Homework Title / No. : _Assignment-1_______________ _Assignment-1_____________________Course ______Course Code :
ECE 30_______ 30_______
Course Instructor: _Miss Ritu Gupta___________ Gupta___________ Course Tutor Tutor (if applicable) : ___do_________ Date of Allotment: ___________________ _____________________ __ Date of submission :
26/02/2010________
Student’s Roll No._54_______________ No._54______________________ _______ Section No. : _____H6802__________ _____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 + 1000πt + cos 4000 πt.
Solution:
(i)
Given, X(t)= 5cos 1000πt ×cos 4000 πt.
= 5/2 5/2 cos cos 100 1000π 0πt× t× cos cos 40 400 00 πt. πt. =5/2[cos (4000π+1000π) t + cos(4000π+1000π)t =5/2[cos5000πt+cos30 =5/2[cos5000πt+cos3000πt] 00πt] .............................. (1) The standard eqn is, is, x (t) =A1 cosw1t +A2 cosw2t .................. (2) Comparing both eqn (1) and (2):Let,
w1=5000π
Since,
w1=5000π 2πf 1=5000π Hence, f 1=2500
Again, Hence,
w2=3000π 2πf 2 =3000π f 2=1500
and
w2=3000π
So, the maximum frequency component present in the given signal is, f 1= 2500 Hence, Nyquist rate =2×f m =2×2500 =5000Hz Nyquist interval = 1/2f m =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, =2f m=2×50=100 And the nyquist interval is, Ts=1/100=0.01 m sec
iii
.
solution:
=1/(πt)2[sin2(100πt)] =1/(πt)2[1-cos(2×100πt)]/2 =1/2(πt)2- 1/2(πt)2 cos (200πt)
Or, Or,
From the term present in the modify input input signal, signal, Cos(200πt).............1 Comparing 1 to the standard eqn x(t)= Acos (wt) Since , w= 200π 2πf=200π f= 100 Hz So, the nyquist rate is,=2f m=2×100 =200Hz and the nyquist interval is ,T s=1/200=5×10-3
i ˅
The standard eqn is,
x(t) = 5cos 1000πt + 1000πt + cos 4000 πt.................(1) x (t) =A 1 cosw1t +A2 cosw2t .................. (2)
Comparing eqn (1) and eqn (2) :-
and
w1=1000π w2=4000π
Since , Or, Hence,
w1=1000π 2πf 1= 1=1000π f 1=500 Hz
Again ,
Or, Hence,
w2= 4000 π 2 πf 2=4000 π f 2= 2= 2000Hz
So, the nyquist rate is,=2f m=2×2000
=4000 And the nyquist interval is ,Ts=1/4000 =2.5×10-4 sec
Q2 A µ-law comparator comparator uses a compressor 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= v i/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 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= Given,µ= 255 Y=
When, x= 0.2,
Or,
Y=
Or,
Y=
where
Y=
= 0.7126
x=0.4,
Y= +
=0.8358
x=0.6, x=0.6,
Y=
= 0.9083 0.9083
x=0.8, x=0.8,
Y=
x=1,
Y=
x=-0.2,
= 0.9599 0.9599 =1 y=
=-0.7126
x=-0.4,
y=
=-0.8358
x=-0.6,
y=
=-0.9083
x=-0.8,
y=
=-0.9599
x=-1, Plot:
,
y=
x versus y graph:-
= -1
x=Vi/V
(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, or,
1/127=
0.043662814 =ln(1+255 x1) ⃓
or,
e0.043662814=1+255 x1
or,
0.044630061=255 x1
⃓
⃓
⃓
⃓
⃓
x1 =1.750×10-3 So, the smallest affective separation between levels is, min=V*x= 40*1.750*10 40*1.750*10-3= 7×10-3 ⃓
Again,
⃓
let x127 is the value of x for which,
y=1-(1/127) y=1-(1/127)
ln ((1+255x127)/ln256) =126/127 ⃓
or, or,
⃓
ln (1+255x127) =.992125984×5.545177444=5.5015463 ⃓
⃓
1+255x1=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 theorem:: The band pass signal x(t) whose maximum bandwidth is 2f m can be completely sampled and recovered from its sample if it is sampled at the minimum rate of twice the bandwidth. Here f m is the maximum frequency component present in the signal. Hence, if the bandwidth of the signal is 2f m, then the minimum sampling rate must be 4f m band pass signal. Proof: Consider a band pass signal whose band width is 2f m and its spectrum is centered on f c.
Let, 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 (2πf ct) and sin(2πf ct) and then suppressing the sum frequencies by means of low pass filters. Thus x I and xQ component contain only low frequency component.
After few mathematical mathematical manipulation, x(t)=
......(1)
and x(t) =
..............(2)
Comparing 1st equation with 2nd eqn (interpolation eqn of low pass signal) , we observed that that x(t) x(t) is repl repla aced ced by by x( x( Here,
x( And
Ts=
)
) = x(nTs) = sampled version of band pass signal
Thus, if 4f m samples per second are taken, then the band pass signal of bandwidth 2f m can be completely recovered from its samples. Hence, for band pass signal of bandwidth 2f m, Minimum sampling rate = Twice of bandwidth =4f m samples per second.
Part- B Q4 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 transmission. 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 f m Hz. This low pass filter blocks all the frequency component above the cut-off frequency. Now signal x(t) is band limited to f m Hz. 2. Sampler: The sampling and hold circuit samples the signals at the freq f s. sampling freq is selected sufficiently above to the nyquist rate, f s
2f m
Quantizer: 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 i s then converted to parallel digital words for each sample. Decoder: 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 samples each of two stereo signals with with a 16 bit analog to digital converter (ADC) at 44.1kb/s. (a) The bit stream stream of digitized digitized data is augmented augmented by the addition addition of error correcting correcting bits, bits, clock extractio extraction n 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 record an hour’s worth of music. music. Determine Determine the number number of bits recorded recorded on a CD. (c) For a comparison, 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 rate,, =2×44.1×10 3×16=1.411 Mb/sec
Due to the 100% overhead output rate is , =2×1.411×10 6b/s =2.822Mb/sec
(b) The no. Of bits recorded on the CD, =2.822×10 6×(3600 sec) =10.16×10 9 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 samples are quantize quantized d into 1024 levels, levels, determin determinee the number number of binary pulses pulses required required to encode each sample. (c) Determine Determine the binary binary pulse pulse rate of the binarybinary- coded coded signal, signal, and the minimum minimum bandwidth bandwidth require required d 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 f s
=9
Hence, f s=45MHZ (ii) The no. of bits or train pulses given by, q = 2v,
Since,
Where q = no. of
levels ,
v=
bits in PCM Hence, 1024 = 2v Or,
Hence
210
=2v
v=10
(iii) The bit rate , r ≥ v fs ≥ 10× 9 × 10 6 = 9 × 107 bits per sec. Ans. So, the transmission bandwidth is , BW ≥ ½ r≥ ½ 9 × 107 ≥45 MHz Ans.
