TYPICAL QUESTIONS & ANSWERS

AE06/AC04/AT04 SIGNALS & SYSTEMS TYPICAL QUESTIONS & ANSWERS PART– I OBJECTIVE TYPE QUESTIONS Each Question carries 2 marks. Choose the correct or ...
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AE06/AC04/AT04

SIGNALS & SYSTEMS

TYPICAL QUESTIONS & ANSWERS PART– I

OBJECTIVE TYPE QUESTIONS Each Question carries 2 marks. Choose the correct or best alternative in the following: Q.1

n

The discrete-time signal x (n) = (-1) is periodic with fundamental period (A) 6 (B) 4 (C) 2 (D) 0 Ans: C Period = 2

Q.2

The frequency of a continuous time signal x (t) changes on transformation from x (t) to x ( α t), α > 0 by a factor 1 (A) α . (B) .

α

(C) α . 2

(D)

α.

Transform Ans: A x(t) α>1 α 0 compression in t, expansion in f by α. expansion in t, compression in f by α.

Q.3

A useful property of the unit impulse δ (t) is that (B) δ (at) = δ (t) . (A) δ (at) = a δ (t) . 1 (C) δ (at) = δ (t) . (D) δ(at ) = [δ(t )]a . a Ans: C Time-scaling property of δ(t): δ(at) = 1 δ(t), a > 0 a

Q.4

The continuous time version of the unit impulse δ (t) is defined by the pair of relations

1

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SIGNALS & SYSTEMS

∞ t=0 (B) δ (t) = 1, t = 0 and ∫ δ (t) dt = 1 . t ≠0 . -∞ ∞ 1, t ≥ 0 (C) δ (t) = 0, t ≠ 0 and ∫ δ (t) dt = 1 . (D) δ(t ) =  . -∞ 0, t < 0 1 (A) δ (t) =  0

Ans: C δ(t) = 0, t ≠ 0 → δ(t) ≠ 0 at origin +∞

∫ δ(t) dt = 1 → Total area under the curve is unity. -∞

[δ(t) is also called Dirac-delta function]

Q.5

.

Two sequences x1 (n) and x2 (n) are related by x2 (n) = x1 (- n). In the z- domain, their ROC’s are (A) the same. (B) reciprocal of each other. (C) negative of each other. (D) complements of each other. z Ans: B x1(n) X1(z), RoC Rx z Reciprocals x2(n) = x1(-n) X1(1/z), RoC 1/ Rx

Q.6 The Fourier transform of the exponential signal e jω 0 t is (A) a constant. (B) a rectangular gate. (C) an impulse. (D) a series of impulses. Ans: C Since the signal contains only a high frequency ωo its FT must be an impulse at ω = ωo Q.7 If the Laplace transform of f (t ) is (A) cannot be determined. (C) is unity.

(s

ω

2

, then the value of Lim f (t ) t →∞ + ω2 (B) is zero. (D) is infinity.

)

L

Ans: B f(t)

ω s2 + ω2

Lim f(t) = Lim t ∞ s 0 =

Lim s

0

s F(s)

[Final value theorem]

sω s2 + ω2

=0

Q.8 The unit impulse response of a linear time invariant system is the unit step function u (t ) . For t > 0, the response of the system to an excitation

e −at u (t ), a > 0, will be (A) ae −at .

(B) 2

1 − e −at . a

AE06/AC04/AT04

SIGNALS & SYSTEMS

(C) a (1 − e − at ) .

(D) 1 − e − at .

Ans: B h(t) = u(t); x(t) = e-at u(t), a > 0  System response y(t) = L−1  .  s s + a 1

1

1 1 1  = L−1  − a  s s + a 

= 1 (1 - e-at) a 0

∑ δ(n − k ) has the following region of convergence

Q.9 The z-transform of the function

k = −∞

(A) z > 1

(B) z = 1

(C) z < 1

(D) 0 < z < 1 0

Ans:

x(n) =

C

∑ δ(n-k) k = -∞

0

x(z) =

∑ z-k = …..+ z3 + z2 + z + 1 (Sum of infinite geometric series)

k = -∞

= 1 , 1–z

z < 1

Q.10 The auto-correlation function of a rectangular pulse of duration T is (A) a rectangular pulse of duration T. (B) a rectangular pulse of duration 2T. (C) a triangular pulse of duration T. (D) a triangular pulse of duration 2T. Ans: D T/2

RXX (τ) = 1 ∫ x(τ) x(t + τ) dτ T -T/2

triangular function of duration 2T.

Q.11 The Fourier transform (FT) of a function x (t) is X (f). The FT of dx (t ) / dt will be (A) dX(f ) / df . (B) j2πf X(f ) . (C) jf X (f ) . (D) X (f ) / ( jf ) . ∞

Ans: B (t) = 1 ∫ X(f) ejωt dω 2π - ∞ ∞

d x = 1 ∫ jω X(f) ejωt dω dt 2π - ∞ ∴ d x ↔ j 2π f X(f) dt

Q.12 The FT of a rectangular pulse existing between t = − T / 2 to t = T / 2 is a (A) sinc squared function. (B) sinc function. (C) sine squared function. (D) sine function. 3

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SIGNALS & SYSTEMS

Ans: B x(t) = 1, 0, +∞

-T ≤ t ≤ T 2 2 otherwise

X(jω) = ∫ x(t) e-jωt dt = -∞

+T/2

∫ e-jωt dt = e-jωt -T/2 jω

+T/2

-T/2

= - 1 (e-jωT/2 - ejωT/2) = 2 jω ω

ejωT/2 - e-jωT/2 2j

= 2 sin ωT = sin(ωT/2) .T ω 2 ωT/2 Hence X(jω) is expressed in terms of a sinc function.

Q.13

An analog signal has the spectrum shown in Fig. The minimum sampling rate needed to completely represent this signal is (A) 3 KHz . (B) 2 KHz . (C) 1 KHz . (D) 0.5 KHz .

Ans: C

For a band pass signal, the minimum sampling rate is twice the bandwidth, which is 0.5 kHz here.

Q.14 A given system is characterized by the differential equation: d 2 y(t ) dy(t ) − − 2 y(t ) = x (t ) . The system is: dt dt 2 (A) linear and unstable. (B) linear and stable. (C) nonlinear and unstable. (D) nonlinear and stable.

Ans:A

d2y(t) – dy(t) – 2y(t) = x(t), x(t) x(t) y(t) h(t) dt2 dt system The system is linear . Taking LT with zero initial conditions, we get s2Y(s) – sY(s) – 2Y(s) = X(s) 1 = 1 or, H(s) = Y(s) = 2 X(s) s – s – 2 (s –2)(s + 1) Because of the pole at s = +2, the system is unstable.

Q.15 The system characterized by the equation y(t ) = ax (t ) + b is (A) linear for any value of b. (B) linear if b > 0. (C) linear if b < 0. (D) non-linear. 4

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SIGNALS & SYSTEMS

Ans: D The system is non-linear because x(t) = 0 does not lead to y (t) = 0, which is a violation of the principle of homogeneity. Q.16 Inverse Fourier transform of u (ω) is 1 1 (A) δ(t ) + . 2 πt 1 (C) 2δ(t ) + . πt

(B)

1 δ(t ) . 2

(D) δ(t ) + sgn (t ) .

FT X(jω) = π δ(ω) + Jω

Ans: A x(t) = u(t)

2π x(-ω)

Duality property: X(jt) u(ω)

1

1 δ(t) + 1 2 πt

Q.17 The impulse response of a system is h (n ) = a n u (n ) . The condition for the system to be BIBO stable is (A) a is real and positive. (B) a is real and negative. (C) a > 1 . (D) a < 1 . +∞

Ans: D Sum S =

+∞

∑ h(n) = n = -∞ +∞



∑  an u(n) 

n = -∞

∑ a n

(

u(n) = 1 for n ≥ 0 )

n=0



1 1- a

if a < 1.

Q.18 If R1 is the region of convergence of x (n) and R 2 is the region of convergence of y(n), then the region of convergence of x (n) convoluted y (n) is (A) R1+ R 2 . (B) R1−R 2 . (C) R 1 ∩ R 2 . (D) R1∪R 2 . z Ans:C x(n) X(z), RoC R1 z y(n) Y(z), RoC R2 z x(n) * y(n) X(z).Y(z), RoC at least R1 ∩ R2 Q.19

( )

The continuous time system described by y(t ) = x t 2 is (A) causal, linear and time varying. (B) causal, non-linear and time varying. (C) non causal, non-linear and time-invariant. (D) non causal, linear and time-invariant.

5

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SIGNALS & SYSTEMS

Ans: D y(t) = x(t2) y(t) depends on x(t2) i.e., future values of input if t > 1. System is anticipative or non-causal α x1(t) → y1(t) = α x1(t2) β x2(t) → y2(t) = β x2(t2) α x1(t) + β x2(t) → y(t) = αx1(t2) +β x2(t2) = y1(t) + y2(t) System is Linear System is time varying. Check with x(t) = u(t) – u(t-z) → y(t) and x1(t) = x(t – 1) → y1(t) and find that y1(t) ≠ y (t –1).

Q.20

If G(f) represents the Fourier Transform of a signal g (t) which is real and odd symmetric in time, then G (f) is (A) complex. (B) imaginary. (C) real. (D) real and non-negative. FT

Ans: B g(t)

G(f)

g(t) real, odd symmetric in time G*(jω) = - G(jω); G(jω) purely imaginary.

Q.21

For a random variable x having the PDF shown in the Fig., the mean and the variance are, respectively, (A) 1 and 2 . 2 3 (B) 1 and 4 . 3 2 (C) 1 and . 3 (D) 2 and 4 . 3 +∞

Ans:B Mean = µx(t) = ∫ x fx(t) (x) dx -∞

3

= ∫ x 1 dx = 1 x2 3 = 9 – 1 1 = 1 -1 4 4 2 -1 2 2 4 +∞

Variance = ∫ (x - µx)2 fx (x) dx -∞

3

=

∫ (x - 1)2 1 d(x-1) -1 4

6

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SIGNALS & SYSTEMS = 1 (x - 1)3 3 = 1 [8 + 8] = 4 -1 12 3 4 3

Q.22

If white noise is input to an RC integrator the ACF at the output is proportional to − τ   −τ  . (A) exp (B) exp .   RC   RC  (D) exp(- τ RC ) . (C) exp( τ RC ) .

Ans: A RN(τ) = N0 exp -  τ  4RC RC

Q.23

x (n ) = a , a < 1 is (A) an energy signal. (B) a power signal. (C) neither an energy nor a power signal. (D) an energy as well as a power signal. n

Ans: A +∞ ∞ Energy = ∑ x2(n) = ∑ a2 n=-∞

n=-∞





= ∑ (a2)

= 1+ 2 ∑ a2

n=-∞

n=1

= finite since a < 1 ∴This is an energy signal.

Q.24

The spectrum of x (n) extends from − ω o to + ωo , while that of h(n) extends from − 2ω o to + 2ω o . The spectrum of y(n ) =



∑ h(k ) x(n − k ) extends k = −∞

from

(A) − 4ω o to + 4ω o . (C) − 2ω o to + 2ω o .

(B) − 3ω o to + 3ω o . (D) − ω o to + ω o

. Ans: D Spectrum depends on H( ejω)

Q.25

The signals x1 (t ) and x 2 (t ) are both bandlimited to (− ω1 , + ω1 ) and (− ω 2 , + ω 2 ) respectively. The Nyquist sampling rate for the signal x1 (t ) x 2 (t ) will be (A) 2ω1 if ω1 > ω 2 . (B) 2ω 2 if ω1 < ω 2 . (C) 2 (ω1 + ω 2 ) . (D) (ω1 + ω 2 ) . 2

Ans: C Q.26

X( ejω) Smaller of the two ranges.

Nyquist sampling rate = 2(Bandwidth) = 2(ω1 – (-ω2)) = 2(ω1 + ω2)

(

If a periodic function f(t) of period T satisfies f (t ) = −f t + T series expansion,

7

2

), then in its Fourier

AE06/AC04/AT04

SIGNALS & SYSTEMS

(A) the constant term will be zero. (B) there will be no cosine terms. (C) there will be no sine terms. (D) there will be no even harmonics. Ans: T

T/2

T

T/2

1 ∫ f(t) dt = 1 ∫ f(t) dt + ∫f(t) dt = 1 T 0 T 0 T/2 T

T/2

∫ f(t) dt + ∫ f(τ + T/2)dτ = 0 0

0

Q.27 A band pass signal extends from 1 KHz to 2 KHz. The minimum sampling frequency needed to retain all information in the sampled signal is (A) 1 KHz. (B) 2 KHz. (D) 4 KHz. (C) 3 KHz. Ans: B Minimum sampling frequency = 2(Bandwidth) = 2(1) = 2 kHz Q.28

The region of convergence of the z-transform of the signal 2 n u (n ) − 3n u (− n − 1) (A) is z > 1 . (B) is z < 1 .

(C) is 2 < z < 3 . Ans: 2nu(n) 3n u(-n-1)

(D) does not exist.

1 , z > 2 1 –2 z -1 1 , z < 3 1 – 3z -1

ROC is 2 < z < 3.

Q.29

The number of possible regions of convergence of the function

(A) 1. (C) 3.

(e −2 − 2)z is (z − e −2 )(z − 2)

(B) 2. (D) 4.

Ans: C Possible ROC’s are z > e-2 , z < 2 and e-2 < z < 2 Q.30

The Laplace transform of u(t) is A(s) and the Fourier transform of u(t) is B( jω) . Then 1 1 (A) B( jω) = A (s ) s = jω . (B) A(s ) = but B( jω) ≠ . s jω 1 1 1 1 (C) A(s ) ≠ but B( jω) = . (D) A(s ) ≠ but B( jω) ≠ . s jω s jω L Ans: B u(t) A(s) = 1 s 8

AE06/AC04/AT04

SIGNALS & SYSTEMS F.T

u(t)

B(jω) = 1 + π δ(ω) jω

A(s) = 1 but B(jω) ≠ 1 s jω

Q.31

Given a unit step function u(t), its time-derivative is: (A) a unit impulse. (B) another step function. (C) a unit ramp function. (D) a sine function.

Ans: A Q.32 The impulse response of a system described by the differential equation d2y + y( t ) = x ( t ) will be dt 2 (A) a constant. (B) an impulse function.. (C) a sinusoid. (D) an exponentially decaying function. Ans: C Q.33

sin(πu ) is denoted by: (πu ) (A) sin c(πu). (C) signum.

The function

(B) sin c(u). (D) none of these.

Ans: C Q.34

The frequency response of a system with h(n) = δ(n) - δ(n-1) is given by (A) δ(ω) - δ(ω - 1). (B) 1 - ejω. (C) u(ω) – u(ω -1). (D) 1 – e-jω.

Ans: D Q.35

The order of a linear constant-coefficient differential equation representing a system refers to the number of (A) active devices. (B) elements including sources. (C) passive devices. (D) none of those.

Ans: D Q.36

z-transform converts convolution of time-signals to (A) addition. (B) subtraction. (C) multiplication. (D) division.

Ans: C Q.37

Region of convergence of a causal LTI system (A) is the entire s-plane. (B) is the right-half of s-plane. (C) is the left-half of s-plane. (D) does not exist.

9

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SIGNALS & SYSTEMS

Ans: B Q.38

The DFT of a signal x(n) of length N is X(k). When X(k) is given and x(n) is computed from it, the length of x(n) (A) is increased to infinity (B) remains N (C) becomes 2N – 1 (D) becomes N2

Ans: A Q.39

The Fourier transform of u(t) is 1 (A) . j2πf 1 (C) . 1 + j2πf

(B) j2πf. (D) none of these.

Ans: D Q.40

For the probability density function of a random variable X given by

f x ( x ) = 5e − Kx u ( x ) , where u(x) is the unit step function, the value of K is 1 1 (A) (B) 5 25 (C) 25 (D) 5 Ans: D Q.41

The system having input x(n) related to output y(n) as y(n) = log10 x (n ) is: (A) nonlinear, causal, stable. (B) linear, noncausal, stable. (C) nonlinear, causal, not stable. (D) linear, noncausal, not stable.

Ans: A Q.42

To obtain x(4 – 2n) from the given signal x(n), the following precedence (or priority) rule is used for operations on the independent variable n: (A) Time scaling → Time shifting → Reflection. (B) Reflection → Time scaling → Time shifting. (C) Time scaling → Reflection → Time shifting. (D) Time shifting → Time scaling → Reflection.

Ans: D Q.43

The unit step-response of a system with impulse response h(n) = δ(n) – δ(n – 1) is: (A) δ(n – 1). (B) δ(n). (C) u(n – 1). (D) u(n).

Ans: B

10

AE06/AC04/AT04 Q.44

SIGNALS & SYSTEMS

If φ(ω) is the phase-response of a communication channel and ωc is the channel dφ(ω) frequency, then − ω = ωc represents: dω (A) Phase delay (B) Carrier delay (C) Group delay (D) None of these.

Ans: C Q.45

Zero-order hold used in practical reconstruction of continuous-time signals is mathematically represented as a weighted-sum of rectangular pulses shifted by: (A) Any multiples of the sampling interval. (B) Integer multiples of the sampling interval. (C) One sampling interval. (D) 1 second intervals.

Ans: B Q.46

ℑ  dx ( t )  If x (t ) ↔ X (s), then ℑ  is given by:  dt 

(A)

dX(s) . ds

(B)

(C) sX (s) − x (0 − ) .

X (s) x −1 (0) − . s s

(D) sX(s) – sX(0).

Ans: C Q.47

The region of convergence of the z-transform of the signal x(n) ={2, 1, 1, 2} ↑ is n=0 (A) all z, except z = 0 and z = ∞ (B) all z, except z = 0. (C) all z, except z = ∞. (D) all z.

Ans: A Q.48

When two honest coins are simultaneously tossed, the probability of two heads on any given trial is: 3 (A) 1 (B) 4 1 1 (C) (D) 2 4

Ans: D Q.49

Let u[n] be a 1, (A) x[n] =  0, 1, (C) x[n] =  0,

unit step sequence. The sequence u[ N − n] can nN 1, (D) x[n] =  otherwise 0,

11

be described as n≤N otherwise n≥N otherwise

AE06/AC04/AT04

SIGNALS & SYSTEMS

1, n ≤ N Ans (B) x[n] =  0, otherwise Here the function u(-n) is delayed by N units.

Q.50

A continuous-time periodic signal x(t ) , having a period T, is convolved with itself. The resulting signal is (A) not periodic (B) periodic having a period T (C) periodic having a period 2T (D) periodic having a period T/2

Ans (B) periodic having a period T Convolution of a periodic signal (period T) with itself will give the same period T. Q.51

If the Fourier series coefficients of a signal are periodic then the signal must be

(A) continuous-time, periodic (C) continuous-time, non-periodic

(B) discrete-time, periodic (D) discrete-time, non-periodic

Ans B) discrete-time, periodic This is the property of the discrete-time periodic signal. Q.52

Q.53

The Fourier transform of a signal x(t ) = e 2t u (−t ) is given by 1 2 (A) . (B) 2 − jω 1 − jω 1 2 (C) (D) j2 − ω j2 − ω 1 Ans (A) 2 − jω 1 1 . Therefore, FT of u(-t) = . If a function x(t) is multiplied FT u(t) = jω − jω by e 2t , then its FT will be F ( jω ) jω → jω − 2 . Hence the answer.

1

For the function H ( jω ) =

(A) 1 (C) 2

2 + 2 jω + ( jω )

2

, maximum value of group delay is

(B) 1/2 (D) 3

Ans None of the given answers is correct. Q.54

A continuous-time signal x(t ) is sampled using an impulse train. If X ( jω ) is the Fourier transform of x(t ) , the spectrum of the sampled signal can be expressed as ∞

(A)



∑ X ( jω + kω )δ (ω )

(B)

s

k = −∞ ∞

(C)

∑ X ( jkω ) * δ (ω + kω ) s

k = −∞ ∞

∑ X ( jω ) * δ (ω + kω )

(D)

s

k = −∞

∑ X ( jω )δ (ω + kω ) s

k = −∞

12

AE06/AC04/AT04

SIGNALS & SYSTEMS ∞

Ans (A)

∑ X ( jω + kω )δ (ω ) s

k = −∞

Since the spectrum consists of various harmonics k = −∞ to ∞ and discretely spread at an interval of fundamental frequency fs. Hence the answer.

Q.55

The region of convergence of a causal finite duration discrete-time signal is

(A) (B) (C) (D)

the entire z-plane except z = 0 the entire z-plane except z = ∞ the entire z-plane a strip in z-plane enclosing jω -axis

Ans (A) The entire z-plane except z = 0 X ( z) =

n2

∑ x[n]z

−n

. This sum should converge provided each term in the sum is

n = n1

finite. However, if there is a non-zero causal component for n2>0, then X(z) will have a term involving z-1 and thus ROC cannot include z = 0.

Q.56

( )

Let H e jω be the frequency response of a discrete-time LTI system, and H I (e jω ) be the frequency response of its inverse. Then,

(A) H (e jω )H I (e jω ) = 1 (C) H (e jω ) * H I (e jω ) = 1

(B) H (e jω )H I (e jω ) = δ (ω ) (D) H e jω * H I e jω = δ (ω )

( )

( )

( ) ( )

Ans (A) H e jω H I e jω = 1 Since H (e jω ) and H I (e jω ) are the inverse of each other, their product should equal 1. Q.57

The transfer function of a stable system is H ( z ) = Its impulse response will be n n (A) (0.5) u[n] + (2 ) u[n]

(C)

1 1 . + −1 1 − 0. 5 z 1 − 2 z −1

(B) − (0.5) u[− n − 1] + (2 ) u[n] n

(0.5)n u[n] − (2)n u[−n − 1]

n

(D) − (0.5) u[− n − 1] − (2 ) u[−n − 1] n

n

Ans (C) (0.5) u[n] − (2 ) u[− n − 1] (A) and (C) are the possible IFTs of the given system function. However, the system is stable; therefore (C) is the only correct answer. n

Q.58

n

The probability cumulative distribution function must be monotone and (A) increasing (B) decreasing (C) non-increasing (D) non-decreasing

Ans (D) non-decreasing The probability cumulative distribution function increases to 1 monotonically and there after remains constant. Q.59

The average power of the following signal is

13

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SIGNALS & SYSTEMS

A2 2 (C) AT12

(B) A 2

(A)

(D) A 2T1

Ans: (D) W=

Q.60



T1 / 2

−T1 / 2

x(t ) 2 dt = A 2 T1

Convolution is used to find: (A) The impulse response of an LTI System (B) Frequency response of a System (C) The time response of a LTI system (D) The phase response of a LTI system Ans: (C) k =∞

y ( n ) = x ( n ) * h( n ) =

Time response

∑ x(k )h(n − k )

k = −∞

Q.61

The Fourier Transform of a rectangular pulse is (A) Another rectangular pulse (B) Triangular pulse (C) Sinc function (D) Impulse. Ans: (C) This can be seen by putting the value of pulse function in the definition of Fourier transform.

Q.62

The property of Fourier Transform which states that the compression in time domain is equivalent to expansion in the frequency domain is (A) Duality. (B) Scaling. (C) Time Scaling. (D) Frequency Shifting. Ans: (B) Substituting the square pulse function f(t) in F ( jω ) =





−∞

f (t )e jωt dt

gives the sinc function.

Q.63

What is the Nyquist Frequency for the signal

x(t) =3 cos 50πt +10 sin 300πt – cos100πt ?

(A) 50 Hz (C) 200 Hz

(B) 100 Hz (D) 300 Hz 14

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SIGNALS & SYSTEMS

Ans: (D) Here the highest frequency present in the signal is ω m = 300π or f m = 150 Hz. Therefore the Nyquist frequency f s = 2 f m = 300 Hz.

Q.64

The step response of a LTI system when the impulse response h(n) is unit step u(n) is (A) n+1 (B) n (C) n-1 (D) n 2 Ans: (A) ∞

y ( n ) = x ( n ) * h ( n) = u ( n) * u ( n ) =



6

u ( k )u ( n − k ) =

k = −∞

∑ u (k )u(n − k ) k =0

y (0) = 1, y (1) = 2, y (2) = 3, ...., y (n) = (n + 1) y (n) = (n + 1) .

Q.65

The Laplace transform of u(t) is

(A) (C)

1 s 1

(B) s2 (D) s

s2

Ans: (A) ∞

Substituting f (t ) = u (t ) in the relation F ( s ) = ∫ f (t )e − st dt gives the answer. o

Q.66

The function which has its Fourier transform, Laplace transform, and Z transform unity is (A) Gausian (B) impulse (C) Sinc (D) pulse Ans: (B) Substituting f(t) = δ (t ) in the definitions of Fourier, Laplace and Z-transform, we get the transforms in each case as 1.

Q.67

The Z transform of δ (n − m ) is (A) z − n

(C)

(B) z − m

1 z−n

(D)

1 z−m

Ans: (B) The Z-transform of a delayed function f(n-m) is z-m times the Z-transform of the function f(n).

Q.68

1 4

If the joint probability pdf of f (x, y ) = , 0 ≤ x, y ≤ 2, P(x + y ≤ 1) is 1 8 1 (C) 4

1 16 1 (D) 2

(A)

(B)

15

AE06/AC04/AT04

SIGNALS & SYSTEMS

Ans: (A) P( x + y ) =

Q.69

1 1− y

∫∫

0 0

1 1 dxdy = 4 4

1

1− y

∫ x0 0

dy =

The period of the signal x(t ) = 10 sin 12πt + 4 cos 18πt π (A) (B) 4 1 (C) (D) 9

1 4

1

1

∫ (1 − y )dy = 8 . 0

is 1 6 1 3

Ans: (D) There are two waveforms of frequencies 6 and 9, respectively. Hence the combined frequency is the highest common factor between 6 and 9,i.e., 3. Hence period is 1/3.

Q.70

The autocorrelation of a rectangular pulse is (A) another rectangle pulse (C) Triangular pulse

(B) Square pulse (D) Sinc pulse

Ans: (C) Autocorrelation involves the integration of a constant which gives a ramp function. Hence the triangular pulse.

Q.71

If the Fourier series coefficients of a signal are periodic then the signal must be (A) continuous-time, periodic (B) discrete-time, periodic (C) continuous-time, non periodic (D) discrete-time, non perodic Ans: (B) It is the property of the discrete-time periodic signal. ∞

Q.72

The area under the curve

∫ δ (t ) dt is

−∞

(A) ∞ (C) 0

(B) unity (D) undefined

Ans: (B) ∞

∫ δ (t ) = 1

By definition of delta function,

Q.73

−∞

A transmission is said to be _____________ if the response of the system is exact replica of the input signal.

(A) LTI (C) Distortionless

(B) Distorted (D) Causal

Ans: (C) Since y (n) = x(n) .

16

AE06/AC04/AT04 Q.74

SIGNALS & SYSTEMS

Laplace Transform of tn is always equal to n n! (A) n (B) n s s n! (C) n +1 (D) All s Ans: (C) tn =

£

Q.75





0

t n e − st dt =

n! s n +1

For a stable system

(A)

z 1

(D)

z ≠1

Ans: (A) For the system to be stable, the ROC should include the unit circle.

Q.76

The region of convergence of a causal finite duration discrete time signal is (A) The entire ‘z’ plane except z = 0 (B) The entire ‘z’ plane except z = ∞ (C) The entire ‘z’ plane (D) A strip in z-plane Ans: (A) The ROC of the causal finite duration will have negative power of z. The ROC is the entire z-plane except z = 0.

Q.77

The CDF for a certain random variable is given as

−∞< x≤0 0,  2 F X ( x ) = kx , 0 < x ≤ 10 100k , 10 < x < ∞  The value of k is (A) 100 (B) 50 (C) 1/50 (D) 1/100 Ans: (D) From the given F(x), we get dF ( x ) = 0 + 2kx + 0 = 2kx dx 10



∴ 2kxdx = 1 0

or 100k = 1 → k = 1 / 100

Q.78

The group delay function τ(ω) is related to phase function φ(ω) as −d d (A) τ (ω ) = φ (ω ) (B) τ (ω ) = φ (ω ) dω dω 2

17

AE06/AC04/AT04

SIGNALS & SYSTEMS

(C) τ (ω ) =

d2 φ (ω ) dω 2

(D) τ (ω ) =

d2 φ (ω ) dω

Ans: (A): By definition.

Q.79

Two sequences x1 (n ) and x 2 (n ) are related by Z-domain, their ROCs are (A) same (C) negative of each other

x 2 (n ) = x1(− n ) . In the (B) reciprocal of each other (D) complement of each other

Ans: (B)

ROC of Z [x2 (n)] is outside the circle of radius r2 while ROC of Z [x1 (−n)] is inside the circle of radius r1 such that r2 = 1/ r1.

Q.80

The autocorrelation of a sinusoid is (A) Sinc pulse (C) Rectangular pulse

(B) another sinusoid (D) Triangular pulse

Ans: (B) ∞

φ XX (t ) = ∫ x(τ ) x(τ − t )dτ =

−∞ ∞



−∞

A sin ωτ × A sin ω (τ − t )dτ

A2 ∞ [(cos t − cos 2ωτ . cos t − sin 2ωτ . sin t ]dτ 2 −∞ A2  π = K [(cos t − cos 2ωτ . cos t − sin 2ωτ . sin t ]dτ  2  −π  2 π A = K  [cos t ]dτ  = K ' cos t 2  −π  . =







Thus the autocorrelation is a sinusoid.

Q.81

Which of the following is true for the system represented by y (n ) = x(− n )

(A) Linear (C) Causal

(B) Time invariant (D) Non Linear

Ans.: (A) The given function is of the form y = mx . Hence linear.

Q.82

The Fourier transform of impulse function is (A) δ(ω) (B) 2πω (C) 1 (D) sinc f Ans: (C) ∞

FT of

Q.83

δ (t ) = ∫ δ (t )e − jωt dt = 1 −∞

Convolution is used to find (A) amount of similarity between the signals 18

AE06/AC04/AT04

SIGNALS & SYSTEMS

(B) response of the system (C) multiplication of the signals (D) Fourier transform Ans: (B) Convolution of the input signal x(n) and the impulse response h(n) is given by ∞

∑ x ( k ) h(n − k )

y ( n ) = x ( n) * h( n ) =

, where y (n) is the response of the system.

k = −∞

Q.84

The final value of x(t ) = 2 + e − 3t u (t ) is (A) 2 (B) 3

[

]

(C) e − 3t

(D) 0

Ans: (A)

[

]

−3t Final value = Lt t →∞ x(t ) = Lt t →∞ 2 + e u (t ) = 2 .

Q.85

Discrete time system is stable if the poles are (A) within unit circle (C) on the unit circle

(B) outside unit circle (D) None

Ans: (A) The ROC should include the unit circle.

Q.86

The z transform of − u (− n − 1) is 1 (A) 1− z 1 (C) 1 − z −1

(B) (D)

z 1− z z

1 − z −1

Ans: (C) −∞

z[− u (−n − 1)] = − ∑ [u (−n − 1)]z − n = −[ z + z 2 + z 3 + ...] = n = −1



Q.87

The area under Gaussian pulse

∫e

−π t2

z 1 = z − 1 1 − z −1

dt is

−∞

(A) Unity (C) Pulse

(B) Infinity (D) Zero

Ans: (A)





2 e − πt dt =

−∞

Q.88





−∞

e − x 2π

x

π

dx = 2 π





−∞

The spectral density of white noise is (A) Exponential (C) Poisson

19

x e − x dx =1.

(B) Uniform (D) Gaussian

AE06/AC04/AT04

SIGNALS & SYSTEMS

Ans: (B) The distribution of White noise is homogeneous over all frequencies. Power spectrum is the Fourier transform of the autocorrelation function. Therefore, power spectral density of white noise is uniform.

20

AE06/AC04/AT04

SIGNALS & SYSTEMS

PART – II

NUMERICALS & DERIVATIONS Q.1. Determine whether the system having input x (n) and output y (n) and described by n relationship : y (n) = ∑ x (k + 2) is (i) memoryless, (ii) stable, (iii)causal (iv) k =-∞ linear and (v) time invariant. (5) Ans: y(n) =

∑ x(k + 2) k = -∞

(i) Not memoryless - as y(n) depends on past values of input from x(-∞) to x(n-1) (assuming)n > 0) (ii) Unstable- since if x (n) ≤ M, then y(n) goes to ∞ for any n. (iii) Non-causal - as y(n) depends on x(n+1) as well as x(n+2). (iv) Linear the principle of superposition applies (due to ∑ operation) (v) Time – invariant - a time-shift in input results in corresponding time-shift in output.

Q.2.

Determine whether the signal x (t) described by x (t) = exp [- at] u (t), a > 0 is a power signal or energy signal or neither.

(5)

Ans: -at x(t) = e u(t), a > 0 x(t) is a non-periodic signal. +∞



∞ - at

- at

Energy E = ∫ x2(t) dt = ∫ e 2 dt = e 2 -∞ 0 -2a

= 1 (finite, positive) 2a 0

The energy is finite and deterministic. x(t) is an energy signal.

Q.3. Determine the even and odd parts of the signal x (t) given by A e - α t t>0 x (t) =  0 t0 t 0, A > 0, -∞ < t < ∞ Even part xe(t) = Odd part

xo(t) =

x(t) + x(-t) 2 x(t) - x(-t) 2

21

AE06/AC04/AT04

SIGNALS & SYSTEMS x(t) A

Ae

-αt

t 0 x(-t) Ae

+αt

A t 0 xe(t) A/2 0

t

xo(t) A/2 t 0 -A/2

Q.4.

Use one sided Laplace transform to determine the output y (t) of a system described by d2y dy dy + 3 + 2y (t) = 0 where y (0 − ) = 3 and =1 (7) 2 dt dt t =0− dt Ans: d2y + 3 dy + 2 y(t) = 0, y(0-) = 3, dy =1 2 dt dt dt t = 0 s2 Y(s) – s y(0) – dy + 3 [s Y(s) – y(0)] + 2 Y(s) = 0 dt t = 0 (s2 + 3s + 2) Y(s) = sy(0) + dy + 3 y(0) dt t = 0 (s2 + 3s + 2) Y(s) = 3s + 1 + 9 = 3s + 10 Y(s) =

3s + 10 = 3s + 10 s + 3s + 2 (s + 1)(s + 2) 2

= A + B s+1 s+2

22

AE06/AC04/AT04

SIGNALS & SYSTEMS

A = 3s + 10 = 7; s + 2 s = -1 Y(s) =

B = 3s + 10 = -4 s + 1 s = -2

7 - 4 s+1 s+2

y(t) = L-1 [Y(s)] = 7e-t – 4e-2t = e-t( 7 - 4e-t) The output of the system is y(t) = e-t( 7 - 4e-t) u(t)

Q. 5. Obtain two different realizations of the system given by y (n) - (a+b) y (n – 1) + aby (n – 2) = x (n).Also obtain its transfer function.

(7)

Ans: y(n) – (a + b) y(n-1) + ab y(n-2) = x(n) Y(z) – (a+b) z-1 Y(z) + ab z-2 Y(z) = X(z) Transfer function H(z) = Y(z) = 1 X(z) 1 – (a+b) z-1 + ab z-2 y(n) = x(n) + (a + b) y(n-1) - ab y(n-2) Direct Form I/II realization

Q. 6.

Alternative Realisation

An LTI system has an impulse response h (t) = exp [ -at] u (t); when it is excited by an input signal x (t), its output is y (t) = [exp (- bt) -exp (- ct)] u (t) Determine its input x (t). (7)

Ans: -at h(t) = e u(t) for input x(t) Output y(t) = (e

-bt

L h(t)

-ct

- e ) u(t) L

H(s), y(t)

L Y(s), x(t)

23

X(s)

AE06/AC04/AT04 H(s) =

SIGNALS & SYSTEMS 1 ; Y(s) = 1 - 1 = s+c–s–b = c-b s+a s+b s+c (s + b)(s + c) (s + b)(s + c)

As H(s) = Y(s) , X(s) = Y(s) X(s) H(s) X(s) = (c - b)(s + a) = (s + b)(s + c)

A + B s+b s+c

A = (c - b)(s + a) = (c – b)(-b + a) = a - b (s + c) s = -b (-b + c) B=

(c - b)(s + a) = (c – b)(-c + a) = c - a (s + b) s= -c (-c + b)

X(s) = a – b + s+b x(t) = (a – b) e

-bt

c-a s+c

+ (c – a) e

-ct

The input x(t) = [(a – b) e

-bt

-ct

+ (c – a) e ] u(t)

Q.7. Write an expression for the waveform f (t ) shown in Fig. using only unit step function and powers of t. (3) Ans:

f(t) = E [ t u(t) – 2(t – T) u(t – T) + 2(t – 3T) u(t – 3T) – (t – 4T) u(t – 4T)] T

Q.8. For f(t) of Q7, find and sketch f ′(t ) (prime denotes differentiation with respect to t). Ans: f(t) = E [ t u(t) – 2(t – T) u(t – T) + 2(t – 3T) u(t – 3T) – (t – 4T) u(t – 4T)] T

24

(3)

AE06/AC04/AT04

SIGNALS & SYSTEMS

f ΄(t) E/T t 0

T

2T

3T

4T

-E/T f ΄(t) = E [u(t) – 2 u(t - T) + 2 u(t – 3T) –u(t – 4T)] T

Q.9.

.

Define a unit impulse function δ(t ) .

(2)

Ans: Unit impulse function δ(t) is defined as: δ(t) = 0, t ≠ 0 +∞

∫ δ(t) dt = 1 -∞

It can be viewed as the limit of a rectangular pulse of duration a and height 1/a when a 0, as shown below.

Q.10. Sketch the function g (t ) =

g (t ) → δ(t ) as ∈→ 0 .

3 3



(t − ∈)2 [u (t ) − u (t − ∈)] and show that

(6)

Ans: As ε

g(t) 3/ε

0, duration ε ∫ g(t) dt = 1

0, amplitude



0

0

Q.11.

ε

t

t Show that if the FT of x (t) is X ( jω) , then the FT of x   is a X ( jaω) . a Ans: FT x(t) X(jω)

25

(6)

AE06/AC04/AT04

SIGNALS & SYSTEMS

FT X1(jω) , then

Let x t a +∞

X1(jω) = ∫ x -∞ +∞

=

∫ x(α) e

t e-jωt dt a -jωaα

-∞ +∞

- ∫ x(α) e

Let t = α a

dt = a dα

a dα if a> 0

-jωaα

-∞ Hence X1(jω) = a

a dα if a < 0 +∞ -

α

∫ x(α) e jωa dα = a x (jωa) -∞

Q.12.

Solve, by using Laplace transforms, the following set of simultaneous differential (14) equations for x (t).

Ans: 2 x ′(t ) + 4 x (t ) + y ′(t ) + 7 y(t ) = 5u (t ) x ′(t ) + x (t ) + y ′(t ) + 3y(t ) = 5δ(t ) The initial conditions are : x (0 − ) = y(0 − ) = 0 . 2 x΄(t) + 4 x(t) + y΄(t) + 7 y(t) = 5 u(t) x΄(t) + x(t) + y΄(t) + 3 y(t) = 5 δ(t) L L L L s X(s), δ(t) 1, u(t) x(t) X(s), x΄(t)

1 s

(Given zero initial conditions) 2 sX(s) + 4 X(s) + sY(s) + 7 Y(s) = 5 s sX(s) + X(s) + sY(s) + 3 Y(s) = 5 (2s + 4) X(s) + (s+7) Y(s) = 5 s (s + 1) X(s) + (s+3) Y(s) = 5 X(s) = 5 s+7 S 3 5 s+3 2s+4 s+7 s+1 s+3 Or, X(s) = 5s + 35 – 5 – 15/s 2s2 + 6s + 4s + 12 - s2 – 8s – 7 = - 5s2 + 30s –15 = -5 2 s(s + 2s + 5) s

s2 + 6s – 3 = s2 + 2s + 5

Then A (s2+ 2s +5) + B s2 +Cs = -5(s2 + 6s – 3)

∴ A +B = -5 2A + C = -30

26

A + Bs+ C s s2+ 2s +5

AE06/AC04/AT04

SIGNALS & SYSTEMS

5A =15 Thus A = 3, B = -8, C = -36 and we can write X(s) = 3 – 8 s +1 – 14 2 2 2 (s + 1)2 + 22 s (s + 1) + 2 -t -t x(t) = (3 – 8 e cos 2t – 14 e sin 2t) u(t)

Q.13.

Find the Laplace transform of t sin ω0 t u (t ) .

(6)

Ans: L sin (ω0t)

ω0 s2 + ω02 L

Using t f(t)

- d [F(s)], ds

L [ t sin (ω0t) u(t) ] = - d ds 0 - ω0(2s) (s2 + ω02)2

=

Q.14.

ω0 s2 + ω02

2ω 0 s (s + ω02)2

=

2

Find the inverse Laplace transform of

s−2 s(s + 1)3

.

(8)

Ans: F(s) = A=

s-2 = A s(s+1)3 s

+

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

D (s+1)3

A(s+1)3 + Bs(s+1)2 + Cs(s+1) + Ds = s-2

s-2 = -2 3 (s+1) s=0

s3 : A+B = 0 D=

s-2 = 3 s s = -1 A = -2

s2 : 3A + 2B + C = 0 D=3

F(s) =

-2 + 2 + 2 + 3 s s+1 (s+1)2 (s+1)3 -t -t -t f(t) = -2 + 2 e + 2 t e + 3 t2 e 2 -t f(t) = [-2 + e ( 3 t2 + 2t + 2 ) ] u(t) 2

27

B=2 C=2

AE06/AC04/AT04 Q.15.

SIGNALS & SYSTEMS

Show that the difference equation y(n ) − αy(n − 1) = −αx (n ) + x (n − 1) represents an all-pass transfer function. What is (are) the condition(s) on α for the system to be stable? (8)

Ans: y(n) – α y(n-1) = - α x(n) + x(n-1) Y(z) – α z-1 Y(z) = - α X(z) + z-1 X(z) (1-α z-1) Y(z) = (-α + z-1) X(z) H(z) = Y(z) = -α + z-1 = 1- α z z- α X(z) 1- α z-1 Zero : z = 1 α

As poles and zeros have reciprocal values, the transfer function represents an all pass filter system.

Pole : z = α Condition for stability of the system : For stability, the pole at z = α must be inside the unit circle, i.e. α < 1.

Q.16. Give a recursive realization of the transfer function H (z ) = 1 + z −1 + z −2 + z −3

(6)

Ans: H(z) = 1 + z-1 + z-2 + z-3 = 1 – z –4 Geometric series of 4 terms –1 1–z First term = 1, Common ratio = z –1 As H(z) = Y(z) , we can write X(z) (1 – z –1) Y(z) = (1 – z –4) X(z) or Y(z) = X(z) (1 – z –4) = W (z)(1 – z –4) (1 – z –1) The realization of the system is shown below.

Q.17

Determine the z-transform of x 1 (n ) = α n u (n ) and x 2 (n ) = −α n u (− n − 1) and indicate their regions of convergence.

28

(6)

AE06/AC04/AT04

SIGNALS & SYSTEMS

Ans: x1(n) = αn u(n) X1(z) =

x2(n) = -αn u(-n-1)

and

1 RoC αz-1 < 1 i.e., z > α -1 1-αz -1

X2(z) =

∑ - αn z-n n=-∞ ∞

= - ∑ α-n zn = -( α-1z + α-2z2 + α-3z3 + ………) n=1

= - α-1z ( 1 + α-1z + α-2z2 + ……..) - α-1z 1- α-1z

=

=

z z-α

=

1 ; 1 - α z-1

RoC

α -1 z < 1 i.e., z < α

Q.18. Determine the sequence h (n ) whose z-transform is 1 H (z ) = , r 0 -0.5, t 7 y(7) = 1 x(7) = -1 = y(-7) 2 2 y(6) = 1 x(6) = 0 = y(-6) 2 y(5) = 1 x(5) = 1 = y(-5) 2 2 y(4) = 1 x(4) = 2 = y(-4) 2 y(3) = 1[x(3) + x(-3)] = 0 = y(-3) 2 y(2) = 1[x(2) + x(-2)] = 0 = y(-2) 2 y(1) = 1[y(1) + y(-1)] = 1 = y(-1) 2 y(0) = 1[ y(0) + y(0)] = 2 2 (v) Parseval’s theorem: π

2



∫ X(e ) dω = 2π -π

Q.45



2

∑ x(n) = 2π(1 + 1 + 4 +1 + 1 + 4 + 1 + 1) = 28π n = -∞

If the z-transform of x (n) is X(z) with ROC denoted by R x , find the

42

AE06/AC04/AT04

SIGNALS & SYSTEMS n

∑ x (k ) and its ROC.

z-transform of y(n ) =

(4)

k = −∞

Ans: z x(n) y(n) =

X(z),

RoC Rx

n

0



∑ x(k) =

∑ x(n-k) =

∑ x(n-k)

k = -∞

k=∞

k=0



Y(z) = X(z)

∑ z-k =

X(z) , RoC at least Rx ∩ (z > 1) 1 - z-1

k=0

Geometric series

Q.46 (i) x (n) is a real right-sided sequence having a z-transform X(z). X(z) has two poles, one of which is at a e jφ and two zeros, one of which is at r e − jθ . It is also known that

∑ x (n ) = 1 .

Determine X(z) as a ratio of polynomials in z −1 .

(6)

(ii) If a = 1 , r = 2, θ = φ = π 4 in part (b) (i), determine the magnitude of X(z) on the 2 unit circle. (4)

Ans: z .

(i) x(n) : real, right-sided sequence -jθ



; ∑x(n) = X(1) = 1

X(z) = K (z- re )(z- re ) jΦ

(z- ae )( z- ae jθ

X(z)

-jΦ

)

-jθ

= K z2 –zr (e +e ) + r2 jΦ jΦ z2 –za (e + e ) + a2 = K 1 – 2r cosθ z-1 + r2 z-2 = K. N(z-1) 1 – 2a cosΦz-1 + a2 z-2 D(z-1) where K. 1 – 2r cosθ + r2 = X(1) = 1 1 – 2a cosΦ + a2 i.e., K =

1– 2a cosΦ + a2 1 – 2r cosθ + r2

(ii) a = ½, r = 2, θ = Φ = π/4 ; K = 1 – 2(½).(1/√2) + ¼ 1 – 2(2) (1/√2) + 4 X(z) = (0.25) . 1 – 2(2) (1/√2) z-1 + 4z-2 1 – 2(½).(1/√2) z-1 + ¼ z-2

43

= 0.25

AE06/AC04/AT04

SIGNALS & SYSTEMS jω

1 - 2√2 z-1 + 4z-2 1 – (1/√2) z-1 + ¼ z-2

= (0.25)



X(e ) =

(0.25)

-jω

- jω

1 - 2√ 2 e +4e2 -jω - jω 1 – (1/√2) e +¼e2

-jω

= - 2√ 2 + e + 4 e jω -jω -2√2+ 4e + e jω

X(e )

=1

Q.47 Determine, by any method, the output y(t) of an LTI system whose impulse response h(t) is of the form shown in fig(a). to the periodic excitation x(t) as shown in fig(b). (14)

Ans:

Fig(a)

Fig(b)

h(t) = u(t) – u(t-1) => H(s) =

1- e -s s

First period of x(t) , xT(t) = 2t [u(t) – u(t- ½) ] = 2[ t u(t) – (t-1/2) u(t-1/2) –1/2 u(t-1/2)] ∴ XT(s) = 2[1/s2 – e-s/2 / s2 – 1/2 e-s/2 / s ] X (s) = XT(s) / 1 – e-s/2 Y(s) =

=

1 − e−s 1 . s 1 − e −s / 2

2 s

=

3

2 s3

=2



1− e 2 

−s / 2



− 0.5se − s / 2   s2 

(1 + e−s / 2 )[1 − e−s / 2 − 0.5 s e−s / 2 ] (1 − e

1 − e −s s3

−s



− 0.5s (e − s / 2 + e − s )

)

e −s / 2 + e −s s2

 1  1 Therefore y(t) = t u(t) – (t-1)2 u(t-1) –  t − u  t +  − ( t − 1) u ( t − 1)  2  2 2

This gives y (t) = t2 0< t < 1/2 t2 –t +1/2 1/2 < t < 1 1/2 t >1

44

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(not to scale)

Q.48

Obtain the time function f(t) whose Laplace Transform is F(s ) =

s 2 + 3s + 1

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

.

(14)

Ans: F(s) = s2+3s+1 = A + B + C + D + E (s+1)3(s+2)2 (s+1) (s+1)2 (s+1)3 (s+2) (s+2)2 A(s+2)2(s+1)2 + B(s+2)2(s+1) + C(s+2)2 + D(s+1)3(s+2) +E(s+1)3 = s2+3s+1 = 1-3+1 = -1 C = s2+3s+1 (s+2)2 s= -1 1

C = -1

= 4-6+1 = 1 E = s2+3s+1 3 (s+1) s= -2 -1

E=1

A(s2+3s+2)2 + B(s2+4s+4)(s+1) + C(s2+4s+4) + D(s3+3s2+3s+1)(s+2) + E(s3+3s2+3s+1) = s2+3s+1 A(s4+6s3+13s2+12s+4) + B(s3+5s2+8s+4) + C(s2+4s+4) + D(s4+5s3+9s2+7s+2) + E(s3+3s2+3s+1) = s2+3s+1 s4 :

A+D = 0

3

s : 6A+ B+ 5D +E = 0

;

A+B+1 = 0

as 5(A+D) = 0, E = 1

s2 : 13A+5B+C+9D+3E = 1

; 4A+5B+1 = 0

as 9(A+D) = 0, C = -1, E = 1

s1 : 12A+8B+4C+7D+3E = 3 ; 5A+8B-4 = 0

as 7(A+D) = 0, C = -1, E = 1

s0 : 4A+4B+4C+2D+E = 1 A+B = -1 ; 4(A+B)+B+1 = 0 or –4+B+1 = 0 or B=3 A=-4 A = -1-3 = - 4 D=4

A+D = 0 or D = -A = 4 F(s) = - 4 + 3 + -1 + 4 + 1 (s+1) (s+1)2 (s+1)3 (s+2) (s+2)2

f(t) = L-1[F(s)] = - 4e-t + 3t e-t – t2 e-t + 4e-2 t + t e-2t = [e-t(-4 + 3t - t2) + e-2 t(4 + t)] u(t)

45

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f(t) = [e-t(-4 + 3t - t2) + e-2 t(4 + t)] u(t)

Q.49 Define the terms variance, co-variance and correlation coefficient as applied to random variables.

(6)

Ans: Variance of a random variable X is defined as the second central moment E[(X-µX)]n, n=2, where central moment is the moment of the difference between a random variable X and its mean µX i.e., +∞

σX² = var [X] = ∫ -∞

(x- µX)² fx(x) dx

Co-variance of random variables X and Y is defined as the joint moment: σXY = cov [XY] = E[{X-E[X]}{Y-E[Y]}] = E[XY]-µ Xµ Y where µ X = E[X] and µ Y = E[Y]. Correlation coefficient ρXY of X and Y is defined as the co-variance of X and Y normalized w.r.t σXσY : ρXY = cov [XY] = σXY σXσY σXσY Q.50 Determine the total energy of the raised-cosine pulse x(t), shown in Fig.1 defined by: (8) π π 1   (cos ω + 1), − ≤ t ≤ x(t) =  2 ω ω .  0, otherwise

t

π − ω

π ω Fig.1

Ans: +∞

Energy E =

∫x

−∞

2

π ω

( t )dt =

1 3π 2 ∫ 4 (cos ωt + 1) dt = 4ω units.

− ωπ

FT

Q.51 State the sampling theorem, given x ( t ) ↔ X (ω) . For the spectrum of the continuous-time signal, shown in Fig.2, consider the three cases f s = 2f x ; f s > 2f x ; f s < 2f x and draw the spectra, indicating aliasing. (8)

46

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SIGNALS & SYSTEMS

− fx

fx

Fig.2

Ans: FT

Sampling theorem: Given x ( t ) ↔ X (ω) , if X (ω) = 0 for ω > ωm , and if ωs > 2ωm ,

2π , Ts = Sampling interval, then x(t) is uniquely Ts determined by its samples x ( nTs ) → where n = 0, ± 1, ± 2, ….. ( ωs > 2ω m ⇒ Nyquist rate.) where sampling frequency ωs =

X δ (f ) f s X ( 0)

-fx

0

fx

f

f fs = f x

-fx

0

fx

f

f f s > 2f x

(2fx) Guardband -fx

0

fx

2fx

f f s < 2f x

overlap ⇒ alia sin g -fx

Q.52

0

fx

Consider a continuous-time signal x(t). FT

f

f

(8)

Show that X ( t ) ↔ 2π x (−ω) , using duality (or similarity) property of FT s. 1 Find x(t) from X (ω) = , using the convolution property of FTs. (1 + jω) 2

(i) (ii)

Ans: +∞

(i)

f

x(t) =

1 X(ω)e jωt dω 2π ω=∫−∞

Using duality property of FTs, t ↔ ω ,

47

AE06/AC04/AT04

SIGNALS & SYSTEMS +∞

x (ω) =

+∞

1 1 X ( t )e jtωdt , or , x ( −ω) = X ( t )e − jtω dt ∫ 2π t =−∞ 2π t =∫−∞ +∞

∴ 2π x (−ω) =

FT

− jtω ∫ X( t )e dt , i.e., X(t ) ↔ 2πx (−ω) .

−∞

(ii) Find x(t) from X (ω) = X (ω) =

1 (1 + jω) 2

=

1 (1 + jω) 2

, using the convolution property of FTs.

FT 1 1 1 , and, e − t u ( t ) ↔ . . 1 + jω 1 + jω 1 + jω FT

Convolution property of FTs ⇒ x ( t ) = x1 ( t ) * x 2 ( t ) ↔ X (ω) = X1 (ω)X 2 (ω) . +∞

∴ x(t) =

∫e

−τ

1, 0 < τ < t > 0 u (τ)u ( t − τ) =  0, t < 0.

u (τ)e −( t −τ) u ( t − τ)dτ .

−∞

= t.e − t , t > 0 ∴ x(t) = t e −t u (t ) .

Q.53 Find the difference equation describing the system represented by the block-diagram shown in Fig.3, where D stands for unit delay. (8)

Σ

Σ



1 2



1 4 Fig.3

Ans: Intermediate variable f(n) between the summers: 1 1 f (n ) = x (n − 1) − f (n − 1) − f (n − 2) 2 4 y(n ) = 2 x (n ) + f (n ) , or, f (n ) = y(n ) − 2 x (n ) f ( n − 1) = y( n − 1) − 2 x ( n − 1) f ( n − 2) = y ( n − 2) − 2 x ( n − 2) 1 1 1 y(n ) + y(n − 1) + y(n − 2) = 2 x (n ) + 2 x (n − 1) + x (n − 2) . 2 4 2 Q.54 For the simple continuous-time RC frequently-selective filter shown in Fig.4, obtain the (8) frequency response H(ω). Sketch its magnitude and phase for -∞ < ω < ∞.

48

AE06/AC04/AT04

SIGNALS & SYSTEMS

Fig.4

Ans: FT dy( t ) + y( t ) ↔ X(ω) = RCjωY(ω) + Y(ω) dt Y (ω) 1 1 H (ω) = = = . X (ω) 1 + jωCR  ω  1 + j  ω0 

KVL ⇒ x ( t ) = RC or,

1

H (ω) =

 ω  1 +   ω0 

2

 ω arg H(ω) = − tan −1    ω0  magnitude spectrum

π 2

phase spectrum

3 dB 0 −

- ω0

Q.55

0

ω0

ω

π 2

ω

Consider the signal x ( t ) = e − t u ( t ) + e −2 t u ( t ) . Express its Laplace Transform in the N (s) form: X (s) = K. , K = system constant. Identify th region of convergence. D(s) Indicate poles and zeros in the s-plane. (8)

Ans: L

x ( t ) = e − t u ( t ) + e − 2 t u ( t ) ↔ X (s ) =

1 1 + s +1 s + 2

3  s +  2s + 3 N(s) 2 , K = 2. X (s) = =1  =K (s + 1)(s + 2) D(s) s 2 + 3s + 2

(

)

49

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SIGNALS & SYSTEMS

3 1 Zero ⇒ s = − . 2 2 Poles ⇒ s = −1,−2. ROC ⇒ R e {s} > −1. R e {s} > −2. ∴ Common R o C is R e {s} > −1.

Q.56 Given input x(n) and impulse response h(n), as shown in Fig.5, evaluate (8) y(n ) = x (n ) * h (n ) , using DTFTs.

Fig.5

Ans: DTFT

y(n ) = x (n ) * h (n ) ↔ Y (e jΩ ).H (e jΩ ) .

( ) n = −1 n = 0 n =1 ( ) n = −1 n = 0 n =1 2 ∴ Y (e jΩ ) = a (e jΩ + 1 + e − jΩ ) = a (e − j2Ω + 2e jΩ + 3+ 2e − jΩ + e − j2Ω ) H e jΩ = e + jΩ + 1 + e − jΩ ; X e jΩ = ae + jΩ + a + ae − jΩ

DTFT

As δ(n − n 0 ) ↔ e − jΩn o , y(n) = aδ(n+2) + 2aδ(n+1) + 3aδ(n) + 2aδ(n-1) + aδ(n-2). y(n) = {a, 2a, 3a, 2a, a} ↑ n=0

Q.57 Determine

the

inverse

( ) = e− j2Ω − 5e− jΩ + 6 .

Xe

Ans:

DTFT,

by

partial

fraction

6

jΩ

( )

expansion,

of

(8)

3 −2 + . − jΩ − jΩ 1 − jΩ 1 − jΩ − jΩ 2 − jΩ e − 3 e − 2 1− e 1− e e −5e +6 3 2 n n n n  1 1 1 1  ∴ x (1n ) = −2  u (n ) + 3  u (n ) = − 2  + 3   u (n ) .  3 2  2     3  X e jΩ =

(

6

) (

)

=

(

50

6

)(

)

=

AE06/AC04/AT04

SIGNALS & SYSTEMS

Q.58 State the initial-value and final-value theorems of Laplace Transforms. Compute the L 3s + 4 initial-value and final-values for x ( t ) ↔ X(s) , where x (s) = . (8) s(s + 1)(s + 2) 2 Ans: Initial-value theorem: If f(t) and its first derivative are Laplace transformable, then the initial value of f(t) is: f (0 + ) = lim f ( t ) = lim sF(s) . t →0 +

s →∞

Final-value theorem: If f(t) and its first derivative are Laplace transformable, and f(t) is not a periodic function, then the final value of f(t) is: lim f ( t ) = lim sF(s) . t →∞

s →0

4  3 +  s  Initial value ⇒ x (0 + ) = lim sX(s) = lim = 0. 1 s →∞ s →∞  2 1 + (s + 2 )  s (3s + 4) = 1 . Final value ⇒ lim x ( t ) = lim sX(s) = lim t →∞ s →0 s →0 (s + 1)(s + 2 )2 Q.59 Find, by Laplace Transform method, the output y(t) of the system described by the dy( t ) differential equation: + 5 y( t ) = x ( t ) where input x ( t ) = 3e −2 t u ( t ) and the initial dt condition is y(0) = -2. (8) Ans: dy( t ) + 5 y( t ) = 3e − 2 t u ( t ), y(0) = -2. dt L L 1 y( t ) ↔ Y(s), u ( t ), e − 2 t ↔ . s+2 3 ∴ sY (s) − y(0 + ) + 5Y (s) = . s+2 −2 3 A B 2 + . Y (s) = = + − (s + 2)(s + 5) (s + 5) s + 2 s + 5 s + 5 1 3 3 = − . A= s = −2 = 1 s+2 s+5 s+5 3 B= ∴ y( t ) = e − 2 t − 3e −5 t u ( t ) s = −5 = −1 . s+2

(

)

Q.60 An LTI system is characterised by the difference equation: x(n – 2) – 9x(n – 1) + 18x(n) = 0 with initial conditions x(-1) = 1 and x(-2) = 9. Find x(n) by using z-transform and state the properties of z-transform used in your calculation. (8) Ans: x(n – 2) -9x(n – 1) + 18x(n) = 0 By using z

x (n − n 0 ) ↔ z −n 0 X (z) + x (−n 0 ) + z −1x (−n 0 + 1) + z − 2 x (−n 0 + 2) + ..... + z −( n 0 −1) x (−1) 51

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We get [ x ( −2) + x ( −1) z −1 + z −2 X( z)] − 9[ x ( −1) + z −1X ( z)] + 18X ( z) = 0 . 123 123 123 9

1

1

   z −1  1 2  . X (z) = = −  1 −1 1 − 2  18  1 − 1 z −1 1 − 1 z −1  181 − z + z    3  6  18  2  n −1 n −1 1  1  1  ∴ x (n ) =   − 2   u (n − 1) . 18  16   3    − z −1

z

Q.61 Determine the discrete-time sequence x(n), given that x (n ) ↔ X (z) =

z2 + z z 3 − 3z 2 + 3z − 1 (8)

.

Ans: Assume that x(n) is casual. Then

z 3 − 3z 2 + 3z − 1 z 2 + z

z −1 + 4z −2 + 9z −3 + 16z −4 + ......

z 2 − 3z + 3 − z −1 4z − 3 + z −1 4z − 12 + 12z −1 − 4z −2 9 − 11z −1 + 4z − 2 9 − 27 z −1 + 27 z −2 − 9z −3 16z −1 − 23z − 2 + 9z − 3 16z −1 − 48z −2 + 48z −3 − 16z −4 25z − 2 − 39z −3 + 16z − 4 z δ(n − n ) ←→ z−n 0  0   ∴ x ( n ) = δ( n − 1) + 4δ(n − 2) + 9δ( n − 3) + 16δ( n − 3) + ..... x(n) = {0, 1, 4, 9, 16, …..} ↑ n=0

∴ X ( y) = z −1 + 4z −2 + 9z −3 + 16z −4 + .....

Q.62 Explain the meaning of the following terms with respect to random variables/processes: (i) Wide-sense stationary process. (ii) Ergodic process. (iii) White noise. (iv) Cross power spectral density. (8) Ans: (i) Wide-sense stationary process. For stationary processes, means and variances are independent of time, and covariance depends only on the time-difference if in addition, the N-fold joint p.d.f. depends on the time origin, such a random process is called wide- sense stationary process. (ii) Ergodic process. Ergodic process is one in which time and ensemble averages are interchangeable. 52

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SIGNALS & SYSTEMS

For ergodic processes, all time and ensemble averages are interchangeable, not just the mean, variance and autocorrelation function. (iii) White noise. White noise is an idealised form of noise, the power spectral density of which is independent of frequency. “White” is in parlance with white light that contains all frequencies within the visible band of electromagnetic radiation.. (iv) Cross power spectral density. Cross power spectral density of two stationary random processes is defined as the FT of their cross-correlation function FT

R xy (τ) ↔ Sxy (f ), where, R xy (τ) = E{X ( t ).Y( t + τ)} . Q.63

A random variable X is characterised by probability density function shown in Fig.6:  x 1 − , 0 ≤ x ≤ 2 f x (x) =  2 f X (x )  0, otherwise Compute its: Probability distribution function; Probability in the range 0.5< x ≤1.5; Mean value between 0 ≤ x ≤ 2; and Mean-square value E(x2). Fig.6 (8) Ans: +∞

p.d.f. f X ( x ) =

x

x2  α α α = − f ( ) d 1 d α = x − , 0< x ≤ 2.   ∫X ∫ 2  4 0 −∞

1 . 2

Probability = f X (1.5) − f X (0.5) = (0.5 < X ≤ 1.5) 2

2  x Mean value m X = ∫ x 1 − dx = 2 3 0 

[ ]

2

2  x Mean-squared value E x = ∫ x 2 1 − dx = . 3  2 0 2

Q.64

Determine the fundamental frequency of the signal x[ n] = e

− j 4πn

3

+e

j 3πn

8

.

Ans:

x (n) = e Fundamental Frequency =

Q.65

− j 4πn 3

+e

− j 3πn 8

=e

− j 32πn 24

+e

− j 9πn 24

1 . 24

A CT system is described by y (t ) =

t3

∫ x(λ )dλ . Find if the system is time

−∞

invariant and stable.

(6)

Ans: 53

AE06/AC04/AT04

SIGNALS & SYSTEMS t 3

∫ x(λ )dλ .

y(t) =

−∞

Let x(t) be shifted by t0 then the corresponding output yi(t) will be t 3

∫ x (λ − t

yi(t) =

0

t −t0 3

∫ x(λ ′)dλ ′ where λ ′ = λ − t

) dλ =

0

.

−∞

−∞

Original output shifted by t0 sec is t −t0 3

∫ x ( λ ) dλ

y0(t) =

−∞

Hence the system is time-invariant. If x(t) is bounded, output will be bounded. Hence the system is stable. Q.66

Let x(t ) be a real signal and x(t ) = x1 (t ) + x 2 (t ) . Find a condition so that ∞





x(t ) dt =

−∞

x1 (t ) dt + 2

−∞ ∞

Ans:



2



x(t ) dt = 2

−∞ ∞





∫ x (t )

2

2

(6)

dt

−∞

2

∫−∞ ( x1 (t ) + x2 (t ) ) dt ∞

x(t ) dt = 2

−∞ ∞





x1 (t ) dt + 2

−∞







x 2 (t ) dt + 2 ∫ x1 (t ) x 2 (t ) dt

−∞

−∞

2

The term 2 ∫ x1 (t ) x 2 (t ) dt will become zero if x1 (t ) is the even part and x 2 (t ) −∞

is the odd part of x(t ) or vice-versa. Then ∞



−∞

Q.67

x(t ) dt = 2





x1 (t ) dt + 2

−∞



∫ x (t )

2

2

dt .

−∞

If h1 [n] = δ [n] , h2 [n] = δ [n − 1] + 2δ [n − 2] , h3 [n] = δ [n + 1] + 2δ [n + 2] are the impulse responses of three LTI systems, determine the impulse response of the system shown in Fig.1.

h1 [n]

Fig. 1

h2 [n]

+ h3 [n]

h1(n) = δ(n) h2(n) = δ(n-1)+2δ(n-2) h3(n) = δ(n+1)+2δ(n+2) Impulse response of the system, h(n) = h1(n) + h2(n)*h3(n) = δ(n) + 2δ(n+1) + 5δ(n) + 2δ(n-1) = 6δ(n) + 2δ(n+1) + 2δ(n-1).

Ans:

54

+

AE06/AC04/AT04 Q.68

SIGNALS & SYSTEMS

Given that y (t ) = x(t ) ∗ h(t ) , determine x(at ) ∗ h(at ) in terms of y (t ) . If a is real, for what values of a the system will be (i) causal, (ii) stable? (10) Ans: y(t) = x(t)*h(t) Thus, Y(s) = X(s)H(s) Now x(at) has Laplace transform (1/a) X(s/a) . Similarly h(at) has Laplace transform (1/a) H(s/a) . Thus Laplace transform of x(at)*h(at) = (1/a2)X(s/a) H(s/a) = (1/a) (1/a) Y(s/a) = Laplace transform of (1/a)y(at) Assuming the original system to be causal and stable, (i) to maintain only causality, a can take any value, (ii) to maintain stability, a > 0.

Q.69

One period of a continuous-time periodic signal x(t ) is as given below.  t , −1 < t < 1 x (t ) =   0, 1 < t < 2

(10)

Determine Fourier series coefficients of x(t ) , assuming its period to be 3.  t , −1 < t < 1  0, 1 < t < 2

Ans: x(t ) =  1 a0 = T

ak =

=

1 T

1  0  1  − tdt + tdt  = −1  T 0





 − t 2  0  t 2 1  1   +  =  2  −1  2  0  T

1  1  0  1  − jkω t − jkω t − jkω t  x(t )e 0 dt  =  − te 0 dt + te 0 dt  T T   T −1 0







  e jk ω 0 e jk ω 0 − 1   e − jk ω 0 e − jk ω 0 − 1   + + − +     k 2 ω 0 2   jk ω 0 k 2 ω 0 2     jk ω 0

 2  e jk ω 0 − e − jk ω 0   2  e jk ω 0 + e − jk ω 0    − 1     + 2 2  2j 2  k ω 0    k ω 0     2  sin k ω 0 cos k ω 0 − 1  =  +  2 T  k ω 0  k 2ω 0 cos 2 π k / T − 1   sin 2 π k / T = +T  πk 2π 2 k 2   1 a0 T =3 = 3 sin 2 k / 3 cos 2 π k / 3 − 1  π  a k T =3 =  +3  πk 2π 2 k 2   =

Q.70

1 T

Determine Fourier series coefficients of the same signal x(t ) as in Q69, but now, assuming its period to be 6. What is the relationship between the coefficients determined in Q69 & Q70? (6) Ans:Let the period be T2 = 2T1 = 6. Following the above procedure, we get 55

AE06/AC04/AT04

SIGNALS & SYSTEMS 2  0  4 2 = = 2a01  − tdt + tdt  = −2  T2 T1 0 0 2 1   ak 2 =  − te − jkω 0 t dt + te − jkω 0 t dt  T2 − 2  0 cos 4 π k / T 2 − 1   sin 4 π k / T 2 = 2 + T2  πk 4π 2 k 2   sin 2 k / T cos 2 k / T − 1 π π   1 1 = 2 + T1  = 2ak1 2 2 π k 2 π k  

a02 =

1 T2









Thus the Fourier coefficients are doubled when the period is doubled. The function with higher period will have all the harmonics present in the lower period function as even harmonics. Q.71

The Fourier transform of a signal x(t) is described as  0.5π − 1 < ω < 0  ω , −1 < ω < 1  and arg X ( jω ) = − 0.5π 0 < ω < 1 X ( jω ) =   0,

otherwise

 

0

otherwise

Determine whether x(t) is real or complex.

Q.72

Ans:The magnitude response is symmetric and the phase response is antisymmetric. So x(t ) is real. 4 Determine the inverse Fourier transform of X ( jω ) = 2 sin 2 ω using the

ω

convolution property of the Fourier transform. Ans: X ( jω ) =

4

2

 2

(4)



sin 2 ω =  sin ω  sin ω  =P(jω).P(jω) ω2 ω  ω  Multiplication in the frequency domain is equivalent to convolution in the time domain. sin ω Inverse Fourier transform of P(jω) = 2 is a pulse ω 1, p (t ) =  0,

t ≤1 t > 1.

Therefore 2 − t , t ≤ 2 x (t ) = p(t ) ∗ p (t ) =  t >2 0,

Q.73

A system is described by the difference equation y[n] = x[n] + ay[n − 1] . Find the impulse response of the inverse of this system. From the impulse response, find the difference equation of the inverse system. (8) Ans: y[n] = x[n] + ay[n − 1] H ( z) =

Y ( z) 1 = X ( z ) 1 − az −1

For Inverse System:

56

AE06/AC04/AT04

SIGNALS & SYSTEMS

1 = 1 − az −1 H ( z) Impulse Response: h1(n) = 1δ ( n) − aδ ( n − 1) Difference equation: y ( n) = x( n) − ax( n − 1) Determine the autocorrelation of the sequence {1, 1, 2, 3} .

Transfer function H 1 ( z ) =

Q.74

(8)

Ans:x(n) = (1, 1, 2, 3) ∞

rxx (k ) =

Since

∑ x ( m) x ( m − k ) = r

xx ( − k )

m = −∞

r (0) = x (0) x(0) + x (1) x(1) + x(2) x(2) + x(3) x(3) = 15 r (1) = x (0) x(1) + x (1) x (2) + x( 2) x(3) = 9 = r (−1) r ( 2) = x(0) x (2) + x(1) x(3) = 5 = r ( −2) r (3) = x(0) x (3) + x(1) x(3) = 3 = r (−3) r (≥ 4) = 0 Thus

Q.75

r (n) = [3, 5, 9, 15, 9, 5, 3] ↑ Determine the cross correlation of the processes x1 (t ) = A cos(2πf c t + θ ) and x 2 (t ) = B sin( 2πf c t + θ ) , where θ is an independent random variable uniformly distributed over the interval (0,2π ) . (8) Ans:x1(t) = A cos (2πfct + θ) and x2(t) = B sin (2πfct + θ) Rxx(τ) = E{A cos {2πfc(t + τ)+θ}B sin (2πfct + θ)} AB E[sin {2πfc(-τ)} + sin {2πfc(2t + τ + 2θ)}] 2 π AB AB 1 =sin(2πfcτ) + sin{2πfc(2t +τ+2θ)}dθ 2 2 2π −π

=



=-

Q.76

AB sin (2πfcτ) 2

∞ sin πt  n is sampled by p (t ) = ∑ δ  t − . Determine and πt 2 n = −∞  (8) sketch the sampled signal and its Fourier transform.

A signal x(t ) =

Ans: x(t ) =

∞ sin πt  n sampled by p (t ) = ∑ δ  t − . Thus, the sampled signal πt 2 n = −∞ 

is n sin π   2 . x (n) = n π  2

DTFT of x( n) is a pulse

57

AE06/AC04/AT04

SIGNALS & SYSTEMS

π  2, ω ≤ X (ω ) =  2 0, otherwise Q.77

Determine the Fourier transforms of (i) x1[n ] = sin nω0 and (ii) x2 [n] = (sin nω0 )u[n ] e jω0 n − e − jω0 n 1 jω0 n = e − e − jω0 n 2j 2j 1 [δ (ω + ω 0 ) − δ (ω − ω 0 )] X (ω ) = 2j

[

x1 [n ] = sin nω 0 =

Ans: (i)

(8)

]

e jω 0 nu ( n) − e − jω 0 nu ( n) 2j 1  1 1  X (ω ) = −  jω0 − jω − jω0 − jω  2 j 1 − e e 1− e e  1 jω  1 1  = e  jω − jω jω0 − jω0  2j e −e e − e 

x2 [n] = (sin nω0 )u ( n) =

(ii)

  sin ω 0 = e jω   jω j 2ω − 2e cos ω 0  1 + e

=

Q.78

 sin ω 0 1   2  cos ω − cos ω 0 

Find the inverse Laplace transform of X (s ) = possible ROCs. Ans:

s 4 + 3s 3 − 4 s 2 + 5s + 5 for all s 2 + 3s − 4 (8)

s 4 + 3s 3 − 4 s 2 + 5 s + 5 s 2 + 3s − 4 5s + 5 5s + 5 3 2 = s2 + = s2 + . = s2 + 2 + ( s + 4)( s − 1) s + 3s − 4 s + 4 s −1

H(s) =

d 2δ (t ) + 3e − 4t u (t ) + 2e t u (t ), ROC σ > 1 dt 2 d 2δ (t ) h (t ) = + 3e − 4t u (t ) − 2e t u (−t ), ROC − 4 < σ 1. Q.81

Determine the ROC of aX ( z ) + bY ( z ) , given that

0.25 z , z > 0.5 . (z − 0.5)(z − 1.5) (z − 0.25)(z − 0.5) For what relationship between a and b the ROC will be the largest? (8) X (z ) =

z

, 0.5 < z < 1.5, Y ( z ) =

z , 0.5 0.5 ( z − 0.25)( z − 0.5)

Ans:X(z) =

0.25a + 0.375b   z (a + 0.25b) z −  a + 0.25b   aX(z) + bY(z) = ( z − 0.5)( z − 1.5)( z − 0.25)

Since ROC is decided by the three poles, ROC of aX(z) + bY(z) is 0.50 1, Sgm (t ) =  − 1, t < 0 ∞

F (ω ) = Lt a →0 ∫ e

−a t

Sgm(t )e − jωt dt

−∞ FT [Sgm] 0 ∞   = Lt a →0  ∫ − e −( a + jω )t dt + ∫ e − ( a + jω )t dt  0 −∞  1 1 2 = + = jω jω jω 2 ∴ F (ω ) = ω Unit Step Function

1 t ≥ 0 u (t ) =  0 t < 0

Sgn (t) = 2u (t ) − 1 ,

Since

1 [sgn(t ) + 1] 2  1 1 2 ∴ FT of u (t ) =  + 2πδ (ω ) = + πδ (ω ) 2  jω  jω u (t )

Q.88

  

=

Determine the Fourier transform a two-sided exponential function −t (8) x(t ) = e and draw its magnitude spectrum. Ans:

f (t ) = e

−t

62

AE06/AC04/AT04

SIGNALS & SYSTEMS ∞

F (ω ) =

∫e

−t



0

e

− jωt

∫e

dt =

−∞

+ ∫ e −(1+ jω )t dt

(1− jω ) t

−∞

0

1 1 2 = + = 1 − jω 1 + jω 1 + ω 2 2 F (ω ) = 1+ ω2 ∴ Q.89

ind the Discrete Fourier transform of the following sequences. (i) x(n ) = a n , 0 < a < 1 (Find N point DFT) π (Find 4 point DFT) (ii) x(n ) = cos n

(8)

4

Ans: N −1

∑ x ( n) e

X (k ) =

− j 2πnk / N

,

k = 0,1,..., N − 1

n =0

(i)

N −1

∑a e

=

n − j 2πnk / N

,

n=0 N −1

∑ (ae

=

)

− j 2πnk / N n

=

n=0

= (ii) N = 4 x(n) = cos 0, = 1, N −1

1− aN , 1 − ae − j 2πk / N

cos

π

4 0707,

∑ x ( n) e X (k ) =

,

cos

n =0

,

∑ x ( n) e

Q.90

N

π 2

3π 4 -0,707

,

cos

k = 0,1,..., N − 1

3

∴ X (k ) = {X (0), = {1,

)

k = 0, 1, ...., N − 1

0,

− j 2πnk / N

=

(

1 − ae − j 2πk / N 1 − ae − j 2πk / N

− j 2πnk / N

,

k = 0, 1, 2, 3

n =0

X (1), 1 − j1.414,

X (3)} 1 + j1.414}

X (2), 1,

(i) Find the circular convolution of the following sequence (rectangular) 1, 0 ≤ n ≤ N − 1 x1 (n) = x 2 ( n) =  0, otherwise

(ii) Compute the DFT of a) x (n ) = δ (n ) b) x(n ) = δ (n − n 0 )

(4) N −1

X 1 ( k ) = X 2 (k ) =

∑W n =0

Ans: (i)

63

nK N

N , k = 0 = 0therwise 0

AE06/AC04/AT04

SIGNALS & SYSTEMS N 2 , k = 0  elsewhere X 3 ( k ) = X 1 ( k ) X 2 (k ) = 0,

x3 ( n ) = N , 0 ≤ n ≤ N − 1 X ( k ) = DFT [δ ( n)] = 1

(ii) a)

− jωno X ( jω ), we get b) Since x[n − no ] ↔ e − jωmo DFT [δ (n − n0 ] = e

Q.91

Find the Nyquist frequency of the following signals. (ii) Sa 2 (100 t ) (iv) 10 sinc(2t )

(i) Sa (100 t ) (iii) 25 cos(500 πt ) x=

(8)

sin x x

Ans: Sa (i) This function will have frequency response a rectangular pulse with 100 100 fm = , fs = 2 fm = 2π π Hz maximum frequency (ii) The sin2(2πft) = (1-cos 4πft)/2, the maximum frequency will be 2f. 200 100 fm = , fs = 2 fm = π π Hz 500 fm = , f s = 2 f m = 500 Hz 2π (iii) sin πt sin 2πt sinc t = . 10 2πt . Its frequency response will π t (iv) Thus 10 sinc(2t) =

be a rectangular pulse X ( f ) such that the maximum frequency f m = 1 Hz. Hence sampling frequency f s = 2 f m = 2 Hz. Q.92

Define ideal low pass filter and show that it is non-causal by finding its (8) impulse response. Ans:

e − j 2πft0 ,− B ≤ f ≤ B H( f ) =  0, f > B  B B e j 2πf (t −t0 ) j 2πf ( t − t 0 ) = h(t ) = − B ∫ e df j 2πf (t − t 0 ) − B −B

[

]

1 e j 2πB (t −t0 ) − e − j 2πB (t −t0 ) j 2π (t − t 0 ) sin 2πB (t − t 0 ) = 2B = 2 BSinc[2 B (t − t 0 )] 2πB (t − t 0 ) =

The system is non-causal as there is some response before t = 0. Q.93

Obtain the Laplace transform of the square wave shown. 64

(8)

AE06/AC04/AT04

SIGNALS & SYSTEMS

Ans: 2T

[ f (t )] = ∫ f (t )e £

4T

− st

dt + ∫ f (t )e dt +.... +

0

− st

2T

( n + 2 )T

∫ f (t )e

− st

dt

nT

 2T  − 2 sT  2T  − st   = ∫ f (t )e dt + ∫ f (t )e dt .e + ... +  ∫ f (t )e − st dt .e − nsT 0 0  0  2T   = 1 + e − 2 sT + e − 4 sT + .... + e − nsT  ∫ f (t )e − st dt  0  2T T 2T     1 1 − st − st − st = f ( t ) e dt = 0 . 5 e dt + 0 . 5 e dt     − 2 sT ∫ ∫ 1 − e − 2 sT  ∫0 T  = 1− e 0  1 0 . 5 0 . 5   = − e − sT − 1 + e − 2 sT − e − sT  s 1 − e − 2 sT  s  2 1 0 . 5  (1 − e −sT )  = − 2 sT  s 1− e   − sT 0.5 1 − e  =   s 1 + e − sT  2T

− st

(

)

(

Q.94

)

(

)

Find the inverse Laplace transforms of the following functions. (i)

 s +1    s+ 2

s2

(s + a )

2

+b

(ii) ln 

2

(8)

Ans: −1

(i) £

s2

(s + a )

2

  s+a a b = s − 2 2 2 2  b ( s + a) + b  +b  (s + a) + b 2

a   d f (t ) = e − at cos bt − e − at sin bt  u (t ) f (t ) + f (0 + ) b   = dt where

    a 2 − at − at − at − at  − be sin bt − ae cos bt + − ae cos bt + e sin bt  + δ (t )u (t )   b    =  2 a  =  sin bt − 2a cos bt − b sin tbt e − at u (t ) + δ (t )  b  s +1 = F ( s ) = (ii) ln s + 2 ln ( s + 1) − ln ( s + 2)

(

)

65

AE06/AC04/AT04

SIGNALS & SYSTEMS dF 1 1 = − ds s + 1 s + 2  dF  ∴   = e −t − e − 2t u (t )  ds  ∴

(

)

Hence, −1

 dF    t £  ds  1 − e −t − e − 2t u (t ) = t

[F (s)] = − 1

£

−1

[

Q.95

]

Obtain the z transforms and hence the regions of convergence of the following sequences. (i) x(n ) = [u (n ) − u (n − 10 )] 2 − n (ii) x (n ) = cos (π n ) u (n ) (8) x (n) = [u (n) − u (n − 10)]2 − n

Ans: (i)

9

Z [x( n)] = ∑ 2 − n Z − n n=0 n

 1  1−    2z  −n = 9  1  1 1 = ∑  z > 1 − 2 n =0  2 z  2 z , ROC is e jπn + e − jπn x(n) = cos(πn )u (n) = u ( n) 2

(ii) ∴ Z [x (n)] =

9 1  9 jπn − n 1  ∞ jπ −1 − jπn − n  e z + e z = e z    2  n=0 n =0  2  n =0



∑(



1 1 1  + jπ −1 − jπ −1   2 1 − e z 1− e z  1 1 1  z =  + = −1 −1  2 1 + z 1+ z  z +1,



) + ∑ (e n

− jπ

n =0

n z −1  

)

=

Q.96

ROC is

z >1

A second order discrete time system is characterized by the difference equation y (n ) − 0.1 y (n-1)-0.02 y (n-2 ) = 2 x(n ) − x(n − 1) . Find y (n ) for n ≥ 0 when x(n) = u(n) and the initial conditions are given as y(–1) = –10, y(–2) = 20 (8) Ans:

Since

y ( n) − 0.1 y ( n − 1) − 0.02( n − 2) = 2 x( n) − x( n − 1) ,

[

]

[

Y (z ) − 0.1 Y (−1) + z −1Y (z) − 0.02 Y (−2) + z −1Y(−1) + z −2 Y (z) −1

= 2X ( z ) − z X ( z ) Substituting the initial values and rearranging, we get

[

]

Y ( z ) 1 − 0.1z −1 − 0.02 z − 2 + 1 − 0.4 + 0.2 z −1 = 66

2z 1 − z −1 z −1

]

AE06/AC04/AT04

SIGNALS & SYSTEMS ( z − 0.2)( z − 0.1) 2 z − 1 0.2 = − − 0.6 z −1 z z2 (2 z − 1) z 2 0.2 0.6 Y ( z) = − − ( z − 1)( z − 0.2)( z − 0.1) z ( z − 0.2)( z − 0.1) ( z − 0.2)( z − 0.1) = Y1 ( z ) + Y2 ( z ) + Y3 ( z )

Y ( z) =

Now Y1 ( z ) ( 2 z − 1) z = z ( z − 1)( z − 0.2)( z + 0.1)

By partial fraction expansion, we get 1.13 z 0.5 z 0.36 z Y1 ( z ) = + + ( z − 1) ( z − 0.2) ( z + 0.1) Similarly 2 z 2 z Y2 ( z ) = +− 3 ( z − 0.2) 3 ( z + 0.1)

Y3 ( z ) = 0.4

z z + 0.2 ( z − 0.2) ( z + 0.1)

Thus,

Y (z) = Hence

1.13 z 0.56 z 0.83 z − + ( z − 1) ( z − 0.2) ( z + 0.1)

[

]

y (n) = 1.13 − 0.56(0.2) + 0.83(0.1) u ( n)

Q.97

n

n

A continuous random variable has a pdf f (x ) = Kx 2e− x ; x ≥ 0 . Find K, and mean and variance of the random variable. (8) Ans: By the property of PDF ∞

∫ Kx

2 −x

e dx = 1

0

Or, 2K = 1 → K = ½ Mean value of x is E ( x) =





2 −x 3 −x ∫ xf ( x)dx = ∫ xKx e dx = 0.5∫ x e dx = 3 0

0

Rx

Now E(x 2 ) =

∫x

2

f ( x )dx =





0

x 2 Kx 2 e − x dx =0.5





0

x 4 e − x dx = 12

Rx

Variance of x is 2 V ( x) = E ( x 2 ) − {E ( x)} = 12 - 9 = 3.

Q.98

Find the autocorrelation of the following functions: (i) g (t ) = e − at u (t ) t  T

t  T

(ii) g(t) = A Π   where A Π   is a rectangular pulse with period T and 67

AE06/AC04/AT04

SIGNALS & SYSTEMS

magnitude A.

(8)

Ans: g (t ) = e − at u (t )

(i) R g (τ ) =





−∞





−∞

g (t ) g (t − τ )dt

e − at e − a ( t −τ )u (t )dt = e aτ

0

e − 2 at dt =

e − aτ 2a

t   g (t ) = A Π T

(ii) R g (τ ) =









−∞

g (t ) g (t − τ )dt

=

R g (−τ )

For τ < T R g (τ )

T

= ∫τ

A 2 dt = A 2 (T − τ )

For τ ≥ T R g (τ )



g (t ) g (t − τ )dt 0, τ =∫ = T

≥T

 A 2 (T − τ ), τ < T = τ ≥ T. 0

Q.99

Find the Fourier series of the following periodic impulse train.

(8)

Ans: A0 =

2I T0 2I Bn = T0 An =

1 T0

I T0 T0 / 2 2πnt 2I ∫−T0 / 2 δ (t ) cos T0 dt = T0 T0 / 2 2πnt ∫−T0 / 2 δ (t ) sin T0 dt =0

I 2I ∴ x(t ) = + T0 T0

Q.100



T0 / 2

−T0 / 2

δ (t ) dt =

2πnt I = cos ∑ T0 T0 n =1 ∞



∑e

2πnt T0

n = −∞

The Magnitude and phase of the Fourier Transform of a signal x(t) are shown in the following figure. Find the signal x(t).

Ans: 68

AE06/AC04/AT04

SIGNALS & SYSTEMS X ( jω ) = π , − W ≤ ω ≤ W ω 0

and

πe jπ / 2 , − W ≤ ω ≤ 0 X ( jω ) =  − jπ / 2 , 0 ≤ω ≤W πe

Thus

x(t ) = F −1 [X ( jω )] =

1 2π





−∞

X ( jω )e jωt dω

W 1  0 jπ / 2 jωt πe e dω + ∫ πe − jπ / 2 e jωt dω  ∫  0 2π  −W W W jωt ) 1 =  ∫ e j (π / 2−ωt ) dω + ∫ e − j (π / 2− dω    0 0  2

=

=

1 2 

W



0

 e j (π / 2−ωt ) + e − j (π / 2− jωt ) dω  = Wcos(π / 2 − ωt )dω = W(sin ωt )dω 0    0





2 1 1  Wt  W t  Wt  = [1 − cos Wt ] = 2 sin 2   = 2 sinc 2   t t  2  2π  2π 

Q.101

Find the Discrete Time Fourier Transforms of the following signals and draw (8) its spectra. (i) x1 (n ) = a

n

a 2 h2 (t ) = 2e −3t u (t ) − e 2t u ( −t ), - 3 < σ < 2 h3 (t ) = −2e −3t u ( −t ) − e 2t u (−t ), σ < −3 h1(t) is causal but unstable due to the pole at s = 2, h2(t) is non-causal due to the pole at s = 2 but stable, and h3(t) is also non-causal due to both the poles but stable. Thus, the system cannot be both causal and stable simultaneously. Q.108

Determine the inverse Z transform of the following X(z) by the partial fraction expansion method. (8) X ( z) =

z+2 2z 2 − 7 z + 3

if the ROCs are (i) z > 3 , (ii) z
3 all poles are interior, x(n) is causal. n

2 1 1 x(n) = δ (n) −   u (n) + (3) n u (n) 3 3 2

Therefore, (ii) When z
. 3 1 1 1 X ( z) 2 1 − z −1 + z − 2  z −  z −  4 8 2  4  H ( z) z 2 1 = = − 1 1 1  1 z  z−  z −  z −  z − 2 4 2  4  Y (z) −

  1 n  1 n  ∴ h(n) = 2  −    u (n)   2   4   (iii) Here

Q.110

z , z >1 z −1 z3 Y ( z) = X ( z)H ( z) = , z >1 1 1 ( z − 1)( z − )( z − ) 2 4 8 z 1 z z = −2 + , z >1 1 3 1 3 z −1 z− z− 2 4 n n 8  1  11  y (n) =  − 2  +    u (n) 3  4    3  2  X ( z) =

A random variable X has the uniform distribution given by 1  , for 0 ≤ x ≤ 2π f X ( x ) =  2π 0, otherwise Determine mean, mean square, Variance. ∞

Ans:

Mean m x =

∫ xf x ( x)dx =

−∞



1

∫ x 2π dx = π 0



Mean square X 2 == E ( X 2 ) =

2 ∫ x f x ( x)dx =

−∞

Variance:

(10)

σ x 2 = E( X 2 ) − mx 2

72

1 2π



∫x 0

2

4 dx = π 2 3

4 1 = π 2 −π 2 = π 2 3 3

AE06/AC04/AT04

SIGNALS & SYSTEMS

π

σx = Q.111

3

Discuss the properties of Gaussian PDF.

(6)

Ans: Property 1: The peak value occurs at x = m, i.e., mean value 1 at x = m (mean) f x ( x) = σ 2π Property 2: Plot of Gaussian PDF exhibits even symmetry around mean value, i.e., f x (m − σ ) = f x ( m + σ )

Property 3: The mean under PDF is 1 / 2 for all values of x below mean value and ½ for all values of above mean value, i.e., 1 P ( X ≤ m) = P ( X > m) = 2 Q.112

A stationary random variable x(t) has the following autocorrelation function

R x (τ ) = σ 2 e

where σ 2 , µ are constants. Rx (t ) is passed through a

−µ τ

filter whose impulse response is h(τ ) = αe −ατ u (τ ) where α is a constant, u (τ ) is unit step function. (i) Find power spectral density of random signal x(t). (ii) Find power spectral density of output signal y(t). (8) S X (ω ) = FT[RX (t )] =

Ans: (i)

=





−∞





−∞



−∞

σ 2e µ τ e − jωτ dτ =

H (ω ) = FT[h(τ )] =

(ii)



R X (τ )e − jωτ dτ

2 µσ 2 µ2 +ω2

α α + jω

αe −ατ u (τ )e − jωτ dτ = 2

2

S y (ω ) = H (ω ) S X (ω ) =

Q.113

Determine the convolution of the following two continuous time functions. (8) x(t ) = e − at u (t ) , a > 0 and h(t ) = u (t ) Ans: y (t ) = h(t ) * u (t ) =





−∞

=−

Q.114

α α 2 µσ 2 S X (ω ) = 2 α + jω α +ω2 µ2 +ω2

x (λ ) h(t − λ ) dλ =

[ ]





−∞

[

e −aλ u (λ )u (t − λ )dλ =

t

∫e o

− aλ



]

1 −aλ 1 1 e = 1 − e −at . 0 a a

Determine signal energy and power of the following signals (i) x(n ) = u (n ) (ii) x(t ) = e −3t Ans:

73

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AE06/AC04/AT04

SIGNALS & SYSTEMS ∞

E=

(i)

1 2N

P = lt N →∞ T

E=

(ii)



n = −∞ N 2





∑ u [n] = 1 2

n = −∞

x [n ] = lt N →∞

n=− N

1 2N

N

∑ u [n] = 1. 2

n=− N

T

1 ∫ [e ] dt = ∫ [e ]dt = − 6 [e − 3t 2

−6 t

−T

− 6T

]

− e 6T = ∞

−T

P = LtT → ∞

Q.115

x 2 [n] =

[

]

1 1 − 6T E = LtT → ∞ − e − e 6T = ∞ 2T 12T

Find the DTFT of the sequence x(n) = u(n).

(4)

Ans: ∞



n = −∞

n =0

DTFT x(n) = X (e jω ) =

∑ x(n)e − jωn = ∑1.e − jωn =

1 1 − e − jω

It is not convergent for ω = 0.

( )

∴ X e jω =

e jω / 2 2 sin

Q.116

ω

, ω ≠ 0.

2

Find the inverse Fourier transform of δ (ω ) . Ans: ∞ 1 1 j ωt −1 F [δ (ω )] = e δ (ω )e jωt dω = ∫ 2π − ∞ 2π

[ ]

(4)

ω =0

=

1 2π

Q.117 Check whether the following signals are energy or power signal and hence find the corresponding energy or power. (6) −α (t ) (i) x(t ) = Ae ⋅ u (t ), α > 0 2 (ii) x(t ) = cos ω 0 t Ans: (i) x(t ) = Ae −αt u (t ), α > 0 ∞



E=

Then

2

x(t ) dt = A2

−∞

2 e −2αt ∞ A = − 2α 0 2α

Since 0 < E < ∞, x(t ) is an energy signal. (ii) Since x(t ) = cos 2 ω o t is a periodic function, it is a power signal. P = Lt To → ∞

1 To

1 = Lt T → ∞ To

= Lt T → ∞

1 To

To / 2



x 2 (t )dt

− To / 2

To / 2

∫ [

− To

]

1 cos ωot dt = Lt T → ∞ To /2 2

2

To / 2

To / 2

∫ [cos

− To / 2

3 1 ∫ 8 [3 + 4 cos 2ω t + cos ω t ]dt = 8 o

− To / 2

74

o

4

]

ωot dt

AE06/AC04/AT04 Q.118

SIGNALS & SYSTEMS

Find the convolution of two rectangular pulse signals shown below.

(10)

Ans: For − ∞ ≤ t ≤ 4 and t ≥ 10 the output is 0. t

For

4 ≤ t ≤ 6 , y (t ) = ∫ 2dt = 2(t − 4) 4 8

For

6 < t ≤ 8 , y (t ) = ∫ 2dt = 2(8 − 6) = 4 6 10

For

8 < t ≤ 10 , y (t ) = ∫ 2dt = 2(10 − t ) t

Thus y(t) is as shown in the figure. 2

Q 119 Given the Gaussian pulse x(t ) = e −π t , determine its Fourier transform.

(8)

Ans: X (ω ) =





−∞

x(t )e − jωt dt =



2

e −πt e − jωt dt =

−∞

 



 

2 π

πt 2 + jωt =  π t +

Expressing



2    

+

ω2 4π





e − (πt

2

+ j ωt )

−∞

dt

,

we have X (ω ) =





 



 

2 π

− πt + e

−∞

ω2 ω2 − − e 4π dt = e 4π

u = πt +

Let

ω2 2 X (ω ) = e 4π π −

Then

2    





0

e

jω 2 π





e

−∞



 

2 π

2    

dt

,

− (u )2

ω2 2 π − πf 2 du = e 4π =e π 2 −

Q.120 Find the exponential Fourier series of the following signal.

75

 

−  πt +

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AE06/AC04/AT04

SIGNALS & SYSTEMS

Ans: Here T = 1sec, ω o = 2π , x(t ) =

10 t = 10t T 1

T

Cn =

1 x(t )e − jnωot dt = ∫ 10te − jn 2πt dt ∫ T 0 0

 te − j 2πnt 1 1 e − j 2πnt  = 10  −∫ dt   − j 2πn 0 0 − j 2πn  1  e − j 2πnt e − j 2πnt   = 10  + 2 2  − j 2πn 4π n 0  5 = j πn ∞ ∞ 5 x(t ) = ∑ C n e − j 2πnt = ∑ j e j 2πnt ; n = −∞ πn n = −∞ π

Now

n≥0



θ n = tan −1 ∞ =  2

π

−  2

n≤0

Q.121 State and prove the following properties of DTFT. (i) Time shifting, frequency shifting (ii) Conjugate symmetry (iii) Time reversal.

(6)

Ans: x ( n ) ↔ X (e jω )

(i) Time shfting: Frequency shifting: (ii)

x ( n − no ) ↔ e − j

ωno

X (e jω )

e jωo n x(n) ↔ X (e j (ω −ωo ) ) x * (n) ↔ X * (e − jω ) X (e jω ) = X * (e − jω ) , x (n) real

[

(iii)

]

Even [x( n)] ↔ Re X (e jω ) Odd [x(n)] ↔ j Im X (e jω ) x ( − n ) ↔ X (e − jω )

[

]

Q.122 Consider a stable causal LTI system whose input x(n ) and output y (n ) are related through second order difference equation 1 3 y (n ) −   y (n − 1) + y (n − 2 ) = 2 x(n ) . 8 4 n

1 Determine the response for the given input x(n ) =   u (n ) 4 Ans: 76

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AE06/AC04/AT04

SIGNALS & SYSTEMS n

3 1 1 y (n) − y (n − 1) + y (n − 2) = 2 x(n) = 2  u (n) 4 8 4 Taking DTFT on both sides 3 1 2 Y (e jω ) − e − jω Y (e jω ) + e − 2 jω Y (e jω ) = 1 − jω 4 8 1− e 4 2 jω Y (e ) =  3 − j ω 1 − 2 j ω   1 − jω  1 − 4 e + 8 e  1 − 4 e  2 = − j ω  1   1 − jω   1 − jω  1 e −  2  1 − 4 e  1 − 4 e  2 = 2  1 − jω   1 − jω  1 − 2 e  1 − 4 e  8 4 2 = − − − j − j ω ω  1   1   1 − jω  2 1 1 − e − e  2   4  1 − e   4  Taking inverse DFT n

n

n

1 1 1 y (n) = 8  u (n) − 4  u (n) − 2(n + 1)  u (n) 2 4 4

Q.123

A continuous time signal is x(t ) = 8 cos 200πt

(8)

(i) Determine the minimum sampling rate. (ii) If fs = 400 Hz, what is discrete time signal obtained after sampling? (iii) If fs = 150 Hz, what is discrete time signal obtained after sampling?

Ans:Here ω = 200π → f = 100 Hz (i) Minimum sampling rate = 2 f = 2 × 100 = 200 Hz f 100 1 (ii) = = f s 400 4 1 πn ∴ x(n) = 8 cos 2πfn = 8 cos 2π n = 8 cos 4 2 (iii) Here

∴ x(n) = 8 cos 2πfn = 8 cos 2π

f 100 2 = = f s 150 3

2 4πn 2π  4πn  n = 8 cos = 8 cos 2π − n = 8 cos 3 3 3  3 

Q.124 State and prove Parseval’s theorem for continuous time periodic signal.

77

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AE06/AC04/AT04

SIGNALS & SYSTEMS

Ans: Parseval’s theorem: The Parseval’s theorem states that the energy in the time-domain representation of a signal is equal to the energy in the frequency domain representation normalized by 2π. Proof: The energy in a continuous time non-periodic signal is 2



E = ∫ x(t ) dt. −∞

2

Since x(t ) = x(t ) x * (t ) , from the Fourier series we get 1 ∞ X * ( jω )e − jωt dω . 2π − ∞ ∞  1 ∞  E= x (t )  X * ( jω )e − jωt dω  dt −∞  2π − ∞ 



x * (t ) =



Hence,



Now interchanging the order of the integrations, we get E=

1 2π

X * ( jω )  −∞ 





1 x (t )e − jωt dt  dω = −∞ 2π 









−∞

X * ( jω ) X ( j ω ) d ω

1 ∞ 2 X ( j ω ) dω . ∫ − ∞ 2π Thus, the energy in the time-domain representation of the signal is equal to the energy in the frequency-domain representation normalized by 2π. =

Q.125 Compute the magnitude and phase of the frequency response of the first order discrete time LTI system given by equation (10) y (n ) − Ay (n − 1) = Bx(n ) where A < 1 . Ans: y ( n) = Ay ( n − 1) + Bx( n )

h(n) = BA n u (n) H (e jω ) = 1 − Ae

Since



∑ h ( n) e

− j ωn

=

B jω

1 − Ae − e = (1 − A cos ω ) + jA sin ω

n = −∞ − jωn

(1 − A cos ω ) + ( A sin ω ) = 1 + A Angle (1 − Ae ) = tan 1 −AAsincosωω

1 − Ae − jωn =

2

2

− jωn

∴ H ( e jω ) =

2

− 2 A cos ω

−1

B 1 + A − 2 A cos ω 2

and angle H (e jω ) = Angle  B − tan −1 

A sin ω  . 1 − A cos ω 

Q.126 Determine the Fourier transform of unit step x(t ) = u (t ). Ans: Fourier transform of x(t) is

X (ω ) =





−∞

Thus Fourier transform of u(t) is

78

x(t )e − jωt dt

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AE06/AC04/AT04

SIGNALS & SYSTEMS X (ω ) =





u (t )e − jωt dt =

−∞





0

e − jωt dt =

1 . jω

Q.127 By using convolution theorem determine the inverse Laplace transform of the following functions. (8) 1 1 (i) (ii) s2 s2 − a2 s 2 (s + 1) Ans: 1 (i) F (s) = 2 2 = F1 ( s ) F2 ( s ) s (s − a 2 )

(

)

1 s2

where

F1 ( s ) =

Thus

f1 (t ) = t

F2 ( s) =

and

f 2 (t ) =

and

1 s − a2 2

1 sinh (at ) a

t

£ −1 F ( s ) = ∫ f 1 (t − τ ) f 2 (τ )dτ

Now

0 t

1 = (t − τ ) sinh (aτ )dτ a 0



1 [sinh at − 1] a3 1 F (s) = 2 = F1 ( s) F2 ( s) s ( s + 1) 1 1 F1 ( s ) = 2 and F2 ( s ) = ( s + 1) s =

(ii) where

f 1 (t ) = t

Thus

and

f 2 (t ) = e − t

t

£ F ( s ) = ∫ f1 (t − τ ) f 2 (τ )dτ −1

Now

0 t

t

t

0

0

0

= ∫ (t − τ )e −τ dτ = ∫ te −τ dτ − ∫ τe −τ dτ = (−te −t + t ) + (te −t + e −t − 1) = t + e −t − 1

Q.128 Check the stability & causality of a continuous LTI system described as (s − 2) H (s ) = (s + 2)(s − 3) Ans: s−2 1 4 1  . =  + ( s + 2)( s − 3) 5  s + 2 s − 3  The system has poles at s = -2, s = 3. Thus, the response of the system will be

Given

H (s) =

4 − 2t 1 e u (t ) + e3t u (t ), σ > 3 5 5 4 − 2t 1 h2 (t ) = e u (t ) − e3t u (−t ), - 2 < σ < 3 5 5

h1 (t ) =

79

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AE06/AC04/AT04

SIGNALS & SYSTEMS

4 1 h3 (t ) = − e − 2t u (−t ) − e 3t u ( −t ), σ < −2 5 5

Note that the response h1(t) is unstable and causal, h2(t) is stable and non-causal, h3(t) is stable and non-causal. Thus, the system cannot be both stable and causal simultaneously.

Q.129

Find the z -Transform X ( z ) and sketch the pole-zero with the ROC for each of the following sequences. (8) n

n

n

n

1 1 (i) x(n ) =   u (n ) +   u (n ) 2  3

1 1 (ii) x(n ) =   u (n ) +   u (− n − 1)  3 2 Ans: n

n

1 1 (i) x(n) =   u (n) +   u (n) 2  3 n

z 1 ,   u ( n) ↔ 1 2 z− 2 n z 1 ,   u (n) ↔ 1  3 z− 3 Thus,

Z >

z >

1 2

1 3

5  2 z z −  z z 12   X (z) = + = , 1 1  1  1 z− z−  z −  z −  2 3  2  3 n

(ii)

Z >

1 2

n

1 1 x(n) =   u (n) +   u (− n − 1)  3 2 n

z 1 ,   u (n) ↔ 1  3 z− 3

z >

1 3

n

z 1 1 , z <   u (−n − 1) ↔ − 1 2 2 z− 2 z z 1 z ∴ X (z) = − =− , 1 1 1  1 6 z− z−  z −  z −  3 2 2  3 

Q.130 Determine the inverse z transform of x(z ) = convergence re (i) z > 1, (ii) z
1 ∴ x ( n) =

(ii) ROC is z