INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
SAMPLE PROBLEMS
PROBLEM ONE: Exponential Modulation The spectrum of the signal x t
0( ) in the block diagram below is given by
Otherwise ,0
,)(0
mm
fff
f
fX
The spectrum of the signal x t1( ) in the block diagram below is given by
Otherwise ,0
,1)(1
mm
fff
f
fX
The signal )(2 tx , obtained through the block diagram given below, is mixed by an
exponential modulator with carrier signal given by the tfj me
6 to produce the exponentially
modulated signal tfj metxts
62 )()(
.
a).- (15 points) Obtain spectrum of tfj metxts
62 )()(
. Show all work.
b).- (10 points) Plot the spectrum obtained in part a) above.
Remarks: You can use the following Fourier transformations without proof:
)(2
atfj
ffeF a
)(2
1)(
2
1)2cos()( aaa ffXffXtftxF
)(2
1)(
2
1
2
1
2
12cos(
22aa
tfjtfja ffffeeFtfF aa
xx0 0 (t)(t)
xx11(t)(t)
Cos fmt4
xx22(t)(t)
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
SAMPLE PROBLEMS FORM MIDTERM EXAM
2
PROBLEM ONE – Solution: Exponential Modulation
a).- tfj meFfXfS6
2
fXfffffXfXtfFfXfX mmm 10102 22
12
2
14cos
fXffXffXfX mm 1002 22
12
2
1
mm ffXfffXfS 33 22
mmm ffXffXffXfS 32
1
2
15
2
1100
b).-
S(f)
f 0 fm 2fm 3fm 4fm 5fm 6fm
1/2
1
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
SAMPLE PROBLEMS FORM MIDTERM EXAM
3
PROBLEM TWO: Ideal Bandpass
Obtain the impulse response of and ideal, linear phase,
bandpass filter starting from the impulse response of an
ideal, linear phase, lowpass filter whose frequency
response is given by the following expression:
m
mftj
Lff
ffefH
d
,0
,)(
2
PROBLEM TWO - SOLUTION: Ideal Bandpass
We proceed to obtain the impulse response of the low-pass
filter given by the following frequency response:
m
mftj
Lff
ffefH
d
,0
,)(
2
m
m
dd
ff
ff
fttj
f
f
tfjftjLL dfedfeefHth
)(2 2 2)()(
m
m
d
m
m
d
f
f
fttj
d
ff
ff
fttje
ttjdfe
)(2 )(2
)(2
1
)(2)(2
1)(
)(2 )(2 )(2
d
fttjfttjf
f
fttj
dL
ttj
eee
ttjth
mdmdm
m
d
)(
)(2sin
)(2
)(2 )(2
d
md
d
fttjfttj
Ltt
ftt
ttj
eeh
mdmd
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
SAMPLE PROBLEMS FORM MIDTERM EXAM
4
)(22
)(2
)(2sin2dmm
md
mdmL ttfSincf
ftt
fttfh
)()(2cos()(2)( 111cLcLcLB ffHffHFtfthFFfHF
)()()()2cos()(2 1 thffHffHFtfth BcLcLcL
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
SAMPLE PROBLEMS FORM MIDTERM EXAM
5
PROBLEM 03: Linear Time Invariant Systems - Continuous Filters The linear system T below is called a tapped-delay line or transversal filter.
a).- Show that the linear system T is indeed a filter.
b).- Plot the input x t u t u t( ) ( ) ( ) 1 and output y t T x t( ) ( ) .
Note: S x t x t( ) ( ) 1
SS S
+1 -1 +1 -1
x(t) y(t)
T
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
SAMPLE PROBLEMS FORM MIDTERM EXAM
6
PROBLEM 03: Solution - Linear Time Invariant Systems - Continuous Part a).- The output of this linear system is given by
y t x t x t x t x t( ) ( ) ( ) ( ) ( ) 1 2 3
To determine if the system is a filter, we proceed to test for the condition of time invariance:
Given that the system's equation is given by
y t T x t( ) ( ) ,
the system is time-invariant if the following identity is satisfied:
y t t T x t t( ) ( ) 0 0
We proceed to obtain the left hand side of this identity from the system's equation:
y t t x t t x t t x t t x t t( ) ( ) (( ) ) (( ) ) (( ) ) 0 0 0 0 01 2 3
For the right hand side of the identity, we set g t x t t( ) ( ) 0 . We then have
T g t g t g t g t g t( ) ( ) ( ) ( ) ( ) 1 2 3
Substituting for g t x t t( ) ( ) 0 , we get
T x t t x t t x t t x t t x t t( ) ( ) (( ) ) (( ) ) (( ) ) 0 0 0 0 01 2 3 We thus prove that the two
sides of the equation y t t T x t t( ) ( ) 0 0 , establishing in this manner the time-invariance
identity condition.
PROBLEM 03: Solution - Linear Time Invariant Systems - Continuous
Part b).- If the input to the filter is given by x t u t u t( ) ( ) ( ) 1 , the output is then given by
y t x t x t x t x t( ) ( ) ( ) ( ) ( ) 1 2 3
A plot of
x t u t u t( ) ( ) ( ) 1
and
x t u t u t( ) ( ) ( )1 1 2
is given below. Observe that the graph of x t( )1 can be obtained from the graph of x t( ) by
delaying the graph of x t( ) by one and then multiplying by 1 .
The output
y t x t x t x t x t( ) ( ) ( ) ( ) ( ) 1 2 3
can be obtained in the same manner. The output can also be obtained by substituting the input
x t u t u t( ) ( ) ( ) 1 into the above output equation, obtaining the following result:
y t u t u t u t u t
u t u t u t u t
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
1 1 2
2 3 3 4
Simplifying, we obtain
)4()3(2)2(2)1(2)()( tutututututy
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
SAMPLE PROBLEMS FORM MIDTERM EXAM
7
SS S
+1 -1 +1 -1
x(t) y(t)
T
t
y(t)=x(t)-x(t-1)+x(t-2)-x(t-3)
1
4
-1
t
-x(t-1)=-(u(t-1)-u(t-2))
2
-1
1
t
x(t)=u(t)-u(t-1)
1
1
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
SAMPLE PROBLEMS FORM MIDTERM EXAM
8
PROBLEM 04: Convolution Operation with Impulse Functions
a).- A filter has an impulse response h t T t t( ) ( ) ( ) 3 . Obtain the output
y t T x t x t h t( ) ( ) ( ) ( ) for input signal x t t( ) ( ) 5 .
PROBLEM 04: Solution - Convolution Operation with Impulse Functions
y t T x t x t h t x h t d t d t( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
5 3 8
PROBLEM 05: Fourier Transform of Impulse Functions a).- Obtain the Fourier transform of y t T x t x t h t( ) ( ) ( ) ( ) for Problem One above.
b).- Take the product F x t F h t( ) ( ) for Problem 04 above and show that it is equal to part the
part a) of this Problem 05. Proceed to state the time-domain linear convolution theorem.
PROBLEM 05: Solution - Fourier Transform of Impulse Functions
a).- F y t F x t h t t e dt ej ft j f( ) ( ) ( ) ( )
8 2 2 8
b).- F x t t e dt ej ft j f( ) ( )
5 2 2 5 F h t t e dt ej ft j f( ) ( )
3 2 2 3
F x t F h t e e ej f j f j f( ) ( ) 2 5 2 3 2 8
REMARK: The time-domain convolution theorem
states that the Fourier transform of the convolution of two
signals is equal to the product of the Fourier transforms of
each of the individual signals.
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
SAMPLE PROBLEMS FORM MIDTERM EXAM
9
PROBLEM 06: Ideal Zero-Phase Low-Pass Filter Obtain the output of the ideal low-pass filter T with frequency response given by
H ff
L ( ),
1 800
0
, Otherwise
if it has as its input the signal x t t t( ) cos( ) cos( ) 2 1200 1400 .
Use F f t f f f fcos( ) ( ) ( )21
2
1
20 0 0 and the time-domain linear convolution theorem to
arrive at your answer. Please, read carefully and explain your answer.
b).- Obtain the Fourier transform of y t T x t x t h t( ) ( ) ( ) ( ) for part a) above.
PROBLEM 06: Solution - Ideal Zero-Phase Low-Pass Filter
a).- F f t f f f fcos( ) ( ) ( )21
2
1
20 0 0 F t f fcos( ) ( ) ( )2 1200
1
21200
1
21200
F t F t f fcos( ) cos( ) ( ) ( ) 1400 2 7001
2700
1
2700
F t t f f f fcos( ) cos( ) ( ) ( ) ( ) ( )2 1200 14001
21200
1
21200
1
2700
1
2700
y t F X f H f F f f f f H f tL L( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) cos( )
1 1
1
21200
1
21200
1
2700
1
2700 2 700
b).- F t F t f fcos( ) cos( ) ( ) ( ) 1400 2 7001
2700
1
2700
f+1200
HL(f)
-1200
X(f)
Y(f)=HL(f). X(f)
+700-700
f
Y(f)
+700-700
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
SAMPLE PROBLEMS FORM MIDTERM EXAM
10
PROBLEM 7: Ideal Channel
An ideal channel in a communication system is described
by the following input output equation )()()( fYfHfY ciLco
Obtain the output of the channel if the input is the signal )6002cos()2002cos()( tttyci and the ideal filter is a zero-
phase filter with cut-off frequency Hzfm 500 .
PROBLEM 7 – SOLUTION: Ideal Channel
The input/output equation is )()()( fYfHfY ciLco . We first
obtain the spectrum of the input to the channel: )6002cos()2002cos()()( ttFtyFfY cici
We use the following trigonometric identities: )sin()sin()cos()cos()cos( BABABA
)sin()sin()cos()cos()cos( BABABA
)cos()cos(2)cos()cos( BABABA
)cos()cos()cos()cos(21
21 BABABA
Let 6002A and 2002B . Thus,
)4002cos()8002cos()6002cos()2002cos()(21
21 tttttyci ,
and
400400800800)(41
41 fffffYci
Thus, the spectral output )()()( fYfHfY ciLco becomes:
400400800800)(41
41 fffffHfY Lco
Finally, the temporal output is: )4002cos()(41 ttyco
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
SAMPLE PROBLEMS FORM MIDTERM EXAM
11
PROBLEM 8: DSB-SC SYSTEM
The spectrum of the modulating signal )(txm in a DSB-SC
system is given by the following expression:
Otherwise ,0
,1)( m
mmff
f
f
fX
Design a DSB-SC communication system for a channel
with no noise or distortion by providing all the all its basic
components in block diagram form and get and plot the
spectrum of its demodulator.
PROBLEM 8 – SOLUTION: DSB-SC SYSTEM
A block diagram of a DSB-SC communications system is
composed of five basic blocks. The transmitter is
composed of two blocks, namely, a linear modulator and
a bandpass filter. The receiver is composed of two
blocks, namely, a linear demodulator and a lowpass
filter. The fifth block describes de channel of the system
which is the medium between the transmitter and receiver.
Assuming that no noise or distortion is experienced at the
channel, the input to demodulator is the signal ).()( tyty cico
The signal )()( txty cci is essentially the output of the linear
modulator; that is, )2cos()()()()( tftxtctxty cmmci .
Thus, the spectrum at the input to demodulator is the
signal:
ccmccico fffffXfXfYfY 21
21)()()()(
Using the linear property of the convolution, we obtain:
cmcmco fffXfffXfY )()()(21
21 ,
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
SAMPLE PROBLEMS FORM MIDTERM EXAM
12
where
)()()( cmcmcm ffXdffXfffX
Thus,
)()()(21
21
cmcmco ffXffXfY
The spectral output of the demodulator, then becomes: )()()()( ccoccocccod ffYffYfffffYfY
Substituting for coY , we obtain:
)()2()()2()(21
21
21
21 fXffXfXffXfY mcmmcmd
We obtain the following final result:
)2()()2()(21
21
cmmcmd ffXfXffXfY
A plot of the spectrum of the output of the demodulator
is given in the graph below.
-2fc +2fc
f
Yd(f)
1/2 1/2
1
+fm -fm
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
SAMPLE PROBLEMS FORM MIDTERM EXAM
13
Problem 9: Realizable Channel Models We present the following sample problem to review the concept of linear, time-invariant (LTI )
systems, which are called “filters”. The basic operation of a filter is the convolution operation.
The filters presented in this problem are realizable filters used to model communication channels
as low-pass filters. A realizable filter is one which can be constructed with passive circuit elements.
This problem assumes that there is no noise in the channel. The basic operation performed by the
channel to obtain the output signal, again, is the convolution operation, an integral operation
performed with two signals or functions, the input signal to the channel and the channel’s impulse
response function which we always denote by using the notation ( )h t :
tdthxty ,)()()(
The impulse response of the RC-filter below is given by
0
1( ) ( ) ( ); , t
oh t T t h e u t hRC
Its step response is given by
0( ) ( ) 1 ( )h ty t T u t e u t
Obtain the output )(ty if the inputs are the following signals:
a) 1
( ) ( ) ( ); x t t t
b) 1
( ) ( ) ( ); x t u t u t
c) tnfj
etx 02)(
Remark: The response of a basic channel filter, with no
noise, to a complex exponential signal is the same
complex exponential signal times a constant, known
as an eigenvalue of the channel filter. The constant is the
frequency response of the system evaluated at the
frequency of the complex exponential input frequency, in
this case, 0
f .
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
SAMPLE PROBLEMS FORM MIDTERM EXAM
14
Problem 9 – SOLUTION: Realizable Channel Models
Part a).-
T t t T t T t h t h t y t ( ) ( ) ( ) ( ) ( ) ( ) ( )
y t h e u t h e u t h e u t eu to
t
o
t
o
t( ) ( ) ( ) ( ) ( )( )
Part b).-
T u t u t T u t T u t y t( ) ( ) ( ) ( ) ( )
y t e u t e u tt t( ) ( ) ( )( ) 1 1
Rearranging,
y t u t e u t u t e eu tt t( ) ( ) ( ) ( ) ( )
y t u t u t e u t eu tt( ) ( ) ( ) ( ) ( )
Part c).-
)()()()( 02tythtxeTtxT
tnfj
dthxty )()()(
dtueety thnfj
)()( )(2 00
t
nfjtht
nfjthdeedeeety
020020)(
t
nfj
nfjeety
02
02
1)(
t
nfj
nfjety
02
02
1)(
)()()( 0
2
2
1 0
0txfHetxTty
tnfj
nfj
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
SAMPLE PROBLEMS FORM MIDTERM EXAM
15
PROBLEM 10: Frequency Response Function The frequency response of an ideal, linear-phase, low-pass filter is given by the following equation:
,0)(
,2 0
m
ffftj
ff
efH
m
, where Sec 2.00 t , Hz 500mf
1a.- Get its real fHR and imaginary fH I parts.
1b.- Sketch its magnitude fH .
1c.- Sketch its phase ftf 02 .
1d.- Obtain its impulse response function )(th .
REMARK:
f
f
ftj dfefHth 2)()(
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
SAMPLE PROBLEMS FORM MIDTERM EXAM
16
PROBLEM 10 – SOLUTION :Freq. Response Function
a.- Get its real fHR and imaginary fH I parts:
,0)(
,2 0
m
ffftj
ff
efH
m
)(sin)()(cos)(
)(sin)(cos)()()()( )(
ffHjffH
fjffHefHfHfH
LL
L
fj
LL
From these two equations we conclude that:
Hz500 ,2)( ;1)( 0 mL ffftf θfH . Thus,
ftfHL 02cos)(real , and ftfHL 02sin))(imag
b.- Sketch its magnitude fH :
c.- Sketch its phase ftf 02 :
Hz500 ,5002.022)( ;1)( 0 mmmmL ffftf θfH
Hz500 ,2002)( ;1)( 0 mmmmL ffftf θfH
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
SAMPLE PROBLEMS FORM MIDTERM EXAM
17
d.- Obtain its impulse response function )(th :
m
m
m
m
f
f
fttj
f
f
ftjftjftjLL dfedfeedfefHth 00 2222)(
m
m
m
m
ff
fttj
f
f
fttjL e
ttjdfeth
00 2
0
2
)(2
1)(
0
0
0
222sin
)(2)(
00
tt
ttf
ttj
eeth m
fttjfttj
L
mm
51
0 1000100022)( tSincttfSincfth mmL
%drsinc
clear all close all Fs=5000; %Sampling frequency
Ts=1/Fs; %Sampling Time
ni=950; %Initial Signal Sample
nf=1050; %Final Signal Sample
Tv=(nf-ni+1)*Ts; %Signal Duration in Seconds
T0=(1/5); %Time Delay or Time Shift fm=500; %Cut Off Frequency
t=(ni-1)*Ts:Ts:(nf-1)*Ts; %Time discretization
hL=2*fm*sinc(2*fm*(t-T0)); %Impulse Response Function
figure
plot(t,hL,t,hL,'o')
grid
axis([min(t) max(t) min(hL) max(hL)]); xlabel('Time in Seconds') ylabel('Amplitude') title('Impulse Response Function')
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
SAMPLE PROBLEMS FORM MIDTERM EXAM
18
Problem 11: Fading channel Show that the continuous-time fading channel given by
)()()()( dttxtxtxTty is a filter.
Problem 11 – SOLUTION: Fading channel
The system )()( txTty is linear, if it satisfies:
)()()()( 2121 txbTtxaTtbxtaxT
r.h.s.:
)()()( 111 dttxatxatxaT
)()()( 222 dttxbtxbtxbT
)()()()()()( 221121 dd ttxbtxbttxatxatxbTtxaT
l.h.s.:
Let )()()( 21 tbxtaxtg
)()()( dttgtgtgT
Substituting for )()()( 21 tbxtaxtg :
)()()()()()( 212121 dd ttbxttaxtbxtaxtbxtaxT
The r.h.s. is equal to l.h.s. and the system is LINEAR.
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
SAMPLE PROBLEMS FORM MIDTERM EXAM
19
The system is time-invariant or TI, if it satisfies:
)()( 00 ttyttxT
r.h.s.:
)()()( 000 dtttxttxtty
l.h.s.:
Let )()( 0ttxtg
)()()( dttgtgtgT
Substituting for )()( 0ttxtg :
)()()( 000 tttxttxttxT d
The r.h.s. is equal to the l.h.s. and the system is TI. Since the system is LINEAR and TI, the system is a FILTER.
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
SAMPLE PROBLEMS FORM MIDTERM EXAM
20
Problem 12: DC Level Shifting System Show that the time-invariant system given by
RdcdtcxtxTty , ;)()()( , is NOT a filter.
Problem 12 – SOLUTION: DC Level Shifting System
The system is )()( txTty linear, if it satisfies:
)()()()( 2121 txbTtxaTtbxtaxT
r.h.s.:
adtacxtxaT )()( 11
bdtbcxtxbT )()( 22
bdtbcxadtacxtxbTtxaT )()()()( 2121
bdadtbcxtacxtxbTtxaT )()()()( 2121
l.h.s.:
Let )()()( 21 tbxtaxtg
dtcgtgT )()(
Substituting for )()()( 21 tbxtaxtg :
dtbxtaxctbxtaxT )()()()( 2121
dtcbxtcaxtbxtaxT )()()()( 2121
The r.h.s. is not equal to the l.h.s. and T is NOT a FILTER
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
SAMPLE PROBLEMS FORM MIDTERM EXAM
21
Problem 13: Fading Channels
Fig. 1: Modeling a Fading Channel with Doppler Effects and AWGN
Fig. 2: Model of a Fading Channel with Doppler Effects and AWGN
Problem: SISO Channel Characterization for 5G Systems
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
SAMPLE PROBLEMS FORM MIDTERM EXAM
22
1a.- 40 Points: The linearity property of a continuous-time system can be expressed as two simpler properties:
i) Additivity property: 𝑇{𝑥1(𝑡) + 𝑥2(𝑡)} = 𝑇{𝑥1(𝑡)} + 𝑇{𝑥2(𝑡)}. ii) Homogeneity property: 𝑇{𝑎𝑥(𝑡)} = 𝑎𝑇{𝑥(𝑡)}.
Proceed to use the additivity property to prove that the Adder System depicted in Fig. 2 above IS NOT a LINEAR system. 1a.- 40 Points: SOLUTION 𝑦𝑐𝑜(𝑡) = 𝑇{𝑦𝑏(𝑡)} = 𝑦𝑏(𝑡) + 𝑛(𝑡) r.h.s.:
𝑦𝑐𝑜1(𝑡) = 𝑇{𝑦𝑏1
(𝑡)} = 𝑦𝑏1(𝑡) + 𝑛(𝑡)
𝑦𝑐𝑜2(𝑡) = 𝑇{𝑦𝑏2
(𝑡)} = 𝑦𝑏2(𝑡) + 𝑛(𝑡)
∴ 𝑇{𝑦𝑏1(𝑡)} + 𝑇{𝑦𝑏2
(𝑡)} = 𝑦𝑏1(𝑡) + 𝑛(𝑡) + 𝑦𝑏2
(𝑡) + 𝑛(𝑡)
l.h.s.:
𝑦𝑏3
(𝑡) = 𝑦𝑏1(𝑡) + 𝑦𝑏2
(𝑡)
𝑇{𝑦𝑏1(𝑡) + 𝑦𝑏2
(𝑡)} = 𝑇{𝑦𝑏3(𝑡)} = 𝑦𝑏3
(𝑡) + 𝑛(𝑡)
∴ 𝑇{𝑦𝑏3(𝑡)} = 𝑇{𝑦𝑏1
(𝑡) + 𝑦𝑏2(𝑡)} = 𝑦𝑏1
(𝑡) + 𝑦𝑏2(𝑡) + 𝑛(𝑡)
The Adder System is NOT a LINEAR system.
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
SAMPLE PROBLEMS FORM MIDTERM EXAM
23
1b.- 30 Points: Provide 𝑦𝑐(𝑡) for the Delay System in Fig. 2 if:
𝑦𝑐𝑖(𝑡) = 𝑥(𝑡), 𝐾 = 3, 𝐴0 = 1, 𝐴1 = 0.5, 𝐴2 = 0.25 ,
𝜏0 = 1 𝑆𝑒𝑐., 𝜏1 = 2 𝑆𝑒𝑐., 𝜏2 = 3 𝑆𝑒𝑐. 1b.- 30 Points: SOLUTION
𝑦𝑐(𝑡) = 𝑇{𝑦𝑐𝑖(𝑡)} = 𝑇{𝑥(𝑡)} = ∑ 𝐴𝑘
2
𝑘=0
𝑥(𝑡) ∗ 𝛿(𝑡 − 𝜏𝑘)
𝑦𝑐(𝑡) = 𝑇{𝑦𝑐𝑖(𝑡)} = 𝑇{𝑥(𝑡)} = ∑ 𝐴𝑘
2
𝑘=0
𝑥(𝑡 − 𝜏𝑘)
𝑦𝑐(𝑡) = 𝐴0𝑥(𝑡 − 𝜏0) + 𝐴1𝑥(𝑡 − 𝜏1) + 𝐴2𝑥(𝑡 − 𝜏2)
𝑦𝑐(𝑡) = 1.0𝑥(𝑡 − 1.0) + 0.5𝑥(𝑡 − 2.0) + 0.25𝑥(𝑡 − 3.0)
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
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1c.- 20 Points: Plot 𝑦𝑐(𝑡) for the Delay System in Fig. 2 if:
𝑦𝑐𝑖(𝑡) = 𝑥(𝑡), 𝐾 = 3, 𝐴0 = 1, 𝐴1 = 0.5, 𝐴2 = 0.25 ,
𝜏0 = 1 𝑆𝑒𝑐., 𝜏1 = 2 𝑆𝑒𝑐., 𝜏2 = 3 𝑆𝑒𝑐.
Assume that 𝑥(𝑡) is a square pulse of unit amplitude and
duration of one second.
1c.- 20 Points: SOLUTION
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
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1d.- 10 Points: Plot 𝑦𝑐(𝑡) for the Delay System in Fig. 2 if:
𝑦𝑐𝑖(𝑡) = 𝑥(𝑡), 𝐾 = 3, 𝐴0 = 1, 𝐴1 = 0.5, 𝐴2 = 0.25 ,
𝜏0 = 1 𝑆𝑒𝑐., 𝜏1 = 1.5 𝑆𝑒𝑐,, 𝜏2 = 2.5 𝑆𝑒𝑐.
Assume that 𝑥(𝑡) is a square pulse of unit amplitude
and duration of one second.
1d.- 10 Points: SOLUTION
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
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Problem 14: DSB-SC Transmitter System The spectrum (Fourier transform) of the modulating signal
𝒙𝒎(𝒕), with its one-sided bandwidth equal to 𝒇𝒎, is given by:
𝑿𝒎(𝒇) = {𝟏 −
|𝒇|
𝒇𝒎, |𝒇| ≤ 𝒇𝒎
𝟎, |𝒇| > 𝒇𝒎
The DSB-SC wireless communication system with an AWGN noisy channel is presented in block diagram form in Fig. 1 below.
Fig. 1: Complete Wireless DSB-SC Communication System
1a.- (50 points) Obtain and plot 𝒀𝒄𝒊(𝒇) if 𝒉𝑩(𝒕) = 𝜹(𝒕). 1b.- (50 points) Obtain 𝒀𝒄(𝒇) and plot its magnitude if the
impulse response of the channel’s filter is 𝒉𝑪(𝒕) = 𝜹(𝒕 − 𝟏
𝟒).
1a.- Solution: The spectrum or Fourier transform of the output of the modulator is given by the following expression:
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
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𝑿𝒄(𝒇) = 𝑿𝒎(𝒇) ∗ 𝑪(𝒇) = 𝑿𝒎(𝒇) ∗ [𝟏
𝟐𝜹(𝒇 − 𝒇𝒄) +
𝟏
𝟐𝜹(𝒇 + 𝒇𝒄)]
𝑿𝒄(𝒇) =𝟏
𝟐𝑿𝒎(𝒇) ∗ 𝜹(𝒇 − 𝒇𝒄) +
𝟏
𝟐𝑿𝒎(𝒇) ∗ 𝜹(𝒇 + 𝒇𝒄)
𝑿𝒄(𝒇) =𝟏
𝟐𝑿𝒎(𝒇 − 𝒇𝒄) +
𝟏
𝟐𝑿𝒎(𝒇 + 𝒇𝒄)
The spectrum of the channel input signal is the product of the spectrum of the
output of the modulation times the frequency response of the bandpass filter:
𝒀𝒄𝒊(𝒇) = 𝑿𝒄(𝒇) ⋅ 𝑯𝑩(𝒇) = [𝟏
𝟐𝑿𝒎(𝒇 − 𝒇𝒄) +
𝟏
𝟐𝑿𝒎(𝒇 + 𝒇𝒄)] ⋅ 𝑯𝑩(𝒇)
Since the frequency response function of the bandpass filter is equal to one
for all frequencies, 𝑯𝑩(𝒇) = 𝟏, the spectrum of the channel input signal is
equal to the spectrum of the output of the modulator: 𝒀𝒄𝒊(𝒇) = 𝑿𝒄(𝒇).
Fig. 1a: Spectrum of Noisy Channel Input Signal )(tyci
1b.- Solution: The spectrum or Fourier transform of the impulse response function of the
channel’s filter is known as the frequency response of the filter as is given by:
𝑯𝑪(𝒇) = 𝑭{𝒉𝑪(𝒕)} = 𝑭 {𝜹 (𝒕 −𝟏
𝟒)} = 𝒆
−𝒋𝟐𝝅(𝟏𝟒)𝒇
= 𝒆−𝒋(
𝝅𝟐)𝒇
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
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𝒀𝒄(𝒇) = 𝒀𝒄𝒊(𝒇) ⋅ 𝑯𝑪(𝒇) = [𝟏
𝟐𝑿𝒎(𝒇 − 𝒇𝒄) +
𝟏
𝟐𝑿𝒎(𝒇 + 𝒇𝒄)] ⋅ 𝒆
−𝒋(𝝅𝟐)𝒇
|𝒀𝒄(𝒇)| = |𝒀𝒄𝒊(𝒇) ⋅ 𝑯𝑪(𝒇)| =𝟏
𝟐𝑿𝒎(𝒇 − 𝒇𝒄) +
𝟏
𝟐𝑿𝒎(𝒇 + 𝒇𝒄)
Thus, we obtain the same plot as in Part 1a of the problem:
Fig. 1b: Spectrum of Channel Filter Output Signal 𝒚𝒄(𝒕)
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
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Problem 15: Scaled 5G Fading or “Ghost” Channel Model A scaled version of a fading or "ghost" channel, used in 5G communications
technology, is modeled using the equation )()()( kcicico ttytyty ,
where , are constants and kt is a constant parameter representing the
“ghost” time-delay. Set 15.0 , 85.0 , and 5kt milliseconds.
1a).- (20 points) Obtain the impulse response function 𝒉(𝒕) of this filter
model using the equation of the system and the following fact:
If 𝑻{𝒚𝒄𝒊(𝒕)} = 𝒚𝒄𝒐(𝒕); then, 𝑻{𝜹(𝒕)} = 𝒉(𝒕)
1b).- (60 points) Obtain the following channel output signal 𝒚𝒄𝒐(𝒕):
𝒚𝒄𝒐(𝒕) = 𝑻{𝒚𝒄𝒊(𝒕)} = 𝑻{𝒄𝒐𝒔(𝟐𝝅𝟓𝟎𝒕) + 𝒄𝒐𝒔(𝟐𝝅𝟏𝟎𝟎𝒕)}
1c).- (20 points) It is desired to plot the channel’s input signal 𝒚𝒄𝒊(𝒕) and
output signal 𝒚𝒄𝒐(𝒕) for a total time duration of 𝑽 = (𝟐𝟎) ∙ 𝟏𝟎−𝟑 𝑺𝒆𝒄. and
a total number of 𝑵 = 𝟖𝟎 sample points. Write a small MATLAB script to
plot the input and output signals in order to meet the desired objective.
Fig. 1: Toy Model of a Ghost Channel Created by a Low Flying Aircraft .
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
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1a).- (20 points) Obtain the impulse response function 𝒉(𝒕) of this filter
model using the equation of the system and the following fact:
If 𝑻{𝒚𝒄𝒊(𝒕)} = 𝒚𝒄𝒐(𝒕); then, 𝑻{𝜹(𝒕)} = 𝒉(𝒕).
1a.- Solution:
𝑻{𝒚𝒄𝒊(𝒕)} = 𝜶𝒚𝒄𝒊(𝒕) + 𝜷𝒚𝒄𝒊(𝒕 − 𝒕𝒌)
𝒉(𝒕) = 𝑻{𝜹(𝒕)} = 𝜶𝜹(𝒕) + 𝜷𝜹(𝒕 − 𝒕𝒌)
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
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1b).- (60 points) Obtain the following channel output signal 𝒚𝒄𝒐(𝒕):
𝒚𝒄𝒐(𝒕) = 𝑻{𝒚𝒄𝒊(𝒕)} = 𝑻{𝒄𝒐𝒔(𝟐𝝅𝟓𝟎𝒕) + 𝒄𝒐𝒔(𝟐𝝅𝟏𝟎𝟎𝒕)}
1b.- Solution:
Using linearity:
𝒚𝒄𝒐(𝒕) = 𝑻{𝒚𝒄𝒊(𝒕)} = 𝑻{𝒚𝒄𝒊𝟏(𝒕) + 𝒚𝒄𝒊𝟐(𝒕)} = 𝑻{𝒚𝒄𝒊𝟏(𝒕)} + 𝑻{𝒚𝒄𝒊𝟐(𝒕)}
𝒚𝒄𝒐𝟏(𝒕) = 𝑻{𝒚𝒄𝒊𝟏(𝒕)} = 𝑻{𝒄𝒐𝒔(𝟐𝝅𝟓𝟎𝒕)}
𝑻{𝒄𝒐𝒔(𝟐𝝅𝟓𝟎𝒕)} = 𝜶𝒄𝒐𝒔(𝟐𝝅𝟓𝟎𝒕) + 𝜷𝒄𝒐𝒔[𝟐𝝅𝟓𝟎(𝒕 − 𝒕𝒌)]
𝒚𝒄𝒐𝟐(𝒕) = 𝑻{𝒚𝒄𝒊𝟐(𝒕)} = 𝑻{𝒄𝒐𝒔(𝟐𝝅𝟏𝟎𝟎𝒕)}
𝑻{𝒄𝒐𝒔(𝟐𝝅𝟏𝟎𝟎𝒕)} = 𝜶𝒄𝒐𝒔(𝟐𝝅𝟏𝟎𝟎𝒕) + 𝜷𝒄𝒐𝒔[𝟐𝝅𝟏𝟎𝟎(𝒕 − 𝒕𝒌)]
𝒚𝒄𝒐(𝒕) = 𝒚𝒄𝒊𝟏(𝒕) + 𝒚𝒄𝒊𝟐(𝒕)
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
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1c).- (20 points) It is desired to plot the channel’s input signal 𝒚𝒄𝒊(𝒕) and
output signal 𝒚𝒄𝒐(𝒕) for a total time duration of 𝑽 = (𝟐𝟎) ∙ 𝟏𝟎−𝟑 𝑺𝒆𝒄. and
a total number of 𝑵 = 𝟖𝟎 sample points. Write a small MATLAB script to
plot the input and output signals in order to meet the desired objective.
1c.- Solution:
%Fading or "ghost" channel.
clear all
close all
alpha=0.15;
beta=0.85;
tk=5*10^-3;
fa=50;
fb=100;
N=80;
V=20*10^-3;
Ts=V/N;
t=0:Ts:V-Ts;
yci=cos(2*pi*50*t)+cos(2*pi*100*t);
ayci=alpha*(cos(2*pi*50*t)+cos(2*pi*100*t));
bycid= beta*(cos(2*pi*50*(t-tk))+cos(2*pi*100*(t-tk)));
yco= ayci + bycid;
plot(t, yci, t,yci, '.', t,yco,t,yco,'.')
xlabel('Time in Seconds')
ylabel('Amplitude')
title('Input yci(t) "BLUE" / Output yco(t) "ORANGE"')
grid
INEL4301 - COMMUNICATIONS THEORY I Prof. Domingo Rodríguez
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