MAY/JUNE 2006(4TH SEM)1)what is meant by aliasing? how can it be avoided?

The superimposition of high frequency component on the lower frequency component is known as aliasing.

EX:A band limited signal x(t) having no frequency component for|Ω|>Ωm.If we sample this signal with a sampling frequency F<2fm,the periodic continuation of X(jΩ) results in spectral overlap & we cant recover the spectrum X(jΩ) using lowpass filter.This s known as aliasing effect.

TO AVOID THIS:The sampling frequency must be greater than twice the highest frequency present in the signal.(i.e)F≥2fm.

2)Is the system y(n)=lnx(n) is linear and time invariant?NOTE:if a system s linear if and only if,

T[a1x1(n)+ a2x2(n)]= a1T[x1(n)]+ a2T[x2(n)],where y1(n)=T[x1(n)], y2(n)=T[x2(n)].Sol:

y1(n)=ln x1(n)= T[x1(n)], y2(n)=ln x2(n)= T[x2(n)] then the LHS is,a1T[x1(n)]+ a2T[x2(n)]→ a1 ln x1(n)+ a2 ln x2(n)-------------------------(1)

T[a1x1(n)+ a2x2(n)] →ln a1x1(n)+ a2x2(n) --------------------(2)(1)≠(2).So it s non linear system…..ii)if the input is delayed by k units in time ,then

y(n,k)=lnx(n-k)------------------------------------------(1)if the output is delayed by k units in time then,

y(n-k)=lnx(n-k)------------------------------------------(2)(1)=(2).therefore this s a time-invariant system.

3)Define DFT pair.The discrete fourier transform of finite duration sequence x(n) obtained by sampling the fourier transform X(ejw) at N equally spaced points over the interval 0≤w≤2Π with a spacing of 2Π/N.This s denoted by,

X(k)=X(ejw)|w=2Πk/N where,0≤k≤N-1.4)Differenciate b/w DIT and DIF algorithms.

DIT(Decimation –In-Time) DIF(Decimation –In-Frequency)1)Decimation occurs in Time domain 1)Decimation occurs in Frequency domain2)O/P sequence x(n) is partioned into two sequence each of length N/2 samples.

2)I/P sequence is partioned into two sequence each of length N/2 samples.

3)for DIT,input is in bit reversal while the output is in natural order.

3)for DIF,ouput is in bit reversal while the input is in natural order.

4)two inputs Xm(p) & Xm(q) are combined to give the output Xm+1(p)& Xm+1(q) via ,the operation is, Xm+1(p)= Xm(p)+WN

kXm(q) Xm+1(q)= Xm(p) -WN

kXm(q)WN

k----twiddle factor

4)two inputs Xm(p) & Xm(q) are combined to give the output Xm+1(p)& Xm+1(q) via ,the operation is, Xm+1(p)= Xm(p)+ Xm(q) Xm+1(q)= [Xm(p)- Xm(q)] WN

k

WNk----twiddle factor

5)Find the transfer function of normalized Buttorworth filter of order 1 by determining the pole values.

Actually the transfer function for normalized filter,H(s)H(-s)=1/1+(-s2)N

For order N=1,H(s)=1/s+1Pole value is (-1)6)What does frequency warping (warping effect)means?

The relationship between the analog and digital frequencies in bilinear transformation is given by,

Ω=(2/T)tan(ω/2) For smaller values of ω there exist linear relationship b/w ω and Ω.But for large values of

ω the relationship is non linear. This non linearity introduces distortion in the frequency axis. This is known as warping effect.

This effect compresses the magnitude & phase response at high frequencies. The warping effect can be eliminated by prewarping the analog filter.

7)State the advantages of FIR filter over IIR filter.FIR FILTER ADVANTAGES IIR FILTER DISADVANTAGES1)FIR filters have exact linear phase. 1)These filters don’t have linear phase.2)FIR filters are always stable. 2)physically realizable and stable

IIR filters cant have linear phase.3)FIR filters can be realized in both recursive and non-recursive.

3)IIR filters are easily realized recursively.

4)Greater flexibility to control the shape of their magnitude response.Filters wit any arbitrary magnitude response can be tackled using FIR filters.

4)Less flexibility,usually limited to specific kind of filters.

5)Errors due to round off noise are less severe in FIR filters, mainly becoz feedback is not used.

5)The round off noise in IIR filters are more.

EXTRA POINTS FOR ADVANTAGES OF FIR FILTER:FIR filters are free of limit cycle oscillations, when implemented on a finite-word length

digital system.8)List out the different forms of structural realizations available for realizing a FIR system.

Traversal structure(or)direct form realization Cascade realization Linear phase realization Lattice structure Polyphase realization

9)Bring out the difference b/w fixed point & floating point arithmetic.FIXED POINT FLOATING POINT1)Fast operation 1)slow operation2)Relatively economical 2)more expensive3)Small dynamic range 3)Increased dynamic range4)Round off errors occurs only for addition 4)Round off errors occurs with both addition

and multiflication.

5)Overflow occurs in addition. 5)Overflow doest arise.6)Used in small computers. 6)Used in large &general purpose computers.

10)How will u avoid limit cycle oscillations due to overflow in addition?OVERFLOW OSCILLATATIONS:

An overflow in addition of two or more binary numbers occurs when the sum exceeds the word size available in the digital implementation of the system.

Let us consider two positive numbers N1, N2N1=0.111→7/8N2=0.110→6/8N1+N2=1.101→ -5/8 in sign magnitude . this s wrong …

This problem is eliminated by modifying the adder characteristics .When an overflow is detected, the sum of adder is set equal to the minimum value.

APRIL/MAY 2008(4 TH SEM) 1)Determine Z transform for x(n)= -nanu(-n-1).

First find z transform for -anu(-n-1).STEP 1: -1 ∞

X(z)= -∑ anz-n= - ∑ (a-1z)-n n=-∞ n=1

\

∞

=-∑ (a-1z)-n -1. n=0

The above series converges to |b-1z|<1,i.e ,for |z|<b

X(z)= - (1/1-a-1z)-1 =z/z-a ROC:|z|<a

u(n)=1 for n≥0 =0 for n<0

u(-n-1)=0 for n≥0 =1 for n≤1

∞X(z)=∑x(n)z-n

n= -∞

∞∑ rn=1+r+r2+r3…………..=1/1-r if |r|<1n=0

Now we have to find, x(n)= -nanu(-n-1). That is nx(n).

So, -nanu(-n-1) will be,

=-z d/dz(z/z-a )= za/(z-a)2

2)Find whether the signal y=n2x(n) is linear.NOTE:if a system s linear if and only if,

T[a1x1(n)+ a2x2(n)]= a1T[x1(n)]+ a2T[x2(n)],where y1(n)=T[x1(n)], y2(n)=T[x2(n)].

Solution:y1(n)=T[x1(n)]= n2x1(n);y2(n)=T[x2(n)]= n2x2(n);

a1T[x1(n)]+ a2T[x2(n)]→ a1 n2x1(n)+ a2 n2x2(n)---------------(1)RHS

T[a1x1(n)+ a2x2(n)] → n2 [a1x1(n)+ a2x2(n)] → a1 n2x1(n)+ a2 n2x2(n)-------------------(2)LHS

The system is a linear system.3)what are the advantages of bilinear mapping(or )bilinear transformations?

Advantages:1. The bilinear transformation provides one-to-one mapping.2. Stable continuous systems can be mapped into realizable,stable,

digital systems3. There is no aliasing.

DisAdvantages:a. The mapping is highly non-linear producing frequency

compressions at high frequencies.b. Neither the impulse response nor the phase response of the

analog filter is preserved in a digital filter obtained by bilinear transformation.

4)How many multiplication and addition is needed for radix/2 FFT?The number of multiplications and additions required to compute N-point DFT

using radix-2 FFT are

Z[nx(n)]=-zd/dzX(z)

d/dz(x/y)=(x'y-y

'x)/y2

(MULTIPLICATIONS)

Nlog2N (ADDITIONS)

respectively.Example:

FFT algorithm with 32-point sequence,N=32Required multiplications is=(32/2) log2(32)=16*5=80.Required addition is =32*5=160.

EXTRA POINTS TO REMEMBER:The no of complex multiplications using direct computation is ,

N2=322=1024Then the speed improvement factor=using Direct computation/Using FFT

=1024/80=12.8

5)Find the DFT for x(n)=1,-1,1,-1.Sol:

Using FFT,First we have to find the bit reversal order…

x(0)=x(00)=x(00)=x(0)=1x(1)=x(01)=x(10)=x(2)=1

x(2)=x(10)=x(01)=x(1)=-1x(3)=x(11)=x(11)=x(3)=-1

So the DFT is 0,0,4,0

6)Define Parsevals Theorem.If x1(n) & x2(n) are complex valued sequences ,then parsevals relation states that,

Where the contour of integration must be in the overlap of the regions of convergencen of X1(v) and X2

*(v)(1/v*).7)advantages of FIR filter:

1)FIR filters have exact linear phase.2)FIR filters are always stable.3)FIR filters can be realized in both recursive and non-recursive.4)Greater flexibility to control the shape of their magnitude response.Filters wit any arbitrary magnitude response can be tackled using FIR filters.

5)Errors due to round off noise are less severe in FIR filters, mainly becoz feedback is not used.

8)Define phase delay.The phase delay is that it should satisfy the following condition.

h(n)=-h(N-1-n)9)Define truncation error for sign magnitude representation and for 2’s complement representation.

Truncation is the process of discarding all the bits less significant than least significant bits is retained.

1)for magnitude the error is satisfies -2 2-b<ε<0 for all x

Here ,ε = xT-x

,xT is the truncated value.

2)for 2’s complement:x is 0≥xT-x> -2-b

10)what are the types of limit cycle oscillation?i. Zero limit cycle oscillation

ii. Overflow limit cycle oscillationZero limit cycle oscillation:

For IIR filter, implemented with infinite precision arithmetic, the o/p should approach zero in the steady state if the input is zero, and it should approach a constant value if the input is a constant. However, with an implementation using finite length register an output can occur even with zero input if there is non-zero initial condition on one of the registers. The output may be a fixed value or it may oscillate b/w finite positive and negative values. This effect is referred to as zero-input limit cycle oscillations and is due to the non-linear nature of the arithmetic quantization.Overflow limit cycle oscillation

The addition of two fixed point arithmetic numbers cause overflow when the sum exceeds the word size available to store the sum. This overflow caused by adder make the filter output to oscillate b/w maximum amplitude limits. Such limit have been referred to as overflow oscillations.

November/DECEMBER-20084TH SEM

1)State the sampling theorem.

Let xa(t) is s band limited signal with Xa(jΩ)=0 for

Ω<Ωm. Then xa(t) is uniquely determined from its samples x(n)= xa(nT) if the

sampling frequency F≥2fm i.e sampling frequency must be at least the highest frequency in the signal.

2)State any two properties of LTI system?a. An LTI system is causal if and only if its impulse response

is zero for negative values of n

b. An LTI s/y is stable if it produces a bounded o/p sequence for every bounded i/p sequence. If for some bounded input sequence x(n),the o/p is unbounded(infinite),the system is classified as unstable. So the necessary & sufficient condition for stability is,

3)Show the saving in time in performing FFT as against DFT.

DFT FFT

N2 number of complex multiplication required to perform DFT

number of complex multiplication required to perform FFT

N2-N number of complex addition is required. N log2N number of complex addition is required.

4)Draw the basic butterfly of the radix 4 DIT algorithm.

h(k)=0 for k<0

∞ ∑ h(n) < ∞

n = -∞

5)Use backward difference for the derivative and the convert the analog filter to digital filter given H(s)= 1

(refer mam also)

s2+16SOLUTION:

H(s)=

6)State the relationship b/w the analog & digital frequencies when converting an analog filter to digital filter using bilinear transformation.

Replace s by ---à

Ω=2 tan (ω/2) T

Ω àanalog frequency

àdigital frequency

7)Compare FIR & IIR Filters.FIR FILTER ADVANTAGES IIR FILTER DISADVANTAGES1)FIR filters have exact linear phase. 1)These filters don’t have linear phase.2)FIR filters are always stable. 2)physically realizable and stable

IIR filters cant have linear phase.3)FIR filters can be realized in both recursive and non-recursive.

3)IIR filters are easily realized recursively.

4)Greater flexibility to control the shape of their magnitude response.Filters wit any arbitrary magnitude response can be tackled using FIR filters.

4)Less flexibility,usually limited to specific kind of filters.

5)Errors due to round off noise are less severe in FIR filters, mainly becoz feedback is not used.

5)The round off noise in IIR filters are more.

EXTRA POINTS EXTRA POINTS6)it has a finite duration unit sample response 6)it has a infinite duration unit sample response7)It depends on present and past input only. 7)It depends on present input as well as past

inputs and outputs.8)Draw the direct form I structure of the FIR filter.

9)What is quantization? The process of converting a discrete time continuous amplitude signal x(n)

into a discrete-time discrete amplitude signal xq(n) is known as quantization. This is done by rounding off each sample in x(n) to the nearest quantization level.

Then each sample in xq(n) is represented by a finite number of digits using a decoder.

q = range of signal = R Number of quantization levels

2b+1

Where qàquantization step size.Common methods of quantization is, 1.Truncation 2.Rounding.10)limit cycle oscillations: In the recursive system,the finite precision arithmetic operation causes periodic oscillation in the o/p. These oscillation is called limit cycle oscillation.Types of limit cycle oscillation

I)Zero limit cycle oscillation

II)Overflow limit cycle oscillationZero limit cycle oscillation:

For IIR filter, implemented with infinite precision arithmetic, the o/p should approach zero in the steady state if the input is zero, and it should approach a constant value if the input is a constant. However, with an implementation using finite length register an output can occur even with zero input if there is non-zero initial condition on one of the registers. The output may be a fixed value or it may oscillate b/w finite positive and negative values. This effect is referred to as zero-input limit cycle oscillations and is due to the non-linear nature of the arithmetic quantization.Overflow limit cycle oscillation

The addition of two fixed point arithmetic numbers cause overflow when the sum exceeds the word size available to store the sum. This overflow caused by adder make the filter output to oscillate b/w maximum amplitude limits. Such limit have been referred to as overflow oscillations.