmful interference whichmay be caused by these applications.

2. Fourier series coefficients and discrete amplitude spectrum

(a) x1(t) = cos(2t) + cos(4t)

First, determine the period of the signal. The fundamental frequency f1 of cos(2t)is f1 = 1 and therefore its period T1 =

1f1

= 1. Similarly, the period of cos(4t) is

T2 =12 . The period of x1(t) is found as the least common multiple (lcm) of T1 and

T2, or T0 = lcm(T1, T2) = lcm(1,12) =

12 lcm(2, 1) = 1. Now use Eulers formula and

express the signal as

x1(t) =

1

2ej2t +

1

2ej2t

+

1

2ej4t +

1

2ej4t

= 12

ej2t + 12

ej2t + 12

ej2(2)t + 12

ej2(2)t.

As a result,

xn =

12 , n = 1,2;

0, otherwise.

A sketch of the amplitude |xn| of the discrete spectrum (the phase is zero because thecoefficients are real) is shown in the figure below:


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n

0 1 2-3 -2 -1 3 4-4

|xn|

1/2 1/2 1/2 1/2

A sketch ofx1(t) is shown in the figure below. The figure was produced using the matlabscript:

t=-1:0.01:1;

x1=cos(2*pi.*t)+cos(4*pi.*t);

plot(t,x1)

axis tight

grid on

xlabel(t)

ylabel(x_1(t))

1 0.5 0 0.5 1

1

0.5

0

0.5

1

1.5

2

t

x1

(t)

(b) x2(t) = cos(2t) cos(4t + /3)

In this case, similar to part (a), we have T0 = 1 and

x2(t) = 12

ej2t +1

2

ej2t12

ej(4t+/3) +1

2

ej(4t+/3)=

1

2ej2t +

1

2ej2t

1

2ej/3ej2(2)t

1

2ej/3ej2(2)t,

with

xn =

12 , n = 1;

12 ej/3, n = 2;

0, otherwise.

The discrete amplitude spectrum |xn| is the same as in part (a).


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(c) x3(t) = 2 cos(2t) sin(4t)

As before, T0 = 1 and

x3(t) =

ej2t + ej2t

1

2jej4t

1

2jej4t

=1

2ej2t +

1

2ej2t +

j

2ej2(2)t

j

2ej2(2)t,

where

xn =

12 , n = 1;

j2 , n = 2;

0, otherwise.

Again, the discrete amplitude spectrum |xn| is the same as in part (a). It is left as anexcercise to plot the signal x3(t) using Matlab.

3. Fourier transform and amplitude spectrum

(a) x(t) = (t 3) + (t + 3)

X(f) = F{x(t)} = sinc(f)

ej2(3)f + ej2(3)f

,

where we have used the Fourier transform pair

(t) sinc(f)

and the propertyx(t t0) X(f)e

2t0f

Now, using the fact that 2 cos() = ej + e

j , we obtain the result

X(f) = sinc(f) cos(6f).

To obtain the sketch of the amplitude spectrum, note that the signal cos(6f) is periodicin f with period 13 .

The following figure was produced with the Matlab script:

f=-6:0.01:6;

X=abs(sinc(f).*cos(6*pi.*f));

plot(f,X)axis([ -6 6 0 1])

grid on

xlabel(f)

ylabel(X(f))


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6 4 2 0 2 4 60

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

f

X(f)

(b) x(t) = t2 1

Write the signal as

x(t) =

1

2(t 2)

=

t

2

t=t2

.

Using (in this order) first the scaling property

x(at) 1

|a|X

f

a

and then the time-shift property

x(t t0) X(f)ej2t0f

the result is obtained:X(f) = 2 sinc(2f)e4t,

and |X(f)| = |sinc(f)|. The sinc function was sketched in class.

(c) x(t) = 4t4

cos(2f0t). The Fourier transform is obtained by using the scaling property

and the Fourier transform pair

x(t)cos(2f0t) 1

2[X(f + f0) + X(f f0)]

which give

X(f) = 4 42

sinc [4(f + f0)] + 4 42

sinc [4(f f0)]

= 8 sinc [4(f + f0)] + 8 sinc [4(f f0)] .

The spectrum thus consists of two copies of a sinc function, centered at f0. The zerocrossings of the sinc function occur at nonzero multiples of 14 . For f0

14 , this signal is

an example of a bandpass signal.

The amplitude spectrum for f0 = 3 is shown in the figure below and was produced withthe Matlab script:


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f=-8:0.01:8;

f0=3;

X=abs(8*sinc(4*(f+f0)) + 8*sinc(4*(f-f0)));

plot(f,X)

axis([ -6 6 0 8])

grid on

xlabel(f)

ylabel(X(f))

6 4 2 0 2 4 60

1

2

3

4

5

6

7

8

f

X(f)

sol homework 1

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