# SSC Examination ECE,Computer

**SSC Examination ECE,Computer**

**1.** Find the z-transform of a^{n} cos nq.

(a) (z-a cos q) / z^{2} – 2za cos q

(b) z(z – a) / z^{2} – 2za + a^{2}

(c) z(z – a cos q) / z^{2} – 2za cos q + a^{2 }**(Ans)**

(d) None of these

** _
Explanations : **Since, z(cos nq) = z(z – cos q) / z

^{2}– 2z cos q + 1 = u(z)

∴ By change of scale property, we have

__

z(a^{n} cos nq) = u (z/a)

= z/a (z/a – cos q) / z^{2}/a^{2} – 2z/a cos q + 1

= z(z – a cos q) / z^{2} – 2za cos q + a^{2}

**2.** Matrix will become singular, if value of x is

(a) 4** (Ans)**

(b) 6

(c) 8

(d) 12

**Explanations : **The given matrix will become singular, if

**⇒ ** 8(-12) – x (-24) = 0

**⇒ ** -96 + 24x = 0

**⇒ ** x = 4

**3.** The number of vertices of odd degree is always

(a) even number ^{ }**(Ans)**

(b) odd number

(c) prime number

(d) zero

**Explanations : **We know that the graph of even number of vertices always have even degree.

i.e., ∑ d(v) = even (integer) …..(i)

v ∈v_{even}

Also know that,

Sum of all degrees is twice the number of edges,

**⇒ ** ∑ d(v) = 2e (∴ e number o edges)

**⇒ ** ∑ d(v) + ∑ d(v) = 2e

v ∈v_{odd }v ∈v_{even}

**⇒ ** ∑ d(v) = 2e (even integer)

v ∈v_{odd }

_{ –} ∑ d(v) (even integer)

v ∈v_{even}

**⇒ ** ∑ d(v) = even integer

v ∈v_{odd }

(∵as x-y is even when x,y ∈ even)

Hence, *G* has even number of odd degree vertices.

**4.** There are 9 objects and 9 boxes. Out of 9,5 objects cannot fit to be put in 3 small boxes. How many arrangements can be made, such that each object can be put in one box only?

(a) 17280 ^{ }**(Ans)**

(b) 17820

(c) 17208

(d) 17028

**Explanations : **Since, 5 objects cannot fit to be put 3 small boxes, these 5 objects can be put in the remaining 6 boxes in^{ 6}P_{5} ways.

The remaining 4 objects can be put in ^{4}P_{4} ways.

∴ Total number of arrangements = ^{6}P_{5} x ^{4}P_{4}

= (6 x 5 x 4 x 3 x 2) x (4 x 3 x 2 x 1)

= 720 x 24

= 17280

**5.** What values of *A, B, C* and *D* satisfy the following simultaneous Boolean equations?

__

*A* + *AB* = 0, *AB* = *AC*

_ ___

*AB* + *AC + CD = CD*

(a) *A* = 1, *B* = 0, *C* = 0, *D* = 1

(b) *A* = 1, *B* = 1, *C* = 0, *D* = 0

(c) *A* = 1, *B* = 0, *C* = 1, *D* = 1

(d) *A* = 1, *B* = 0, *C* = 0, *D* = 0 ^{ }**(Ans)**

** **__

**Explanations : **Given, *A* + *AB* = 0 and *AB* = *AC*

Let *A* = 1 and *B* = 0

__ __

Then, *A* + *BA* = 1 + 1.0 = 0 + 0 = 0

And also, *AB *= 1.0 = 0

*AC *= 1.0 = 0, then *C* = 0

We have, also given

* _ _
AB + AC + CD = * 1.0 + 1.0 + 0.

*D*

= 0 + 1.1 + 0

= 0 + 1 + 0 = 1

___

So, CD should be 1.

For this, let *D* = 0

__ __

⇒ *CD* = 0.0

_

= 0 = 1

Hence, *A *= 1, *B* = 0, *C* = 0, *D* = 0

**6.** The functionally complete set is

(a) {⌉,^,v}

(b) {↓^}

(c) {↑}^{ }**(Ans)**

(d) {→,^}

**Explanations : **We know that, {^,⌉} and {v,⌉} are functionally complete.

For NAND (↑)

*P* ↑ *Q* = 7 (*P* ^ *Q*)

7*P* = *P* ↑ *P*

*P *v *Q* = (*P* ↑ *P*)↑(*Q *↑ *Q*)

*P *v *Q* = (*P* ↑ *Q*)↑(*P* ↑ *Q*)

So {↑} are functionally complete.

**7****.** The function* C*(t) = e^{at} u(t)* e^{bt} u(t) is

(a) 1/a – b (e^{at }– e^{bt}) u(t)**(Ans)**

(b) 1/a + b (a^{at }– e^{bt}) u(t)

(c) 1/a – b (e^{-at }– e^{-bt}) u(t)

(d) 1/a + b (e^{-at }+ e^{-bt}) u(t)

**Explanations : **Given, *C*(t) = e^{at} u(t)* e^{bt} u(t)

Taking Laplace transform,

*C*(s) = 1/(s-a) . 1/(s-b)

[From time convolution property]

*C*(s) = 1/a – b [1/(s – a) – 1/(s – b)]

Taking inverse Laplace transform,

*C*(t) = 1/a – b [e^{at} – e^{bt}] u(t)

**8.** For a distortionless transmission,

1. amplitude response is constant.

2. phase response is linear with frequency.

Which of the following is/are correct?

(a) Only 1

(b) Only 2

(c) Both 1 and 2 **(Ans)**

(d) Neither 1 nor 2

**Explanations : **

Frequency response of distortionless transmission.

**9****.** Hilbert transform is known as

(a) – p phase shifter

(b) – p/2 phase shifter **(Ans)**

(c)** **p** **phase shifter

(d) 2p phase shifter

**Explanations : **

Transfer function of Hilbert transform, *i.e., *every frequency component is shifted by – p/2.

**10****.** The VSWR can have any value between

(a) 0 and 1

(b) -1 and + 1

(c) 0 and ¥

(d) 1 and ¥ **(Ans)**

**Explanations : **The VSWR

* s *= 1 + t

1 – t

where t is reflection coefficient value between 0 and 1.

Hence, s varies from 1 to ¥.

**11.** If the closed-loop transfer function of a control system is given as T(*s*) = (*s *– 5) , then it is

(*s *+ 2)(*s *+ 3)

(a) an unstable system

(b) an uncontrollable system

(c) a minimum phase system

(d) a non-minimum phase system **(Ans)**

**Explanations : **If closed-loop transfer function has any zero in the right half of s-plane, then the system is non-minimum phase system.

**12. **The open-loop DC gain of a unity negative feedback system with closed-loop transfer function s + 4 is

s^{2 }+ 7s + 13

(a) 4/13

(b) 4/9** (Ans)**

(c) 4

(d) 13

**Explanations : **We know that for unity feedback system, closed transfer fuction

*T*(*s*) = *G*(*s*)/ 1 +* G*(*s*) = *s *+ 4 / *s ^{2 }*+ 7s + 13

1 +* G*(*s*)/ *G*(*s*) = *s ^{2 }*+ 7s + 13 /

*s*+ 4

1/* G*(*s*) = *s ^{2 }*+ 7s + 13 /

*s*+ 4 – 1 =

*s*+ 6s + 9 /

^{2 }*s*+ 4

*G*(*s*) =* s *+ 4 / *s ^{2 }*+ 6s + 9

For DC gain; *s *=* *0, thus

*G*(0) =* *4/9

**13.** The equivalent inductive reactance of the circuit is

(a)* j* 15 Ω

(b) *j* 20 Ω

(c) *j* 10 Ω

(d) *j* 12 Ω** (Ans)**

Hence, the equivalent inductive reactance

Z = *j *(3 + 5 + 6) –* j*2 – *j*3 +* j*4

= *j *12 Ω

**14****.** In series *R-L-C* circuit at resonance frequency, which of the following statements is correct?

(a) Voltage across capacity is greater than voltage across inductor

(b) Voltage across inductor is greater than voltage across capacitor

(c) Voltage across inductor is equal to voltage across capacitor ** (Ans)**

(d) Current is minimum

**15.** The Fourier transform of a function is equal to its two sided Laplace transform evaluated

(a) on the real axis of the *s*-plane

(b) on a line parallel to the real axis of the *s*-plane

(c) on the imaginary axis of *s*-plane ** (Ans)**

(d) on a line parallel to the imaginary axis of the *s*-plane

**Explanations : **The Fourier transform is the continuous sum of exponential of the form e^{j}^{wt},where frequencies re restricted to the imaginary axis.

The Laplace transform is more generalized, where the frequency variable is s = s + jw.

If we evaluate the Laplace transform in the imaginary axis of the *s*-plane (s = 0).

**16****.** A 220 V three-phase, two pole, 50Hz induction motor is running at a slip of 5%. Then speed of rotor in revolution per min, is

(a) 3000 rev/min

(b) 150 rev/min

(c) 285 rev/min

(d) 2850 rev/min** (Ans)**

**Explanations : ** The speed of magnetic field is

*n*_{sync }= 120 f_{e} / *P *= 120(50 Hz) / 2 = 3000 rev/min) = 2850 rev/min

**17.** The polar plot is shown below. The type of system is

(a) type 0

(b) type 1

(c) type 2** (Ans)**

(d) type 3

**18.** The polar plot of* G*(j w) for different damping ratio for the system is shown below.

Which of the following is correct ?

(a) x_{1}> x_{2}

(b) x_{1 }> x_{2}** (Ans)**

(c) x_{1 }= x_{2}

(d) Can’t be determined

**19.** The root locus of the system *G*(s) *H*(s) = * K
s*(

*s*+ 2)(

*s*+ 3)

has the breakaway point located at

(a) (-0.5, 0)

(b) (-2.548, 0)

(c) (-4, 0)

(d) (-0.784, 0)** (Ans)**

**Explanations : ** We have,

1 + *G*(s) *H*(s) = 0

or 1 + * K
s*(

*s*+ 2)(

*s*+ 3)

or *K *= –*s *(*s ^{2} *+ 5

*s*+ 6

^{2}*s*)

*dK / ds = –*(3*s ^{2 }*+ 10

*s +*6) = 0

which gives, *s = *– 10 ± √100 – 72

6

= – 0.784, -2.548

The location of poles on *s*-plane is

Since, breakpoint must lie on root locus. So, *s* = – 0.784 is possible.

**20.** Convolution of x(t + 5) with impulse function d(t -7) is equal to

(a) x(t – 12)

(b) x(t + 12)

(c) x(t – 2)** (Ans)**

(d) x(t + 2)

**Explanations : ** If x(t)* h(t) = y(t)

Then, x(t – T_{1})* h(t – T_{2}) = y(t – T_{1 }– T_{2})

Thus, x(t + 5)* d(t – 7) = x(t + 5 – 7) = x(t – 2)