CALIFORNIA STATE UNIVERSITY, BAKERSFIELD (CSUB)DEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERING & COMPUTER SCIENCE

ECE 423: DIGITAL COMMUNICATIONS

Homework 4Solution

QUESTION 1:(25 POINTS)You are given the baseband signals (i) m(t ) = cos1000πt ; (ii) m(t ) = 2cos1000πt+sin2000πt ;(iii) m(t ) = cos1000πt cos3000πt ; (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 )cos10,000πt .

c) Identify the upper sideband (USB) and the lower sideband (LSB) spectra.

Answer: Chapter 4

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

(i) M(j)

USB

\

-500 500 /(Hz) -5.5 -4.5

(ii) MagnitudeIM(j)I

USB

Modulated signal spectrum

LSB LSB

\

4.5

Modulated magnitude spectrum

LSB 1/2

l/4

LSB

USB

5.5 /(kHz)

USB

/(kHz)

-1000 -500 500 1000 -6 -5.5 -4.5 -4 4 4.5 5.5 6 Modulated signal spectrum

,-114 USB LSB LSB USB

1/8(kHz)

-2000 -1000 1000 2000 -7 -6

(ii) Phase:

L.M(f)

7r /2 )

500

-500 /(Hz) -5.5

-rr/2

-4 -3

LS(!! rr/2

-4.5

) -'Tr /2 --

Fig. 84.2-la

3 4

'

4.5

6 7

5.5 f (kHz)

C)

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

50

1/2

1

1/2

l/4

1

Chapter 4

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

(i) M(j)

USB

\

-500 500 /(Hz) -5.5 -4.5

(ii) MagnitudeIM(j)I

USB

Modulated signal spectrum

LSB LSB

1/2 \

4.5

Modulated magnitude spectrum

LSB 1/2

l/4

LSB

USB

5.5 /(kHz)

USB

/(kHz)

-1000 -500 500 1000 -6 -5.5 -4.5 -4 4 4.5 5.5 6 Modulated signal spectrum

,-114 USB LSB LSB USB

1/8(kHz)

-2000 -1000 1000 2000 -7 -6

(ii) Phase:

L.M(f)

7r /2 )

500

-500 /(Hz) -5.5

-rr/2

-4 -3

LS(!! rr/2

-4.5

) -'Tr /2 --

3 4

'

4.5

6 7

5.5 f (kHz)

C)

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

50

1/2

1

2

L_M(f) = { -7r /2, f = 5007r /2,

0, f = -500 else

(iii) m(t) = coswm ,1t·coswm ,2t = cos 10001rt·cos30001rt = ! (cos21rfm.1t + cos21rfm .2t) = ! (cos20001rt + cos 40001rt)--+f m ,I = lOOOHz f m ,2 = 2000Hz

(iv) Since m (t) = e-lOJtl, we have

.J-0

20 F(m(t)) = M(f) = 100+41r2f2

·20

/(Hz)

20 JO

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

0 f (Hz)

51

3

QUESTION 2:(25 POINTS)Two signals m1(t ) and m2(t ), both band-limited to 5000 Hz, are to be transmitted simultane-ously over a channel by the multiplexing scheme shown in the following figure. The signal atpoint b is the multiplexed signal, which now modulates a carrier of frequency 20,000 Hz. Themodulated signal at point c is transmitted over a channel.

EE411 Homework 7 

 

1. (4.2‐1)Given the baseband signal (i)  ttm π1000cos)( = ; (ii)  ||10)( tetm −= . For each one, do the 

following: (a) Sketch the spectrum of m(t). (b) Sketch the spectrum of the DSB‐SC signal  ttm π10000cos)( . 

(c) Identify the upper sideband (USB) and the lower sideband (LSB) spectra. 

 

 

2. (4.2‐7)Two signals m1(t) and m2(t), both band‐limited to 5000 Hz, are to be transmitted simultaneously over a channel by the multiplexing scheme shown in Fig. 1. The signal at point b is the multiplexed signal, which now modulates a carrier of frequency 20,000 Hz. The modulated signal at point c is transmitted over a channel. (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 m1(t) and m2(t) from the modulated signal at point c.                                      

                

M1(f)

f5000-5000 0

5000-5000 0 f

M2(f) Σ

2cos(20000πt)

m1(t)

m2(t) a

b c

2cos(40000πt)

   

                                                                                                               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 m1(t ) and m2(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 30000Hz (from 5000 Hz to 35000 Hz).

4

(c) The following figure shows the receiver diagram to recover both m1(t ) and m2(t ) from themodulated signal at point c.

5

QUESTION 3:(25 POINTS)An amateur audio scrambler/descrambler pair is shown in the following figure.QUESTION 3:(25 POINTS)

An amateur audio scrambler/descrambler pair is shown in the following figure.EE456–Digital Communications (Fall 2015) ASSIGNMENT

( )m t

2cos(40,000 )tπ

LPF0-20 kHz

( )y tLPF

0-10 kHz

( )x t

( )z t

02cos( )tωA

( )M f

f10 kHz10 kHz− 0

Figure 4: An audio scrambler considered in Question 6.

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

(b) Show whether or not you can descramble z(t) to recover m(t).

University of Saskatchewan Page 3

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.

c) Show whether or not you can descramble z(t ) to recover m(t ).

3

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.

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 •··� /�/(lt�z)

-30 -20 -10 10 20 30

f',, I '··

-10 10 /(kHz)

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 ) = cos100πt +2cos300πt

b) m(t ) = sin100πt sin500πt

In each case:

i. Sketch the spectrum of m(t ).

ii. Find and sketch the spectrum of the DSB-SC signal 2m(t )cos1000πt .

iii. From the spectrum obtained in part (ii), suppress the LSB spectrum to obtain the USBspectrum.

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 generatethe signals of the same relative magnitude, it is convenient to multiply m(t ) by 2cos(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. Simi-larly, to obtain the USB spectrum, we suppress the LSB spectrum (between −ωc andωc ) fromthe DSB-SC spectrum. The following figures show the three cases.

8

I 1 f fi

� I I-ISO -SO so 150 f

__ J_tJTL .. -4SO -JSO o 350 450

I I

f IP-1/1

�J -6SO -5SO -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 <PLsB (t) = 2 cos (7001rt) + cos (9001rt) and <l>usB (t) = cos (I I001rt) + 2 cos ( 13001rt).

(b) We can express:

<i>LSB (t) = ! [cos (4007Tt) + cos (6007Tt)] and <l>usa (t) = ! [cos (14001rt) + cos (16007Tt)].

r�·I l '" 11O>,,..(/)

1 I I , 1 1 1 I I 1 � ·'.!00 -200 0 200 JOO

I -800 -700 -300 -200 0 200 JOO 700 IIOO I

O>,.,.(J)

l '"a>,,.(J)

l l I 114 I l .. 1 1 I I � .)00 -200 0 200 � I -MOO -700 u 700 800 I

<hsa (t) = [cos (100,rt) + 2 cos (3007Tt)] cos (lOOOt) + [sin ( I001rt) + 2 sin (3001rt)] sin (lOOOt) = cos (1000 - 100,r)t + 2cos (1000 - 3001r)t

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

66

9