trigonometric fourier series of an even function does not have ? answers and solution given in last.

Unit Exercise – 1

(1 Mark Questions)

1. The trigonometric Fourier series of an even function does not have the

(a) DC term

(b) cosine terms

(c) sine terms

(d) odd harmonic terms

1. If the unit step response of a network is (1 – e-at), then its unit impulse response is

(a) ae-at

(b) a-1 e-at

(c) (1 – a-1) e-at

(d) (1 – a) e-at

1. A system is defined by its impulse response h(n) = 2n u(n – 2). the system is

(a) stable and causal

(b) causal but not stable

(c) stable but not causal

(d) unstable and non-causal

1. The trigonometric fourier series for the waveform f(t) shown below contains

(a) only cosine terms and zero value for the DC component

(b) only cosine terms and a positive value for the DC component

(c)  only cosine terms and a negative value for the DC component

(d) only sine terms and a negative value for the DC component

1. Consider the z-transform x(z) = 5z2 + 4z-1 + 3, 0 <|z|< 0. the inverse z-transform x|n| is

(a) 5 [n + 2] + 3[n] + 4[n – 1]

(b) 5 [n – 2] + 3 [n] + 4 [n + 1]

(c) 5 [n + 2] + 3 u[n] + 4u [n – 1]

(d) 5u [n – 2] + 3u [n] + 4u [n + 1]

1. For an n-point FFT algorithm with n = 2m, which one of the following statements is true?

(a) it is not possible to construct a signal flow graph with both input and output in normal order

(b) the number of butterfiles in the mth stage is n/m

(c) in – place computation requires storage of only 2n node data

(d) computation of a butterfly requires only one complex multiplication

1. The fourier series of a real periodic function has only

P : Cosine terms if it is even

Q : Sine terms if it is even

R : Cosine terms if it is odd

S : Sine terms if it is odd

which of the above statements are correct?

(a) p and s

(b) p and r

(c) q and s

(d) q and r

1. A function is given by f(t) = sin2 t + cos 2t. which of the following is true?

(a) f (t) has frequency components at 0 and 1/2 Hz

(b) f (t) has frequency components at 0 and 1/ Hz

(c) f (t) has frequency components at 1/2 and 1/ Hz

(d) f (t) has frequency components at 0, 1/2 and 1/ Hz

1. The ROC of z-transform of the discrete time sequence x(n) = (1/3)n u(n) – (1/2)n u (-n – 1) is

(a) 1/3 < |z| < 1/2

(b) |z| > 1/2

(c) |z| < 1/3

(d) 2 < |z|< 3

1. The input and output of a continuous time system are respectively denoted by x(t) and y(t). which of the following descriptions corresponds to a causal system?

(a) y(t) = x(t – 2) + x(t + 4)

(b) y(t) = (t – 4) x(t + 1)

(c) y(t) = (t + 4) x(t – 1)

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

1. The impulse response h(t) of a linear time-invariant continuous time system is described by h(t) = exp (at) u(t) + exp (bt) u(-t), where u(t) denotes the unit step function and a and b are real constants. this system is stable if

(a) a is positive and b is positive

(b) a is negative and b is negative

(c) a is positive and b is negtive

(d) a is negative and b is positive

1. If the laplace transform of a signal y(t) is y(s) = 1/s(s – 1) then, its final value is

(a) -1

(b) 0

(c) 1

(d)  unbounded

1. A low-pass filter having a frequency response H(j) = A(0) ej does not produce any phase distortion if

(a) A (0) – c2 , (0) = k2

(b) A = C2 , (0) = K

(c) A = C , (0) = K2

(d) A = C, (0) = K-1

1. Let x(t) – x(j) be fourier transform pair. the fourier transform of the signal x(5t – 3) in terms of x(j) is given as

(a) 1/5 e-j3/5 x (j/5)

(b) 1/5 ej3/5 x (j/5)

(c) 1/5 e-j3 x (j/5)

(d) 1/5 ej3 x (j/5)

1. The dirac delta function (t) is defined as

(a) (t) = {1,  t = 0  0, otherwise

(b) (t) = {t = 0  0, otherwise

(c) (t) = {1,   t = 0  0, otherwise   and (t) dt = 1

(d) (t) = {t = 0   0, otherwise and (t) dt = 1

1. If the region of convergence of x1[n] + x2[n] is 1/3 < |z| < 2/3, then the region of convergence of x1[n] – x2 [n] includes

(a) 1/3 < |z| < 3

(b) 2/3 < |z| < 3

(c) 3/2 < |z| < 3

(d) 1/3 < |z| < 2/3

1. Choose the function f(t) – < t < + 0, for which a fourier series cannot be defined.

(a) 3 sin (25t)

(b) 4 cos (20t + 3) + 2 sin (10t)

(c) exp (-|t|) sin (25t)

(d) 1

1. The function x(t) is shown in fighre.

Even and odd parts of a unit step function u(t) are, respectively.

(a) 1/2 , 1/y x(t)

(b) -1/2, 1/2 x(t)

(c) 1/2, – 1/2 x(t)

(d) -1/2, -1/2 x(t)

1. The region of convergence of z-transform of the sequence (5/6)n u(n) – (6/5)n u(-n – 1) must be

(a) |z| < 5/6

(b) |z| > 6/5

(c) 5/6 < |z| < 6/5

(d) 6/5 < |z| < 0

1. Which of the following can be impulse response of a causal system?
2. Let x(n) = (1/2)n u(n), = x2 (n), and y(ej) be the fourier transform of y(n). the y (ejo) is

(a) 1/4

(b) 2

(c) 4

(d) 4/3

1. The power in the signal

s(t) = 8 cos (20t – 2) + 4 sin (15 t) is

(a) 40

(b) 41

(c) 42

(d) 82

1. The impulse response h[n] of a linear time-invariant system is given by

where, h[n] = u[n + 3] +u[n – 2] -2 u[u – 7]

where, u[n] is the unit step sequence. the above system is

(a) stable but not causal

(b) stable and causal

(c) causal and unstable

(d) ustable and non-causal

1. The z-transform of a system is h(z) = z/z – 0.2. if the ROC is e|z| < 0.2, then the impulse response of the system is

(a) (0.2) n u[n]

(b) (0.2)n u[n – 1]

(c) -(0.2)n u[n]

(d) -(0.2)n u[-n – 1]

1. The fourier transform of a conjugate symmetric function is always

(a) imaginary

(b) conjugate anti-symmetric

(c) real

(d) conjugate symmetric

unit Exercise – 1

1. (c)

trigonometric fouier series is represented by

f(t) = ao + [an cos nt + bn sin nt],

0o = 2/t

here,    a0 = 1/t  f(t) dt

an = 2/t  f(t) cos (0o nt) dt

bn = 2/t  f(t) sin (not) dt

since, f(t) is even here

hence, bn = 2/t  f(t) sin nt dt = 0

hence, does not contain sine term.

1. (a)

unit step response y(t) = 1 – e-at

taking laplace transform,

y(s) = 1/s – 1/s + a = a/s(s + a)

here, x(t) = u(t). hence,  x(s) = 1/s

so,  transfer function is

h(s) = y(s)/x(x) = a/s(s + a) – 1/s

= a/s + a

now, for impulse response,

y(s) = a/s + a

hence, impulse response

y(t) = ae-at

1. (b)

h(n) = 2n u(n – 2)

here, for n < 0 h[n] = 0

hence, system is causal.

h[n] is a continuously increasing function. hence, unstable.

1. (c)

the function f(t) is an even function and has negative DC offset vlue.

f(-t) = f(t)

hence, all sine coefficient (bn = 0) becomes zero.

therefore, fourier series of the waveform f(t) contains only cosine term and negative value for DC component.

1. (a)

given,  x(z) = 5z2 + 4z-1 + 3

we know that

[n] – 1

from time shifting property, if

x[n] = x(z)

x[n – k] = z-k x(z)

hence, 5[n + 2] = 5z2

4[n – 1] = 4z-1

3 [n] = 3

hence, inverse transform of x(z) is

x[n] = 5[n + 2] + 4[n – 1] + 3[n]

1. (b)

on FFT to create one sample we require N complex multiplication and it can be represented conveniently using signal flow graph known butterflies and number of butterflies in the mth stage is n/m.

1. (a)
2. (b)

given, f(t) = sin2 t + cos 2 t

= 1 – cos 2t/2 + cos 2 t

= 1/2 + 1/2 cos 2 t

hence, f(t) has DC component 1/2, i.e., frequency component of zero.

the cosine term has 0 = 2 rad/s

i.e., 2f = 2

f = 1/Hz

hence, f(t) has component at frequency 1/ Hz.

1. (a)

x(n) = (1/3)n  u(n) – (1/2)n  u(-n – 1)

x1 [n] = (1/3)n  u(n)

x1 (z) = 1/1- (1/3) z-1

with ROC |z| > |1/3|

x2[n] = – (1/2)n  u(-n – 1)

x2(z) = 1/1 – (1/z) z-1

with ROC |z|< |1/2|

hence, ROC of x[n]

1/3 < |z| < 1/2

1. (c)

A linear time-invariant system is said to be causal if the output is dependent only on the current and/or past value of the input signal.

hence,  y(t) = (t + 4) x(t – 1) is the causal system. other system depends on future values of input signal.

1. (b)

An LTI continuous time system is stable if and only if its impulse response is absolutely integrable.

|h|d < 0

h(t) = eat u(t) + ebt u(t)

[ea u + eb u d] < 0

ea d + eb d] < 0

ea d < 0

eb d < 0

the above conditions satisfy if a is negative and b is positive.

1. (d)

final value theorem exists if sF (s) is analytic on the imaginary axis and in the right-half of the s-plane, i.e., pole lies in the left-half plane.

y(s) = 1/s(1 – s)

pole lies in the right-half of s-plane. hence, the system is unstable and output is unbounded.

1. (c)

given,

h(0) = A(0) ej(0)

for low-pass filter doesn’t produce any phase distortion,

<H(0) is linear with 0.

<H (0)

(0) = K

1. (a)

x(t) – x(j)

from time scaling property,

x(at) – 1/|a| x (j/|a|)

x(5t) – 1/5 x (j/5)

from time shifting property,

x(t – td) = e-jotd  x(j)

x(5t – 3) = x (5(t – 3/5))

x(5(t – 3/5) – 1/5 e-j3/5 x (j/5)

1. (d)

the dirac delta function has infinity (undefined) value at t = 0 and 0 for t = 0. the area under this function is unity.

(t) = {0,  t = 0  0, otherwise

(t) dt = 1

1. (d)

the region of convergence defines the value of z for which z-transform converges.

ax [n] + by[n] = ax(z) + by(z)

ROC = ROC x(n) ROC (y)

hence, if x1 [n] + x2 [n] has ROC 1/3 < |z| < 2/3

then,  x1 [n] – x2[n] also has ROC 1/3 < |z| < 2/3

1. (c)

For existence of fourier series, the function is absolutely integrable over one period and have finite number of maxima and minima in one period and finite number of finite discontinuities over one period. hence,we can say that periodic waveform is valid and sufficient condition for existence of fourier series, i.e., for periodic function and constant, fourier series can be defined.

f(t) = e-|t| sin (25 t)

is not a periodic function.

1. (a)

even part of signal x(t)

xe(t) = 1/2 [x(t) + x(-t)]

ue(t) = 1/2 [u(t) + u(-t)] = 1/2

odd part of signal x(t)

x0 (t) = 1/2 [x(t) – x(-t)]

u0 (t) = 1/2 [u(t) – u(-t)]

= 1/2 x(t)

1. (c)

x[n] = (5/6)n  u[n] – (6/5)n  u[- n – 1]

x1[n] = (5/6)n u[n]  1/1 – 5/6 z-1

with ROC |z|< |5/6|.

x2[n] – (6/5)n u[n – 1] – 1/1 – 6/5 z-1

with ROC |z|<|6/5|.

hence, ROC of x[n] is ROC (x1[n]) ROC (x2[n])

5/6 <|z|< 6/5

1. (b)

A system is causal if and only if its impulse response is zero for t less than 0.

h(t) = 0 for t < 0

1. (d)

given,  x(n) = (1/2)n u(n)

y(n) = x2(n)

= (1/2)2n u(n)

the fourier transform of y(n)

y(ejo) = y(n) e-jon

putting = 0

y(ejo) = y(n)

= (1/2)2n  u(n)

= (1/4)n

= 1/1 – 1/4 = 4/3

1. (a)

s(t) = 8 cos (20 t – 2) + 4 sin (15 t)

s(t) = 8 sin 20 t + 4 sin 15 t

power = 82/2 + 42/2 = 40

1. (a)

the system to be stable

|h[k]|< 0

|u[k + 3] + u[k – 2] -2u [k – 7] |< 0

u[k + 3] + u[k – 2] – 2 u[k – 7] < 0

h[k] = 15 < 0

hence, system is stable.

for causal system, the impulse response

h[n] = 0 for n < 0

h[n] = u [n + 3] + u[n – 2] – 2u [n – 7]

for n < 0, h[n] = 0, system is non-causal.

1. (d)

h(z) = z/z – 0.2

= 1/1 – 0.2 z-1 with ROC |z| < 0.2

As |z| < 0.2: ROC is the area inside of a circle, where 0.2 is the innermost pole. hence, x[n] is the left sided signal.

by inspection,

h[n] = – (0.2)n u[-n – 1]

1. (c)

signal x(t) is real, then its fourier transform is conjugate symmetric function and vice-versa.