Hw4_solution.pdf

C ALIFORNIA S TATE U NIVERSITY, B AKERSFIELD (CSUB) D EPARTMENT OF E LECTRICAL & C OMPUTER E NGINEERING & C OMPUTER S CIENCE ECE 423: D IGITAL C OMMUNICATIONS

Homework 4 Solution

QUESTION 1:(25 POINTS) You are given the baseband signals (i) m(t ) = cos 1000πt ; (ii) m(t ) = 2 cos 1000πt +sin 2000πt ; (iii) m(t ) = cos 1000πt cos 3000πt ; (iv) m(t ) = exp(−10|t |). For each one, do the following.

a) Sketch the spectrum of m(t ). b) Sketch the spectrum of the DSB-SC signal m(t ) cos 10, 000πt . c) Identify the upper sideband (USB) and the lower sideband (LSB) spectra. Chapter 4

Answer:

(i) m (t) = COS Wm t = cos 21rfm t = cos 10007rt -+ fm = 500Hz. M(f) = 0.M(f - 500) + 0.5o(f + 500).

M(j)

(i)

Modulated signal spectrum

500

-500 (ii) Magnitude

IM(j)I

LSB

USB

\

1/2 /(Hz)

l/4

-5.5

LSB

USB

500 1000

-2000

-1000

1/2 l/4

5.5 /(kHz)

USB

LSB

4 4.5

-6 -5.5 -4.5 -4

Modulated signal spectrum

USB

,-114

4.5 Modulated magnitude spectrum

1/2 -500

USB

\

-4.5

1

-1000

LSB

5.5

LSB

LSB

6

USB

1/8

1000 2000

-7 -6

-4 -3

/(kHz)

3 4

6

7

(kHz)

(ii) Phase:

L.M(f)

7r /2

LS(!! rr/ 2

)

'

500 -500

-rr/2

/(Hz)

-5.5

)

-4.5

4.5 -'Tr /2

--

5.5 f (kHz) C)

1

USB  LSB


,-114 -2000

-1000

1000 2000

-7 -6

LSB

USB

1/8

-4 -3

3 4

6

7

(kHz)

(ii) Phase:

L.M(f)


7r /2




500 -500

/(Hz)

-rr/2







   




LS(!! rr/ 2

)

#
'

      








-5.5

)

66f 




-4.5

 -'Tr /2

--

5.5 f (kHz) 





4.5

C)



    

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(ii) m (t) = 2 cos Wm,1t + sin Wm,2t = 2cos 21rJm ,1t + sin 21rfm,2t = 2cos 10001rt + sin 20001rt 
 
 - 500) +D:f 
 _. M(f) = o(f - 1000) + o(f + 1000) - o.5jo(f o.5jo(f + 500)  + o(f + 1000) + o.M(f - 500) + o.M(f + 500) IM(J)I = o(f - 1000) 





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2

L_M(f) = {

/2, 0,

(iii) m(t) = coswm ,1t·coswm ,2t = cos 10001rt·cos30001rt = f m ,I = lOOOHz f m ,2 = 2000Hz (iv) Since m(t) = e-lOJtl we have ,

f = 500 f = -500 else

-7r /2,

7r

F(m(t)) = M(f)

! (cos21rf

=

m .1t

!

+ cos21rfm .2t) = (cos20001rt + cos 40001rt)--+

20 100+41r2f2

/(Hz) .J-0

20

·20

JO

Spectrum of e .. 1111•1 co� ( 100001TI.)

0

f (Hz)

51

3

(b) Sketch the spectrum of the DSB‐SC signal  m(t ) cos 10000πt .  (c) Identify the upper sideband (USB) and the lower sideband (LSB) spectra.      2. (4.2‐7)Two signals m (t) and m (t), both band‐limited to 5000 Hz, are to be transmitted 

2 QUESTION 2:(25 POINTS) 1 simultaneously over a channel by the multiplexing scheme shown in Fig. 1. The signal at point b  Two signalsis the multiplexed signal, which now modulates a carrier of frequency 20,000 Hz. The modulated  m 1 (t ) and m 2 (t ), both band-limited to 5000 Hz, are to be transmitted simultaneously over asignal at point c is transmitted over a channel.  channel by the multiplexing scheme shown in the following figure. The signal at (a) Sketch signal spectra at points a, b, and c.  point b is the multiplexed signal, which now modulates a carrier of frequency 20, 000 Hz. The (b) What must be the bandwidth of the distortionless channel?  modulated (c) signal at point c is transmitted over a channel. Design a receiver to recover signals m (t) and m (t) from the modulated signal at point c.                                      1

2

  M1(f)

m1(t) -5000

0

5000

f

Σ

M2(f)

m2(t) -5000

                

0

5000

b

c

a 2cos(40000πt)

f

2cos(20000πt)

   

                                                                                                               Fig. 1                                                  

a) Sketch signal spectra at points a, b, and c.   b) What must be the bandwidth of the distortionless channel?     c) Design a receiver to recover signals m 1 (t ) and m 2 (t ) from the modulated signal at point c.  

Answer:

(b) From the spectrum at point c, it is clear that the channel bandwidth must be at least 30000 Hz (from 5000 Hz to 35000 Hz).

4

(c) The following figure shows the receiver diagram to recover both m 1 (t ) and m 2 (t ) from the modulated signal at point c.

5

QUESTION 3:(25 POINTS) QUESTION POINTS) An amateur audio 3:(25 scrambler/descrambler pair is shown in the following figure. An amateur audio scrambler/descrambler EE456–Digital Communications (Fall 2015) pair is shown in the following figure. ASSIGNMENT M( f )

A

−10 kHz

LPF 0-20 kHz

m( t )

0

10 kHz

f

x (t )

2 cos(ω0t )

2 cos(40,000π t )

z (t )

LPF 0-10 kHz

y (t )

Figure 4: An audio scrambler considered in Question 6.

a) Graphically find and show the spectra of signals x(t ), y(t ), and z(t ) when ω0 = 20, 000π.

(b) Graphically find and show the spectra of signals y(t) and z(t) when ω0 = 30, 000π. a) Graphically find and show the spectra of signals x(t ), y(t ), and z(t ) when ω0 = 20, 000π.

b) Graphically findwhether and show theyou spectra of signals z(t) y(t ),toand z(t ) m(t). when ω0 = 30, 000π. (b) Show or not can descramble recover

b) Graphically andorshow the of signals y(t ), m(t and). z(t ) when ω0 = 30, 000π. c) Showfind whether not you canspectra descramble z(t ) to recover c) Show whether or not you can descramble z(t ) to recover m(t ). Answer: (a) X(/)

_,/

·20

10

·10

·'

20

/(kHz)

Y(/1

/o • ··� ��

-30

-20

-10

10

/�/

20

(lt�z)

30

If',,'·· -10 University of Saskatchewan

10

/(kHz)

Page 3

3

6

(b)

(c) We are able to recover m(t ) from z(t ) in part (a) as shown in the following figure.

But we are not able to recover m(t ) from z(t) in part (b).

7

QUESTION 4(25 POINTS) A modulating signal m(t ) is given by:

a) m(t ) = cos 100πt + 2 cos 300πt b) m(t ) = sin 100πt sin 500πt In each case:

i. Sketch the spectrum of m(t ). ii. Find and sketch the spectrum of the DSB-SC signal 2m(t ) cos 1000πt . iii. From the spectrum obtained in part (ii), suppress the LSB spectrum to obtain the USB spectrum.

iv. Knowing the USB spectrum in part (ii), write the expression φU SB (t ) for the USB signal. v. Repeat part (iii) and (iv) to obtain the LSB signal φLSB (t ). Answer: To generate a DSB-SC signal from m(t ), we multiply m(t ) by cos(ωc t ). However, to generate the signals of the same relative magnitude, it is convenient to multiply m(t ) by 2 cos(ωc t ). This also avoids the nuisance of the fractions 1/2, and yields the DSB-SC spectrum M (ω − ωc ) + M (ω + ωc ) We suppress the USB spectrum (above We and below −ωc ) to obtain the LSB spectrum. Similarly, to obtain the USB spectrum, we suppress the LSB spectrum (between −ωc and ωc ) from the DSB-SC spectrum. The following figures show the three cases.

8

I 1 �I I

f IP-1/1

�J

fi

-ISO -SO

fso

150

__J_tJTL .. -4SO -JSO

o I

350 450

-6SO -5SO

f

I

-1.50 -350 0

JSO -4SO

5SO

6SO

f j O>,-(Jl

l_J____ --- - ��________J_l

-6SO -5SO

-4SO -JSO O

JSO 4SO

(a) We can express usB (t) = cos (I I001rt) + 2 cos ( 13001rt). (b) We can express:

LSB (t) =

! [cos (4007Tt)+ cos (6007Tt)]

and usa (t) =

1 I r�·I I ,

·'.!00 -200

ll

.)00

-200

0

I

0

200

JOO

I

! [cos (14001rt) + cos (16007Tt)].

11 1I

-800 -700

O>,.,.(J)

11

4

I l ..

200

�

I

11

-MOO

-700

-300 -200

l '" 11

'l "

200 JOO

0

u

O>,,..(/)

I1

700

IIOO

I

�

a>,,.(J)

II

700

800

I

�

9 = [cos (100,rt)+ 2cos (3007Tt)] cos (lOOOt)+ [sin (I001rt)+ 2sin (3001rt)] sin (lOOOt) = cos (1000 - 100,r)t+ 2cos (1000 - 3001r)t (t) usa , = [cos (1001rt) + 2cos (3007Tt)] cos (IOOOt) - (sin (l001rt)+ 2sin (3001rt)) sin (lOOOt) = cos (1000 + 100,r)t + 2cos (1000 + 300,r)t
66