Code: A-06/C-04/T-04

Code: AE-06/AT-04/AC-04 Subject: SIGNALS & SYSTEMS

PART – I, VOL – I

TYPICAL QUESTIONS & ANSWERS

OBJECTIVE TYPE QUESTIONS

Each Question carries 2 marks.

Choose the correct or best alternative in the following:

Q.1 The discrete-time signal x (n) = (-1)ⁿ 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 (t), > 0 by a factor

(A) . (B) .

(C) ². (D) .

Transform

Ans: A x(t) x(αt), α > 0

α > 1 compression in t, expansion in f by α.

α < 1 expansion in t, compression in f by α.

Q.3 A useful property of the unit impulse is that

(A) . (B) .

(C) . (D) .

Ans: C Time-scaling property of δ(t):

δ(at) = 1 δ(t), a > 0

Q.4 The continuous time version of the unit impulse is defined by the pair of relations

(A) (B) .

(C) . (D) .

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 R_x

z Reciprocals

x₂(n) = x₁(-n) X₁(1/z), RoC 1/ R_x

Q.6 The Fourier transform of the exponential signal 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 w_o its FT must be an impulse at w = w_o

Q.7 If the Laplace transform of is , then the value of

(A) cannot be determined. (B) is zero.

(C) is unity. (D) is infinity.

Ans: B f(t) ω

s² + ω²

Lim f(t) = Lim s F(s) [Final value theorem]

t ∞ s 0

= Lim sω = 0

s 0 s² + ω²

Q.8 The unit impulse response of a linear time invariant system is the unit step

function . For t > 0, the response of the system to an excitation

will be

(A) . (B) .

(C) . (D) .

Ans: B

h(t) = u(t); x(t) = e^-at u(t), a > 0

System response y(t) =

= 1 (1 - e^-at)

Q.9 The z-transform of the function has the following region of convergence

(A) (B)

(C) (D)

Ans: C x(n) = ∑ δ(n-k)

k = -∞

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

k = -∞

= 1 , ׀z׀ < 1

1 – z

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

R_XX (τ) = 1 ∫ x(τ) x(t + τ) dτ triangular function of duration 2T.

T -T/2

Q.11 The Fourier transform (FT) of a function x (t) is X (f). The FT of will be

(A) . (B) .

(C) . (D) .

∞

Ans: B (t) = 1 ∫ X(f) e^jωt dω

2π - ∞

∞

dx = 1 ∫ jω X(f) e^jωt dω

dt 2π - ∞

\ dx « j 2π f X(f)

Q.12 The FT of a rectangular pulse existing between t = to t = T / 2 is a

(A) sinc squared function. (B) sinc function.

(C) sine squared function. (D) sine function.

Ans: B x(t) = 1, -T ≤ t ≤ T

2 2

0, otherwise

+∞ +T/2 +T/2

X(jω) = ∫ x(t) e^–jωt dt = ∫ e^–jωt dt = e^–jωt

–∞ –T/2 –jω

–T/2

= - 1 (e^-jωT/2 - e^jωT/2) = 2 e^jωT/2 - e^-jωT/2

jω ω 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) .

(B) .

(C) .

(D) .

Ans: C For a band pass signal, the minimum sampling rate is twice the

bandwidth, which is 0.5kHz here.

Q.14 A given system is characterized by the differential equation:

The system is :

(A) linear and unstable. (B) linear and stable.

(C) nonlinear and unstable. (D) nonlinear and stable.

h(t)

Ans:A d²y(t) – dy(t) – 2y(t) = x(t), x(t) x(t) y(t)

dt²dt system

The system is linear . Taking LT with zero initial conditions, we get

s²Y(s) – sY(s) – 2Y(s) = X(s)

or, H(s) = Y(s) = 1 = 1

X(s) s² – 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 is

(A) linear for any value of b. (B) linear if b > 0.

(C) linear if b < 0. (D) non-linear.

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 is

(A) . (B) .

(C) . (D) .

Ans: A x(t) = u(t) X(jω) = π δ(ω) + 1

Jω

Duality property: X(jt) 2π x(-ω)

u(ω) 1 δ(t) + 1

2 πt

Q.17 The impulse response of a system is . The condition for the system to

be BIBO stable is

(A) a is real and positive. (B) a is real and negative.

(C) . (D) .

+∞ +∞

Ans: D Sum S = ∑ ׀h(n)׀ = ∑ ׀ aⁿ u(n) ׀

n = -∞ n = -∞

+∞

£ ∑ ׀a׀ⁿ ( u(n) = 1 for n ≥ 0 )

n = 0

£ 1 if ׀a׀ < 1.

1- ׀a׀

Q.18 If is the region of convergence of x (n) and is the region of convergence of

y(n), then the region of convergence of x (n) convoluted y (n) is

(A) . (B) .

(C) . (D) .

Ans:C x(n) X(z), RoC R₁

y(n) Y(z), RoC R₂

x(n) * y(n) X(z).Y(z), RoC at least R₁∩ R₂

Q.19 The continuous time system described by 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.

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

a x₁(t) ® y₁(t) = a x₁(t²)

b x₂(t) ® y₂(t) = b x₂(t²)

a x₁(t) + b x₂(t) ® y(t) = ax₁(t²) +b x₂(t²) = y₁(t) + y₂(t)

System is Linear

System is time varying. Check with x(t) = u(t) – u(t–1) ® y(t) and

x₁(t) = x(t – 1) ® y₁(t) and find that y₁(t) ¹ 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.

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) .

(B) 1 and .

(C) 1 and .

(D) 2 and .

+∞

Ans:B Mean = μ_x(t) = ∫ x f_x(t) (x) dx

-∞

= ∫ x 1 dx = 1 x² 3 = 9 – 1 1 = 1

-1 4 4 2 -1 2 2 4

+∞

Variance = ∫ (x - μ_x)² f_x (x) dx

-∞

= ∫ (x - 1)² 1 d(x-1)

-1 4

= 1 (x - 1)³ 3 = 1 [8 + 8] = 4

4 3 -1 12 3

Q.22 If white noise is input to an RC integrator the ACF at the output is proportional to

(A) . (B) .

(C) . (D) .

Ans: A

R_N(τ) = N₀ exp - ׀ τ ׀

4RC RC

Q.23 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.

Ans: A +∞ ∞ ∞ ∞

Energy = ∑ x²(n) = ∑ a² = ∑ (a²) = 1+ 2 ∑ a²ⁿ

n=-∞ n=-∞ n=-∞ n=1

= finite since ׀a׀ < 1

\This is an energy signal.

Q.24 The spectrum of x (n) extends from , while that of h(n) extends

from . The spectrum of extends

from

(A) . (B) .

(C) . (D)

Ans: D Spectrum of y(n) is H( e^jω).

X( e^jω) exists out the smaller of the two ranges.

Q.25 The signals and are both bandlimited to and

respectively. The Nyquist sampling rate for the signal

will be

(A) . (B) .

(C) . (D) .

Ans: C

Nyquist sampling rate = 2 x [highest frequency in ] = 2(ω₁ + ω₂)

Q.26 If a periodic function f(t) of period T satisfies , then in its Fourier

series expansion,

(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: A

T T/2 T T/2 T/2

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

T 0 T 0 T/2 T 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.

(C) 3 KHz. (D) 4 KHz.

Ans: B

Minimum sampling frequency = 2(Bandwidth) = 2 kHz

Q.28 The region of convergence of the z-transform of the signal

(A) is . (B) is.

(C) is . (D) does not exist.

Ans:

2ⁿu(n) 1 , ׀z׀ > 2

1 –2 z ^-1

3ⁿ u(-n-1) 1 , ׀z׀ < 3

1 – 3z ^-1

ROC is 2 < ׀z׀ < 3.

Q.29 The number of possible regions of convergence of the function is

(A) 1. (B) 2.

(C) 3. (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 .

Then

(A). (B) .

(C) . (D) .

Ans: B u(t) A(s) = 1

F.T

u(t) B(jω) = 1 + π δ(ω)

jω

A(s) = 1 but B(jω) ≠ 1

s jω

PART – II, VOL – I

NUMERICALS

Q.1. Determine whether the system having input x (n) and output y (n) and described by relationship :

is (i) memoryless, (ii) stable, (iii)causal (iv) linear and (v) time invariant.

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.

Ans:

x(t) = e^-at u(t), a > 0

x(t) is a non-periodic signal.

+ ∞ ∞ ∞

Energy E = ∫ x²(t) dt = ∫ e^-2^atdt = e^-2^at = 1 (finite, positive)

- ∞ 0 -2a 2a

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

Ans:

Assumption : α > 0, A > 0, -∞ < t < ∞

Even part x_e(t) = x(t) + x(-t)

Odd part x_o(t) = x(t) - x(-t)

x(t)

A Ae^-αt

x(-t)

Ae^+αt A

x_e(t)

A/2

0 t

x_o(t)

A/2

-A/2

Q.4. Use one sided Laplace transform to determine the output y (t) of a system described by where y = 3 and

Ans:

d²y + 3 dy + 2 y(t) = 0, y(0-) = 3, dy = 1

dt² 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

s² + 3s + 2 (s + 1)(s + 2)

= A + B

s + 1 s + 2

A = 3s + 10 = 7 ; B = 3s + 10 = -4

s + 2 s = -1 s + 1 s = -2

Y(s) = 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.

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 Alternative Realisation

Q. 6. 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).

Ans:

h(t) = e^-at u(t) for input x(t)

Output y(t) = (e^-bt - e^-ct) u(t)

L L L

h(t) H(s), y(t) Y(s), x(t) X(s)

H(s) = 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) = A + B

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

s + b s + c

x(t) = (a – b) e^-bt + (c – a) e^-ct

The input x(t) = [(a – b) e^-bt + (c – a) e^-ct] u(t)

Q.7. Write an expression for the waveform shown in Fig. using only unit step function

and powers of 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)]

Q.8. For f(t) of Q7, find and sketch (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)]

f ΄(t)

E/T

0 T 2T 3T 4T t

–E/T

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

Q.9. Define a unit impulse function .

Ans:

. Unit impulse function δ(t) is defined as:

δ(t) = 0, t ≠ 0

∫ δ(t) dt = 1

0–

It can be viewed as the limit of a rectangular pulse p(t)of duration a and height 1/a when

a 0, as shown below.

Q.10. Sketch the function and show that

Ans:

g(t) As ε 0, duration 0, amplitude ∞

3/ε ε

∫ g(t) dt 1

0 ε t

Q.11. Show that if the FT of x (t) is , then the FT of is .

Ans:

x(t) X(jω)

Let x t X₁(jω) , then

+∞

X₁(jω) = ∫ x t e^-jωt dt Let t = α dt = a dα

- ∞ a a

+∞

= ∫ x(α) e^-jωaα a dα if a> 0

-∞

+∞

- ∫ x(α) e^-jωaα a dα if a < 0

-∞ +∞

Hence X₁(jω) = ׀a׀ ∫ x(α) e^-jωa^α dα = ׀a׀ X (jωa)

-∞

Q.12. Solve, by using Laplace transforms, the following set of simultaneous differential equations for x (t).

Ans:

The initial conditions are : .

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

x(t) X(s), x΄(t) s X(s), δ(t) 1, u(t) 1

(Given zero initial conditions)

2 sX(s) + 4 X(s) + sY(s) + 7 Y(s) = 5

sX(s) + X(s) + sY(s) + 3 Y(s) = 5

(2s + 4) X(s) + (s+7) Y(s) = 5

(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 s² + 6s – 3 = A + Bs+ C

s(s² + 2s + 5) s s² + 2s + 5 s s²+ 2s +5

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

\ A +B = -5

2A + C = -30

5A =15

Thus A = 3, B = -8, C = -36 and we can write

X(s) = 3 – 8 s +1 – 14 2

s (s + 1)² + 2² (s + 1)² + 2²

x(t) = (3 – 8 e^-t cos 2t – 14 e^-tsin 2t) u(t)

Q.13. Find the Laplace transform of .

Ans:

sin (ω₀t) ω₀

s² + ω₀²

Using t f(t) - d [F(s)],

L [ t sin (ω₀t) u(t) ] = - d ω₀

ds s² + ω₀²

= 0 - ω₀(2s) = 2ω₀s

(s² + ω₀²)² (s² + ω₀²)²

Q.14. Find the inverse Laplace transform of .

Ans:

F(s) = s-2 = A + B + C + D

s(s+1)³ s s+1 (s+1)²(s+1)³

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

(s+1)³ s=0

s³ : A+B = 0

D = s-2 = 3

s s = -1 s² : 3A + 2B + C = 0

F(s) = -2 + 2 + 2 + 3

s s+1 (s+1)²(s+1)³

f(t) = -2 + 2 e^-t + 2 t e^-t + 3 t² e^-t

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

Q.15. Show that the difference equation represents an all-pass transfer function. What is (are) the condition(s) on for the system to be stable?

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

X(z) 1- α z^-1 z- α

Zero : z = 1 As poles and zeros have reciprocal values, the transfer function

α 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

Ans:

H(z) = 1 + z^-1 + z^-2 + z^-3 = 1 – z ^–4 Geometric series of 4 terms

1 – z ^–1 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 and and indicate their regions of convergence.

Ans:

x₁(n) = αⁿ u(n) and x₂(n) = -αⁿ u(-n-1)

X₁(z) = 1 RoC ׀αz^-1׀ < 1 i.e., ׀z׀ > α

1-αz^-1

-1

X₂(z) = ∑ - αⁿz^-n

n=-∞

∞

= - ∑ α^-nzⁿ = -( α^-1z + α^-2z² + α^-3z³ + ………)

n=1

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

= - α^-1z = z = 1 ; RoC ׀α ^-1 z׀ < 1 i.e., ׀z׀ < ׀α׀

1- α^-1z z-α 1 - α z^-1

Q. 18. Determine the sequence whose z-transform is .

Ans:

. H(z) = 1 , ׀r׀ < 1

1-2r cosθ z ^-1 + r² z ^-2

= 1 , ׀r׀ < 1

(1-r e^jθ z^-1) (1-r e ^-jθ z ^-1)

= A + B , ׀r׀ < 1

(1-r e^jθ z^-1) (1-r e ^-jθz ^-1)

where A= 1 = 1

(1-r e^–jθ z^-1) r e^jθ z^-1=11- e ^-j2θ

B = 1 = 1

(1-r e^jθ z^-1) r e^-jθ z^-1=11- e^j2θ

\ h(n) = 1 ( r e^jθ )ⁿ+ 1 ( r e^-jθ )ⁿ

1- e^-2jθ1- e^2jθ

\h(n) = rⁿ u(n)

              = rⁿ e ^{j(n +
1)θ}  - e^- ^{j(n +
1)θ}   u (n)

e^jθ- e^-jθ

= rⁿ sin(n+1) θ u (n)

sinθ

Q.19. Let the Z- transform of x(n) be X(z).Show that the z-transform of x (-n) is .

Ans:

x(n) X(z) Let y(n) = x(-n)

∞ ∞ ∞

Then Y(z) = ∑ x(-n)z^-n = ∑ x(r) z^+r = ∑ x(r) (z^-1 ) ^-1 = X (z^-1 )

n = -∞ r = -∞ r = -∞

Q.20. Find the energy content in the signal .

Ans:

x(n) = e^-0.1ⁿsin 2πn

+∞ +∞ 2

Energy content E = ∑ ׀x²(n)׀ = ∑ e^-0.2ⁿ sin 2πn

n = - ∞ n = - ∞ 4

+∞

E = ∑ e^-0.2n sin² nπ

n = - ∞ 2

+∞

E = ∑ e^-0.2n 1-cosnπ

n = - ∞ 2

+∞

= 1 ∑ e^-0.2n [1 – (-1)ⁿ]

2 n = - ∞

Now 1 – (-1)ⁿ= 2 for n odd

0 for n even

∞ ∞

Also Let n = 2r +1 ; then E = ∑ e^{-.2(2r +1
)} = ∑ e^-.4re^{- .2}

r = - ∞ r = - ∞

∞ ∞

= e^-0.2∑ e^-.4r+∑ e^.4r The second term in brackets goes to infinity . Hence

r = 0 r = 1 E is infinite.

Q.21. Sketch the odd part of the signal shown in Fig.

Ans:

Odd part x_o(t) = x(t) – x(-t)

Q.22. A linear system H has an input-output pair as shown in Fig. Determine whether the

system is causal and time-invariant.

Ans:

System is non-causal the output y(t) exists at t = 0 when input x(t) starts only at

t = +1.

System is time-varying the expression for y(t) = [ u (t) – u (t-1)(t –1) + u (t –3) (t – 3) – u (t-3) ] shows that the system H has time varying parameters.

Q.23. Determine whether the system characterized by the differential equation is stable or not.

Ans:

d²y(t) - dy(t) +2y(t0 = x(t)

dt² dt

L L

y(t) Y(s); x(t) X(s); Zero initial conditions

s² Y(s) – sY(s) + 2Y(s) = X(s)

System transfer function Y(s) = 1 whose poles are in the right half plane.

X(s) s² – s + 2

Hence the system is not stable.

Q.24 Determine whether the system is invertible.

Ans:

y(t) = ∫ x(τ) dτ

- ∞

Condition for invertibility: H^-1H = I (Identity operator)

H Integration

H^-1 Differentiation

x(t) y(t) = H{x(t)}

H^-1{y(t)} = H^-1 H{x(t)} = x(t)

The system is invertible.

Q.25 Find the impulse response of a system characterized by the differential equation

Ans:

y΄(t) + a y(t) = x(t)

L L L

x(t) X(s), y(t) Y(s), h(t) H(s)

sY(s) + aY(s) = X(s), assuming zero initial conditions

H(s) = Y(s) = 1

X(s) s + a

The impulse response of the system is h(t) = e^-at u(t)

Q.26. Compute the Laplace transform of the signal .

Ans:

y(t) = (1 + 0.5 sint) sin1000t

= sin 1000t + 0.5 sint sin 1000t

= sin 1000t + 0.5 cos 999t – cos 1001t

= sin 1000t + 0.25 cos 999t – 0.25 cos 1001t

Y(s) = 1000 + 0.25 s - 0.25 s

s² + 1000²s² + 999² s² + 1001²

Q.27. Determine Fourier Transform of the signal and

determine the value of .

Ans:

We assume f(t) = e^-αt cos(ωt + θ) u (t) because otherwise FT does not exist

FT +∞

f(t) F(ω) = ∫ e^-αt e^{j(ωt + θ)} + e^{-j(ωt
+ θ)} e^-jωt dt

0 2

+∞

F(ω) = 1 ∫ [e^-αt e^-jωt e^{jωt
+ jθ} + e^-αt e^-jωt e^{-jωt –
jθ}] dt

2 0

+∞

= 1 ∫ [e^{-αt + jθ} + e^{– jθ}e^{–(α + 2jω)t}] dt

2 0

F(ω) = 1 (α + 2jω) e^jθ+α e^{– jθ}

2 α (α + 2jω)

= 1 2α cos θ+ 2jωe^jθ

2 α (α + 2jω)

= α cos θ + j ω cos θ – j ω sin θ

α (α + 2jω)

F(ω) ² = α² cos ²θ+ ω² – 2αω sin θ cos θ

α² (α² + 4ω²)

ω²+ α² cos²θ – αω sin2 θ

α² (α² + 4ω²)

Q.28. Determine the impulse response h(t) and sketch the magnitude and phase

response of the system described by the transfer function

Ans:

H(s) = s² + ω₀²

s² + ω₀s + ω₀²

H(jω) = (jω)² + ω₀² = ω₀² - ω²

(jω)² + ω₀ (jω) + ω₀² ω₀² - ω² + j ω ω₀

Q Q

H(jω) = ω₀² - ω²

[(ω₀² - ω²)² + ω²ω ₀ ²]^{1/ 2}

Arg H(jω) = - tan^-1 ω ω₀

ω₀² - ω²

Ω	H(jω)	Arg H(jω)
0	1	0
∞	1	0
ω_0-	0	- π/2
ω₀₊	0	+ π/2

H(jω)

Magnitude

₀ ω₀ ω

arg H(jω)

Phase

+ π/2

0 ω₀ ω

- π/2

Q.29. Using the convolution sum, determine the output of the digital system shown in

Fig..

Ans:

Assume that the input sequence is and that the system is

initially at rest.

x(n) = { 3, -1, 3 }, system at rest initially (zero initial conditions)

n = 0

x(n) = 3δ(n) - δ(n-1) + 3δ(n-2)

X(z) = 3 - z^-1 + 3z^-2

Digital system: y(n) = x(n) + 1 y(n-1)

Y(z) = X(z) = 3 - z^-1 + 3z^-²= -10 -6z^-1 + 13

1 – 1 z^-1 1 – 1 z^-1 1 – 1 z^-1

2 2 2

by partial fraction expansion.

Hence y(n)= -10 d (n) – 6 d (n-1) +

Q.30. Find the z-transform of the digital signal obtained by sampling the analog signal at intervals of 0.1 sec.

Ans:

x(t) = e^-4^t sin 4t u(t), T = 0.1 s

x(n) = x(t nT) = x(0.1n) = ( e^-0.4 )ⁿ sin(0.4n)

z α = e^-0.4 = 0.6703, 1 = 1.4918

x(n) X(z) α

x(n) = sin Ωn u(n) z sin Ω Ω = 0.4 rad = 22.92˚

z² –2z cos Ω + 1

sin Ω = 0.3894; cos Ω = 0.9211

αⁿ x(n) X (z/α)

X(z) = 1.4918z (0.3894)

(1.4918)² z² – 2(1.4918)z(0.9211) + 1

X(z) = 0.5809z

2.2255 z² – 2.7482z + 1

Q.31. An LTI system is given by the difference equation .

i. Determine the unit impulse response.

ii. Determine the response of the system to the input (3, -1, 3).

Ans:

y(n) + 2y(n-1) + y(n-2) = x(n)

Y(z) + 2z^-1 Y(z) + z^-2 Y(z) = X(z)

(1 + 2z^-1 + z^-2)Y(z) = X(z)

(i). H(z) = Y(z) = 1 = 1 ( Binomial expansion)

X(z) 1 + 2z^-1 + z^-2(1 + z^-1)²

= 1- 2z^-1 + 3 z^-2 - 4 z^-3 + 5 z^-4 - 6 z^-5 + 7 z^-6 - …….

h(n) = δ(n) - 2δ(n-1) + 3δ(n-2) -…..

= {1,-2,3,-4,5,-6,7,….} is the impulse response.

n=0

(ii). x(n) = { 3,-1,3 }

n=0

= 3δ(n) - δ(n-1) + 3δ(n-2)

X(z) = 3 - z^-1 + 3z^-2

Y(z) = X(z).H(z) = 3 - z^-1 + 3z^-2 = 3(1 + 2z^-1 + z^-2) – 7z^-1

1 + 2z^-1 + z^-21 + 2z^-1 + z^-2

= 3 – 7 z^-1

(1 + z^-1)²

y(n) = 3δ(n) + 7nu(n) is the required response of the system.

Q.32. The signal x(t) shown below in Fig. is applied to the input of an

(i) ideal differentiator. (ii) ideal integrator.

Sketch the responses.

Ans:

x(t) = t u(t) – 3t u(t-1) + 2t u(t-1.5)

x(t)

(i) 0 < t < 1

1 t

y(t) = ∫ t dt = t²y(1)= 0.5

0 2

0 1 1.5 (ii) 1 < t < 1.5

dx(t ) t

dt y(t) = y(1) + ∫(3-2t)dt

1 1

= 0.5 + (3t – t²) = 0.5 + 3t – t² – 3 + 1

0 t = 3t – t² – 1.5

For t =1: y(1) = 3 – 1 – 1.5 = 0.5

(iii) t ≥ 1.5 : y(1.5) = 4.5 – 2.25 –1.5 = 0.75

-2

∫x(t)dt

0.5

0 1 1.5 t

Q.33. Sketch the even and odd parts of

(i) a unit impulse function (ii) a unit step function

(iii) a unit ramp function.

Ans:

Even part x_e(t) = x(t) + x(-t)

Odd part x_o(t) = x(t) - x(-t)

(i)unit impulse (ii)unit step (iii)unit ramp

function function function

Q.34. Sketch the function .

Ans:

f(t)

1 f(t) = 1 0 < t < T, 2 T < t 3 T

-1 T< t < 2T, 3T < t <4T, ……

….. -T 0 T ….. t

-1

Q.35. Under what conditions, will the system characterized bybe

linear, time-invariant, causal, stable and memory less?

Ans:

y(n) is : linear and time invariant for all k

causal if n₀ not less than 0.

stable if a > 0

memoryless if k = 0 only

Q.36. Let E denote the energy of the signal x (t). What is the energy of the signal

x (2t)?

Ans:

Given that

E = dt

To find E΄=

Let 2t =r then E΄=

Q.37. x(n), h(n) and y(n) are, respectively, the input signal, unit impulse response and

output signal of a linear, time-invariant, causal system and it is given that

where * denotes convolution. Find the possible sets

of values of and .

Ans:

. y(n-2) = x(n-n₁) * h(n-n₂)

z^-² Y(z) = z^-n1 X(z) . z^-n2 H(z)

z^-² H(z) X (z) = z^-(n₁⁺ⁿ₂⁾ X(z) H(z)

n₁+n₂ = 2

Also, n₁, n₂ ≥ 0, as the system is causal. So, the possible sets of values for n₁and n₂are:

{n_1, n₂} = {(0,2),(1,1),(2,0)}

Q.38. Let h(n) be the impulse response of the LTI causal system described by the difference

equation and let . Find .

Ans:

y(n) = a y(n-1) + x(n) and h(n) * h₁(n) = δ(n)

Y(z) = az^-¹ Y(z) + X(z) and H(z) H₁(z) =1

H(z) = Y(z) = 1 and H₁(z) = 1

X(z) 1-az^-¹ H(z)

H₁(z) = 1-az^-¹ or h₁(n) = δ(n) – a δ(n-1)

Q.39. Determine the Fourier series expansion of the waveform f (t) shown below in terms of

sines and cosines. Sketch the magnitude and phase spectra.

Ans:

Define g(t) = f(t) +1. Then the plot of g(t) is as shown below and,

w = 2p/2p = 1

because T =2p

g(t) = 0 - π< t < - π/2

2 - π/2< t < π/2

0 π/2< t < π

Let g(t) = a₀+ S (a_n cos nt + b_nsin nt)

n=1

Then a₀= average value of f(t) =1

a_n = = ^π/2 = [2 /(n π)] . 2sin n π/2

_-π/2

= [4 /(n π)] . sin n π/2

= 0 if n= 2,4,6 ……

4 /(n π) if n= 1,5,9 ……

- 4 /(n π) if n= 3,7,11……

Also, b_n = = ^π/2 = [4 /(n π)] [ cos n π /2 - cos n π /2] = 0

_–π/2

Thus, we have f(t) = -1 + g(t)

= (4/)[ cost – (cos3t)/3 + (cos5t)/5 …..]

spectra :

(Magnitude) X π/4

1/3

1/5

1/7

_{-7 -6 -5 -4 -3 -2 -1 0 1 2
3 4 5 6 7} a

Phase

-7 -3 3 7

-π

Q.40. Show that if the Fourier Transform (FT) of x (t) is , then

FT .

Ans:

x(t) X(ω)

+∞

i.e., x(t) = 1 ∫ X(ω) e^jωt dω

2π -∞

+∞

d [x(t)] = 1 ∫ X(ω) jω e^jωt dω

dt 2π -∞

d [x(t)] FT jω X(jω)

Q.41. Show, by any method, that FT .

Ans:

+∞

x(t) = 1 ∫ X(jω) e^jωt dω

2π -∞

+∞

x(t) = 1 ∫ π δ(ω) e^jωt dω = 1

2π -∞ 2

1 FT π δ(ω)

Q.42 Find the unit impulse response, h(t), of the system characterized by the relationship :

Ans:

h(t) = ∫ d (τ) dτ = 1, t ≥ 0

-∞ 0, otherwise

= u(t)

Q.43. Determine the frequency response of u(t).

Ans:

As shown in the figure, u(t) = (1/2) + x(t)

where x(t) = 0.5, t >0

–0.5, t <0

\dx/dt = d (t) and FT[d (t)] = jwX(w)

\X(w) = 1/(jw). Also FT[1/2] = pd (w)

Therefore FT [u(t)] = pd(w) + 1(jw).

Q.44. Let denote the Fourier Transform of the signal x (n) shown below .

Without explicitly finding out , find the following :-

(i) X (1) (ii)

(iii) X(-1) (iv) the sequence y(n) whose Fourier

Transform is the real part of .

(v) .

Ans:

∞

X(e^jω) = ∑ x(n) e^-jωn

n = -∞

+∞

(i) X(1) = X(e^j0) = ∑ x(n) = –1 + 1 + 2 + 1 + 1 + 2 + 1 –1 = 6

-∞

+π π

(ii) x(n) = 1 ∫ X(e^jω) e^jωn dω ; ∫ X(e^jω) dω = 2π x(0) = 4π

2π -π -π

+∞

(iii) X(-1) = X(e^jπ) = ∑ x(n) (-1)ⁿ = 1+ 0-1+2-1+0-1+2-1+0+1 =2

n = -∞

(iv) Real part X(e^jω) x_e(n) = x(n) + x(-n)

y(n) = x_e(n) = 0, n < -7, n > 7

y(7) = 1 x(7) = -1 = y(-7)

2 2

y(6) = 1 x(6) = 0 = y(-6)

y(5) = 1 x(5) = 1 = y(-5)

2 2

y(4) = 1 x(4) = 2 = y(-4)

y(3) = 1[x(3) + x(-3)] = 0 = y(-3)

y(2) = 1[x(2) + x(-2)] = 0 = y(-2)

y(1) = 1[y(1) + y(-1)] = 1 = y(-1)

y(0) = 1[ y(0) + y(0)] = 2

(v) Parseval’s theorem:

π 2 ∞ 2

∫ X(e^jω) dω = 2π ∑ x(n) = 2π(1 + 1 + 4 +1 + 1 + 4 + 1 + 1) = 28π

-π n = -∞

Q.45 If the z-transform of x (n) is X(z) with ROC denoted by , find the

z-transform of and its ROC.

Ans:

x(n) X(z), RoC R_x

n 0 ∞

y(n) = ∑ x(k) = ∑ x(n-k) = ∑ x(n-k)

k = -∞ k = ∞ k = 0

∞

Y(z) = X(z) ∑ z^-k = X(z) , RoC at least R_x∩ (׀z׀ > 1)

k = 0 1 - z^-¹

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 and two zeros, one of which is at . It is also known that . Determine X(z) as a ratio of polynomials in .

(ii) If in part (b) (i), determine the magnitude of X(z) on the unit circle.

Ans: z

. (i) x(n) : real, right-sided sequence X(z)

X(z) = K (z- re^-jθ)(z- re^jθ) ; ∑x(n) = X(1) = 1

(z- ae^jΦ)( z- ae^-jΦ)

= K z² –zr (e^jθ+e^-jθ) + r²

z² –za (e^jΦ+ e^jΦ) + a²

= K 1 – 2r cosθ z^-1 + r²z^-2 = K. N(z^-1)

1 – 2a cosΦz^-1 + a²z^-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 = 0.25

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

= (0.25) 1 - 2√2 z^-1 + 4z^-2 X(e^jω) = (0.25) 1 - 2√2 e^-jω + 4 e^-2^jω

1 – (1/√2) z^-1 + ¼ z^-2 1 – (1/√2) e^-jω + ¼ e^-2^jω

= - 2√2 + e^jω + 4 e^-jω

-2√2+ 4e^jω + e^-jω

X(e^jω) = 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).

Ans:

Fig(a) Fig(b)

h(t) = u(t) – u(t-1) => H(s) =

First period of x(t) , x_T(t) = 2t [u(t) – u(t- ½) ]

= 2[ t u(t) – (t-1/2) u(t-1/2) –1/2 u(t-1/2)]

\ X_T(s) = 2 [(1/s² )– (e^-s/2 / s² )– (1/2) e^-s/2 / s ]

X (s) = X_T(s) / 1 – e^-s/2

Y(s) =. 2

= 2

Therefore y(t) = t² u(t) – (t-1)² u(t-1) –

This gives y (t) = t² 0< t < 1/2

t² –t +1/2 1/2 < t < 1

1/2 t >1

(not to scale)

Q.48 Obtain the time function f(t) whose Laplace Transform is .

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

C = s²+3s+1 = 1-3+1 = -1

(s+2)² s= -1 1

E = s²+3s+1 = 4-6+1 = 1

(s+1)³ s= -2 -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

s³ : 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

A = -1-3 = - 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)

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.

Ans:

Variance of a random variable X is defined as the second central moment

E[(X-μ_X)]ⁿ, n=2, where central moment is the moment of the difference between a random variable X and its mean μ_X i.e.,

+ ∞

σ_X² = var [X] = ∫ (x- μ_X)² f_x(x) dx

-∞

Co-variance of random variables X and Y is defined as the joint moment:

σ_XY = 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 ρ_XY of X and Y is defined as the co-variance of X and Y normalized w.r.t σ_Xσ_Y :

ρ_XY = cov [XY] = σ_XY

σ_Xσ_Y σ_Xσ_Y