Midterm Hw123problems Hw123sol

7/30/2019 Midterm Hw123problems Hw123sol

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EECE4572 Summer 2010

Communication Systems I Prof. Salehi

MIDTERM SOLUTIONS

Problem 1:

The power spectral density of a WSS information source is shown below (the unit of power

spectral density is Watts/Hz). The maximum amplitude of this signal is 200.

-

6

JJJJJ

f (Hz)

SX(f )

30003000 20002000

2

1. What is the power in this process?

2. Assume that this signal is transmitted using a uniform PCM system with 512 quantiza-

tion levels, what is the resulting SQNR in decibels and what is the minimum required

transmission bandwidth if in sampling the signal a guard band of 1 KHz is used?

3. If the available transmission bandwidth is 47 KHz, design a PCM system that achieves the

highest possible SQNR. What is the resulting SQNR and the guard band?

Solution:

1. P =

SX(f)df = 4000 2 + 2

12 2 1000 = 10000 Watts.

2. = log2 N = 9, then SQNR = 4.8+69+10log10100002002

52.8 dB, and Bt = W +2WG =

9 3000 + 4.5 1000 = 31500 Hz

3. W + 12WG 47000 or 3000 +12WG 47000. The largest integer that satisfies this

is = 15 which gives WG = 266.6 Hz and SQNR = 4.8+615+10log10100002002

88.8 dB.

Problem 2: 35 points

In the block diagram shown below X(t) denotes a zero-mean WSS (wide-sense stationary) and

white random process with power spectral density SX(f ) =N02

.

-

- 2 ddt

m+?

6

- LPF: [W , W ] -X(t) Y (t) = X(t) + 2X(t) Z(t)


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The block denoted by LPF represents an ideal lowpass filter that passes all frequencies in

the frequency range from W to W and blocks all other frequencies. Answer the following

questions, your answers will be in terms of N0.

1. What is the power spectral density and the mean ofY(t)?

2. What is the power spectral density ofZ(t)?

3. Is Z(t) a WSS random process? Why?

4. What is the variance ofZ(t) ifW = 4?

5. What is the power in Y(t)?

Solution:

1. h(t) = (t) + 2(t), hence H(f) = F[h(t)] = 1 + j4f. We have mY = mXH(0) =

0 1 = 0 and SY(f ) = SX(f )|H(f)|2 =

N02 |1 + j4 f|

2 =N02 (1 + 16

2f2).

2. For the LPF H1(f ) = (f /2W ), henceSZ(f ) = SY(f )|H1(f )|2 =N02 (1+16

2f2)(f /2W ).

3. Since input is WSS and system is LTI the output will be WSS.

4. E[Z2(t)] = E[Z(t)Z(t)] = RZ(0) =

SZ(f)df =N02

44(1 + 16

2f2) df = N0(4 +13

210 2).

5. PY =

SY(f ) =

N02 (1 + 16

2f2) df = .

Problem 3: 35 points

A discrete-memoryless source X has the alphabet X = {x1, x2, x3, x4, x5, x6}, with corre-

sponding probabilities 116 ,

14 ,

18 ,

14 ,

116 ,

14

.

1. What is the entropy of this source?

2. Design a Huffman code for this source, what is the average codeword length of the Huff-

man code?

3. Can you design a more efficient Huffman code by using the second extension of this

source (i.e., designing a Huffman code for sequences of two outputs)? Why?

4. If you are asked to assign new probabilities to X (different from the ones given above)

such that the entropy is maximized, what probabilities would you assign? What is the

resulting entropy?

Solution:

1. H(X) = 6

i=1 pi log2 pi = 34 log2

14

18 log2

18

216 log2

116 = 2.375 bits/symbol.

2. Designing the Huffman code gives codewords {1110,00,110,01,1111,10} (or some equiv-

alent code) with R =

pili = 2.375 binary symbols/source symbol.

3. Since already R = H(X) no improvement is possible.

4. Equiprobable distribution has the highest entropy, hence pi = 1/6 and H(X) = log2 6 =

2.58.


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Homework 1 Problems


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Homework 2 Problems

Homework 3 Problems


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EECE4572 Summer 2012

Communications I Prof. Salehi

Homework 1 Solution

Note: You need to use Fourier transform properties and F.T table in this HW. To prepare for this

HW read Chapters 2.

Problem 1

1. Using scaling, shift, and modulation properties of the F.T., determine the F.T. ofx(t), where

x(t) =

cos(t), 0 t 40, otherwise.

2. Derive the magnitude spectrum of this signal usingMatlab

and plot it (magnitude spectrumis the magnitude of the F.T. of a signal, i.e., |X(f)|). In your HW solutions include both the

Matlab code and the resulting plot. Is your plot symmetric (even)? Why? (Note: If you do not

know how to use Matlab to find F.T. look at Chapter 1 of the Matlab recommended book and

in particular Illustrative Problem 1.5 on page 21. The Matlab fundamentals handout posted

on the blackboard site is also useful to refresh your memory.)

3. Now let x1(t) and x2(t) be defined as

x1(t) =

cos(2 f0t), 0 t 40, otherwise. x2(t) =

cos(t), 0 t T0, otherwise.

Note that in x1(t) the width of the pulse is kept constant at 4 but the frequency f0 can change.In x2(t) the width of the pulse can change but the frequency is kept at

12

. Plot |X1(f )| for

f0 = 1, 2, 4 and |X2(f ) for the pulse durations T = 8, 16 and explain how changing f0 and T

change the magnitude spectrum of the signal.

Solution

Since the highest frequency is 4 we choose the parameter fs to be 20 which is well above twice the

highest frequency. The frequency resolution df determines how accurate you want your graph to

be, we choose df=0.001. The maximum time is 16 we choose the representation from 32 to 32 to

represent the signal precisely, the listing of the program is given below.

df=0.001;

fs=20;

ts=1/fs;

t=[-32:ts:32];

x1=zeros(size(t));

x1(641:721)=cos(pi*t(641:721));


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x2=zeros(size(t));

x2(641:721)=cos(2*pi*t(641:721));

x3=zeros(size(t));

x3(641:721)=cos(4*pi*t(641:721));

x4=zeros(size(t));

x4(641:721)=cos(8*pi

*t(641:721));

x5=zeros(size(t));

x5(641:801)=cos(pi*t(641:801));

x6=zeros(size(t));

x6(641:961)=cos(pi*t(641:961));

[X1,x11,df1]=fftseq(x1,ts,df);






X11=X1/fs;

X21=X2/fs;

X31=X3/fs;

X41=X4/fs;

X51=X5/fs;

X61=X6/fs;

f=[0:df1:df1*(length(x11)-1)]-fs/2;

plot(f,fftshift(abs(X11)))






The plots denoted by X1 through X6 are shown on the next page. X1 is the original plot for T = 4

and f0 = 1/2. The following table gives T and f0 for various Xs.

X T f0

X1 4 1/2

X2 4 1

X3 4 2

X4 4 4

X5 8 1/2

X6 16 1/2

As seen increasing f0 moves the peaks away from zero and locates them at the corresponding

frequencies, increasing T makes the peaks of the signals higher and more like impulses (note the

vertical scale).


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10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 100

0.5

1

1.5

2

2.5

X1

10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 100

0.5

1

1.5

2

2.5

X2

10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 100

0.5

1

1.5

2

2.53

10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 100

0.5

1

1.5

2

2.54

10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 100

0.5

1

1.5

2

2.5

3

3.5

4

4.55

10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6

7

8

9

X6


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Problem 2

Problem 2.10 parts 2, 3.

Solution

2) Using the time shift theorem, we have

F[x(t)] = F[(t 3) +(t + 3)]

= sinc(f)ej2 f3 + sinc(f)ej2f3

= 2sinc(f ) cos(23f )

3) Using time-scaling and time-shift theorems we have

F[x(t)] = F[(2t + 3) +(3t 2)]

= F[(2(t +3

2)) +(3(t

2

3)]

= 12 sinc2( f2 )e

j f3 + 13 sinc2( f3 )e

j2 f2

3

Problem 3

Problem 2.12 parts a, b, e.

Solution

a) We can write x(t) as x(t) = 2( t4 ) 2(t2 ). Then

F[x(t)] = F[2(t

4)] F[2(

t

2)] = 8sinc(4f ) 4sinc2(2f )

b)

x(t) = 2(t

4) (t) F[x(t)] = 8sinc(4f ) sinc2(f )

e) We can write x(t) as x(t) = (t + 1) +(t) +(t 1). Hence,

X(f) = sinc2(f)(1 + ej2 f + ej2f) = sinc2(f)(1 + 2cos(2f))

Another approach is to note that x(t) = 2

t2

(t), hence X(f) = 4sinc2(2f ) sinc2(f ).

Problem 4

Problem 2.26 part 5.

Solution

5) Using the convolution theorem we obtain

Y(f) = (f )(f )

Y(f) is shown below.


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12

12

12

1

Y(f)

f

We notice that we can write Y(f) = 12(f ) +12(2f ), and hence y(t) =

12

sinc(t) + 14 sinc2

t2

.

Problem 5

Problem 2.17.

Solution

(Convolution theorem:)

F[x(t) y(t)] = F[x(t)]F[y(t)] = X(f)Y(f)

Thus

sinc(t) sinc(t) = F1[F[sinc(t) sinc(t)]]

= F1[F[sinc(t)] F[sinc(t)]]

= F1[(f )(f)] = F1[(f)]

= sinc(t)

Problem 6Problem 4.10, parts 1,2.

Solution

1) The random variable X is Gaussian with zero mean and variance 2 = 108. Thus p(X > x) =

Q( x) and

p(X > 104) = Q

104

104

= Q(1) = .159

p(X > 4 104) = Q

4 104

104

= Q(4) = 3.17 105

p(2 104 < X 104) = 1 Q(1) Q(2) = .8182

2)

p(X > 104X > 0) = p(X > 104, X > 0)

p(X > 0)=

p(X > 104)

p(X > 0)=

.159

.5= .318


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Problem 7

X is a Gaussian random variable with mean 3 and variance 4. Find the following probabilities.

1. P(X > 0).

2. P(X < 8).

3. P (2 < X < 2).

4. P (4 < (X + 1)2 < 16).

Solution

Obviously m = 3 and = 2. We have

1. P(X > 0) = Q

032

= Q(1.5) = 1 Q(1.5) = 1 0.0668 = 0.9332.

2. P(X < 8) = Q

382

= Q(2.5) = 1 Q(2.5) = 1 0.00621 = 0.99379.

3. P (2 < X < 2) = Q32

2

Q3(2)

2

= Q(0.5) Q(2.5) = 0.308 0.0062 = 0.3018.

4. P (4 < (X+1)2 < 16) = P (2 < |X+1| < 4) = P (2 < X+1 < 4)+P (2 < X1 < 4) = P (1 < X

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