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UNIT I

SIGNALS AND SYSTEMS

1. What are the basic elements of DSP and its requirements

Ans. The basic elements of digital signal processing system are shown in fig. below.

is measured as

The different blocks of this system as follows.

1 I/P Signal It is the signal generated from some transducer or from some communication system It may be biomedical signal like ECG or

EEC Generally input signal is analog in nature It is denoted as x(t) 2. Anti Aliasing filter: It is basically a low pass filter. It is used for the

following purposes. (a) It removes the high frequency noise contain in input signal.

(b) It avoids aliasing effect that means it is used to band limit the signal.

3. Sample and Hold circuit. It takes samples of I/P signal. It keeps the

voltage level of I/P signal relatively constant which is the requirement of ADC

4 Analog to digital converter It is used to convert analog signal into digital form This is required because digital signal processor accepts

the signal which is digital in nature. 5. Digital Signal Processor : It processes input signal digitally. Input

signal making modifying the signal as per requirement. For this purpose DSP processors like ADSP 2100 or TMS 320 are used.

6 Digital to analog converter (DAC) The output of digital signal processor is digital in nature. But the required final output is analog in

nature. So to convert digital signal into analog signal DAC is used. 7 Reconstruction Filter Output signal of DAC is analog that means it is

continous signal. But it may contain high frequency components. Such high frequency components are unwanted. To remove these

components reconstruction filter is used.

2. Compute convolution of y (n) of the signals.

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Ans.

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3. Define the stability conditions for a linear time invarient

system. Determine the range of values of a for which the LTI system with impulse response h(n) as defined below is stable.

Ans. LTl system is stable if its inpulse response is absolutely summable.

Consider a linear time invarient system having impulse response h(n). Let x(n) be input applied to this system. Now according to the

definition of convolution, the output of such system is expressed as

The input x(n) will be bounded if I x(n) is less than some finite number. Let us denote this finite number by Mx. Thus for input signal

to be bounded

Here Mx is finite number, so its values should be less than infinity. Thus eq. (2) can be written as

Now taking absolute value of both sides of eq. (1)

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We will read R.H.S. of eq. (4) as absolute value of summation of

terms. If we take sign outside then the term become

This summation of absolute Thus:

values of terms. Always absolute values of sum of terms is less than or equal to the sum of absolute value of terms.

Here x(n k) is delayed input signal. If input is bounded then its delayed version is also bounded. This is because delay or qlding is

related to time shifting operations By performing these operations the magrjitu4e is not changed. Now for bounded input,

We know that Mx is finite number. We want 0/P y(n) to be bounded.

That means I y(n) should be finite. So eq. (8) to obtain finite output we have

Here h(k)= h(n) is the impulse response of LTI system. Thus eq. (9) gives the conditions

of stability in terms of impulse response of the system.

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From above, it is abvious that given system is stable if a

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5. An LTI system is described by . Find the response of this system for an input of x(n) = 10 cos (0.05

3m).

Ans. Given

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6. Find the signal energy of the signal x(t) = U(t) U(10 t)?

Ans. Energy of input sequence x(t) is given by

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7. Determine the impulse response for the cascade of two

linear time invariant systems having impulse responses.

8. List various properties of z-transform.

Ans. The various properties of z-transform are given below.

1. Linearity

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11. Parsevals theorem:

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1. Shifting:

1. Multiplication (by nm). [or differentiation in z-domain]

4. Scaling in z-domain: [Multiplication by an ]

5.Time Reversal:

6. Conjugation:

7. Convolution:

8. Initial value:

9. Final value:

10. Correlation of two sequence.

where the contour of integration must be overlap of the regions of convergence of

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9. Determine the z-fransform of the signal

Ans. By using Eulers identity the signal x (n) can be expressed as

By simply linearity property.

After some simple algebraic manipulations we obtain the desired result namely.

10. Find the z-transform and region of convergence for the following sequence.

Apply initial value theorem and check the z-transform whether

it is correct or not.

Ans.

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Thus calculated z transfer is corrects.

11. Find the inverse z transform of

Ans. Given:

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By using long division method.

12. Find all possible inverse z-transform of the following.

Ans. Given:

In partial fraction expansion form,

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13. Consider a signal x (n) given by

Ans.

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The sum in convergences then

For convergence of X (z) both sum must converge

which requires that

1

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Thus the ROC is because the ROC is the common region where

both sum are finite. The fig shows the ROC of the z-transform of each of the

individual terms & for combined signal.

14. Find the one sided z-transform of discrete sequence

generated by mathematically sampling the continuous time

function

Ans. Given

The discrete sequence is generated by replacing t by kT, where T is

the sampling period.

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15. Determine the inverse z-transform of

Ans.(a) Since the ROC is exterior of a circle, we except x (n) to be a

causal sequence. This we divide so as to obtain a series in negative power of z. Carrying out the long division, we obtain

(b)When the ROC is interior of the circle, the signal x (n) is anticausal

signal. Thus we divide so as to obtain a series in power of z as follows.

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16. Find the inverse Z-transform of the function,

Ans. Given

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From above equation we can say that if a complex sinusoidal signal is given as input signal to LTI system, then the output is a sinusoid of

the same frequency modified by H(w). Hence H(w) is called the frequency response of the system. The H(w) produces a change in the

amplitude and phase of the input signal. The frequency response H(w) of LTI system is also given by the ratio

of Fourier transform of output to Fourier transform of input.

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17. Determine the inverse z-transforms of

Ans. Given

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18. Perform circular, convolution of two sequences

Ans. Circular convolution is

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UNIT II

FREQUENCY TRANSFORMATIONS

1. Discuss various properties of DFT.

Ans. Properties of the DFT.

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

then for any two constants a and b,

2. Periodicity

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3. Circular shift of a sequence : This property is analogous to the time shifting

property 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

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

Using equations (6.6) and (6.7) in (6.8), we have,

The brackets term in eqn. (6.9) has the form,

If we substitute the eqn. (6.11) into eqn. (6.12), we have,

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The equation (6.12) 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. (6.12), the second sequence is

circularly time reversed and circularly shifted w.r.t. of first. The equation (6.12) 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.

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If we change the index from n to m by defining m = N - n, then,

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7. Circular time shift:

Proof.

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8. Multiplication of two sequences:

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

Equation (6.18) can be written as

The proof of this property is similar to circular convolution.

2. State and prove Circular correlation and parsevals theorem 1. Circular correlation: For complex valued sequence x (n) and y (n).

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

Proof.

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

multiplication of the their DFTs and from complex conjugate property

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

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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.

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Now consider the second summation that means

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

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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

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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 DFTs given by eq (17) & (18) Let us consider an example of 8 pomt OFT That

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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.

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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.

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Since we have divided x(n) into two parts, wq can write separate

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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,

summation for even and odd sequences as follows:

The 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).

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Comparing each summation with the definition of DFT,

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

Now 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,

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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,

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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.

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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.

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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

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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.

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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

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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.

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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.

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5. Draw a 8 point radius 2 FF1 DIT flow graphs and obtain DF1 of the following

sequence

Ans.

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6. 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

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

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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 DFTs of x1 (n) and x2(n). The four point DFT of x1 (n) is

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

Ans. We know that the inverse DFT is expressed as

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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 DFTs. 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

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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.

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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.

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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.

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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: The 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.

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The butterfly computations of third stage are shown below

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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.

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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 exceed L +

M - 1. Therefore, a DFT of size , is required to represent (y (n)) in the frequency domain.

Now if

then ..(3)

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where {X (k)) and {H (k)) are the N-point DFIs 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 especiffy 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

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Then the N-point IDFT yields the result

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duration M. The input data sequence is segmented into blocks of L

points, where, by assumption, L>> M without loss of generality. Overlap-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..

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

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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 DFTs 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

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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 fig.

11. At 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

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domain. This computational efficiency is the basic advantage of using

the DFT to compute the output of an FIR filter.

UNIT III 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

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Parallel realization is shown in fig. below.

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2. Obtain the cascade realization of the system characterized

by transfer function.

Ans.

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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.

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Taking inverse z-transform

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

Direct from-I realization is shown below

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Direct form II realization is shown below.

4. Find the block diagram representations of parallel realization

of the system with transfer function.

Ans. Given

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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

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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

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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:

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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

The parallel form structure is shown below:

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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 = 4

The 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).

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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.

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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:

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6 Calculation of system function

7 Calculation of transfer function of digital filter

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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

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Using bilinear transformation

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(b) Impulse Invariance Method:

Taking Inverse Laplace transform we obtain

Let

Taking z-Transform

Assume T =1 sec.

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UNIT IV

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 and

These 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

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The realization is shown in fig below:

3. Discuss signal flow graph representation and lattice form structures for FIR

systems.

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,

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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,

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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.

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Here 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. What is fixed point representation? Ans. In fixed point representation the bits allowed for integer part and fractional part and so the position of binary point is fixed. The main

drawback of this representation is that, due to the fixed integer and fraction part, too large and to small values cannot be represented. The

bit to the right represent the fractional part of the number and those to the left represent the integer part. For Example: The binary no. 010.11100 has the value 2.875 is decimal. The negative numbers are represented in three different form for fixed

point arithmetic 1. Sign-magnitude form. 2. Ones-complement form 3. Twos-complement form. 1. Sign Magnitude form: In this form, the MSB is used to represent the given no. is positive or negative. Let N be the length of binary bits, then (N-i) bit will represent magnitude and MS represent sign. For example:

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2. Ones complement form: In this form the positive number is represented as in the sign magnitude notation. But the negative number is obtained by complementing all thebits of the positive

number. For example:

3. Twos complement form: In this form positive numbers are represented as in sign magnitude and ones complement. The negative number is obtained by complementing all the bits of the +ve number

and adding one to the least significant bit. For example:

5. What is floating point representation? Ans. In floating point representation, a positive number is represented as

Where M is called mantissa and it will be in binary fraction format. The

value of M will be in the range of and E is called exponent and it is either a positive or negative integer. In this form, both mantissa and exponent uses one bit for representing

sign. Usu - ally the LSB is mantissa and exponent is used to represent the sign.

A I in the LSB represent negative sign and a 0 in the LSB represent positive sign. The floating point representation is explained by considering a five bit mantissa and three bit exponent with a total size of eight bits. In

mantissa the LSB is used to represent the sign and other four bits are used to represent a binary fraction number. In exponent the LSB is

used to represent the sign and the other two bits are used to represent a binary integer number.

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6. 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

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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.

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Antisymmetric Condition:

For this

Then

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

impulse response is antisymmetric about

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The filters that satisfy the above conditions and have a delay of Ni 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

7. 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.

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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

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Step 6. To find the magnitude response of

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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,

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8. 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)

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For 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

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9. 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,

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Step3. To find the Hamming window sequence.

Step 4. To find the filter co-efficents

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

sequence.

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

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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

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Step 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.

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

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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,

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So But we know that

Step 4. To find the filter transfer function.

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Step 5. To find the realizable filter transfer function.

From the realizable filter transfer function, we have

Step 6. To find the magnitude frequency response

11. Design a low pass FIR filter using hamming window to

meet the following specifications.

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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

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The frequency response is given by

UNIT V

APPLICATIONS

1.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(ej). The z-transform

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is defined as andX(ej) is nothing but X(z) evaluated on the unit circle z = ej. 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 inputsingle 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 inputmultiple output (MIMO) systems and they are characterized by a (possibly rectangular) matrix

transfer function H(z).

2. Explain Application in channel equalization In 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.

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The discrete-time equivalent of an MIMO digital communication system with a symbol spaced equalizer (SSE) [39] is shown in Fig.2.3(a). 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 [57], it can be used with fat channel transfer functions, and it often allows for FIR solutions while SSE does

not.

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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 discrete time, this is

modeled as shown in Fig.2.3(b). 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

and the simplified scheme is shown in Fig.2.3(c). Note that the noise also needs to be modified, but this is not

the main point of discussion here. We recall from Section 2.3 that a zero-forcing FSE H(z) in Fig.2.3(c) 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 [57]; 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

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4. Design a two stage decimator for the following specification

Decimating factor=100 Passband=0

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5. Design a two stage decimator for the following specification Decimating factor=50

Passband=0

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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

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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

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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 /2, the quadrature frequency

8. Multirate applications in digital communications

The block diagram of the communication system that is the focus of this thesis is shown

Even though the figure title reads general, numerous variations and extensions are possible. The system focuses on a single user with the corresponding discrete message s(n). This message is to be

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transmitted to the single receiver, only after sustaining the

perturbations introduced by the transmission medium. These perturbations are modeled by the continuous-time LTI channel cc(t)

and the appropriate additive noise at the input of the receiver. The design challenge amounts to ensuring that the received sequence s(n)

resembles the original message under some criteria To this end, the receiver often introduces some redundancy combined with the

appropriate pre-processing. The goal of this block is to facilitate the signal reconstruction at the

transmitter (more about this in the following subsection). The obtained signal x(n) with rate 1/T is converted to analog and sent through the

medium to the receiver, only after the pulse-shaping. The combined effect of the pulse-shaping and the physical channel is often referred

to as the equivalent channel and denoted by fc(t). At the receiver, the corrupted signal is first sampled and digitized,

according to the sampling rate q/T. In this thesis we mostly focus on

the case when q > 1 which corresponds to acquiring more information than absolutely necessary about the signal and the channel. This further facilitates the signal recovery. The corresponding digital equalizer works at the higher rate and that combined with its

increased complexity is the price to pay for the improvement in performance achieved by oversampling. Finally, the signal rate needs

to be reduced back to the rate of s(n). This is usually achieved after decimating by q and removing the redundancy.

All the signals in Fig.1.9 can be scalars or vectors. Correspondingly, the LTI systems can be SISO or MIMO. The unconventional notation in

Fig is chosen to reflect these possibilities. Vector signals arise naturally in situations when there is a number of sensors, but can also be

obtained by blocking scalar signals . The system is further complicated if there is more than one In this thesis we mainly focus on the blocks

for redundancy insertion and pre-processing together with the

corresponding equalization and redundancy removal as well as equalization

For different values of the parameter q in the vector and the scalar

case Also, of special interest will be the system modifications for

application in multiuser communications and the corresponding

equalization algorithms.

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