Homework Title / No. : _2___________________________Course Code : __ECE-304_______

Course Instructor : _Miss RITU_____________________ Course Tutor (if applicable) : ____________

Date of Allotment : ___________ Date of submission:13/03/2010______________

Student’s Roll No.___42__________

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 :

Piyush kamra_

Evaluator’s comments: _____________________________________________________________________

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Content of Homework should start from this page only:

Q1 The TDM system which is used to multiplex the four signals

m1(t)=cos(1000πt), m2(t)=cos(2000πt), m3(t)=cos(3000πt)and m14(t) = cos(4000πt).

(a) If each signal is sampled at the same sampling rate, calculate the minimum sampling rate fs.

(b) What is the commutator speed (in revolutions per second)?(c) What is the commutator which will allow each of the four signals to be

sampled at a rate faster than is required to satisfy the nyquist criterion for individual signal?

Solution:

(a) f1= 1000π/2π Hence, f1 =500Hz f2=2000 π/2π Hence, f2=1000Hz

f3=3000π/2 πHence , f3=1500Hz Since, f4=4000 π/2 π Hence, f4=2000HzSo, the maximum frequency present in the signal is f4=2000Hz. Hence sampling rate fs =2Fm fs = 2*2000=4000Hz(b) Commutator will be equal to the sampling frequencyFs= 2000Hz .ans(c) The minimum commutator speed for nyquist rate to be satisfied is fs =2000Hz

Q2 Consiider the boinary sequence 0100101. Draw the waveforms for the following signaling formats.

(a) Unipolar NRZ signalling format.(b) Bipolar RZ signaling format.(c) AMI (alternate mark inversion) RZ signalling format.

Ans:-

Q3 Compare the PCM, DPCM, DM and ADM.?Solution:

Sr no.

Parameter of Comparison

Pulse code Modulation(PCM)

Delta Modulation(DM)

Adaptive Delta Modulation

Differential Pulse Code Modulation(DPCM)

1. Number of bits

It can use 4, 8 or 16 bits per sample.

It uses only one bit for one sample.

Only one bit is used to encode one sample.

Bits can be more than one but less than PCM.

2. Levels and step size.

The no of levels depends on no of bits. Level size is kept fixed.

Step size is kept fixed and cannot be varied.

According to the signal variation, step size varies (i.e.

Here, fixed no levels are used.

adaptive)3. Quantization

error and distortion.

Quantization error depends on no of levels used.

Slope overload distortion and granular noise are present.

Quantization noise is present but other erres are absent.

Slope overload distortion and quantization noise is present.

4. Transmission bandwidth.

Highest bandwidth is required since no of bits are high.

Lowest bandwidth is required.

Lowest bandwidth is required.

Bandwidth required is lower than PCM.

5. Feedback. There is no feedback in transmitter and receiver.

Feedback exists in transmitter.

Feedback exists.

Here,feedback exists.

6. Complexity of implementation

System is complex.

Simple. Simple. Simple

Part: B

Q4 Explain the Delta modulator (DM). What are the errors in DM? How these errors can be eliminated?Solution:

Delta modulation transmits only one bit per sample. Here the present sample is compared with previous value and this result whether the amplitude is increased or decreased is transmitted. Input signal x(t)is approximated to a step signal by a delta modulator. This step size is kept fixed the difference between the input signal x(t) and its staircase is approximated signal by the delta modulator. Sampled signal x(nets) is applied to the summer and this fed to the one bit quantize that has output b(nets) this output is added with delayed unit impulse u(n-1)Ts) and o/p of delayed cut is fed to the summer cut that produced error signal e(nets) the Delta Modulator transmitter is given below.

The error in Delta Modulation is

1.Slope overload distortion.

2.Granular or Idle Distortion.

(1)Slope overload distortion:-

This distortion arises because of large dynamic range of the input signal. as can be observed from fig ,the rate of rise of input signal(t) is so high that the staircase signal cannot approximately it, the steep size ‘∆’ becomes too small for staircase signal μ(t)to follow the step segment of x(t).hence, there is a large error between the staircase approximated signal and the original input signal x(t).this error or noise is known as slope overload distortion. To reduce this error, the step size must be increased when slope of signal x(t) is high.

Since the step size of delta modulator remains fixed, its maximum or minimum slopes occur along straight lines.therefore,this modulator is also known as linear delta modulator(LDM).

(2)Granular or IDLE Noise

Granular or Idle noise occur when the step size is too large compared to small variations in the input signal.this means that for very small variation in the input signal,the staircase signal is changed by large amount(∆) because of large step size,. Fig shows that when the input signal is almost flat, the staircase signal μ(t) keeps on oscillating by ± ∆ around the signal. The solution to this problem is to make step size small.

Therefore, a large step size is required to accommodate wide dynamic range of the input signal signal(to reduce slope overload distortion) and small steps are required to reduce granular noise.Infact,adaptive delta modulation is the modification to overcome these errors.

Q5 consider a sine wave of frequency fm and amplitude Am, which is applied to a delta modulator of step size ∆. Show that slope –overload distortion will occur if.

Where Ts is the sampling period. What is the maximum power that may be transmitted without slope overload distortion?Ans:-

Let us consider x(t)=Amsin(2 )

It may be noted that the slope of x(t) will be max. When the derivative of x(t)w.r.t. will be

maximum. Maximum slope of =step size/sampling period=

We know that slope overload distortion will take palce if slope of sine wave is greater than the slope of delta modulator that is i.e,

Max|dx(t)/dt|>

Max|dAmsin /dt

Max| Am

Am

Am>

This is the condition for maximum slope overloading.

For, maximum power that may be transmitted without slope overload distortion is = A2

m/2

= ( )2/2

= 2/8π2F2mTs

2 ans.

Q6 consider a low- pass signal with a bandwidth of 3 Khz. A linear delta modulation system, with step size ∆ =0.1V, is used to process this signal at a sampling rate ten times the nyquist rate.(a) Evaluate the maximum amplitude of a test sinusoidal signal of frequency 1 KHz,

which can be processed by the processed by the system without slope-overload distortion.

(b) For the specification is given in part (a), evaluate the output signal-to-noise ratio under (i) pre filtered (ii) post filtered conditions.

Ans:-(a) given, = 0.1v

And BW= 3khz

Maximum allowable amplitude of the i/p signal is, Amax=

wmTs

=o.1×Fs/2π×103

=60×o.1/2π=0.96 volt

(a)Signal to noise ratio(post filtered conditions):

(SNR)0= (S/N)o

=3fs3/8 π2×106

=3*(2*10*3*103)3/8 π2×106

=8210.1=20log(8210.11)=180.26db Ans.