inel4301 - communications theory i prof. domingo …domingo/teaching/inel4301/sample... · inel4301...

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)


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



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



4

)(22

)(2

)(2sin2dmm

md

mdmL ttfSincf

ftt

fttfh

)()(2cos()(2)( 111cLcLcLB ffHffHFtfthFFfHF

)()()()2cos()(2 1 thffHffHFtfth BcLcLcL



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



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



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



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.



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



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



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 ,



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



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 .



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



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)()(



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



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')



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.



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.



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



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

https://www.youtube.com/watch?v=fnP0muG3BT0&t=57s

https://www.youtube.com/watch?v=FXVgiIY3zd4

https://www.mathworks.com/help/comm/ref/awgn.html

https://www.youtube.com/watch?v=fnP0muG3BT0&t=57s

https://www.youtube.com/watch?v=FXVgiIY3zd4

https://www.mathworks.com/help/comm/ref/awgn.html



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.



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)



24

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



25

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



26

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:

https://en.wikipedia.org/wiki/Fourier_transform



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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

https://en.wikipedia.org/wiki/Fourier_transform



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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)



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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 .



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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)



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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)



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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



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