Signals And System.docx

ASSIGNMENT Course Code Course Name Programme Department Faculty

ECC201A Signals and Systems B.TECH COMPUTER SCIENCE AND ENGINEERING FET

Name of the Student

B.PRAVEEN REDDY

Reg. No

17ETCS002043

Semester/Year

4TH / 2ND

Course Leader/s

Dr T. Christy Bobby

i

Declaration Sheet Student Name

B. PRAVEEN REDDY

Reg. No

17ETCS002043

Programme

B.TECH

Course Code

ECC201A

Course Title

Signals and Systems

Course Date

18/02/19

Course Leader

Dr T. Christy Bobby

Semester/Year 4TH / 2ND

to

18/03/19

Declaration The assignment submitted herewith is a result of my own investigations and that I have conformed to the guidelines against plagiarism as laid out in the Student Handbook. All sections of the text and results, which have been obtained from other sources, are fully referenced. I understand that cheating and plagiarism constitute a breach of University regulations and will be dealt with accordingly.

Signature of the Student

Date

18/03/19

Submission date stamp (by Examination & Assessment Section)

Signature of the Course Leader and date

Signature of the Reviewer and date

ii

Faculty of Engineering and Technology Ramaiah University of Applied Sciences Department Semester/Batch

Electronic and Communication Engineering

B. Tech. in CSE

Programme

rd

4 /2017

ECC201A Course Code Course Leader(s) Ms. Prafulla Kumari K. S & Dr T. Christy Bobby

Signals and Systems

Course Title

Assignment – 02 Reg.No.

Se cti on s Pa rt A Pa rt B

17ETCS002043

Name of Student B. PRAVEEN REDDY

Marks Marking Scheme

A1

MM ax ark s

Essay on DCT for Audio Signal Processing

First Examiner Marks

Moderator

5

Part-A Max Marks

5

B 1.1

Formulation and solution of the difference equation

6

B 1.2

Plotting of traffic

2

B 1.3

Comments on the variability of the traffic

2

B.1 Max

10

2

B 2.3

Computation of Laplace Transforms 𝑊(𝑠), 𝐻(𝑠) and 𝑌(𝑠) Computation of the response when only secret signal is sent Recovery of the message signal

B 2.4

Modification for the case with convolved inputs

3

Marks

B 2.1 B 2.2

2 3

B.2 Max Marks

10

Total Assignment Marks

25

iii

Course Marks Tabulation Component-1 (B) Assignment A

First Examiner

Remarks

Moderator

Remarks

B.1 B.2 Marks (Max 25 )

Signature of First Examiner

Signature of Moderator

Please note: 1. Documental evidence for all the components/parts of the assessment such as the reports, photographs, laboratory exam / tool tests are required to be attached to the assignment report in a proper order. 2. The First Examiner is required to mark the comments in RED ink and the Second Examiner’s comments should be in GREEN ink. 3. The marks for all the questions of the assignment have to be written only in the Component – CET B: Assignment table. 4. If the variation between the marks awarded by the first examiner and the second examiner lies within +/- 3 marks, then the marks allotted by the first examiner is considered to be final. If the variation is more than +/- 3 marks then both the examiners should resolve the issue in consultation with the Chairman BoE.

iv

PART A Discrete Cosine Transform The Discrete Cosine Transform (DCT) is a transform that is very common when encoding video and audio tracks on computers. Many "codecs" for movies rely on DCT concepts for compressing and encoding video files. The DCT can also be used to analyze the spectral components of images as well. The DCT is very similar to the DFT, except the output values are all real numbers, and the output vector is approximately twice as long as the DFT output. It expresses a sequence of finite data points in terms of sum of cosine functions. Audio Signal Processing Audio signal processing is at the heart of recording, enhancing, storing and transmitting audio content. Audio signal processing is used to convert between analog and digital formats, to cut or boost selected frequency ranges, to remove unwanted noise, to add effects and to obtain many other desired results. This process can be done on an ordinary PC or laptop, as well as specialized recording equipment. An audio signal is a representation of sound, typically as an electrical voltage. Audio signals have frequencies in the audio frequency range of roughly 20 to 20,000 Hz (the limits of human hearing). Audio signals may be synthesized directly, or may originate at a transducer such as a microphone, musical instrument pickup, phonograph cartridge, or tape head. Loudspeakers or headphones convert an electrical audio signal into sound. Digital representations of audio signals exist in a variety of formats. Discrete Cosine Transform for Audio Signal Processing The discrete cosine transform(DCT) signal processing technique come with many applications on a great number of engineering fields. Here I propose to apply techniques to the compression of audio signals. Using spectral analysis and the properties of the DCT, we can treat audio signals as sparse signals in the frequency domain. This is especially true for sounds representing tones. Here I propose the use of DCT to obtain an efficient representation of audio signals, especially when they are sparse in the frequency domain. By using the DCT as signal preprocessor in order to obtain a sparse representation in the frequency domain, here I show that the subsequent application of the signals with less information than the wellknown sampling theorem. This means that our results could be the basis for a new compression method for audio and speech signals. Conclusion I have proposed an efficient joint implementation of DCT, as a method to obtain a sparse audio signal representation, and the application of the compressive sampling algorithm to this sparse signal. The DCT speech signal representation has the ability to pack input data into as few coefficients as possible. This allows the quantizer to discard coefficients with relatively small amplitudes without introducing audio distortion in the reconstructed signal. Although the compressive sampling technique is used primarily for compression sample images, we achieve reasonable results due to the preprocessing of the audio signal.

PART B

v

Solution to Question No. B1: B1.1 Formulation and solution of the difference equation: The given data for this question is follow: 𝑦(𝑛) = 𝑎1𝑦(𝑛 − 1) + 𝑎2 𝑦(𝑛 − 2) + 𝑥(𝑛); 𝑥(𝑛) = 𝐴𝑐𝑜𝑠(𝜋𝑛) ; 𝑦(0) = 𝑑(0), 𝑦(1) = 𝑑(1) The values of 𝑎1 , 𝑎2 , 𝑎3 , 𝐴 , 𝑑(0) & 𝑑(1) are as follow: University ID

a1 a2

A

Student No:

5

6

-4

d(0) d(1) 4

5

So, The given differential equation is follow:  𝑦(𝑛) = 𝑎1𝑦(𝑛 − 1) + 𝑎2 𝑦(𝑛 − 2) + 𝑥(𝑛) Now, Substitute the value of 𝑎1&𝑎2 in the given equation  𝑦(𝑛) = 5𝑦(𝑛 − 1) − 4𝑦(𝑛 − 2) + 𝑥(𝑛) 𝑦(𝑛) − 5 𝑦(𝑛 − 1) + 4 𝑦(𝑛 − 2) = 𝑥(𝑛) -----------------------(i) Here, the given equation is discrete signals so, the auxiliary equation is of the given form 1 − 5𝜆−1 + 4𝜆−2 = 0 𝜆2 − 5𝜆 + 4 = 0 𝜆2 − 1𝜆 − 4𝜆 + 4 = 0 𝜆(𝜆 − 1) − 4(𝜆 − 1) = 0 (now, take (𝜆 − 1) common in next step) (𝜆 − 1) ∗ (𝜆 − 4) = 0 𝜆 = 1

&

𝜆=4

Therefore, the natural response of the given differential equation of discrete signals are as follow, 𝑦 𝑛 (𝑛) = 𝐶1 (1)𝑛 + 𝐶2 (4)𝑛 --------------------------(ii) Now, we have to find the particular solution first then we can find the forced response of the given system: For Particular solution: we have the given value of 𝑥(𝑛) = 𝐴𝑐𝑜𝑠(𝜋𝑛) Now, we can write 𝑥(𝑛) = 6 ∗ (−1)𝑛 [Note: here replace A by 8 and we can write 𝑐𝑜𝑠(𝜋𝑛) = (−1)𝑛 ] Now, the value of 𝑦 𝑝 (𝑛) = 𝑘 ∗ (−1)𝑛 -----------(iii) Now, put the value of eq (iii) in eq (i) 𝑘(−1)𝑛 − 5𝑘(−1)(𝑛−1) + 4𝑘(−1)(𝑛−2) = 6(−1)𝑛 5𝑘(−1)𝑛 4𝑘(−1)𝑛 ) + ( ) (−1) (−1)2

𝑘(−1)𝑛 − (

= 6(−1)𝑛 [Take (−1)𝑛 common in next step]

(𝑘 + 5𝑘 + 4𝑘) ∗ (−1)𝑛 = 6(−1)𝑛 10 𝑘 = 6 3

𝑘 = 5

vi

3 5

Hence, the solution of particular solution is 𝑦 𝑝 (𝑛) = ( ) ∗ (−1)𝑛 --------------(iv) Now, we have to write the equation of force response which is follow: 𝑦 𝑓 (𝑛) = 𝑦 𝑛 (𝑛) + 𝑦 𝑝 (𝑛) Add eq (ii) and eq (iv) 3

𝑦 𝑓 (𝑛) = 𝐶1 (1)𝑛 + 𝐶2 (4)𝑛 + (5) ∗ (−1)𝑛 ------------------(v) Now, for finding the value of 𝐶1 & 𝐶2 we have to apply the given initial conditions: Initial conditions are as follow:  𝑦(0) = 𝑑(0), 𝑦(1) = 𝑑(1) 𝑦(0) = 4, 𝑦(1) = 5 Now apply the initial conditions in eq (v) we get the value of 𝐶1 & 𝐶2 3 5

𝑦(0) = 𝐶1 (1)0 + 𝐶2 (4)0 + ( ) ∗ (−1)0 3

𝑦(0) = 𝐶1 + 𝐶2 + 5 3 5

𝐶1 + 𝐶2 + = 4 3

𝐶1 + 𝐶2 = 4 − 5 𝐶1 + 𝐶2 =

20−3 5

𝐶1 + 𝐶2 =

17 5

-------------------(vi)

Now, for 𝑦(1) 6

 𝑦(0) = 𝐶1 (1)1 + 𝐶2 (4)1 + (15) ∗ (−1)1 𝑦(1) = 1𝐶1 + 4𝐶2 −

3 5

3 5

1𝐶1 + 4𝐶2 − = 5 1𝐶1 + 4𝐶2 = 5 + 𝐶1 + 4𝐶2 =

25+3 5

𝐶1 + 4𝐶2 =

28 5

3 5

-------------------(vii)

Now, I have to find the value 𝐶1 & 𝐶2 by using the eq (vi) & (vii) Now, subtract the eq (vi) – eq (vii) 𝐶1 + 𝐶2 =

17 5 28

𝐶1 + 4𝐶2 = 5 -------------------------

−3𝐶2 =



−𝐶2 =

17−28 5

−11 15

11

𝐶2 = 15

vii

Now, put the value of 𝐶2 = 𝐶1 + 𝐶2 =

11 in 15

eq (vi)

17 5

𝐶1 +

11 15

17 5

𝐶1 =

17 11 − 5 15

𝐶1 =

51−11 15

=

40

𝐶1 = 15 8

𝐶1 = 3 Now substitute the value of 𝐶1 & 𝐶2 in the forced equation we get the total solution of the differential equation: 6 15

𝑦(𝑛) = 𝐶1 (1)𝑛 + 𝐶2 (4)𝑛 + ( ) ∗ (−1)𝑛 8

11

6

 𝑦(𝑛) = (3) (1)𝑛 + (15) (4)𝑛 + (15) ∗ (−1)𝑛 B1.2 Plotting of traffic: Here, we have to plot the traffic as a function of 𝑛 = 0,1, … . .50. For this we have the total solution of the differential equation 8

11

6

𝑦(𝑛) = (3) (1)𝑛 + (15) (4)𝑛 + (15) ∗ (−1)𝑛 now, we have to plot the discrete signal graph using stem plot in MATLAB: MATLAB CODE: n = [0:50]; y = (8/3).*(1).^n+(11/15).*(4).^n+(6/15).*(-1).^n; fprintf('%f\n',y); stem(n,y,'fill','--') hold on grid on xlabel('value of n = 1..50') ylabel('value of discrete signals') title('Plot of the discrete signal') Command window: 3.800000 5.200000 14.800000 49.200000 190.800000 753.200000 3006.800000 12017.200000

viii

48062.800000 192241.200000 768958.800000 3075825.200000 12303294.800000 49213169.200000 196852670.800000 787410673.200000 3149642686.800000 12598570737.199999 50394282942.799995 201577131761.199980 806308527038.799930 3225234108145.199700 12900936432574.799000 51603745730289.195000 206414982921150.780000 825659931684593.120000 3302639726738366.500000 13210558906953456.000000 52842235627813816.000000 211368942511255260.000000 845475770045021060.000000 3381903080180084200.000000 13527612320720337000.000000 54110449282881348000.000000 216441797131525390000.000000 865767188526101560000.000000 3463068754104406200000.000000 13852275016417625000000.000000 55409100065670500000000.000000 221636400262682000000000.000000 886545601050728000000000.000000 3546182404202912000000000.000000 14184729616811648000000000.000000 56738918467246592000000000.000000

ix

226955673868986370000000000.000000 907822695475945470000000000.000000 3631290781903781900000000000.000000 14525163127615128000000000000.000000 58100652510460510000000000000.000000 232402610041842040000000000000.000000 929610440167368160000000000000.000000 GRAPH: 29

10

Plot of the discrete signal

x 10

9

value of discrete signals

8 7 6 5 4 3 2 1 0

0

5

10

15

20 25 30 value of n = 1..50

35

40

45

50

Figure 1: Discrete plot of the signal obtained as a solution in above question. B1.3 Comments on the variability of the traffic: As we can see in the above graph that traffic is varying with respect to time such that for every increase in time there is some increase in traffic. In the Beginning phases we can see that the increase in traffic with respect to time is gradual but as it reaches the 45 unit mark a very high increase is noticed. Thus beyond this point the increase in traffic/variable in traffic is exponential. According to the values provide to me the variation in traffic results in a positive value. Thus Naturally it is a growth Rate.

x

Question No. B2 Solution to Question No. B2: The data is follows:𝑚(𝑡) = 𝑒 −𝑎𝑡 𝑢(−𝑡) 𝑤(𝑡) = 𝑒 −𝑏𝑡 𝑢(𝑡) ℎ(𝑡) = 𝑐 𝑢(𝑡) & 𝑠(𝑡) = 𝑚(𝑡) + 𝑤(𝑡) The value of 𝑎 , 𝑏 & 𝑐 𝑎𝑟𝑒 𝑎𝑠 𝑓𝑜𝑙𝑙𝑜𝑤: a

b

c

1

4

8

B2.1 Computation of Laplace Transforms 𝑊(𝑠), 𝐻(𝑠) and 𝑌(𝑠): Here, we have to find the Laplace transforms of 𝑊(𝑠), 𝐻(𝑠)& 𝑌(𝑠) Now, first I have to find 𝑊(𝑠)first, For this we have the given eq of 𝑤(𝑡) = 𝑒 −𝑏𝑡 𝑢(𝑡) 1 𝑡>0 Here, 𝑢(𝑡) is unit step function which is defined as 𝑢(𝑡) = { } 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 Graph of u(t) is follow:  𝑤(𝑡) = 𝑒 −𝑏𝑡 𝑢(𝑡) Now, apply Laplace transform ∞

𝐿[𝑤(𝑡)] = ∫−∞ 𝑒 −𝑠𝑡 𝑒 −𝑏𝑡 𝑢(𝑡)𝑑𝑡 0

∞

 𝐿[𝑤(𝑡)] = ∫−∞ 𝑒 −𝑠𝑡 𝑒 −𝑏𝑡 ∗ 0𝑑𝑡 + ∫0 𝑒 −𝑠𝑡 𝑒 −𝑏𝑡 ∗ 1𝑑𝑡 ∞

𝑊(𝑠) = ∫0 𝑒 −(𝑠+𝑏)𝑡 𝑑𝑡 𝑒 −(𝑠+𝑏)𝑡

 𝑊(𝑠) = [ −(𝑠+𝑏) ] 𝑊(𝑠) =

∞ 0

1 [𝑒 −∞ −(𝑠+𝑏)

− 𝑒0]

1

𝑊(𝑠) = 𝑠+𝑏 [Now, replace the b by 3 in next line we get,] 1

 𝑊(𝑠) = 𝑠+4 ---------------(i) 1 𝑠+4

Hence, Laplace transform of 𝑊(𝑠) = Now, I have to find the value of 𝐻(𝑠)

For, this I have the given value of ℎ(𝑡) = 𝑐 𝑢(𝑡) Now, apply Laplace transform ∞

𝐿[ℎ(𝑡)] = ∫−∞ 𝑒 −𝑠𝑡 ∗ 𝑐 ∗ 𝑢(𝑡)𝑑𝑡 0

∞

𝐿[ℎ(𝑡)] = 𝑐 ∗ [∫−∞ 𝑒 −𝑠𝑡 ∗ 0𝑑𝑡 + ∫0 𝑒 −𝑠𝑡 ∗ 1𝑑𝑡 ] ∞

𝐻(𝑠) = 𝑐 ∗ ∫0 𝑒 −𝑠𝑡 𝑑𝑡

xi

𝑒 −𝑠𝑡

 𝐻(𝑠) = 𝑐 ∗ [ −𝑠 ] 𝐻(𝑠) =

𝑐 [𝑒 −∞ −𝑠

∞ 0

− 𝑒 0]

𝑐

𝐻(𝑠) = 𝑠 [Now, replace the c by 9 in next line we get,] 8

 𝐻(𝑠) = 𝑠 --------(ii) Hence, the Laplace transform of 𝐻(𝑠) =

8 𝑠

Now, we have to find the value of 𝑦(𝑠) For finding this we have to draw the L.T.I system and Find 𝑦(𝑡) L.T.I system is follow:

Ʃ

From the above L.T.I system we can calculate 𝑦(𝑡) 𝑦(𝑡) = ℎ(𝑡) ∗ 𝑠(𝑡) 𝑦(𝑡) = ℎ(𝑡) ∗ [𝑚(𝑡) + 𝑤(𝑡)] Now, apply Laplace transform 𝐿[ 𝑦(𝑡)] = 𝐻(𝑠) ∗ [𝑀(𝑠) + 𝑊(𝑠)]  𝑌(𝑠) = 𝐻(𝑠) ∗ 𝑀(𝑠) + 𝐻(𝑠) ∗ 𝑊(𝑠) ----------------(iii) Here, I have to find the value of 𝑚(𝑡) first,  𝑚(𝑡) = 𝑒 −𝑎𝑡 𝑢(−𝑡) Here, I have 𝑢(−𝑡) as a function so the value get reversed in 𝑢(−𝑡) 1 𝑡<0 Here, 𝑢(−𝑡) is unit step function which is defined as 𝑢(−𝑡) = { } 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 Now, take the value of 𝑚(𝑡) = 𝑒 −𝑎𝑡 𝑢(−𝑡) Now, apply Laplace transform ∞

𝐿[𝑚(𝑡)] = ∫−∞ 𝑒 −𝑠𝑡 ∗ 𝑒 −𝑎𝑡 ∗ 𝑢(−𝑡)𝑑𝑡 0

∞

 𝐿[𝑚(𝑡)] = ∫−∞ 𝑒 −𝑠𝑡 ∗ 𝑒 −𝑎𝑡 ∗ 1𝑑𝑡 + ∫0 𝑒 −𝑠𝑡 ∗ 𝑒 −𝑎𝑡 ∗ 0𝑑𝑡 0

𝑀(𝑠) = ∫−∞ 𝑒 −(𝑠+𝑎)𝑡 𝑑𝑡 [Note: here, I Consider – (𝑠 + 𝑎) = 𝐶] 0

 𝑀(𝑠) = ∫−∞ 𝑒 𝐶𝑡 𝑑𝑡 𝑀(𝑠) = [

0 𝑒 𝐶𝑡 ] 𝐶 −∞

1

 𝑀(𝑠) = 𝐶 [𝑒 0 − 𝑒 −∞ ] 1

 𝑀(𝑠) = −(𝑠+𝑎) [1 − 0] [replace 𝐶 𝑏𝑦 – (𝑠 + 𝑎)] 1

 𝑀(𝑠) = −(𝑠+1) [𝑟𝑒𝑝𝑙𝑎𝑐𝑒 𝑎 𝑏𝑦 1] 1

𝑀(𝑠) = −(𝑠+1) ----------(iv) 1

Hence, the value of 𝑀(𝑠) = −(𝑠+1) Now, From eq (iii)

xii

 𝑌(𝑠) = 𝐻(𝑠) ∗ 𝑀(𝑠) + 𝐻(𝑠) ∗ 𝑊(𝑠) Now, put the value of 𝐻(𝑠), 𝑀(𝑠)&𝑊(𝑠) we get the value of 𝑌(𝑠)as 8

1

8

1

 𝑌(𝑠) = 𝑠 ∗ −(𝑠+1) + 𝑠 ∗ 𝑠+4 8

1

1

 𝑌(𝑠) = [ + ] 𝑠 −(𝑠+1) 𝑠+4 8 (𝑠+1)−(𝑠+4)

 𝑌(𝑠) = 𝑠 [ (𝑠+1)(𝑠+4) ] 8

−3

 𝑌(𝑠) = 𝑠 [(𝑠+1)(𝑠+4)] −24

𝑌(𝑠) = [𝑠∗(𝑠+1)∗(𝑠+4)] ----------(v) Hence, we get the value of 𝑌(𝑠) = [

−24 ] 𝑠∗(𝑠+1)∗(𝑠+4)

B2.2 Computation of the response when only secret signal is sent: 𝑣(𝑡) = 𝑤(𝑡) ∗ ℎ(𝑡) Now, apply Laplace transform 𝐿[𝑣(𝑡)] = 𝑊(𝑠) ∗ 𝐻(𝑠) Now, put the value of 𝑊(𝑠) 𝑎𝑛𝑑 𝐻(𝑠) in the above equation: From eq (i) & (ii) 𝑉(𝑠) = 𝑊(𝑠) ∗ 𝐻(𝑠) -------(vi) 𝑉(𝑠) =

1 𝑠+4

∗

8 𝑠

8

𝑉(𝑠) = (𝑠+4)∗𝑠 -----------------(vii) B2.3 Recovery of the message signal: From, equation(iii) we have  𝑌(𝑠) = 𝐻(𝑠) ∗ 𝑀(𝑠) + 𝐻(𝑠) ∗ 𝑊(𝑠) And, from equation (v) we have  𝑉(𝑠) = 𝑊(𝑠) ∗ 𝐻(𝑠) Now, subtract the eq(iii) – eq(vi) we get,  𝑌(𝑠) − 𝑉(𝑠) = 𝐻(𝑠) ∗ 𝑀(𝑠) + (𝐻(𝑠) ∗ 𝑊(𝑠)) − (𝑊(𝑠) ∗ 𝐻(𝑠))  𝑌(𝑠) − 𝑉(𝑠) = 𝐻(𝑠) ∗ 𝑀(𝑠) 𝑀(𝑠) =

(𝑌(𝑠)−𝑉(𝑠)) 𝐻(𝑠)

Now, replace the value of 𝑌(𝑠), 𝑉(𝑠)&𝐻(𝑠) from eq (vii) , (v) & (ii) we get,

xiii

𝑀(𝑠) =

(

−24 8 )−( ) 𝑠∗(𝑠+1)∗(𝑠+4) 𝑠∗(𝑠+4) 8 𝑠

−3

8

8

[take 𝑠 common then the 𝑠 get divided and we get the eq as]

1

 𝑀(𝑠) = ((𝑠+1)∗(𝑠+4)) − ((𝑠+4)) −3−𝑠−1

𝑀(𝑠) = (𝑠+1)∗(𝑠+4) (𝑠+4)

𝑀(𝑠) = − (𝑠+1)∗(𝑠+4) 1

𝑀(𝑠) = −(𝑠+1) Now, we have to find the inverse Laplace of 𝑀(𝑠) to get the value of 𝑚(𝑡) 1

𝐿−1 [𝑀(𝑠)] = 𝐿−1 [−(𝑠+1)] 1

𝑚(𝑡) = −𝑒 −1𝑡 [Note: Laplace inverse of 𝐿−1 [−(𝑠+1)] = −𝑒 −1𝑡 ] Now, we can replace −1 as 𝑢(−𝑡)  𝑚(𝑡) = 𝑒 −1𝑡 ∗ 𝑢(−𝑡) Hence, we get the correct recovery signal. B2.4 Modification for the case with convolved inputs: Here, in this question we have to Analyze the effect of change in operation between the message and the watermark on the response (𝑤(𝑡) and 𝑚(𝑡) are convolved instead of being added. Here we have to draw the L.T.I system to find convolved inputs: The L.T.I system is as follow:

Now, here we have to take 𝑠(𝑡) = 𝑤(𝑡) ∗ 𝑚(𝑡) (as per L.T.I system) Then, From above eq of 𝑦(𝑡) = ℎ(𝑡) ∗ 𝑠(𝑡) Now, apply Laplace transform we get, 𝐿[𝑦(𝑡)] = 𝐿[ℎ(𝑡) ∗ 𝑤(𝑡) ∗ 𝑚(𝑡)] 𝑌(𝑠) = 𝐻(𝑠) ∗ 𝑊(𝑠) ∗ 𝑀(𝑠) Now, put the value of 𝐻(𝑠), 𝑊(𝑠) & 𝑀(𝑠) we get, 8

1

1

 𝑌(𝑠) = 𝑠 ∗ 𝑠+4 ∗ −(𝑠+1) −8

 𝑌(𝑠) = 𝑠∗(𝑠+4)∗(𝑠+1)

xiv

15