Solved University Questions Papers - GL DataScience

Solved University Questions

Papers

B.E/B.Tech Degree Examination Apr/May 2017

Third Semester

Electronics and Communication Engineering

EC6303 - Signals & Systems

(Common to Biomedical Engineering and Medical Electronics Engineering)

(Regulation 2013)

Answer All Questions.

Part - A (10 × 2 = 20 Marks)

1. Find the summation x (n) =∞∑

n=−∞δ (n − 1) sin 2n.

Refer Page No. 1.57, Example 1.30, sum 2

2. Define a linear system.

Refer Page No. 1.77, section 1.11.1

3. What is the condition for the existence of fourier series for a signal?

Refer Page No. 2.5, section 2.2

4. Give the expression for convolution integral.

y (t) =

∞∫

−∞

x (τ)h (t − τ) dτ

This equation called convolution integral.

S.2 Signals and Systems

5. Given h (t) what is the step response of a CT-LTI system.

6. What is the z-transform of a unit step response.

Refer Page No. 2.52, Example 4.4

7. State Parsevals’ theorem for a continuous time aperiodic signal.

Its states that

E =

∞∫

−∞

|x (t)|2 dt

E =1

2π

∞∫

−∞

|X (jω)|2 dω

8. Find x (∞) of the signal for with z-transform is given by

X (z) =z + 1

3 (z − 1) (z + 0.9)

x (∞) = ltz→1

(z − 1

z

)

X (z)

= ltz→1

(z − 1

z

)

×z + 1

3 (z − 1) (z + 0.9)

= ltz→1

(z + 1)

3z (z + 0.9)

=2

3 (1.9)

x (∞) = 0.3508

9. What is the difference between recursive and nonrecursive system?

Refer Page No. 5.59, Section 5.8

10. What is the necessary and sufficient condition on impulse response for sta-

bility of a causal LTI system.

For CT; Condition for stability

∞∫

−∞

h (t) < ∞

Solved University Questions Papers S.3

For DT; Condition for stability

∞∑

k=−∞

h (k) < ∞

Part - B (5 × 13 = 65 Marks)

11. (a) Find out whether the following signals are periodic or not. If periodic

find the period.

(1) x (t) = 2 cos (10t + 1) − sin (4t − 1)Refer Page No. 1.9. Example 1.6, Sum 1

(2) x (n) = cos (0.1πn)

ω = 0.1π

N =2π

ω× m

=2π

0.1π× m

N = 20 × m ∴ m = 1

N = 20 (rational number)

N = 20 samples

So given signal is periodic.

(b) Find out whether the following signals are energy or power signal or

neither power nor energy. Determine power on energy as the case may

be for the signal x (t) = u (t) + 5u (t − 1) − 2u (t − 2).

u(t)

u(t 1)-

5u(t 1)- 5u(t 1)-

1

1 1

1

1

5 5


(Or)

Determine the properties viz linearity, causality, time invariance and

dynamicity of the given systems.

y (t) =d2y (t)

dt2+ 3t

dy (t)

dt+ y (t) = x (t)

Refer page No. 1.86, Example 1.47, sum 1

y1 (n) = x(n2

)+ x (n)

y2 (n) = log10 x (n)

Refer page No. 1.89, Example 1.47, sum 4

12. (a) Obtain the fourier co-efficient and write the quadrature form of a fully

rectified sine wave.


(Or)

(b) Determine the inverse laplace transform of the following

(i) x (s) =1 − 2s2 − 14s

s (s + 3) (s + 4)

1 − 2s2 − 14s

s (s + 3) (s + 4)=

A

s+

B

(s + 3)+

C

(s + 4)

1 − 2s2 − 14s

s (s + 3) (s + 4)=

A (s + 3) (s + 4) + Bs (s + 4) + Cs (s + 3)

s (s + 3) (s + 4)

1 − 2s2 − 14s = A (s + 3) (s + 4) + Bs (s + 4) + cs (s + 3)

sub s = −3

1 − 2 (−3)2 − 14 (−3) = B (−3) (−3 + 4)

1 − 18 + 42 = B (−3)

25 = B (−3)

B =−25

3

sub s = −4

1 − 2 (−4)2 − 14 (−4) = C (−4) (−4 + 3)

1 − 32 + 56 = C (4)

25 = C (4)

C =25

4


sub s = 0

1 = A (3) (4)

A =1

12

X (s) =1

12s+

(−25

3

)1

s + 3+

25

4×

1

(s + 4)

x (t) = L−1X (s)

Applying inverse laplace transform

x (t) =1

12u (t) −

25

3e−3tu (t) +

25

4e−4tu (t)

(ii) x (s) =2s2 + 10s + 7

(s + 1) (s2 + 3s + 2)

2s2 + 10s + 7

(s + 1) (s2 + 3s + 2)=

A

(s + 1)+

B

(s + 1)2+

C

(s + 2)

2s2 + 10s + 7

(s + 1) (s2 + 3s + 2)=

A (s + 2) (s + 1) + B (s + 2) + C (s + 1)2

(s + 1)2 (s + 2)

2s2 + 10s + 7 = A[s2 + s + 2s + 2

]+ Bs + 2B + C

[s2 + 2s + 1

]

= A[s2 + 3s + 2

]+ Bs + 2B + Cs2 + Cs2 + C

= As2 + 3sA + 2A + Bs + 2B + Cs2 + c2s + C

= s2 [A + C] + s [3A + B + 2C] + (2A + 2B + C)

Equating s2

s2 [A + C] = 2

A + C = 2

Equating s

3A + B + 2C = 10

Equating constant

2A + 2B + C = 7


Solving this three equations

A = 7; B = −1; C = −5

X (s) =7

(s + 1)−

1

(s + 1)2−

5

(s + 2)

Taking inverse laplace transform

x (t) = 7e−tu (t) − te−tu (t) − 5e−2tu (t)

13. (a) A causal LTI system having a frequency response H (jΩ) =1

jΩ + 3is producing an output y (t) = e−3tu (t) − e−4tu (t) for a particular

input x (t). Determine x (t).

Solution:

Given

H (jΩ) =1

jΩ + 3

y (t) = e−3tu (t) − e−4tu (t)

Applying fourier transform on both side

H (jΩ) =1

jΩ + 3−

1

jΩ + 4

=jΩ + 4 − jΩ − 3

(jΩ + 3) (jΩ + 4)

=1

(jΩ + 3) (jΩ + 4)

We know that

Y (jΩ) = H (jΩ)X (jΩ)

X (jΩ) =Y (jΩ)

H (jΩ)

=

1(jΩ+4)(jΩ+3)

1(jΩ+3)

=1

(jΩ + 4) (jΩ + 3)× (jΩ + 3)

X (jΩ) =1

(jΩ + 4)


Taking inverse fourier transform

x (t) = e−4tu (t)

(Or)

(b) Realize the given system in parallel form

H (s) =s (s + 2)

s2 + 8s2 + 19s + 12

Solution:

H (s) =s (s + 2)

(s + 1) (s2 + 7s + 12)

=s (s + 2)

(s + 1) (s + 4) (s + 3)

H (s) =A

(s + 1)+

B

(s + 4)+

C

(s + 3)

s (s + 2)

(s + 1) (s + 4) (s + 3)=

A (s + 3) (s + 4) + B (s + 1) (s + 3) + C (s + 1) (s + 4)

(s + 1) (s + 4) (s + 3)

s (s + 2) = A (s + 3) (s + 4) + B (s + 1) (s + 3) + C (s + 1) (s + 4)

sub s = −3

C (−3 + 1) (−3 + 4) = 3

C (−2) = 3

C = −3

2

sub s = −4

B (−3) (−1) = 8

B =8

3

sub s = −1

A (2) (3) = −1

A =−1

5

H (s) =−1

5 (s + 1)+

8

3 (s + 4)+

(−3/2)

(s + 3)


H (s) =−0.2

(s + 1)︸︷︷︸

⇓H1(s)

+2.6

(s + 4)︸︷︷︸

⇓H2(s)

−1.5

(s + 3)︸︷︷︸

⇓H3(s)

H1 (s) =−0.2

(s + 1); H2 (s) =

2.6

(s + 4); H3 (s) =

−1.5

(s + 3)

H1 (s) =Y1 (s)

X1 (s)×

W1 (s)

W1 (s)=

−0.2

(s + 1)

Y1 (s)

W1 (s)= −0.2

Y1 (s) = −0.2W1 (s)

W1 (s)

X1 (s)=

1

(s + 1)

W1 (s) s + W1 (s) = X1 (s)

sW1 (s) = X1 (s) − W1 (s)

S1

1Y (S)1

X (S)1

SW (S)1

-0.2

H2 (s) =2.6

s + 4

H2 (s) =Y2 (s)

X2 (s)×

W2 (s)

W2 (s)=

2.6

s + 4

Y2 (s)

W2 (s)= 2.6

Y2 (s) = 2.6W2 (s)

W2 (s)

X2 (s)=

1

(s + 4)

sW2 (s) + 4W2 (s) = X2 (s)

sW2 (s) = X2 (s) − 4W2 (s)

S1

-4Y (S)2

X (S)2

SW (S)2

2.6


H3 (s) =−0.2

(s + 1)

H3 (s) =Y3 (s)

X3 (s)×

W3 (s)

W3 (s)=

−0.2

(s + 1)

Y3 (s)

W3 (s)= −0.2

W3 (s)

X3 (s)=

1

(s + 1)

sW3 (s) + W3 (s) = X3 (s)

sW3 (s) = X3 (s) − W3 (s)

S1

Y (S)3

X (S)3

SW (S)3

-0.2

Combine all in parallel form

S1

X(S)

-0.2

S1

-1.5

S1

Y(S)

1

-4

-0.2

14. (a) State and prove sampling theorem.


(Or)

(b) State and prove the following properties of DTFT


(i) Differentiation in frequency

Refer Page No. 2.50, Section 4.3.8

(ii) Convolution in frequency domain.

Refer Page No. 2.49, Section 4.3.7

15. (a) Perform convolution to find the response of the system h1 (n) and

h2 (n) for the input sequences x1 (n) and x2 (n) respectively.

(i) x1 (n) = 1, −1, 2, 3, h1 (n) = 1, −1, 3, 1

(ii) x2 (n) = 1, 2, 3, 2, h2 (n) = 1, 2, 2

(i) x1 (n) = 1, −1, 2, 3, h1 (n) = 1, −1, 3, 1

Matrix method

1 2 3

1 2 31

3 6 9

1 2 3

-2 -4 -6

-1

-3

2

-1

-1

1

3

-2

y (n) = 1, −3, 7, −3, −1, 11, 3

Linear method

Step 1:-

The total number of output sequence = No. of sequence in x (n)

+ No. of sequence in h (n) − 1

= 4 + 4 − 1

= 7

Step 2:-

x (n) starts at n1 = 0

h (n) starts at n2 = 0

n = n1 + n2

n = 0

n = 0, 1, 2, 3, 4, 5, 6


Step 3:-

y1 (n) =∞∑

k=−∞

x (k)h (n − k)

n = 0;

y1 (0) =6∑

k=0

x (k)h (−k)

y1 (0) = 1

n = 1;

y1 (1) =6∑

k=0

x (k)h (1 − k)

y1 (1) = −3

n = 2;

y1 (2) =6∑

k=0

x (k)h (2 − k)

y1 (2) = 7

n = 3;

y1 (3) =6∑

k=0

x (k)h (3 − k)

y1 (3) = −3

n = 4;

y1 (4) =6∑

k=0

x (k)h (4 − k)

y1 (4) = −1

n = 5;

y1 (5) =

6∑

k=0

x (k)h (5 − k)

y1 (5) = 11


n = 6;

y1 (6) =

6∑

k=0

x (k)h (6 − k)

y1 (6) = 3

y1 (n) = 1, −3, 7, −3, −1, 11, 3

(ii) x2 (n) = 1, 2, 3, 2, h2 (n) = 1, 2, 2Matrix method

1 3 2

1 3 41

2 6 4

2 6 4

4

4

2

2

2

2

x (n)2

n (n)2

®

¯

y2 (n) = 1, 4, 9, 14, 10, 4

Linear method:-

Step 1:-

The total number of output sequence = No. of sequence in x (n)

+ No. of sequence in h (n) − 1

= 4 + 3 − 1

= 6

Step 2:-

x (n) starts at n1 = 0

h (n) starts at n2 = 0

n = n1 + n2

n = 0

n = 0, 1, 2, 3, 4, 5

Step 3:-

y2 (n) =∞∑

k=−∞

x (k)h (n − k)


n = 0;

y2 (0) =

6∑

k=0

x (k)h (−k)

y2 (0) = 1

n = 1;

y2 (1) =6∑

k=0

x (k)h (1 − k)

y2 (1) = 4

n = 2;

y2 (2) =6∑

k=0

x (k)h (2 − k)

y2 (2) = 9

n = 3;

y2 (3) =6∑

k=0

x (k)h (3 − k)

y2 (3) = 14

n = 4;

y2 (4) =6∑

k=0

x (k)h (4 − k)

y2 (4) = 10

n = 5;

y2 (5) =6∑

k=0

x (k)h (5 − k)

y2 (5) = 4

y2 (n) = 1, 4, 9, 14, 10, 4

(Or)


(b) For a causal LTI system the input x (n) and output y (n) are related

through a difference equation y (n) −1

6y (n − 1) −

1

6y (n − 2) =

x (n).

Determine the frequency response H(ejω

)and impulse response h (n)

of the system.

y (n) −1

6y (n − 1) −

1

6y (n − 2) = x (n)

Taking z-transform

Y (z) −1

6z−1Y (z) −

1

6z−2Y (z) = X (z)

Y (z)

X (z)=

1

1 −1

6z−1 −

1

6z−2

H (z) =1

1 −1

6z−1 −

1

6z−2

Frequency response

Replace z = e−jω

H(ejω

)=

1

1 −1

6e−jω −

1

6e−2jω

B =

(

z +1

3

)

×z

(z − 1

2

) (z + 1

3

)

∣∣∣∣∣sub z=− 1

3

=−1/32 − 5

6

=−1

3×

6

−5

B =2

5H (z)

z=

5/3(

z −1

2

) +2/5

(

z +1

3

)

H (z) =5/3z

(

z −1

2

) +2/5z

(

z +1

3

)


Taking inverse z-transform

H (z) =5

3

(1

2

)n

u (n) +2

5

(−1

3

)n

u (n)

Part - C

16. (a) Using laplace transform determine the response of the system de-

scribed by the equation

d2y (t)

dt2+ 5

dy (t)

dt+ 4y (t) =

dx (t)

dt

with initial conditions y (0) = 0;dy (t)

dt

∣∣∣∣t=0

= 1 for

To obtain impulse response

H (z) =z2

z2 −1

6z −

1

6

H (z) =z2

(

z −1

2

)(

z +1

3

)

Y (z)

X (z)=

z2

(

z −1

2

)(

z +1

3

)

Y (z) =z2X (z)

(

z −1

2

)(

z +1

3

)

Y (z)

z=

z(

z −1

2

)(

z +1

3

)

Y (z)

z=

A(

z −1

2

) +B

(

z +1

3

)

A =

(z − 1

2

)× z

(

z −1

2

)(

z +1

3

)

∣∣∣∣∣∣∣∣sub z= 1

2


A =

1

2(1

2+

1

3

) =

1

25

6

A =5

3

the input x (t) = e−2tu (t).

Solution:d2y (t)

dt2+ 5

dy (t)

dt+ 4y (t) =

dx (t)

dtApplying laplace transform[

s2Y (s) − sY(0)−

dy (0)

dt

]

+[5s − Y

(0)]

+ 4Y (s) = sX (s)

Sub y (0) = 0;dy (0)

dt= 1

(s2Y (s) − 1

)+ [5s − 0] + 4Y (s) = sX (s)

s2Y (s) − 15s + 4Y (s) = sX (s)

Y (s)[s2 + 5s + 4

]= sX (s) + 1

Given x (t) = e−2tu (t)

X (s) =1

(s + 1)

Y (s)[s2 + 5s + 4

]=

s

s + 1+ 1

Y (s) =s

(s + 1) (s2 + 5s + 4)+

1

(s2 + 5s + 4)

Y (s) =s + (s + 1)

(s + 1) (s2 + 5s + 4)

Y (s) =2s + 1

(s + 1) (s2 + 4) (s + 1)

Y (s) =2s + 1

(s + 1)2 (s + 4)

2s + 1

(s + 1)2 (s + 4)=

A

(s + 1)+

B

(s + 1)2+

C

(s + 4)

2s + 1 = A (s + 4) (s + 1) + B (s + 4) + C (s + 1)2


= A[s2 + 5s + 4

]+ Bs + 4B + Cs2 + 2sC + C

(2s + 1) = As2 + 5sA + 4A + Bs + 4B + Cs2 + 2sC + C

Equating s2

s2 [A + C] = 0

A + C = 0

Equating ss [5A + B + 2C] = 2

Equating constant

4A + 4B + C = 1

Solving

A = 0.7; B = −0.3; C = −0.7

Y (s) =0.7

(s + 1)−

0.3

(s + 1)2+

(−0.7)

(s + 4)


y (t) = 0.7e−tu (t) − 0.3te−tu (t) − 0.7e−4tu (t)

(Or)

(b) Determine the steady state response for the system with impulse re-

sponse h (n) = [j0.5]n for an input x (n) = cos πnu (n).


B.E/B.Tech Degree Examination Nov/Dec 2016

Third Semester



(Common to Biomedical Engineering and Medical Electronics Engineering)

(Regulation 2013)


Part - A (10 × 2 = 20 Marks)

1. Give the mathematical and graphical representation of a continuous time

and discrete time unit impulse functions.

Refer Page No. 1.27; Section 1.4.6

2. State the difference between causal and noncausal system.

Refer Page No. 1.84; Section 1.11.5

3. Find the fourier series representation of the signals x (t) =cos 2πt

3and

determine the fourier series coefficients.

x (t) =ej 2π

3t + e−j 2π

3t

2(1)

x (t) = a0 + a1ejω0t + a−1e

−jω0t (2)

Compare (1) & (2)

a0 = 0

a1 =1

2

a−1 =1

2

4. Find the laplace transform of x (t) = e−atu (t).


5. Convolve the following signals u (t − 1) and δ (t − 1).


6. Given H (s) =s

s2 + 2s + 1. Find the differential equation representation

of the system.

H (s) =Y (s)

X (s)=

s

s2 + 2s + 1

s2Y (s) + 2sY (s) + Y (s) = sX (s)

d2y (t)

dt2+ 2

dy (t)

dt+ y (t) =

dx (t)

dt

7. Find the Nyquist rate of the signal x (t) = sin 200πt − cos 100πt.

x (t) = sin 200πt − cos 100πt

f1 =200 Hz; f2 = 100Hz; (maximum frequency) ; fm = 200 Hz

Nyquist rate = 2W

= 2fm

= 2 × 200

= 400 Hz

Nyquist rate = 400 Hz

8. Find the z-transform of the signal and its associated ROC x (n) = 2, −1, 3, 0, 2.

x (n) =

2, −1, 3

↑n=0

, 0, 2

X (z) =∞∑

n=−∞

x (n) z−n

=2∑

n=−2

x (n) z−n

= x (−2) z2 + x (−1) z1 + x (0) z0 + x (1) z−1 + x (2) z−2

Y (z) = 2z−2 − 1z−1 + 3 + 0z−1 + 2z−2

Y (z) = 2z−2 + z−1 + 3 + 2z−2

9. Convolve the following sequences.

x (n) = 1, 2, 3

h (n) = 1, 1, 2


1 2

1 21

3 6

2 4

3

2

1

1

3

2

h(n)

x(n)

y (n) = 1, 3, 7, 7, 6

10. Given the sequence function H (z) = 2+3z−1 +4z−3 − 5z−4. Determine

the impulse response h (n).

Part - B (5 × 13 = 65 Marks)

11. (a) Determine whether the system is linear, time invariant, causal and

memoryless

y (t) =1

2

t∫

−∞

x (z) dz


(Or)

(b) Sketch the following signals

(i) u (t − 2)

(ii) r (−t + 3)

(iii) 2δ (n + 2) + δ (n) − 2δ (n − 1) + 3δ (n − 3)

(iv) u (n + 2)u (−n + 3)

where u (t), r (t), δ [n], u [n] represent continuous time unit step, con-

tinuous time ramp, discrete time impulse and discrete time step func-

tions respectively.

Solution:

(i) u (t − 2)u(t 2)-

2

1


(ii) r (−t + 3)r( t)-

t t

t

r( t+3)-

(iii) 2δ (n + 2) + δ (n) − 2δ (n − 1) + 3δ (n − 3)

d(n+2)

t

d(n)

n

2 (n+2)+ (n)d d

d -(n 1)

-2

3 (n 3)d -

n

2 (n 1)+3 (n 3)d - d -

2 (n+2)+ (n) 2 (n 1)+3 (n )d d - d - d -3


(iv) u (n + 2)u (−n + 3)

u(-n+3)

u(n+2)u(-n+3)

12. (a) Find the fourier transform of the signal x (t) = cosΩ0tu (t).

(b) State and prove the multiplication and convolution property of fourier

transform.

13. (a) Convolve the following signals

x (t) = e−3tu (t)

h (t) = u (t + 3)

Solution:

y (t) =

∞∫

−∞

x (τ)h (t − τ) dτ

=

∞∫

−∞

e−2τ u (t + 3 − τ) dτ

u (τ)u (t + 3 − τ) = 1

limit from 0 to (t + 3)

=

t+3∫

0

e−3τ · 1 dτ

=

[e−3τ

τ

]t+3

0

=−1

2

e−3(t+3) − e0

⇓1


y (t) =1

2

[

1 − e−3(t+3)]

(b) A system is described by the differential equationd2y (t)

dt2+6

dy (t)

dt+

8y (t) =d

dtx (t) + x (t). Find the transfer function and the output

signal y (t) for x (t) = δ (t).

Solution:

d2y (t)

dt2+ 6

dy (t)

dt+ 8y (t) =

d

dtx (t) + x (t)

Applying laplace transform on both side

s2Y (s) + 6sY (s) + 8Y (s) = sX (s) + X (s)

Y (s)

X (s)=

s + 1

s2 + 6s + 8

Transfer function

H (s) =Y (s)

X (s)=

s + 1

s2 + 6s + 8

Y (s) = X (s) ·s + 1

s2 + 6s + 8

x (t) = δ (t)

X (s) = 1

Y (s) =s + 1

s2 + 6s + 8=

s + 1

(s + 4) (s + 2)=

A

s + 4+

B

s + 2

s + 1 = A (s + 2) + B (s + 4)

A =3

2

B =−1

2

Y (s) =3/2

(s + 4)−

1/2

(s + 2)


y (t) =3

2e−4tu (t) −

1

2e−2tu (t)


14. (a) (i) Discuss the effects of undersampling a signal using necessary di-

agrams. (5)

Refer Page No. 5.36

(ii) Find the z-transform of x (n) = anu (n) − bnu (−n − 1) and

specify its ROC. (8)

Ans:

X (z) =z

z − a+

z

z − b

(b) (i) Give the relation between discrete time fourier transform (DTFT)

and z transform. (5)

Both DTFT and x transform are used to analysis aperiodic dis-

crete time signals.

DTFT

x (n)F.T−−→ X

(ejω

)

X(ejω

) I.F.T−−→ x (n)

X(ejω

)=

∞∑

n=−∞

x (n) e−jωn

x (n) =1

2π

∞∫

−∞

X(ejω

)ejωn dω

z-transform

ejω → z

x (n)Z.T−−→ X (z)

X (z)I.Z.T−−→ x (n)

x [n] = δ [n] +1

6δ [n − 1] −

1

6δ [n − 2]

x [n] = δ [n]6 −2

3δ [n − 1]

(ii) Find the response y (t) of a continuous time system using laplace

transform with transfer function H (s) =1

(s + 2) (s + 3)for an

input x (t) = e−tu (t).

H (s) =Y (s)

X (s)=

1

(s + 2) (s + 3)


Y (s) = X (s) ·1

(s + 2) (s + 3)∴ x (t) = e−tu (t) =

1

s + 1

=1

(s + 1) (s + 2) (s + 3)

Y (s) =1

(s + 1) (s + 2) (s + 3)

Y (s) =A

(s + 1)+

B

(s + 2)+

C

(s + 3)

A =1

2; B = 1; C = −

1

2

Y (s) =

1

2(s + 1)

+1

(s + 2)+

(

−1

2

)

(s + 3)


y (t) =1

2e−tu (t) + e−2tu (t) −

1

2e−3tu (t)

x (z) =

∞∑

n=−∞

x (n)z−n

x (n) =1

2πj

∮

X (z) zn−1 dn

(iii) State and prove the time shifting property and time reversal prop-

erty of z transform. (8)

Refer Page No. 4.28, Section 4.52, Section 4.5.3

15. (a) Convolve the following signals.

x (n) = u (n) − u (n − 3)

h (n) = (0.5)n u (n)

(b) Determine whether the given system is stable by finding H (z) and

plotting the pole zero diagram.

y (n) = 2y (n − 1) − 0.8y (n − 2) + x (n) + 0.8x (n − 1)

Part - C (1 × 15 = 15 Marks)


16. (a) A causal system has input x (n) and output y (n). Find the

(i) System function (4)

(ii) Impulse response h (n) (6)

(iii) Frequency response H(ejω

)(5)


B.E/B.Tech Degree Examination May/June 2016

Third Semester



(Regulation 2013)


Part - A (10 × 2 = 20 Marks)

1. Sketch the following signals

(i) rect

(t + 1

4

)

rect(t)rect(t+1/4)

0.25-0.75

rect(t+1/4)

-3 0.75

(ii) 5 ramp (0.1t)

5

10

15

5 ramp(t)

5

10

15

5 ramp(0.1t)

10 20 30

2. Given g (n) = 2e−2n−3. Write out and simplify the functions

(i) g (2 − n)


(ii) g( n

10+ 4

)

3. What is the inverse fourier transform of

(i) e−j2πft0

(ii) δ (f − f0)

I.F.T of e−j2πft0 = δ (t + t0)

I.F.T of δ (t − t0) = ejf0t

4. Give the laplace transform of x (t) = 3e−2tu (t) − 2e−tu (t) with ROC.

Refer page No. 4.11, Example 2.35.

5. Find whether the following system whose impulse response is causal and

stable.

h (t) = e−2tu (t − 1)


6. Realize the block diagram representing the system

H (s) =s

s + 1

H (s) =Y (s)

X (s)=

s

s + 1

sY (s) + Y (s) = sX (s)

Y (s) = X (s) −Y (s)

s

S1

-1

X(S) Y(S)

7. Write the conditions for existence of DTFT.

The fourier transform doesnot exist for all periodic function the condition

for x (n) to have fourier transform are


(i) x (n) is absolutely integrable over (−∞, ∞) that is

∞∑

n=−∞

|x (n)|dt < ∞

(ii) x (n) has finite number of discontinuous and a finite number of max-

ima and minima in every finite time interval.

8. Find the final value of the given signal

X (z) =1

1 + 2z−1 + 3z−2

9. From discrete convolution sum, find the step response interms of h (n).

10. Define the non recursive system.

Refer page No. 5.59, Section 5.8

Part - B (5 × 16 = 80 Marks)

11. (a) (i) Find whether the following signals are periodic or aperiodic. If

periodic find the fundamental period and fundamental frequency.

x1 (n) = sin 2πt + cos πt

x2 (n) =sin nπ

3·cos nπ

5

Solution:

x1 (n) = sin 2πt + cos πt

T1 =2π

ω=

2π

2π= 1

T2 =2π

ω=

2π

π= 2

T1

T2=

1

2(rational)

∴ Periodic signal.

Fundamental period T = 2T1 (or) T2

T = 2 sec


(ii) Find whether the following signals are power or energy signals.

Determine power and energy of the signals.

g (t) = 5 cos(

17πt +π

4

)

+ 2 sin(

19πt +π

3

)

g (n) = (0.5)n u (n)

E = ltN→∞

N∑

n=−N

|x (n)|2

= ltN→∞

N∑

n=−N

(0.5)2n u (n) u (n) = 1; limit 0 to N

= ltN→∞

N∑

n=0

(0.25)n (1)

= ltN→∞

1

1 − 0.25

= ltN→∞

1

0.75

E =1

0.75

P = ltN→∞

1

(2N + 1)

N∑

n=−N

|x (n)|2

= ltN→∞

1

(2N + 1)×

1

0.75

P = 0

(Or)

(b) Find whether the following systems are time variant or fixed. Also

find whether the systems are linear or nonlinear.

d3y (t)

dt3+ 4

d2y (t)

dt2+ 5

dy

dt+ y2t = x (t) (8)

y (n) = an2X (n) + bnX (n − 2) (8)

12. (a) Obtain the fourier series co-efficient & plot the spectrum for the given

waveform. (16)


(Or)

(b) (i) From basic formula, determine the fourier transform of the given

signals obtain the magnitude and phase spectra of the given sig-

nals.

te−atu (t) , a > 0

e−a|t|, a > 0



(ii) State and prove Rayleigh’s energy theorem.

Refer page No. 2.39, Section 2.7.13

13. (a) (i) Using graphical convolution, find the response of the system whose

impulse response is (8)

h (t) = e−2tu (t)

For an input

x (t) =

A for 0 ≤ t ≤ 20 otherwise

.

(ii) Realize the following a indirect form II. (8)

d3y (t)

dt3+4

d2y (t)

dt2+7

dy (t)

dt+8y (t) = 5

d2x (t)

dt2+4

dx (t)

dt+7x (t)

Solution:

Applying laplace transform

s3Y (s) + 4s2Y (s) + 7sY (s) + 8Y (s) = 5s2X (s) + 4sX (s) + 7X (s)

H (s) =Y (s)

X (s)=

5s2 + 4s + 7

s3 + 4s2 + 7s + 8


Y (s)

W (s)= 5s2 + 4s + 7

Y (s) = 5s2W (s) + 4sW (s) + 7W (s)

W (s)

X (s)=

1

s3 + 4s2 + 7s + 8

s3W (s) + 4s2W (s) + 7sW (s) + 8W (s) = X (s)

s3W (s) = X (s) − 4s2W (s) − 7sW (s) − 8W (s)

S1

S1

S1

-4

-7

-8

Y(S)X(S)

5

4

7

S W(S)3

(Or)

(b) (i) An LTI system is defined by the differential equation (10)

d2y (t)

dt2− 4

dy (t)

dt+ 5y (t) = 5x (t)

Find the response of the system y (t) for an input x (t) = u (t) if

the initial conditions are Y (0) = 1;dy (t)

dt

∣∣∣∣t=0

= 2.

(ii) Determine frequency response and impulse response for the sys-

tem described by the following differential equation. Assume

zero initial conditions. (6)

dy (t)

dt+ 3y (t) = x (t)

Solution:

dy (t)

dt+ 3y (t) = x (t)


Applying Laplace transform

sY (s) + 3Y (s) = X (s)

H (s) =Y (s)

X (s)=

1

(s + 3)

Frequency response

s = jω

H (jω) =1

(jω + 3)

Impulse response

Y (s) =X (s)

(s + 3)

x (t) = δ (t)X (s) = 1

Y (s) =1

(s + 3)


y (t) = e−3tu (t)

14. (a) (i) State and prove sampling theorem. (10)


(ii) What is aliasing? Explain the steps to be taken to avoid aliasing.

(6)

Refer Page No. 4.5

(Or)

(b) State and prove the following theorems

(i) Convolution theorem of DTFT. (8)

Refer Page No. , Section 4.3.7

(ii) Initial value theorem of z transform. (8)

Refer Page No. , Section 4.5.7

15. (a) (i) Realize the following system in cascade form. (10)

H (z) =1 +

1

5z−1

(

1 −1

2z−1 +

1

3z−2

)(

1 +1

4z−1

)


(ii) Convolve

x (n) =

1, 1, 0↑, 1, 1

h (n) = 1, −2, −3, 4

(Or)

(b) A system os governed by a linear constant coefficient difference equa-

tion

y (n) = 0.7y (n − 1) − 0.1y (n − 3) + 2x (n) − x (n − 2)

Find the output response of the system y (n) for an input x (n) =u (n)


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