Homework Title / No. : _Assignment-1_____________________Course Code : ECE 30_______

Course Instructor: _Miss Ritu Gupta___________ Course Tutor (if applicable) : ___do_________

Date of Allotment: _____________________ Date of submission : 26/02/2010________

Student’s Roll No._54______________________ Section No. : _____H6802____________________

Declaration: I declare that this assignment is my individual work. I have not copied from any other student’s work or from any other source except where due acknowledgment is made explicitly in the text, nor has any part been written for me by another person.

Student’s Signature: _

Ramjee prasad

Evaluator’s comments: _____________________________________________________________________

Marks obtained : ___________ out of ______________________

Content of Homework should start from this page only:

Part- A

Q1 Find the Nyquist rate and the Nyquist interval for each of the signals:i) X(t) = 5cos 1000πt cos 4000 πt.

ii)

iii)iv) X(t) = 5cos 1000πt + cos 4000 πt.

Solution: (i) Given, X(t)= 5cos 1000πt ×cos 4000 πt.

= 5/2 cos 1000πt× cos 4000 πt.

=5/2[cos (4000π+1000π) t + cos(4000π+1000π)t =5/2[cos5000πt+cos3000πt] .............................. (1)

The standard eqn is, x (t) =A1 cosw1t +A2 cosw2t .................. (2)

Comparing both eqn (1) and (2):-

Let, w1=5000π and w2=3000π

Since, w1=5000π 2πf1=5000π Hence, f1=2500

Again, w2=3000π

2πf2 =3000πHence, f2=1500

So, the maximum frequency component present in the given signal is, f1= 2500

Hence, Nyquist rate =2×fm

=2×2500 =5000Hz

Nyquist interval = 1/2fm

=1/5000=2×10-4

=0.2 m sec

ii

From the given signal , w =100 π 2 π f =100 π

Hence, f=50

So, the nyquist rate is, =2fm=2×50=100And the nyquist interval is, Ts=1/100=0.01 m sec

iii solution: .

Or, =1/(πt)2[sin2(100πt)]Or, =1/(πt)2[1-cos(2×100πt)]/2

=1/2(πt)2- 1/2(πt)2 cos (200πt) From the term present in the modify input signal, Cos(200πt).............1Comparing 1 to the standard eqn x(t)= Acos (wt) Since , w= 200π

2πf=200π f= 100 Hz

So, the nyquist rate is,=2fm=2×100 =200Hz

and the nyquist interval is ,Ts=1/200=5×10-3

i˅

x(t) = 5cos 1000πt + cos 4000 πt.................(1) The standard eqn is, x (t) =A1 cosw1t +A2 cosw2t .................. (2)

Comparing eqn (1) and eqn (2) :-

w1=1000π and w2=4000π

Since , w1=1000πOr, 2πf1=1000π Hence, f1=500 Hz

Again , w2= 4000 π

Or, 2 πf2=4000 πHence, f2= 2000Hz

So, the nyquist rate is,=2fm=2×2000

=4000And the nyquist interval is ,Ts=1/4000

=2.5×10-4 sec

Q2 A µ-law comparator uses a compressor which output to input by the relation

Here the + sign applies when x is positive and – sign applies when x is negative. Also x= vi/V and y= vo/V where vi and vo are the input and output voltages and the range of allowable volltage is –V to +V. The

parameter determine the degree of compression.

(a) A commonly used value is =255. For this value make a plot of y vs. x from x=-1 to x=+1.

(b) If V=40 volts and 256 quantization levels are employed what is the voltage interval between levels

there is no compression? For =255 what is the minimum and what is the maximum effective

separation between levels.

Solution: (a) Given, µ= 255

Y= , where x=Vi/V

Or, Y=

Or, Y=

When, x= 0.2, Y= = 0.7126

x=0.4, Y= + =0.8358

x=0.6, Y= = 0.9083

x=0.8, Y= = 0.9599

x=1, Y= =1

x=-0.2, y= =-0.7126

x=-0.4, y= =-0.8358

x=-0.6, y= =-0.9083

x=-0.8, y= =-0.9599

x=-1, y= = -1

Plot: x versus y graph:-

(B)

Given,

V=40 volt, quantization level= 256, µ=255

The voltage interval (or step size) with no compression is,

=2V/q = 2*40/256=0.3125

When there is no compression (that is, a nonuniform quantization),the smallest effective separation between levels will be very near to the origin , and the largest affective separation levels will be closet x =1.

Let x1= is that value of x for which y=1/127,

Since, Y=

or , 1/127=

or, 1/127=

or, 0.043662814 =ln(1+255 x1)or, e0.043662814=1+255 x1 or, 0.044630061=255 x1

x1 =1.750×10-3So, the smallest affective separation between levels is,

min=V*x= 40*1.750*10-3= 7×10-3

Again, let x127 is the value of x for which, y=1-(1/127)

ln ((1+255 x127 )/ln256) =126/127

or, ln (1+255 x127 ) =.992125984×5.545177444=5.5015463

or, 1+255 x1 =e5.5015463

x1 = 0.95713

Thus the largest effective separation b/w levels is

= V (1- x127 )=40× ( 1-0.95713)

=1.72

Q3 Explain the bandpass signal sampling theorem. Prove that the minimum sampling frequency (fs) should be twice of bandwidth of signal and sampling frequency should be multiple of bandwidth of signal.

Solution:

Bands pass sampling theorem:

The band pass signal x(t) whose maximum bandwidth is 2fm can be completely sampled and recovered from its sample if it is sampled at the minimum rate of twice the bandwidth. Here fm is the maximum frequency component present in the signal.

Hence, if the bandwidth of the signal is 2fm, then the minimum sampling rate must be 4fm band pass signal.

Proof:

Consider a band pass signal whose band width is 2fm and its spectrum is centered on fc.

Let,

XI(t)= imphase component of x(t)and xQ =quadrature component of x (t)The imphase and quadrature components are obtained by multiplying x(t) by cos (2πfct) and sin(2πfct) and then suppressing the sum frequencies by means of low pass filters. Thus xI and xQ component contain only low frequency component.

After few mathematical manipulation,

x(t)= ......(1)

and x(t) = ..............(2)

Comparing 1st equation with 2nd eqn (interpolation eqn of low pass signal) , we observed that x(t) is replaced by

x( )

Here, x ( ) = x(nTs) = sampled version of band pass signal

And Ts=

Thus, if 4fm samples per second are taken, then the band pass signal of bandwidth 2fm can be completely recovered from its samples.

Hence, for band pass signal of bandwidth 2fm,

Minimum sampling rate = Twice of bandwidth

=4fm samples per second.

Part- BQ4 what is baseband data transmission? Explain the working principles of each blocks of PCM.

Solution:

Whenever a modulating or message signal is impressed upon a carrier signal, the modulated signal is produced. The modulated signal has fixed band of frequencies around carrier frequency. Because the modulated signal is band limited, it is calledband pass or pass band signal. The transmission of such type of modulated signal over achannel is called Band pass data transmission.

Block diagram of PCM system:-

PCM generator

1. Low pass filter:

In PCM generator, the signal x(t)=is applied to a low pass filter of cut off frequency fm Hz. This low pass filter blocks all the frequency component above the cut-off frequency. Now signal x(t) is band limited to fm Hz.

2. Sampler:

The sampling and hold circuit samples the signals at the freq fs. sampling freq is selected sufficiently above to the nyquist rate,

fs 2fm

3. Quantizer:

The o/p of sampler is denoted by X(nTs). This signal X(nTs) is discrete in time and continuous in amplitude. A q-level quantizer compares input x (nTs) with its fixed digital levels. It then assigns any one of the digital level to x(nTs) which result in minimum distortion or error. Thus o/p of quantaizer is a digital level called xq(nTs).

4. Now the quantized signal is level xq(nTs) is given to binary encoder. This encoder converts input signal to ‘v’ binary bits. This encoder is also known as digitlizer.

PCM transmission path:

The path b/w the PCM transmitter and PCM receiver over which the PCM signal travel, is called PCM transmission path. The PCM system use regenerative repeaters for control the distortion produced by the channel. The regenerative repeater performs three basic operations namely quantization, timing and decision making. Hence each repeater actually reproduced the clean noise free PCM signal from the PCM signal distorted by the channel noise.

PCM receiver:

1. Regeneration circuit:

The regeneration circuit receives the signal from the channel and reconstructs them into an original analog signal. The regenerator at the start of PCM receiver reshapes the pulse and removes the noise. This signal is then converted to parallel digital words for each sample.

2. Decoder:

Now the o/p of regeneration circuit goes to the decoder and its convert the digital word into its analog value denoted as xq(t) with the help of a sampler and hold circuit.

3. Reconstruction filters:

The signal from the decoder is passed through a low pass reconstruction filter to get the appropriate original message signal.

Q5 A compact disc (CD) (recording system samples each of two stereo signals with a 16 bit analog to digital converter (ADC) at 44.1kb/s.

(a) The bit stream of digitized data is augmented by the addition of error correcting bits, clock extraction bits, and display and control bit fields. These additional bits represent 100 % overhead. Determine the output bit rate of the CD recording system.

(b) The CD can record an hour’s worth of music. Determine the number of bits recorded on a CD.(c) For a comparison, a high-grade collegiate dictionary may contain 1500 pages, 2 columns per, 100 lines

per column, 8 words per line, 6 letters per word, and 7b per letter on average. Determine the number of bits required to describe the dictionary, and estimate the number of comparable books that can be stored on a CD.

Solution:

(a) The input bit rate, =2×44.1×103×16=1.411 Mb/sec

Due to the 100% overhead output rate is , =2×1.411×106b/s =2.822Mb/sec

(b) The no. Of bits recorded on the CD,=2.822×106×(3600 sec)=10.16×109 bits=10.16 giga bits

(c) No. Of bit reqd to describe dictionary, =1500×2×100×8×6×7

=100.8Mb No. Of comparable books that can be stored on a CD,

=no. of bits recorded on the CD/no. Of bits reqd to describe dictionary

10.16×109×103/100.8×106

= 50.4 =50 books

Q6. A TV signal has a bandwidth of 4.5 MHz. This signal is sampled, quantized and binary coded to obtain a PCM signal.

(a) Determine the signal sampling rate if the signal to be sampled at a rate 20% above the Nyquist rate.(b) If the samples are quantized into 1024 levels, determine the number of binary pulses required to encode

each sample.(c) Determine the binary pulse rate of the binary- coded signal, and the minimum bandwidth required to

transmit this signal.

Solution:

The maximum frequency is given by,

= 4.5 MHz

by using sampling theorem,

signal sampling frequency is,

fs = 2 × 4.5

fs = 9

Hence, fs=45MHZ

(ii) The no. of bits or train pulses given by,

Since, q = 2v, Where q = no. of levels

, v = bits in PCM

Hence,

1024 = 2v

Or, 210 =2v

Hence v=10

(iii) The bit rate ,

r ≥ v fs ≥ 10× 9 × 106 = 9 × 107 bits per sec. Ans.

So, the transmission bandwidth is ,

BW ≥ ½ r≥ ½ 9 × 107 ≥45 MHz Ans.