· Web viewLATHA MATHAVAN ENGINEERING COLLEGE MADURAI-625301 DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING QUESTION BANK WITH ANSWER SUBJECT NAME: EC6502/PRINCIPLES OF

PREPARED BY S.JANARTHANAN AP/ECELATHA MATHAVAN ENGINEERING COLLEGE

MADURAI-625301DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING

QUESTION BANK WITH ANSWER

SUBJECT NAME: EC6502/PRINCIPLES OF DIGITAL SIGNAL PROCESSING

YEAR/SEM: III/V REVISED SYLLABUS (R2013)

TWO MARKS WITH ANSWER

UNIT –I

DISCRETE FOURIER TRANSFORM

1. Define: Discrete Time signalsA discrete signal or discrete-time signalis a function defined only at particular time in stands.it is discrete in time but continuous in amplitude. An example is temperature recorded at regular intervals of time in a day.

2. Define: Discrete time systemA discrete time system is a device or algorithm that operates on a discrete time input signal x(n),According to some well-defined rule, to produce another discrete time signal y(n) called the output signal

3. Whatare the classifications of Discrete time signals? Energy and power signals Periodic and aperiodic signal Symmetric(Even) and Symmetric (Odd) signals Causal and non-causal signals.

4. Whatare the classificationsof Discrete Time systems? Static and dynamic systems Time variant and time in variant systems Linear and non linear Stable and unstable systems Causal and non-causal systems

5. What are the different types of operations performed on discrete time signals? Delay of a signal Advance of a signal Folding or reflection of a signal Time scaling Amplitude scaling Addition of signals Multiplication of signals

6. What is the time invariant system? If the input-Output relation of a system does not vary with time, the system is said to be time invariant or shift invariant.

PREPARED BY S.JANARTHANAN AP/ECE

7.What are energy and power signals?

The energy of a discrete time signal x(n) is defined as E=∑n=∞

∞

¿ x (n )∨¿2¿

A signal x(n) is called an energy signal if and only if the energy obeys the relation 0<E<∞

For an energy signal P=0

For average power of a Discrete time signal x(n) is defined as

P= limN→∞

12N+1 ∑

n=∞

∞

¿ x (n )∨¿2 ¿

8. What is a causal system?A system is said to be causal if the output of the system at any time n depends only on present and past input, but does not depend on future inputs.

9. Define DFTof a discretetime sequence.

The DFTisusedtoconvert a finite discrete time sequence x(n)toan N-point frequencydomain sequence denotedbyX(k). The N-pointDFTof a finiteduration sequencex(n) of lengthL, where L≤ Nisdefined as.

X (K )=∑n=0

N−1

x (n )e− j2 π kn /N k=1 ,2 , .. . ,N−1

10. Define IDFT.

The IDFTisusedtoconvert the N-pointfrequencydomainsequence X(k) to an N-pointtime domainsequence. The IDFTofthesequence X(k)oflengthn is defined as

x (n)= 1N ∑

k=0

N−1

X (K )e j2 π kn /N Where n=0,1,2……..N-1

11. Write twoapplications of DFT.

(a)The DFTis usedfor spectral analysis of signals using a digital computer.

(b)The DFTis used to performfiltering operations on signals using digital computer.

12.WhatisZero Padding. Why itis needed.AppendingZerostoasequenceinordertoincreasethesizeorlengthofthe

sequenceiscalledZeroPadding.Incircularconvolution,whenthetwoinput sequences aredifferentsize, then theyareconverted to equalsize byzero padding.


13.Comparethe OverlapAddandOverlap Save method ofsectioned convolutions.

Overlap Add method Overlap Save method

1)Linear convolution of each sectionoflongersequencewith smaller sequence is performed.

2)ZeroPaddingisnot required.

3)Overlappingofsamplesofinput sectionsare notrequired.

4)Theoverlappedsamplesinthe outputofsectionedconvolutions areaddedtogettheoverall output.

Circularconvolutionofeachsectionof longersequencewithsmallersequenceis performed.(Afterconvertingthemtothe size ofoutput sequence)ZeroPaddingisrequiredtoconvertthe inputsequencestothesizeoftheoutput sequence.

TheN2-1samplesofaninputsectionof longersequenceisoverlappedwithnext inputsection.Dependingonthemethodofoverlapping theinputsamples,eitherthelastN2-1 samplesorthefirstN2-1samplesofthe output sequence of each sectioned convolutionsare discarded

14. WhatisFFT.TheFFTisamethodforcomputingtheDFTwithreducednumberofcalculations.

Thecomputationalefficiencyisachievedbydivideandconquerapproach.Thisis basedon the decompositionof an N–point DFTintosuccessivelysmaller DFT’s.

15. Whatisradix –2FFT.Theradix–2FFTisanefficientalgorithmforcomputingN–pointsequence.In Radix–2FFT theN–

pointsequence isdecimatedinto2–pointsequencesandthe 2–pointDFTforeachdecimatedsequenceiscomputed.Fromtheresultsof2– pointDFT’s, the 4–pointDFT’sare computed.Fromtheresultsof 4–pointDFT’s, the8 – pointDFT’s are computedand so on until we get N –point DFT.

16. WhatisDITradix– 2 FFT.TheDecimationinTime(DIT)radix–2FFT isan efficientalgorithmforcomputing DFT.InDITradix–

2FFT,thetimedomainN–pointsequenceisdecimatedinto2 –pointsequences.Theresultsof2–pointDFT’sareusedtocompute4–point DFT.Twonumbersof2–pointDFT’sarecombinedtoget4–pointDFT.The resultof4–pointDFT’sareusedtocompute8–pointDFT.Twonumbersof4– pointDFT’s are combined to get a8–pointDFT.This processiscontinued untilweget N– pointDFT.

17. Whatisphase factoror twiddle factor.


ThecomplexnumberWNiscalledphasefactorortwiddlefactor.TheWNrepresent a complexnumber e-j2π/n.

It also represent an Nthroot of unity.

18.WhatisDIF radix –2 FFT.TheDecimationinFrequency(DIF)radix–2FFTisanefficientalgorithmfor

computingDFT.InthisalgorithmtheN–pointtimedomainsequenceisconverted intotwonumbersofN/2pointsequences.TheneachN/2pointsequenceis converted to twonumbersofN/4point sequences.This processiscontinued until we getN/2 numbersof 2 –point sequences. Now the2– pointDFT’s of N/2 numbersof 2–pointsequenceswillgivensamples,whichistheN–pointDFTofthetime domainsequence.HeretheequationsformingN/2pointsequences,N/4point sequences etc.; are obtained bydecimation of frequencydomain sequences.

19.Comparethe DIT and DIF radix–2 FFT.

DITradix – 2 FFT. DIF radix– 2FFT.

1.Thetimedomainsequenceis decimated.2.Whentheinputisinbitreversed

order,theoutputwillbeinnormal order andvice versa.

3.Ineachstageofcomputations,the phasefactorsaremultipliedbefore add and subtractoperations.

4.The value of N should be expressed

suchthatN=2mandthisalgorithm consists of m stages of computations.

5.Total number of arithmetic operationsareNlog2Ncomplex additionsand(N/2)log2Ncomplex multiplications.

The frequency domain sequence is decimated.Whentheinputisinbitnormalorder,the outputwillbeinbitreversedorderandvice versa.

Ineachstageofcomputations,thephase factorsaremultipliedafteraddandsubtract operations.

ThevalueofNshouldbeexpressedsuch

thatN=2mandthisalgorithmconsistsofm stages of computations.

Totalnumberofarithmeticoperationsare Nlog2Ncomplexadditionsand(N/2)log2N complexmultiplications.

20.Distinguish between linear convolutionandcircular convolutionof two sequences.

Linear Convolution Circular Convolution1.Ifx(n)isasequenceofLnumberof samplesandh(n)withmnumberof

Ifx(n)isasequenceofLnumberof samplesandh(n)withmnumberof


samples,afterconvolutiony(n)will contain N=L+M –1samples.

2.Linearconvolutioncanbeusedtofind the responseof a linear filter.

3.Zeropaddingisnotnecessarytofind the responseof a linear filter.

samples,afterconvolutiony(n)will contain N=Max(L,M)samples.Circularconvolutioncanbeusedtofind the responseof a linear filter.Zeropaddingisnecessarytofindthe response of alinear filter.

21.List any four properties of DFT.

Let DFT{x (n)} =X (K), DFT{x1 (n)} =X1 (K), DFT{x2 (n)} =X 2(K)

i) Periodicity: X (K+N) =X (K) for all K.ii) Linearity: DFT[a1 x1 (n)+a2 x2(n)]=a1 X1 (K)+a2 X2 (K)iii) DFT of time reversed sequence: DFT[ x(N-n)]=X(N-K)iv) Circular convolution :DFT[x1(n)ּx2(n)]=X1(K) X2(K)

22. How many multiplication and addition are involved in radix- 2 FFT?The total number of complex addition is N log N and the total numbers of complex multiplication

are N/2log N.

23.How many multiplication and addition are involved in DFT?The total numbers of complex additions are N (N-1) and the total number of complex multiplications

are N 2

24.Calculate the number of multiplications needed in calculation of DFT and FFT with 64 point Sequence.

DFTThe number of complex multiplications required using direct computation is

N2=642=4096FFT

The number of complex multiplication required using FFT is N/2logN=64/2log64=19225. What is the difference between DFT and DTFT? (MAY/JUNE 2009)DFT * Obtained by performing sampling operation in both the time and frequency domain * Discrete frequency spectrumDTFT * Sampling is performed only in time domain * Continuous function of frequency spectrum

PREPARED BY S.JANARTHANAN AP/ECEUNIT II

IIR FILTER DESIGN

1. Definean IIR filter.The filtersdesignedbyconsidering allthe infinitesamplesof impulse response are calledIIR filters.

Theimpulseresponse isobtainedbytaking inverse Fouriertransform ofideal frequencyresponse.

2. Compare the digitaland analog filter.Digital Filter Analog Filter

1.Operates on digitalsamplesof the signal.2.Itis governedbylinear difference equation.3.Itconsistsof adders,multipliers and delays implemented in digital logic.

Operates on analogsignals (or actual signals)Itis governedbylinear difference equation.Itconsistsof electrical components like resistors, capacitors andinductors.

3. Whatis impulseinvarianttransformation?The transformation of analog filter todigital filter withoutmodifythe impulse response ofthe

filteriscalledimpulse invarianttransformation.Inthis transformationtheimpulse responseofthedigital filter willbe sampled version of theimpulse responseof the analog filter.

4. Whatis bilinear transformation?The bilineartransformation is conformal mapping that transforms the s– planetoz – plane. In this

mapping the imaginaryaxis ofs–plane is mapped into theunit circleinthe z –plane, theleft halfof s –planeismapped into interiorof unitcircle in z plane and the righthalf ofs– planeismapped into exterior of unit circle in z– plane. The bilinear mappingisaone to one mapping and it is accomplished when

S = 2/T(1 –z-1/1 + z-1)

5. Whatisfrequencywarping.Inbilineartransformation the relationbetweenanalog anddigital frequencies is non-linear. When the s

–planeismapped intoz–plane using bilinear transformation,this non-linear relationshipintroduces distortioninfrequencyaxis, whichiscalled frequency warping.

6.Whatare the advantagesand disadvantages ofbilinear transformation. A d v an ta g esofbi li n e ar tr a nsfor m ation i. The bilineartransformation is one–to-one mapping.ii. There is no aliasing andsothe analog filterneed nothave a bandlimited frequencyresponse.iii. The effect of warping on amplitude response canbe eliminatedbyprewarping theanalogfilter.iv. Itcan beusedtodesigndigital filters withprescribed magnituderesponse with piecewise constantvalues.D i s a dva n t a g e s of b il i neartr a n s f or m ation

i.Thenonlinear relationshipbetween analoganddigital frequencies introduces frequencydistortionwhich is called frequency warping.ii.Usingbilinear transformation, a linearphase analog filter cannotbe transformed tolinear phasedigital filter.

7.Whatisprewarping. Whyit isemployed.

PREPARED BY S.JANARTHANAN AP/ECEInIIR filterdesignusing bilineartransformation, the conversionof thespecified digital frequencies

toanalog frequenciesiscalledprewarping. Itis necessaryto eliminatethe effect of warping on amplitude response.

8.Explainthetechniqueof prewarping.InIIR filterdesignusing bilineartransformation, the conversionof thespecified digital frequencies toanalog frequenciesiscalledprewarping.Usingthe prewar frequencies, the analog filter transfer function is designed andthen itistransformedtodigital filter transfer function.

9. Comparethe impulse invariant andbilinear transformationsImpulse Invariant Transformation Bilineartransformation

i. Itis many-to-one mapping

ii. The relationbetweenanalog and digital frequencyislinear.iii. To preventthe problemof aliasing theanalogfilters should be band limited.iv. The magnitude and phase response ofanalog filter can be preservedbychoosing low samplingtime or high sampling frequency.

Itis one-to-one mapping

The relationbetweenanalog anddigital frequencyis non-linear.Isno problemof aliasingandso the analog filters neednot bebandlimited. Due to the effectof warping, thephase response of analogfilters cannotbe preserved. But the magnitude response canbe preservedbyprewarping.

10. WhatisButterworthapproximation.InButterworthapproximation, the error functionisselected such that the magnitude ismaximallyflatin

the origin (i.e.,at Ω =0) andmonotonicallydecreasing with increasing Ω.

11. WhatisChebyshevapproximation.InChebyshevapproximation, the approximationfunction is selected such that the

errorisminimizedover a prescribedbandof frequencies.

12. Whatare thedifferent typesof structures for realizationof IIRsystems.The differenttypes ofstructures forrealizationof IIRsystemare.

i. Direct – formIstructure,ii. Direct – formIIstructure,iii. Transposed Direct – formIIstructure,iv. Cascade formstructure,v. Parallel formstructure,

13. Why impulse invariant method is not preferred in the design of IIR filters other than low pass filter?

In this method the mapping from s plane to z plane is many to one. i.e. ,all the poles in the s plane between the intervals (2k-1)∏/T to (2k+1)∏/T .Thus there are an infinite number of poles that map to the same location in the z plane, producing an aliasing effect. Due to spectrum aliasing the impulse invariant method is inappropriate in designing high pass filters. That is why the impulse method is not much preferred. in the design of IIR filters other than low pass filter


14. Give any properties of butterworth filters.i)The magnitude response of butterworth filter closely approximates the ideal response as the order N increasesii) The magnitude response of butterworth filter decreases monotonically as the frequency Ώ increases from 0 to ∞iii)the Poles of the butterworth filter lies on a unit circle.

15. What are the disadvantages of IIR filters (compared to FIR filters)? They are more susceptible to problems of finite-length arithmetic, such as noise generated by

calculations, and limit cycles. (This is a direct consequence of feedback: when the output isn't computed perfectly and is fed back, the imperfection can compound.)

They are harder (slower) to implement using fixed-point arithmetic. They don't offer the computational advantages of FIR filters for multirate (decimation and

interpolation) applications.

16.What are the properties of chebyshev filter?1. The magnitude response of the chebyshev filter exhibits ripple either in the stop band or the pass band.

2. The poles of this filter lies on the ellipse

17. Differentiate Butterworth and Chebyshev filter. MAY/JUNE 2006 Butterworth damping factor 1.44 chebyshev 1.06

Butterworth flat response chebyshev damped response

UNIT IIIFIR FILTER DESIGN

1. What are the advantages of FIR filter?1. FIR filter have exact linear phase.2. FIR filters are always stable.3. FIR filter can be realized in both recursive and recursive structures.4. Excellent design methods are available for various kinds of FIR filter.5. FIR filter are free of limit cycle oscillation, when implemented on a finite word length digital system.

2. What are the disadvantages of FIR filter?i) Memory requirement and execution time are very high.

ii The implementation of narrow transition band fir filters is very costly, as it requires considerably more arithmetic operation and hardware components such as multipliers, adders and delay elements.

3. What is the necessary and sufficient condition for the linear phase characteristic of a FIR filter?

http://www.dspguru.com/info/faqs/mrfaq.htm

PREPARED BY S.JANARTHANAN AP/ECEThe phase function should be a linear function of w, in which requires constant group delay and

phase delay.θ(w)=-ζw

for satisfying above conditionh (n)=h(N-1-n)

i.e. The impulse response must be symmetrical about ζ =(N-1)/2

If only constant group delay is desired thenθ(w)=β-ζw

for satisfying above conditionh (n)=-h(N-1-n)

i.e. The impulse response must be symmetrical about ζ =(N-1)/2

4. Distinguish IIR and FIR filters

5. List the well-known design technique for linear phase FIR filter design?1. Fourier series method 2. Window method3. Frequency sampling method6. What is Gibbs phenomenon? OR What are Gibbs oscillations?

One possible way of finding an FIR filter that approximates H d(e j)would be to truncate the infinite Fourier series at n= (N-1/2).Abrupt truncation of the series will lead to oscillation both in pass band and is stop band .This phenomenon is known as Gibbs phenomenon.

7. Give the equation specifying rectangular window function.The weighing function for the rectangular window is given by WR (n) = {1 -(N-1)/2≤n≤(N-1)/2 0 otherwise

8. Give the equation specifying hamming window function.The weighing function for the rectangular window is given by

WH (n) = { 0.54+0.46cos (2 П n /N-1) - (N-1)/2≤n≤ (N-1)/2 0 otherwise

9. Compare Hamming window with Rectangular Window.

FIR IIRImpulse response is finite

They have perfect linear phase

Impulse Response is infinite

They do not have perfect linear phaseNon recursive Recursive

Greater flexibility to control the shape of magnitude response

Less flexibility, usually limited to specific kinds of filters.

PREPARED BY S.JANARTHANAN AP/ECEHamming window Rectangular Window

1.The main lobe width is equal to 8/N and the peak side lobe level is –41dB.

2.The low pass FIR filter designed will have minimum stop band attenuation of –53 dB


2.The low pass FIR filter designed will have minimum stop band attenuation of –21dB

10. Compare Hamming window with Kaiser Window.Hamming window Kaiser window


2.The low pass FIR filter designed will have first side lobe peak of –53 dB

The main lobe width, the peak side lobe level can be varied by varying the parameter and N.

The side lobe peak can be varied by varying the parameter.

11.What is the reason that FIR filter is always stable?

FIR filter is always stable because all its poles are at the origin.

12. What condition on the FIR sequence h (n) are to be imposed N order that this filter can be called a liner phase filter? The conditions are

Symmetric condition h(n)=h(N-1-n)Anti-symmetriccondition h(n)=-h(N-1-n)

13. What is the principle of designing FIR filter using frequency sampling method?In frequency sampling method, a set of sample is determined from the desired frequency response

and are identified as discrete Fourier transform coefficients. The inverse discrete Fourier transform of this set of samples then gives the filter coefficients. The set of sample points used in this procedure can be determined by sampling a desired frequency response Hd(ejw) at N points wk, k=0,1,….N-1 uniformly spaced around the unit circle.

14. Write the steps involved in FIR filter design.i. Choose the desired frequency response Hd(ɷ).ii. Take inverse Fourier transform of Hd(ɷ) to get hd(n).iii. Convert the infinite duration hd(n) to finite duration sequence h(n).iv. Take Z - transform of h(n) to get the transfer function H(z) of the FIR filter.


15. What are the possible types of impulse response for linear phase FIR filters.i. Symmetric impulse response when N is odd.ii. Symmetric impulse response when N is even.iii. Anti-Symmetric impulse response when N is odd.iv. Anti-Symmetric impulse response when N is even.

16. What are the desirable characteristics of the frequency response of window function. The desirable characteristics of the frequency response of window function is i. The width of the main lobe should be small and it should contain as much of the total energy as possible. ii. The side lobes should decrease in energy rapidly as ɷ tends to π.

17.Write the characteristics of rectangular window. i. The main lobe width is equal to 4π/N.ii. The maximum side lobe magnitude is – 13dB.iii. The side lobe magnitude does not decreases significantly with increasing ɷ.

18. List the features of FIR filter designed using rectangular window.

i. The width of the transition region is related to the width of the main lobe of window spectrum.ii. Gibb’s oscillations are noticed in the pass band and stop band. iii. The attenuation in the stop band is constant and cannot be Varied.

19. Write the characteristics features of triangular window.a. The main lobe width is equal to 8π/N.b. The maximum side lobe magnitude is – 25dB.c. The side lobe magnitude slightly decreases with increasing ɷ.

20. List the features of hanning window spectrum.a. The main lobe width is equal to 8π/N.b. The maximum side lobe magnitude is – 31dB.c. The side lobe magnitude decreases with increasing ɷ.

21.What is the advantage in direct form – I structure when compared todirect form – II structure.In direct form – II structure the number of delay element required isexactly half that direct form – I

structure when the number of poles andzeros are equal. Hence it requires less amount of memory

22.What are the difficulties in Cascade realization.The difficulties in Cascade realization.a. Decision of pairing poles and zeros.b. Deciding the order of cascading the first and second ordersections.c. Scaling multipliers should be provided between individualsections to prevent the filter variables from becoming toolarge or too small.

23.What are the advantages and disadvantages of digital filters?

PREPARED BY S.JANARTHANAN AP/ECEAdvantages of digital filters

High thermal stability due to absence of resistors, inductors and capacitors. Increasing the length of the registers can enhance the performance characteristics like accuracy,

dynamic range, stability and tolerance. The digital filters are programmable. Multiplexing and adaptive filtering are possible.

Disadvantages of digital filters The bandwidth of the discrete signal is limited by the sampling frequency. The performance of the digital filter depends on the hardware used to implement the filter.

UNIT IVFINITE WORDLENGTH EFFECTS

1.What is meant by fixed point number?In fixed point number the position of a binary point is fixed. The bit to the right representthe

fractional part and those to the left is integer part.

2. What are the different types of fixed point arithmetic?Depending on the negative numbers are represented there are three forms of fixed pointarithmetic.

They are sign magnitude,1’s complement,2’s complement

3. What is meant by sign magnitude representation?For sign magnitude representation the leading binary digit is used to represent the sign.If it is equal

to 1 the number is negative, otherwise it is positive.

4. What is meant by block floating point representation? What are its advantages?In block point arithmetic the set of signals to be handled is divided into blocks. Eachblock has the

same value for the exponent. The arithmetic operations with in the block uses fixed point arithmetic & only one exponent per block is stored thus saving memory. This representation of numbers is more suitable in certain FFT flow graph & in digital audio applications.

5. What are the advantages of floating point arithmetic?1. Large dynamic range2. Over flow in floating point representation is unlike.

6. What are the three-quantization errors to finite word length registers in digitalfilters?NOV/DEC 20041. Input quantization error2. Coefficient quantization error3. Product quantization error

7. What do you understand by input quantization error?

PREPARED BY S.JANARTHANAN AP/ECEIn digital signal processing, the continuous time input signals are converted intodigital using a b-bit

ACD. The representation of continuous signal amplitude by a fixeddigit produce an error, which is known as input quantization error.

8. How the multiplication & addition are carried out in floating point arithmetic?In floating point arithmetic, multiplication are carried out as follows,Let f1 = M1*2c1 and f2 =

M2*2c2. Then f3 = f1*f2 = (M1*M2) 2(c1+c2) That is, mantissa is multiplied using fixed-point arithmetic and the exponents are added. The sum of two floating-point number is carried out by shifting the bits of the Mantissa of the smaller number to the right until the exponents of the two numbers areequal and then adding the mantissas.

9. What do you understand by 2's complement representation?In two's complement representation positive number are represented as in sign magnitude & 1's

complement. The negative number is obtained by complementing all the bits of the positive number & adding one to the least significant bit. For example (0.5625)10 = (0.100100)2

(-0.5625)10 = 1.011011 0.000001 -------------

(1.011100)2

10. What do you understand by input quantization error? In digital signal processing, the continuous time input signals are converted into digital using a b-bit

ACD. The representation of continuous signal amplitude by a fixed digit produce an error, which is known as input quantization error.

11. What is product quantization error?Product quantization error arises at the output of a multiplier. Multiplication of b bit data with a b bit

coefficient result a product having 2b bits. Since b bit register is used the multiplier output must be rounded or truncated to b bits, which produce an error. This error is known as quantization error.

12. What is coefficient quantization error?The filter coefficient is computed to infinite precision theory. But in digital computation the filter

coefficient are represented in binary and are stored in registers.If b bit register is used the filter coefficient must be rounded or truncated to b bit which produce an

error. Due to quantization of coefficient the frequency response of filter may differ appreciably from the desired response and some time the filter may actually fail to meet the desired specifications. If the poles of desired filter are close to the unit circle, then those of the filter with quantized coefficients may lie just outside the unit circle leading to unstability.

13. What are the different quantization methods? The common methods of quantization are

1. Truncation 2. Rounding


14. What is the relationship between truncation error e(n) and the bits b for representing a decimal into binary? For a 2's complement representation, the error due to truncation for both positive and negative values of x is -2-b≤ e (n) ≤0 Where b is the number of bits

15. What is meant rounding? Discuss its effect on all types of number representation? Rounding a number to b bits is accomplished by choosing the rounded result as the b bit number closest to the original number unrounded. For fixed point arithmetic, the error made by rounding a number to b bits satisfy the inequality -2-b2-b

----- ≤ e (n) ≤ -------- 2 2

for all three types of number systems, i.e., 2's complement, 1's complement & sign magnitude.

For floating point number the error made by rounding a number to b bits satisfy the inequality -2-b≤e (n) ≤ 2-b where e (n) =xT-x

16. What is truncation?Truncation is a process of discarding all bits less significant than least significant than least significant bit that is retained.0.00110011 to 4 bit 0.0011.

17. What is meant by A/D conversion noise? A DSP contains a device, A/D converter that operates on the analog input x (n) to produce xq(n) which is binary sequence of 0s and 1s. At first the signal x(t) is sampled at regular intervals to produce a sequence x(n) is of infinite precision. Each sample x(n) is expressed in terms of a finite number of bits given the sequence xq(n). The difference signal e (n)=xq(n)-x(n) is called A/D conversion noise.

18. What is the effect of quantization on pole location? Quantization of coefficients in digital filters lead to slight changes in their value. This change in

value of filter coefficients modify the pole-zero locations. Some times the pole locations will be changed in such a way that the system may drive into instability.

19. Which realization is less sensitive to the process of quantization?Cascade form.

20. What is meant by quantization step size?Let us assume a sinusoidal signal varying between +1 and -1 having a dynamic range 2. If the ADC

used to convert the sinusoidal signal employs b+1 bits including sign bit, the number of levels available for quantizing x(n) is 2 b+1. Thus the interval between successive levels


q = 2 / 2 b+1 =2-b

Where q is known as quantization step size.

21. How would you relate the steady-state noise power due to quantization and the b bits representing the binary sequence? Steady state noise power

σe2 =2-2b/12

Where b is the number of bits excluding sign bit.

22. What are the two kinds of limit cycle behavior in DSP? 1. Zero input limit cycle oscillations 2. Overflow limit cycle oscillations

23. What is overflow oscillation? The addition of two fixed-point arithmetic numbers cause over flow the sum exceeds the word size

available to store the sum. This overflow caused by adder make the filter output to oscillate between maximum amplitude limits. Such limit cycles have been referred to as over flow oscillations.

24. What is meant by limit cycle oscillation?For an IIR filter, implemented with infinite precision arithmetic the output should approach zero in

the steady state if the input is zero .however the non-linearity due to the finite precision arithmetic operation often cause periodic oscillations to occur in the output. Such oscillation in recursive systems are called zero input limit cycle oscillations.

25. What are the methods used to prevent overflow? There are two methods used to prevent overflow

1. Saturation arithmetic 2. Scaling

26. Define "dead band" of the filter The limit cycle occur as a result of quantization effect in multiplication. The amplitudes of the

output during a limit cycle are confined to a range of values called the dead band of the filter.

27. Explain briefly the need for scaling in the digital filter implementation. To prevent overflow, the signal level at certain points in the digital filter must be scaled so that no

overflow occurs in the adder.

28. Why the limit cycle problem does not exist when FIR digital filter is realized in direct form or cascade form?

In case of FIR filter there are no limit cycle oscillations, if the filter is realized in direct form or cascade form since these structures have no feedback.

29. Why rounding is preferred to truncation in realizing digital filter? 1. The quantization error due to rounding is independent of type of arithmetic. 2. The mean of rounding error is zero.

PREPARED BY S.JANARTHANAN AP/ECE 3. The variance of the rounding error signal is low.

30. What is the steady state variance of the noise in the output due to quantization of the input for the first order filter? y(n)=ay(n-1)+x(n)σe

2 =(2-2b/12)(1/1-a2)

31. Compare the fixed point and floating point arithmetic.

Fixed point arithmetic. Floating point arithmetic.

Fast operation Slow operation

Relatively economical More expensive because of costlier hardware.Small dynamic range Increased dynamic rangeRound off error occur only for addition Round off error occur with both addition and

multiplication.Overflow occur in addition Overflow occur in addition

32. Express the fraction 7/8 and -7/8 in sign magnitude, 2’s complement and 1’s complement.Fraction (7/8) = (0.111)2 in sign magnitude, 2’s complement and 1’s complement.

Fraction (-7/8)= (1.111)2 in sign magnitude (1.000)2 1’s complement. (1.001)2 2’s complement.

UNIT V DSP APPLICATIONS

1. Define sampling rate conversion.

Sampling rate conversion of digital signal can be obtained by converting the digital signal into analog form (D/A) and then re sampling the resulting (A/D) conversion.

2. Define decimation

.The process of reducing the sampling rate by a factor D is known as decimation or down sampling.

3. What is interpolator?[APR/MAY 2010]

Interpolator is also known as up sampler.The process of sampling rate conversion in digital domain can be viewed as linear filtering operation.

4. What is a decimator?

Decimator is also known as down sampler. It reduces sampling rate.

PREPARED BY S.JANARTHANAN AP/ECE5. Mention two applications of multirate signal processing.[APR/MAY 2010]

•Phase shifter

•Transmultiplexers

•Vocoder

6. What do you mean by sub band coding?[NOV/DEC 2010]

Sub band coding is a method where speech signal is sub-divided into several frequency bands and each band is digitally encoded and separated by allocating different bits per sample to the signal of different sub-bands.we can achieve a reduction in a bit rate of the digitalized bit signal. It is very useful to reproduce speech signal efficiently.

7. What is an anti-imaging filter?

The filter which is used to remove the image spectra is known as anti-imaging filter.

8. What is trans multiplexers?

Trans multiplexers is used to convert frequency division multiplexed signals into time division multiplexed signals and vice versa.

9. Mention the two stages of musical sound processing.

First stage: the sound from the singer or the sound from the instrument is recorded on a single track of multi track tape.Second stage: the special audio effects are added to this sound. The special audio effects can be generated by DSP.

10.What are the steps to be followed for the reproduction of the recorded signal?

•Decoding and demodulation

•Error correction and demultiplexing.

11. Define interpolation.

The process of increasing the sampling rate by a factor I is known as interpolation.

12. What is linear filter?

Linear filter process time varying input signals to produce output signals subject to the constraint of linearity. It is also used in statistics and data analysis.

13. Define multirate signal processing systems.[NOV/DEC 2010]

The systems that employ multiple sampling rates in the processing of digital signals are known as multi rate signal processing.

14. What are the two methods used for sampling rate conversion?

First method:The digital signal is converted into analog signal by using DAC. Then analog signal is converted into digital signal using ADC.Secondmethod:sampling rate conversion is performed in digital domain.

PREPARED BY S.JANARTHANAN AP/ECE15. Define aliasing.

Aliasing refers to an effect that causes different signals to become indistinguishable when sampled. It also refers to the distortion when the signal reconstructed from samples is different from the original continuous signal.

16. What are the various advantages of MDSP?

i) Computational requirements are less

ii) Storage for filter co-efficients are less

iii) Finite arithmetic effects are losses

iv)Filter order required in multirate applications are low

UNIT I

DISCRETE FOURIER TRANSFORM

PREPARED BY S.JANARTHANAN AP/ECE16MARKS WITH ANSWER

1. Discuss various properties of DFT.

1. Linearity : The DFT obeys the law of linearity. If

then for any two constants a and b,

2. Periodicity

3. Circular shift of a sequence : This property is analogous to the time shiftingproperty of the DFT, but with some difference.Let us consider a sequence x (n) of length N which is defined for 0 n N - 1. The sample value of such sequence is zero for n <0 and n N. For any arbitrary integer k, the shift sequence x1 (n) = x (n — k) is no longer defined for the range 0 n N - 1. Therefore we need to define another type of shift that will always keep the shifted sequence in the range 0 n N - 1.This shift is known as circular shift that can be represented as the index modulo N. Thus we can write,

where x (n) is represented the circular shift of x (n). Or more generally we candefine


4. Circular convolution and multiplication of two DFTs : Consider two finite duration sequences both of length, with their N - point DFTs and i.e.,

Here we wish to determine the sequence x3 (n) for which the DFT is,

The brackets term has the form,


The equation has the form of convolution sum. However it differs from a linear convolution of x1 (n) and x2 (n) as defined in unit-TI. In linear convolution the computation of the sequence x3 (n) involves multiplying one sequence by a folder and linear shifted version of the other and then summing the values of the product x1 (m) x2 (n - m) for all values of n.Instead the convolution sum in eqn.the second sequence is circularly time reversed and circularly shifted w.r.t. of first. The equation is called circular convolution of two finite duration sequences.Thus we conclude that multiplication of the DFTs of two sequences is equivalent to the circular convolution of the two sequences in time domain. The operation of forming

where, denotes the circular convolution of two N length sequences, x1 (n) and x2 (n).

5. Time reversal of a sequence:

Hence, when the N-point sequence is reverse in time, it is equivalent to reversing the DFT values. The time reversal is illustrated in Fig.

PREPARED BY S.JANARTHANAN AP/ECEIf we change the index from n to m by defining m = N - n, then,

7. Circular time shift:

Proof:

If we change the index from n to m = N + n - 1, then

This Equation can be written as


8. Multiplication of two sequences:

The proof of this property is similar to circular convolution.

2. State and prove Circular correlation and parsevals theorem1. Circular correlation: For complex valued sequence x (n) and y (n).

where is the circular cross-correlation sequence, given as,

Proof.

We know circular convolution of the two sequences is just equal to the multiplication

PREPARED BY S.JANARTHANAN AP/ECEof the their DFTs and from complex conjugate property

2. Parseval’s theorem:For the complex-valued sequence x (n) and y (n), if

3. Develop a Radix-2, 8-point DIF FFT algorithm with neat flow chart.

Ans. Decimation in frequency stands for splitting the sequences in terms of frequency. That means we have split output sequences into smaller subsequences. This decimation is done as follows.

First stage of decimation:

We can divide the stmmation into two parts as follows.

Now consider the second summation that means


Putting this value in 2equatioñ (2), we get

Taking the summation common we get


Here we have no split the sequence in terms of frequency. So we will split X(k) in terms of even numbered and add numbered DFT.co-efficients. Let X(2r) represents even numbered DFT and X(2r + 1) represents odd numbered DFT.

Thus puffing k = 2 r in eq (4), we will get even numbered sequence

By puffing k = 2r + 1 in eq. (4), we will get odd numbered sequence

Here ‘r’ is an integer similar to k and it varies from 0 to N/2 -1

Puffing these values in eq (5) & (6) we get

Puffing these values in eq. (9) & (10) we get


Putting these values in eq. (13) & (14) we get

Note that at this stage we have decimated the sequence of ‘N’ pomt DFT mto two N/2

point DFT’s given by eq (17) & (18) Let us consider an example of 8 pomt OFT That

means N = 8 So considermg (17) & (18) (that means N/2=4) we can obtain N (8-point)

DFT. This is first stage of decimation. Note that eq. (17) indicate 4 (N/2) point OFT of IN\

g(n) and eq (18) indicates 4 (N/2) pomt DFT of h(n) For 8 pomt OFT eq (15) becomes

Here we are computing ‘4’ point DFT, So range of ‘n’ is n=0 to n=3. Putting these

values in eq. (19), we get

Using equations (20) & (22) & eq. (17) & (18) we can draw the flow graph of the


first stage of decimation as shown in fig. below.

Second stage of decimation: In the first stage of decimation we have used 4-point DFT. We can further decimate the sequence by using 2 point DFT. The second stage of decimation is shown in fig below.

Third stage of decimation : In the second stage of decimation we have used 2- point DFT. So further decimation is not possible. Now we will use a butterfly structure to obtain 2-point DFT. Thus the total flow graph of 8 point DIF-FFT is shown below.


4. Discuss DIT and DIF algorithms and also compare the two algorithms.

Ans. Radix-2 Decimation in Time (DIT) Algorithm (DIT FFT) : To decimate means to break into parts. Thus DIT indicates dividing (splitting) the sequence in time domain. The different stages of decimation are as follows:

First stage of decimation Let x(n) be the given input sequence containing ‘N’ samples. Now for decimation in time we will divide. x (n) into even and odd sequences.

Input sequence x(n) has ‘N’ samples. So after decimation;f1(m) and f2(m) will contain N

j- samples.

Now according to the definition of DFT,

Since we have divided x(n) into two parts, wq can write separate summation for even and odd sequences as follows:

PREPARED BY S.JANARTHANAN AP/ECEThe first summation represents even sequence. So we will put n = 2m in first summation. While the second summation represents odd sequence, so we will put n =(2m+1)

(2m +1) in second summation. Since even and odd sequences contain N/2 samples each;

the limits of summation will be from m = 0 to N/2-1

But x (2m) is even sequence, so it is f1(m) and x(2m +1) is odd sequence, so it is f2

(m).

Comparing each summation with the definition of DFT,

We will consider an example of 8 point DFT. That means N = 8.

PREPARED BY S.JANARTHANAN AP/ECENow F1(k) and F2(k) are 4-point (N/2) DFTs. They are periodic with period N/2.

Using periodicity property of DFT we can write,

We are considering an example of 8 point DFT (N = 8). So in Equations (15) and (16), k varies from 0 to 3. Now putting k = 0 to 3 in Equations (15) and (16) we get,

The graphical representation of first stage of decimation shown in Fig.

for 8 point DFT is as


In Fig. input sequence are

That means each sequence contains N/2 samples.

Second stage of decimation:

In the first stage of decimation; we obtained the sequences of length N/2 . That

means for 8-point DFT (N = 8); the length of each sequence is ‘4’ as given by Equations (19) and (20). We discussed that we have to continue this process till we get ‘2’ point sequence.

We can further decimate f1(m) into even and odd samples. Let g11(n)= f1(2m), which contains even samples and let g12(n) = f1(2m +1), which contains odd samples of f1(m).

Note that here range of ‘n’ and ‘m’ is from 0 to N/4-1

Now recall equations (15) and (16). We obtained sequences X(k) and X(k+N/2)

from F1(k) and F2(k). The length of each sequence was N/2. Here in the second stage of

decimation. We are further dividing the sequences into even and odd parts. So similar to Equations (15) and (16) we can write; For F1(k),

Thus, for N = 8 we have the range of K, from K = 0 to K = iHere G11(k) is DFT of g11(n) and G12(k) is DFT of g12(n) Now putting the values of ‘K’ in Equation (21) we get,


Here the values of K are ‘0’ and 1. That means it is 2-point DFT. Thus Equations (23) and (24) shows that we can obtain 4-point DFT by combining two 2-point DFTs. The graphical representation is shown in Fig.

Note that here,

Now similar to Equations (21) and (22) we can write equations for F2(k) as follows:


Similarly from Equation (27), we get

The graphical representation of Equations (28) and (29) is shown in Fig.

Note that here.

Combination of first and second stage of decimation:

Combining Fig. we get the combination of first and second stage of decimation. It is shown in Fig.

At this stage we N/4 have that means 2 point sequences. So further decimation is

not possible. As shown in Fig. we have to compute 2-point DFT.


Computation of 2-point DFT: According to the basic definition of DFT,

We will use Equation (31) to compute 2-point DFT. From Fig. consider the first block of 2-point DFT. It is separately drawn as shown in Fig.

Here input sequences are g11(0) and g11(1). We can denote it by g11(n); where n

varies from 0 to 1. Now the output sequences are G11(0) and G11(1). We can denote it by G11(k); where ‘k’ varies from 0 to 1. Here G11(k) is DFT of g11(n).

Thus for G11(k) we can write Equation (31) as,


Note that this is 2 point DFT, so we have put N = 2.

Now puffing values of k in Equation (32) we get,

Expandin the summation we get,

Expanding the summation we get,

Using Equations (33) and (35), we can represent the computation of 2-point DFT

as shown in Fig. This structure looks like a butterfly. So it is called as FFT butterfly

structure.


Now we know that . Thus we can modify Equation (33) and (35) as

Follows:

This modified butterfly structure is shown in Fig.

Similarly for other 2-point DFTs we can draw the butterfly structure.

Total signal flow-graph for 8-point DIT FF1’

The total signal flow graph is obtained by interconnecting all stages of decimation. In this case, it is obtained by interconnecting first and second stage of decimation. But the starting block is the block used to compute 2-point DFT (butterfly structure). The total signal flow graph is shown in Fig.


5. Draw a 8 point radius 2 FFT’ DIT flow graphs and obtain DFT’ of the following

sequence

Ans.

PREPARED BY S.JANARTHANAN AP/ECE6. Compute 4-point DFT of causal three sample sequence given by.

Ans. By definition of N-point DFT, the kth is complex co-efficient of X (k) for

is given by

Here N = 4, therefore the 4-point DFF is

The values of X (k) can be evaluated for k = 0, 1, 2, 3.


The three sample sequence and its periodic extension are shown in Fig. below.

7. By means of the DFT & IDFT, determine the sequence x3 (n) corresponding to the circular convolution of the sequence x1 (n) and x2(n).

Ans. First we compute the DFT’s of x1 (n) and x2(n).

The four point DFT of x1 (n) is


8. Find the inverse DFT of X(k) = {1, 2, 3, 4}.

Ans. We know that the inverse DFT is expressed as


9. By means of the DFT & IDFT, determine the response of the FIR filter with impulse response.

Ans. The input sequence has length L = 4 and the impulse response has length M = 3. Linear convolution of these two sequences produces a sequence of length N = 6. Consequently, the size of the DFT must be at least six.

For simplicity we compute eight point DFT’s. We should also mention that the efficient computation of the DFT via the Fast Fourier Transform (FFT) algorithm is usually performed for a length N, that is power of 2. Hence eight point DFT of x(n) is


10. An 8-point sequence is given by x(n); x(n) = (2, 2, 2, 2, 1, 1, 1, 1).

Compute 8 point DFT of x(n) by radix -2 DIF-FFT.

Ans. For 8 point DFT by radix 2 FF1 we require 3-stages of computation with 4 butterfly computation in each stage.

The given sequence is the input to the first stage. For other stages of computation, the output of the previous stage will be the input for the current stage.

First stage of computation

The input sequence = {2, 2, 2, 2, 1, 1, 1, 1)

The phase factors involved in first stage of computation are


The output sequence of first stage of computation.

Second stage of computation

The input sequence of 2nd stage

The phase factors involved in 2nd stage of computation are

The butterfly computation of second stage are shown above.


11. An 8-point sequence is given by x (n) = {2, 2, 2, 2, 1, 1, 1, 1). Compute 8-point DFT of X(n) by radix -2 DIT FFT. Also sketch the magnitude and phase spectrum.

Ans. The given sequence is first arranged in the bit reversed order.

For 8-point DFT by radix 2 FFF we require 3 stages of computation wits 4 butterfly computations in each stage. The sequence rearranged in the bit reversed order forms the input to the first stage. For other stages of computation the 0/P of previous stage will be I/P for current stage.

First stage computation:

PREPARED BY S.JANARTHANAN AP/ECEThe I/P time sequence {2, 1, 2, 1, 2, 1, 2, 1)

The butterfly computations of first stage are shown below.

Input sequence = {3, 1, 3, 1, 3, 1, 3, 1)

The phase factors involved in second stage computation are W & W.

The 0/P OFT sequence = {3, 1, 3, 1, 3, 1, 3, 1} Second stage computation are

The butterfly computation are shown below.

The butterfly computations of third stage are shown below


12. Discuss Linear filtering approach for the computation of DFT.


Ans. Linear Filtering Methods Based on the DFT : Since the DFT provides a discrete frequency representation of a finite-duration sequence in the frequency domain, it is interesting to explore its use as a computational tool for linear system analysis and especially, for linear filtering. We have already established that a system with frequency response H (), when excited with an input signal that has a spectrum X (w), possesses an output spectrum Y (w) = X (w) H (w). The output sequence y (n) is determined from its spectrum via the inverse. Fourier transform. Computationally, the problem with this frequency domain approach is that X (w). H (w) and Y (cv) are functions of the continuous variable cv. As a consequence, the computations cannot be done on a digital computer, since the computer can only store and perform computations on quantities at discrete frequencies.

On the other had, the DFT does lend itself to computation on a digital computer. In the discussion that follows, we describe how the DFT can be used to perform linear filtering in the frequency domain. In particular, we present a computational procedure that serves as an alternative to time-domain convolution. In fact, the frequency-domain approach based on the DFT, is computationally more efficient than time-domain convolution due to the existence of efficient algorithms for computing the DFT. These algorithms, which are described in Chapter 6, are collectively called fast Fourier transform (FFT) algorithms.

Use of the DFT in Linear Filtering: In the preceding section it was demonstrated that the product of two DFTs is equivalent to the circular convolution of the corresponding time-domain sequences. Unfortunately, circular convolution is of no use to us if our objective is to determine the output of a linear filter to a given input sequence. In this case we seek a frequency-domain methodology equivalent to linear convolution.

Suppose that we have a finite-duration sequence x (n) of length L which excites an FIR filter of lenth M. Without loss of enerality, let

where h (n) is the impulse response of the FIR filter.

The output sequence y (n) of the FIR filter can be expressed in the time domain as the convolution of x (n) and h (n), that is:

Since h (n) and x (n) are finite-duration sequences, their convolution is also finite in duration. In fact, the duration of y (n) is L + M - 1.

The frequency-domain equivalent to (1) is:

If the sequence y (n) is to be represented uniquely in the frequency domain by samples of its spectrum Y (co) at a set of discrete frequencies, the number of distinct samples must equal or

PREPARED BY S.JANARTHANAN AP/ECEexceed L + M - 1. Therefore, a DFT of size , is required to represent (y (n)) in the frequency domain.

Now if

then ………………..(3)

where {X (k)) and {H (k)) are the N-point DFI’s of the corresponding sequences x

(n) and h (n), respectively. Since the sequences x (n) and h (n) have a duration less than N, we simply pad these sequences with zeros to increase their length to N. This increase in the size of the sequences does not alter their spectra which are continuous spectra, since the sequences are aperiodic. However, by sampling their spectra at N equally spaced points in frequency (computing the N-point DFTs), we have increased the number of samples that represent these sequences in the frequency domain beyond the minimum number (L or M, respectively).

Since the N = L + M - I point DFT of the output sequence y (n) is sufficient to represent y (n) in the frequency domain, it follows that the multiplication of the N-point DFTs X (k) and H (k), according to (3), followed by the computation of the N-Point IDFT, must yield the sequence {y (n)). In turn, this implies that the N-point circular convolution of x (n) with h (n). In other words, by increasing the length of the sequences x (n) and h (n) to N points (by appending zeros), and then circularly convolving the resulting sequences, we obtain the same result as would have been obtained with linear convolution. Thus with zero padding, the DFT can be used to perform linear filtering.

Filtering of Long Data Sequences : In practical applications involving linear filtering of signals, the input sequence x (n) is often a very long sequence. This is especiäffy true in some real-time signal processing applications concerned with signal monitoring and analysis.

Since linear filtering performed via the DFT involves operations on a block of data, which by necessity must be limited in size due to limited memory of a digital computer, a long input signal sequence must be segmented to fixed-size blocks prior to processing. Since the filtering is linear, successive blocks can be processed one at a time via the DFT and the output blocks are fitted together to form the overall output signal sequence.

We now describe two methods for linear FIR filtering a long sequence on a block- by-block basis using the DFT. The input sequence is segmented into blocks and each block is processed via the DFT and IDFT to produce a block of output data. The output blocks are fitted together to form an overall output sequence which is identical to the sequence obtained if the long block had been processed via time-domain convolution.

The two methods are called the overlap-save method and the overlap-add method. For both methods we assume that the FIR filter has duration M. The input data sequence is segmented into blocks of L points, where, by assumption, L>> M without loss of generality.

PREPARED BY S.JANARTHANAN AP/ECEOverlap-save method : In this method the size of the input data blocks is N = L + M -1 and the size of the DFTs and IDFT are of length N. Each data block consists of the last

m - 1 data points of the previous data block followed by L new data points to form a

data sequence of length N = L + M - 1. An N-point DFT is computed for each data block.

The impulse response of the FIR filter is increased in length by appending L — 1 zeros

and an N-point DFT of the sequence is computed once and stored. The multiplication of the two N-point DFTs for the mth block of data yields..

Then the N-point IDFT yields the result

Since the data record is of length N, the first M — 1 points of are corrupted by aliasing and must be discarded. The last L points of are exactly the same as the result from linear convolution and, as a consequence,

To avoid loss of data due to aliasing, the last M — 1 points of each data record are saved and these points become the first M - 1 data points of the subsequent retord, as indicated above. To begin the processing, the first M - 1 points of the first record are set to zero. Thus the blocks of data sequences are

and so forth. The resulting data sequences from the IDFT are given by (8), where the first M - I points are discarded due to aliasing and the remaining L points constitute the desired result from linear convolution. This segmentation of the input data and the fitting of the output data blocks together to form the output sequence are graphically illustrated in fig.

Overlap-add method : In this method the size of the input data block is L points. and the size of the DFTs and IDFT is N = L + M - 1. To each data block we append M - 1 zeros and compute the N-points DFT. Thus the data blocks may be represented as:


and so on. The two N-point DF’Ts are multiplied together to form

The IDFT yields data blocks of length N that are free of aliasing since the size of

the DFTs nd IDFT is N = L + M - 1 and the sequences are increased to N-points by appending zeros to each block.

Since each data block is terminated with M - I zeros, the last M — I points from each output block must be overlapped and added to the first M - I points of the succeeding block. Hence this method is called the overlap-add method. This overlapping and adding yields the output sequence.

The segmentation of the input data into blocks and the fitting of the output data

blocks to form the output sequence are graphically illustrated in following figure.

PREPARED BY S.JANARTHANAN AP/ECEAt this point, it may appear to the reader that the use of the DFT in linear FIR filtering is not only an indirect method of computing the output of an FIR filter, but it may also be more expensive compositionally since the input data must first be convected to the frequency domain vai the DFT, multiplied by the DFT of the FIR filter, and

finally, converted back to the time domain via the IDFr. On the contrary, however, by

using the fast Fourier transform algorithm, as will be shown in Chapter 6, the DFTs and

IDFT require fewer computations to compute the output sequence than the direct

realization of the FIR filter in the time

domain. This computational efficiency is the basic advantage of using the DFT to compute the output of an FIR filter.

UNIT II 16 mark with answer

IIR FILTER DESIGN

1. Determine the cascade and parallel realizations for the system described by the system function


Cascaded Realization:

Ans.

Parallel realization:

In terms of positive powers of z, H(z) can be written as



Parallel realization is shown in fig. below.

2. Obtain the cascade realization of the system characterized by transfer function.

Ans.


3. Obtain the block diagram representations of Direct form I and II realizations of the system with transfer function.

Ans. Given


Realization of H1(z) is shown below.

Taking inverse z-transform

Realization of H2(z) is shown in fig. ahead.

Direct from-I realization is shown below


Direct form II realization is shown below.

4. Find the block diagram representations of parallel realization of the system with transfer function.

Ans. Given


The parallel realization is shown in fig. below.


5. Draw the canonical cascade realization diagram for the system given by:

Ans. Given

Taking z transform of equation given

Canonical realization is obtained using shortcut method as shown below:


6. Obtain direct form-I, direct form-Il, cascade and parallel structure for the system described by:

Ans. We take z-transform both sides of difference equation, we have

1. Direct form-I

the order of

1. Direct form-Il : To obtain direct form-Il realization, we change H1(z) and H2(z), then resulting structure is shown below.


The direct form-Il realization is shown below:

1. Cascade form structure:

The cascade form structure is shown in fig.

1. Parallel form structure : To obtain parallel form structure, H(z) must be expanded in partial fraction i.e.

After some arithmetic calculation, we find that

PREPARED BY S.JANARTHANAN AP/ECEThe parallel form structure is shown below:

7. For given analog filter system function into digita IIR filter by means of bilinear z-transformation. Digital filter is to have resonant

frequency

Ans. The given transfer function is:

……….(1) From eq. (1) we can say that f = 4The value of is given as Now we will find out the value of sampling time (T5) using we relation.

Using bilinear transformation H(z) can be obtained by puffing in the eq. of H (s).


8. A chebyshev low pass filter has the following specifications:(a) Order of the filter = 3(b) Ripple in pass-band = 1 db(c) Cut off frequency = 100 Hz(d) Sampling frequency = 1 kHz.Determine H(z) of the corresponding hR digital filter using bilinear transformation technique. Ans. First we will calculate the values of the edge frequency for analog filter.Given : Order of filter = N= 3

1. Calculation of required design specification of digital filter.

Given


3. For normalized filter 4. Calculation of poles

First we calculate parameter

Now we will calculate the values of and R

5. Calculate poles:

PREPARED BY S.JANARTHANAN AP/ECE6 Calculation of system function

7 Calculation of transfer function of digital filter

9. Convert the analog filter with system functions

into the digital IIR filter by means of the impulse invariance method.

Ans. The partial fraction expansion of is given as:

The corresponding digital filter is then


It should be noted that zero of is not obtained by transforming the zero at s = -z into a zero at

10. Design a chebyshev filter for the following specification using (a) bilinear transformation (b) Impulse invariance method.

Ans. Given


Using bilinear transformation

(b) Impulse Invariance Method:


Taking Inverse Laplace transform we obtain

Let

Taking z-Transform

Assume T =1 sec.

PREPARED BY S.JANARTHANAN AP/ECEUNIT III 16 MARK WITH ANSWER

FIR FILTER DESIGN

1. Obtain a cascade realization using minimum number of multiplications for the system.

Ans. We can consider that H(z) is product of factors andThese two factors are having linear phase symmetry.

The realization is shown in fig below :

2. Realize the system function.

by using direct form structure. Ans. Given that

The realization is shown in fig below:


3. Discuss signal flow graph representation and lattice form structures for FIRsystems. Ans. Lattice Structure for FIR Filters : Consider Mth order FIR system with the transfer function.

Here M denotes the degree of polynomial and is coefficient.Thus when M = 0 we get

Basically is the transfer function which can be written as

Putting Equation (I) in Equation (3) we get.

Taking IZT of both sides we get,

Let M = I then Equation (5) becomes,

In the simplest way Equation (6) can be realized as shown in Fig. (a).


Now the same output can be obtained by using the structure shown in Fig. (b). This

structure is called as single stage lattice structure. This structure provides two outputs

namely A1(n) and B1(n).

A0 and B0 are constant multipliers.

In terms of x (n) we can write,

Similarly, the other output can be written as,

In terms of x (n) we can write,

We know that Equation (7 (b)) is obtained by using single stage lattice structure.

Equations (6) and (7 (b)) are same if,

Similarly, Equation (6) and (7 (a)) are matching if,

Here ‘k’ is called as reflection coefficient.

That means the same output can be obtained using single stage lattice structure.

Now for M = 2; Equation (5) becomes,


As M = 2 we have to cascade two stages to obtain two stage lattice structure as shown in Fig. (c).

From fig. (c) we can write,

Output of first stage:

And the output of second stage is,

Now putting equation (10) in equation (12) we get,

From Equation (11) we can write,

Putting this value in equation (14) we get,

Observe that equations (9) and (15) are matching.

PREPARED BY S.JANARTHANAN AP/ECEHere k1 and k2 are reflection coefficients The same way we can increase the number of stages Finally Mth stage lattice structure is obtained as shown in Fig (d)

4. Describe the magnitude and phase response of FIR filters. How is linear phase FIR filter defined.

Ans. The FIR filter can be characterized by its system function.

Which we consider as polynomial of degree N-1 in the variable .The roots of

this polynomial constitute the zeroes of the filter. The frequency response of eq. (1) is givr’ .by

which is periodic in frequency with period

where is magnitude response and is phase response

For FIR filters with linear phase we can define

where a is a constant phase delay in sample. Symmetric Condition

We can write


which gives us

By taking ratio. (3) to (2), we have

Simplifying the above equation we get

The eq. (4) will be zero when

Therefore, FIR filters will have constant phase and when the impulse response is symmetrical

about , then its phase is piecewise linear. Fig. below shows the property of. eq. (5), for N = 6 and N = 5.

In both the cases when N = 6 and N = 5 in the general case, the unit impulse

response sequence satisfying eq. (5). is symmetrical about For N odd, there is

one sample , that is not matched to any other sample.


Antisymmetric Condition:

For this

Then

Therefore FIR filters have constant group delay and not constant phase delay when

impulse response is antisymmetric about

PREPARED BY S.JANARTHANAN AP/ECEThe filters that satisfy the above conditions and have a delay of N—i samples but

their impulse response are antisymmetric around the centre of the sequence, as opposed to the true linear phase sequence that are symmetric around the centre of the sequence.

Magnitude Specifications : It is shown in fig below:

Hence

Magnitude specification of FIR filter can be written as

5. Design an ideal band pass filter with a frequency response.

Find the values of h(n) for N 7. Find the realizable filter transfer function and magnitude function of

Sol. Step 1. Draw the ideal desired frequency response of bandpass filter.


Form the desired frequency response, we can find that the given response is symmetric N odd

Step 2. To find

Step 3. To find h(n).

For symmetry response


Step 4. To find filter transfer function,

Step 5. To find the realizable filter transfer function

Therefore, the filter co-efficients of the causal filters are,

Step 6. To find the magnitude response of


6. Design an ideal band reject filter with a desired frequency response

Find the value of h(n) for N = 7. Find H(z) and

Sol. Step 1. Draw the ideal desired frequency response of band reject filtr.

From the desired frequency response, we can find that the given response is symmertric, N-odd.

Step 2. To find

In general,

Step 3. To find h(n)

PREPARED BY S.JANARTHANAN AP/ECEFor symetric response,

Step 4. To find filter transfer function,

Step 5. To find the realizable filter transfer function.

Therefore, the filter co-efficients of the causal filters are,

Step 6. To find magnitude function


7. Design an ideal highpass filter with a frequency response

Find the value of h(n) for N = 11 using

(a) Hamming window

(b) Hanning window.

Sol. (a) Hamming Window

Step 1. Draw the desired frequency response of ideal highpass filter.

From the desired frequency response, we can find that the given response is

symmetric.

Step 2. To find

We know that,


Step3. To find the Hamming window sequence.

Step 4. To find the filter co-efficents


Step 5. To find the filter co-efficients using Hamming window sequence.

Step 6. To find the transfer function of the filter.

Step 7. To find transfer function of the realizable filter


The filter co-efficients of causal filters are,

(b) Hanning Window

Step 1. The filter co-efficients can be obtained from part (a), step (2) and step (5)

Step 2. To find the Hanning window sequence

The Hanning window sequence is given by

PREPARED BY S.JANARTHANAN AP/ECEStep 3. To find the filter co-efficients using Hanning window.

The filter co-efficients using Hanning window are

Step 4. To find the transfer function of the filter.

The transfer function of the filter is given by,

Step 5. To find the transfer function of realizable filter.

08. Design an ideal low pass filter with a frequency response

Find the values of h (n) for N 1. Also find the filter transfer and frequency magnitude frequency function.

Sol. Step 1. Draw the desired frequency response:

Given,


From the frequency response, we can find that the given response is a symmetrical

N odd response.

Step 2. To find

In general,

Step 3. To find h(n):

For symmetric response,

So

But we know that


Step 4. To find the filter transfer function.

Step 5. To find the realizable filter transfer function.

From the realizable filter transfer function, we have

PREPARED BY S.JANARTHANAN AP/ECEStep 6. To find the magnitude frequency response

09. Design a low pass FIR filter using hamming window to meet the following specifications.

use 10 tap filter and obtain the impulse response of the desired filter.

Ans. The filter co-efficients are given by :

Given M = 10. The filter co-efficients are:


The hamming window function is

The filter co-efficients of the resultant filter are then

Therefore

The frequency response is given by


UNIT IV FINITE WORLD LENGTH EFFECTS16 MARK WITH NASWER

1.ConsiderthetransferfunctionH(Z)=H1(Z)H2(Z)whereH1(Z)=1/1-A1 Z-

1H2(z)=1/1/1-A2 Z-1 Findtheo/pRoundofnoisepowerAssumea1=0.5anda2=0.6andfindo.proundoffnoisepower.1) Draw the round ofNoise Model. 2) Byusing residue methodfind h1( Z)3)Byusing residue methodfind h2( Z)

2.WhatismeantbyA/Dconversionnoise?Explainindetail?1) ADSP contains adevice,A/D converter that operates on the analog input x(t)to producexq(t)which is binarysequenceof 0s and1s.2) At first the signal x(t)issampled at regular intervalsto produceasequencex(n)is of infinite precision. 3) Each sample x(n)is expressed in terms of afinite number ofbits given the sequence xq(n). Thedifferencesignale(n)=xq(n)-x(n) is called A/Dconversion noise. + derivation.

3.considerthetransferfunctionH(Z)=H1(Z)H2(Z)whereH(Z)=H1(Z)H2(Z)whereH1(Z)=1/1-A1 Z-1 H2(z)=1/1/1-A2 Z-

1Findtheo/pRoundofnoisepowerAssumea1=0.7anda2=0.8andfindo.proundoffnoisepower.1) Draw the round ofNoise Model. 2) Byusing residue methodfind h1( Z)3)Byusing residue methodfind h2( Z)

3. Explainthedifferenttypesofrepresentationsindigitalsystems.Ans:Fixed Point

Floating pointBlock floating point

Fixed Point:Infixed point arithmeticthe position of the binarypoint isfixed. Thebitto the right represent

thefractional part ofthe numbersand those to the left represent the integer part .forexample,the binarynumber 01.1100 the value 1.75 in decimal.depending on thewaynegative numbers are represented,thereare three different forms offixed-point arithmetic. Theyare1. Sign magnitude 2. 1’s complement3. 2’s complement.

PREPARED BY S.JANARTHANAN AP/ECE*It is the fast operation.

*relativelyeconomical

*small dynamic range

*roundoff erroroccurs onlyforaddition

*overflow occurs in addition

*used in small computer.

Floating point:Infloating pointrepresentation a positive number isrepresented asF=2c.Mwherem called

mantissa and cis the exponent can be either positiveornegative.Negativefloating point numbersare generallyrepresented byconsidering the mantissaas a fixed point number.

*slow operation

*more expensive because ofcostlierhardware

*increased dynamicrange

*roundoff errorscanoccurwith the both addition and multiplication.

*overflow does notarise

*used in larger general purpose computer.

Block floating pointIn blockfloating point arithmetic the set ofsignals to behandled is divided in toblocks. Each

block has the same value fortheexponent. The arithmeticoperationswithin the block usefixed point arithmetic and onlyone exponent per block isstored thussaving memory. Thisrepresentationof numbers is mostsuitable in certain FFT flow graphs and in digital audio applications.

6. Disscussthedifferenttypesofquantizationerrors.Ans:

Inputquatization errorProduct quantization error. Coeffient quantization error

The common methods of quantization aretruncation and rounding. Truncation isa process of discarding all bits lesssignificant than least significant bit that isretained.rounding ofa numberof bbits is accomplished bychoosing the rounded result as the b bit number closest to the original number un rounded. The error dueto truncation and rounding is known as quantization error.

Theinput quantization erroris given by

E(n)=Xq(n)-X(n)


Xq(n)=sampled quantized value

X(n)=sampled unquantized value

Theoutput quantizationerroris the variance of the sum of independent random variableis the sum of their variances.If the quantization error are assumed to be independent at different sampling instances.

Infixed point arithmetic the product of two bbit numbersresult in numbers 2b bits long.In digital signal processing applications,itis necessarytoroundthis product to ab bit number,which producean errorknown as product quantization error.

In the design of adigital filterthe coefficientsare evaluated with infiniteprecision. Butwhen theyare quantized, the frequencyresponse oftheactual filter deviatesfrom that which would havebeen obtained with an infinite word lengthrepresentation and the filter mayactuallyfail to meet thedesired specifications.If the poles of the desiredfilter areclose to the unitcircle, then those ofthe filterwith quantized coefficientsmaylie just outside the unit circle.

.

7.ExplainthedifferenttypesoflimitcycleoscillationsandalsothesolutionsAns: Zero input limit cycle oscillations

Overflow input limit cycle oscillations

Zero input limit cycle oscillations:*This limit cycle oscillation has lowamplitude.*This limit cycle occurswhen the input applied tothe system is verylow.

Overflow input limit cycle oscillations*This limit cycle occursbecause of the overflowtaking placein the implementation of

digital filters.

*Inaddition to limit cycle oscillation causing byrounding the result ofmultiplication, thereareseveral typesoflimit cycle oscillation caused byaddition, which make the filter output oscillate between maximum andminimum amplitudes.

Dead band:

It is the rangeof output amplitudes overwhich limit cycle oscillationtakeplace.

Scaling:To prevent overflow .thesignal level at certain pointsin the digital filtersmust be scaled so that

no overflow occursin the adder.

8.ExplaintheconstructionandoperationofchannelvocoderwithblockdiagramAns: The channel vocoder is used in the analysissynthesissystem. We use filter bank to separate the frequencybands. There are about eight to ten filters. The amplitude of thefilters outputs areencodedby level

PREPARED BY S.JANARTHANAN AP/ECEdetectorsand coders.Inaddition to this, pitch and voiceinformationarealso sentalong with them. Also, a widebandexcitation signal is generated atthe receiving end using the transmitted pitch and voicing information.For avoiced signal, the excitation consists of periodicsignal with an appropriate frequency.

Howeverfor unvoiced signal the excitation isa white noise. At the receiver end amatched filter bank isavailable, duetowhich the output level matches the encodevalue.further,individual outputs are combined to producethespeech signal.

UNIT VDSP APPLICATIONS

PREPARED BY S.JANARTHANAN AP/ECE1.Describe about multirate signal processing

The signals of interest in digital signal processing are discrete sequences of real or complex numbers denoted by x(n), y(n), etc. The sequence x(n) is often obtained by sampling a continuous-time signal xc(t). The majority of natural signals (like the audio signal reaching our ears or the optical signal reaching our eyes) are continuous-time. However, in order to facilitate their processing using DSP techniques, they need to be sampled and converted to digital signals. This conversion also includes signal quantization, i.e.,discretization in amplitude, however in practice it is safe to assume that the amplitude of x(n) can be any real or complex number. Signal processing analysis is often simplified by considering

the frequency domain representation of signals and systems. Commonly used alternative representations of x(n) are its z-transform X(z) and the discrete-time Fourier transform X(ejω).

The z-transform is defined as andX(ejω) is nothing but X(z) evaluated on the unit circle z = ejω.

Multirate DSP systems are usually composed of three basic building blocks, operating on a discrete-time signal x(n). Those are the linear time invariant (LTI) filter, the decimator and the expander. An LTI filter, like the one shown in Fig.1.1, is characterized by its impulse response h(n), or equivalently by its z-transform (also called the transfer function) H(z). Examples of the M-fold decimator and expander for M = 2 are shown in Fig.1.2. The rate of the signal at the output of an expander is M times higher than the rate at its input, while the converse is true for decimators. That is why the systems containing expanders and decimators are called ‘multirate’ systems. Fig.1.2 demonstrates the behavior of the decimator and the expander in both the time and the frequency domains. In the z-domain this is described by

The systems shown in Figs.1.1 and 1.2 operate on scalar signals and thus are called single input—single output (SISO) systems. The extensions to the case of vector signals are rather straightforward: thedecimation and the expansion are performed on each element separately. The corresponding vector sequence decimators/expanders are denoted within square boxes in block diagrams. In Fig.1.3 this is demonstrated for vector expanders. The LTI systems operating on vector signals are called multiple input—multiple output (MIMO) systems and they are characterized by a (possibly rectangular) matrix transfer function H(z).

2. Explain Application in channel equalization

PREPARED BY S.JANARTHANAN AP/ECEIn the following, we consider the case where an FIR LBP is used as a MIMO channel equalizer. We will showthat the flexibility in the choice of H(z) can be exploited in order to reduce the undesirable amplification of the channel noise. But, before proceeding to these results, we give a brief overview of some equalization techniques for the vector channels.

The discrete-time equivalent of an MIMO digital communication system with a symbol spaced equalizer (SSE) is shown in FiG. The vector symbol rate at the input x(n) is 1/T . Notice that the equalizer H2(z) works at the same rate (thus the name symbol spaced equalizer). The discrete versions of the pulse shaping filter and the channel, G2(z) and C2(z), respectively, are obtained by sampling the corresponding continuous-time impulse responses also at the rate 1/T . We will refer to their cascade F2(z) = C2(z)G2(z) as the equivalent channel for the SSE case. Therefore, as for the signal x(n), the system from Fig.2.3(a) can be represented as a cascade of the equivalent channel F2(z) and a SSE H2(z).

An ideal equalizer (or a zero-forcing equalizer [39]) H2(z) is then obtained as a left inverse of the equivalent channel F2(z). From this discussion, several drawbacks of symbol spaced equalizers are apparent. The MIMO transfer function F2(z) does not have a left inverse if it is a fat matrix. Even if the matrix is not fat, its invertibility will depend on the rank. Furthermore, if F2(z) is invertible, its inverse is most probably IIR, which often amplifies the noise at the receiver. Finally, it has been observed that the ISI suppression achieved by this equalizer is very sensitive to the phase of the sampling at the receiver.

For all these reasons, a popular alternative is to use a so called fractionally spaced equalizer (FSE). It can be shown to be far less

sensitive to the sampling phase , it can be used with fat channel transfer functions, and it often allows for FIR solutions while SSE does not.

The idea behind an FSE is to let the equalizer work at a higher rate. Because of this additional redundancy, FSEs are both more flexible and more robust than SSEs. In a continuous-time communication system, FSE is realized by sampling the received waveform at M times the symbol rate, and feeding such oversampled signal to the equalizer, which now operates at the rate M/T. In this chapter, the oversampling 24 ratio M is assumed to be an integer. In

PREPARED BY S.JANARTHANAN AP/ECEdiscrete time, this is modeled as shown in FiG. The discrete transfer functions G(z) and C(z) are obtained after sampling the corresponding continuous-time impulse responses at the rate M/T. Thus, the equivalent channel F(z) in this case is such that F2(z) = [F(z)]M

Note that the noise also needs to be modified, but this is not the main point of discussion here. We recall that a zero-forcing FSE H(z) in is nothing but an LBP of the channel matrix F(z). In this section we will exploit the non-uniqueness of this biorthogonal partner with the aim of minimizing the noise power at the receiver. The concept of fractionally spaced equalization is by no means new ; however, the original contribution of this section is the attempt to parameterize the FIR FSE solutions, making the search for the ‘best’ solution analytically tractable. On the other hand, the optimization of MIMO systems of the type shown in

has been considered by several authors in many different contexts (derive the optimal transmitter and receiver for a given channel in the sense of minimizing the overall mean squared error. This MMSE solution clearly outperforms any zero-forcing equalizer, however the price is paid

in terms of complexity: the solution in involves ideal filtering. Here we have taken a simplistic approach of decoupling the problems of ISI and noise suppression.

3. Derivation of poly phase decomposition



4. Design a two stage decimator for the following specification

Decimating factor=100

Passband=0<F<50

Transition band=50<F<345

Input sampling rate=10 khz

Ripple p=1/10, S=1/103


5. Design a two stage decimator for the following specification

Decimating factor=50

Passband=0<F<60

Transition band=50<F<65

Input sampling rate=10 khz

Ripple p=1/10, S=1/103


6. SUBBAND CODING OF SPEECH SIGNALS

Let Speech signal is sampled at Fs samples per sec.A very useful tool in multirate signal processing is the so-called polyphase representation of signals and systems. It facilitates considerable simplifications of theoretical results as well as efficient implementation of multirate systems. Since polyphase representation will play an important role in the rest of the thesis, here we take a moment to formally define it. Consider an LTI system with a transfer function namely, the system for generating y(n) from x(n).

Traditionally, this structure has been called the system for digital interpolation since the rate of y(n) is M times higher than that of x(n). Filter H(z) is usually referred to as the interpolation filter [61]. Suppose the goal is to recover the signal x(n) from y(n). Conceptually the simplest way to achieve this is shown in Fig.1.5(a). Namely, y(n) is first passed through the inverse of the interpolation filter 1/H(z). This recovers the signal at the input of the M-fold expander. The M-fold decimator that follows simply discards the zeros inserted by the expander and the recovery of x(n) is complete. Notice, however, that this is not the only way to reconstruct x(n), simply because the inverse filter forces the discarded samples to be zero, while they can take arbitrary values. Indeed, any filter F(z) with the property that its output preserves the desired samples of x(n) in the appropriate locations, with arbitrary values in between x(n)


7. Two channel Quadrature Mirror Filter bank

•A discrete signal x[n] is first split into a number of subband signals by means of analysis filter bank, the subband signals are processed and then finally combined together by synthesis filter bank

•If up sampling and down sampling factors are equal to the number of subbands then the filter is known as critically sampled filter bank

The up and down sampler are time variant systems and the analysis and synthesis filters are chosen to cancel the effect of aliasing.


When the magnitude of T(z) is constantthe output of the QMF bank is said to be magnitude preserving.

•When the function T(z) is having a linear phasethe filter bank is said to be phase preserving

When the above two conditions are satisfied then the filter bank is called as Perfect reconstruction QMF bank

•Alias free bank is obtained when

•H0(z) –low pass filter

•H1(z) –high pass filter

•The spectrum of H1(z) is a mirror image of H0(z) with respect to π/2, the quadrature frequency

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