: 3 : TEST-10 (Solutions) · 2017. 4. 16. · ACE Engineering Academy...

: 2 : ESE OFFLINE TEST-2017

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01. (a)

Sol: Given data,

20 kVA, 50Hz, 2000/200 V distribution transformer

Power output = V2I2 cos2

= 200 × 100 × 0.8

= 16000 W

Total loss PL = Pi + k2Pc

= 120 + 1 × 300 = 420 W

=L0

L

PPP

1

=42016000

4201

= 97.44%

For maximum efficiency , K =300120

PP

c

i = 0.632

i.e., at 0.632 pu load

max (cos 2 = 0.8) =i0

i

P2PP2

1

=1202632.016000

12021

= 97.68%.

01. (b)

Sol:

1N

0n

*1N

0n

2 )n(x)n(x)n(x*

1N

0k

nkN

2j1N

0n

e)k(XN

1)n(x

1N

0n

nkN

2j1N

0k

* e)n(x)k(XN

1

1N

0k

* )k(X)k(XN

1

1N

0k

2)k(XN

1

: 3 : TEST-10 (Solutions)


01. (c)

Sol: Synchronous impedance,currentSC

voltageOCZs

=100

500= 5

Xs =2a

2s RZ = 22 8.05 = 4.936

E0 =2

s2

a )IXsinV()IRcosV(

2 2(2000 0.8 100 0.8) (2000 0.6 100 4.936)

= 1822 V

% Regulation = 1002000

20001822

= – 8.9%

01. (d)

Sol: In above sequential circuit in initial state v, a ‘0’ inputs generates a ‘0’ outputs and the circuit stays

in state v, an input of 1 produces an output 1 and the circuit will move to next state W. In state W

for the input =0, the output also 0 and the next state is Y and this process is continued.

State table:

Present stateNext state

d=0 d =1

Output

d=0 d=1

V V W 0 1

Y V 0 0

Y V 0 0

Y Z V 0 0

Z V X 0 1

In the above table, for both W and X states their next states and outputs also same so, W and X are

equivalent states so, we can reduce one state by replace ‘X’ with ‘W’

W

X



After reducing X state the minimized state table is

Present stateNext state

d=0 d =1

Output

d=0 d=1

V W 0 1

W Y V 0 0

Y Z V 0 0

V W 0 1

Now again ‘V’ and ‘Z’ are equivalent states So, we can reduce one more state, replace ‘Z’ by ‘V’.

After reducing Z-state , the minimized state table is

Present state Next state

d = 0 d =1

Output

d = 0 d =1

V V W 0 1

W Y V 0 0

Y V V 0 0

Initial given state diagram has 5 states means we require 3 flip-flops, now it reduced to 3 states

means just 2flip-flops required. Here one flip-flop eliminated. So, some complexity will be

reduced.

Z

V W Y

1/0

1/1 0/0

1/0

0/0

0/0

V



01. (e)

Sol: (i) Air-gap power: The power transferred from stator to rotor across the air gap.

i.e., Pgs

RI3 22

r

Where, Pg = Air gap power

Ir = Rotor current per phase

R2 = Rotor resistance per phase

(ii) Internal mechanical power developed (or) Gross mechanical power out

1

s

1RI3 2

2r

(iii) Shaft power is nothing but net mechanical power output.

i.e., shaft power = Gross mechanical output – mechanical losses.

Rotor input: Rotor cu losses: Gross mechanical power output

1

s

1RI3:RI3:

s

RI3

2

2

r2

2

r

22

r

1

s

1:1:

s

1

1 : s : (1s)

Rotor Cu loss = s Rotor input

Gross mechanical power output

= (1s) Rotor Input

Rotor cu loss = Gross mechanical power output s1

s

01. (f)

Sol: Apply DTFT to given difference equation

)e(Xe2

1)e(X)e(Ye

16

3)e(Ye)e(Y jjjjj2jjj

j2j

j

j

jj

e163

e1

e21

1

)e(X

)e(Y)e(H



jj

j

j

e41

1e43

1

e21

1)e(H

jj

j

e41

1

21

e43

1

21

)e(H

)n(u4

15.0)n(u

4

35.0)n(h

nn

01. (g)

Sol: A shunt excited d.c generator may fail to self-excite for any of the following reason.

1. Residual magnetism is absent. This difficulty can be overcome by exciting the field winding

from a separated d.c source for a few seconds with the armature at rest.

2. The field connection to the armature is such that the induced emf due to the residual magnetism

tends to destroy the residual magnetism. This condition can be remedied by reversing the field

connection to the armature.

3. For critical resistance Ry line coincides with the linear portion of the magnetization curve.

4. Speed is less than critical speed. At critical speed, the open circuit characteristic (OCC) is

tangential to the Rf line.

01. (h)

Sol: Given circuit is:

Given circuit has voltage-shunt feedback

To find the input resistance apply some test voltage Vx

Vi–

+Ri gmVi V0

+

–



Apply KCL at node 1

Ix – gm Vi = I1

Ix – gm Vx =i

x

R

V

xmi

xx Vg

R

VI

im

i

x

xin Rg1

R

I

VR

01. (i)

Sol: 240V/120V, 12kVA has rated current of 50 A/100 A. It’s connected as an autotransformer as

shown in figure.

Auto-transformer rating

= 360 × 100 = 36 kVA

It is 3-times then 2-winding connection.

As 2-winding connection Output, P0

= 12 × 1 = 12 kW

Vi–

+Ri gmVi

Ix 1 I1

–gmVi

Vx

–

+

360V

240V

+

+

120V

150A

240V

100A

L

50A



=L0

0

PPP

=

0

L

PP

1

1

= 0.962

From which find full-load loss

1 = 0.962 + 0.962

0

L

PP

(or)0

L

P

P=

962.0038.0

; PL=962.0038.0

12 = 0.474 kW

In auto connection full-load loss remains the same. At 0.85 pf

P0 = 36 × 0.85 = 30.6 kW

=

6.30474.0

1

1

= 0.985 or 98.5%

01. (j)

Sol: y(n) = 2y(n–1) + (n)

y(0) = 2y(–1) + (0)

1)1(y24

2

3)1(y

y(–1) = 2y(–2) + (–1)

)2(y223

43

)2(y

y(1) = 2y(0) + (1) = 2(4) + 0 = 8

y(2) = 2y(1) + (2) = 16

y(3) = 2y(2) + (3) = 32



+

–

0.6

0.2 10

02. (a)

Sol: (i) Resistance of motor = 0.2 + 0.6 = 0.8

Eb1 = 200 – (25 0.8) = 180 V

Let the motor input current be I2 when armature divertor is used.

Series field voltage drop = 0.6 I2

P.D at brushes = 20 – 0.6 I2

Arm. divertor current A10

I6.0200 2

Armature current = I2 –10

I6.0200 2

2a 2

10.6 I 200I

10

As torque in both cases is the same,

1 I1 = 2 I2

22

10.6 I 20025 25 I

10

or

6,250 = 10.6 22I – 200I2

I2 = 35.6 A

P.D at brushes in this case

= 200 – (35.6 0.6)

= 178.6 V

10

2006.356.10I 2a

= 17.74 A

Eb2 = 178.6 – (17.74 0.2) = 175 V



Now2

1

1b

2b

1

2

E

E

N

N

or2

1

1b

2b

1

2

I

I

E

E

N

N

2

1

N 175 25

N 180 35.6

N2 = 314 r.p.m

(ii) Power equation for salient pole synchronous

Machine: (with zero armature Resistance)

1. IdXd = E Vcos

dd X

cosVEI

2. IqXq = V sin

qq X

sinVI

P = (Id voltage component along d-axis) +

(Iq volt along Q-axis)

cosV

X

sinVsinV

X

cosVE

qd

cossinX

Vcossin

X

Vsin

X

EV

q

2

d

2

d

V

Ia

IqVCos

IqXq

Q-axis

IdXd

m

d-axis Id

E

Vsin



2sin

X

1

X

1

2

Vsin

X

EVP

dq

2

d

emd

PsinX

EVpowermagneticElectro

2

relq d

V 1 1sin2 Reluctance power P

2 X X

02. (b) (i)

Sol: x(t) = Asin(t) 0 t 1

x(t) = Asin(t) 0 t 1

2T2

1T 0

1

0

t2jnTt

t

tjnn dte)tsin(A

1

1dte)t(x

T

1C

0

0

0

1

0

t)n21(j1

0

t)n21(j dtedtej2

A

1

0

t)n21(j1

0

t)n21(j

n )n21(j

e

)n21(j

e

j2

AC

)n41(

A2C

2n

A2dt)tsin(A

1

1dt)t(x

T

1C

1

0

T

0

0

0nn

tn2j2

e)n41(

A2A2)t(x

02. (b) (ii)

Sol: EFS coefficient T

0

tjnn dte)t(x

T

1C 0



T

0

00n dt)tnsin(j)tncos()t(xT

1C

T

0

0

T

0

0n dt)tnsin()t(xT

2jdt)tncos()t(x

T

2

2

1C

nnn jba21

C

T

0

tjnn dte)t(x

T

1C 0

T

0

0

T

0

0 dt)tnsin()t(xT2

jdt)tncos()t(xT2

21

nnn jba2

1C

0

T

t

0 adt)t(xT

1C

0

C0 = a0,2

jbaC nn

n

,

2

jbaC nn

n

02. (c) (i)

Sol: Buffer stage using op-amp:

It is also called voltage follower circuit and is shown below

By observing the above circuit we can say that the voltage at the non – inverting input is Ei, the

voltage at the inverting input approaches the voltage at the non – inverting i/p, and the output is at

the same voltage as the inverting input. Hence E0 = Ei and our analysis is complete.

Ei Eo

+ +

+

fig (a)



Fig (b) Equivalent circuit of a buffer amplifier

Eo = Ei

AV = Eo/Ei =1

Since no current flows through Op-Amp, hence the input impedance of the voltage follower is

infinite i.e. Zi = . The output impedance is just that of the ideal operational amplifier itself, i.e

zero Z0 = 0

02. (c) (ii)

Sol:

(A)

Ei

+

-

-++

-Eo

1Ei

+

–

i1

i2

R2

2k

RL=5k

1k

R1 V0

5k

Vs

–

+Vd

R1

1k

2k

R2

Vs

+–

V0

RLAVd

Rs



21

10sd RR

RVVV

V0= AVd

21

10s0 RR

RVVAV

21

10s

0

RR

RVV

A

V

A [ Ideal op-amp]

21

10s RR

RVV

ss1

20 V

K1

K21V

R

R1V

V0= 3Vs

(B) SRdt

dV0

SRdt

dVA i

CL [ AcL Closed loop voltage gain ]

|ACL Vm m| < SR

322fm < 5105

fm < 13.3kHz

fmax = 13.3kHZ

(C) V0= 3Vs

V0 = 32sint = 6sint

mA2.1K5

6I (max)L



03. (a)

Sol: Given p = 1dB, p = 0.2, s = 0.3, s = 15dB

Pre wrapped frequencies: 65.022.0

tan22

tanT2 p

p

02.12

3.0tan2

2tan

T

2 ss

65.0

02.1cosh

110

110cosh

cosh

110

110cosh

N1

1.0

5.11

p

s1

1.0

1.01

p

s

N 3.01

N = 4 (round off to next higher integer)

Calculation of poles: 508.0110 2

11.0 p

17.41 21

2)17.4()17.4(

65.02

a4141N1N1

p

a = 0.237

2

)17.4()17.4(65.0

2b

4141N1N1

p

b = 0.6918

N2)1k2(

2k

, k = 1, 2, 3, 4

1 = 112.5o, 2 = 157.5o, 3 = 202.5o, 4 = 247.5o

S1 = acos1 + jbsin1 = –0.0907 + j0.639

S2 = acos2 + jbsin2 = –0.2189 + j0.2647

S3 = acos3 + jbsin3 = –0.2189 – j0.2647

S4 = acos4 + jbsin4 = –0.0907 – j0.639

The denominator polynomial of H(s) is

H(s) = [(s+0.0907)2 + (0.639)2] [(s+0.2189)2 + (0.2647)2]



H(s) = (s2 + 0.1814s + 0.4165) (s2 + 0.4378s + 0.118)

As N is even

So, the numerator of H(s) is 04381.01

)118.0)(4165.0(2

Transfer function of)1180.0s4378.0s)(4165.0s1814.0s(

04381.0)s(H

22

The digital filter transfer function is

1

1

1

1

z1

z12

z1

z1

T

2s)s(H)z(H

)z6493.0z5548.11)(z8482.0z499.11(

)z1(001836.0)z(H

2121

41

03. (b) (i)

Sol: SAM(t) = Ac[1+kam(t)] cos 2fct

= Ac[1+cos2fmt] cos2fct

= Accos2fct + Ac cos2fmt cos2fct

= Accos2fct + tfcos2tf2cos22A1.0

cmc

= Accos2fct + 0.05Ac[cos2(fc+fm)t+cos2(fc–fm)t]

If lower side band is removed from above signal

tff2cosA05.0tf2cosA)t(S mcccc*AM

= Accos2fct+0.05Ac cos2fct cos2fmt–0.05Ac sin2fct sin2fmt

= [Ac+Ac0.05 cos2fmt] cos2fct – [0.05Ac sin2fmt] sin2fct

Above signal is given to envelop detector. The output of envelop detector is

2mc2

mcc tsinA05.0tcosA05.0A

tcosA1.0A05.0A m2c

2c

2c

tcos1.00025.01A mc

The maximum value of above signal is

05.11025.1Ac (∵given in question)



Ac = 1

cmc

m AAAA

= 1(0.1)

Am= 0.1

03. (b) (ii)

Sol: Standard AM signal is given by

S(t) = Ac [1+kam(t)] cos 2fct

S(t) = Ac cos 2 fct + Ackam(t)cos 2fct

When single tone modulation is considered

Then m(t) = Am cos2fmt

S(t) = Ac cos 2fct +AcKaAm cos 2fmt cos 2fct

Taking kaAm =

S (t) = Ac cos 2 fct tff2cos2

Amc

c

tff2cos2

Amc

c

Comparing the above equation with the given signal

S(t) = 20 cos 2000 t + 5 cos 2200t +5 cos 1800 t

(A) c(t) = 20 cos 2000 t

(B) AC = 20, fc= 1000

52

Ac

= 0.5

(C) 5.020AAAA

cmc

m

Am=1

fc+fm=1100

fm=1100–1000



fm= 1000

m(t) = Am cos2fmt

m(t) = cos200t

(D) power in carrier signal W200

220

2A 22

C

Power in USB W225

2/52

Power in LSB W225

2/52

Power in both side bands W25225

225

8

1

200

25

signalcarrierinpower

bandssideinpower

03. (c)

Sol: (i) Let Is = starting current

Ifl = full load current

2

2

2

2

2

2

s XR

VI

; (s = 1) …….. (i)

2

2

2

f2

2

2

f Xs/R

VI

………….. (ii)

Dividing equation (i) by (ii)

2

f

s

I

I

=

2

2

2

2

2

2

2

f2

XR

Xs/R

= 1ss

ss2

Tmax,2f

2f

2Tmax,

….(iii)

Substituting the values

25 = 1s)04.0(

)04.0(s2

Tmax,2

22Tmax,

(Given that Is = 5Ifl)

or Smax.T = 0.2 or 20%

Tmax = 22

2

s XV5.0

.3

…….. (iv)



Tfl = 2

2

2

f2

f2

2

sX)s/R(

)s/R(V.

3

……….. (v)

Dividing equation (iv) by (v)

f

max

TT

=

f22

22

2f

22

sXRXsR

5.0

=

fTmax,

2f

2Tmax,

ss

ss5.0

=04.02.0

)04.0()2.0(5.0

22

= 2.6

or Tmax = 2.6 pu

(ii) As per equation

f

s

f

s

II

TT

sfl = (5)2 × 0.04 = 1

Ts = 1 pu

03. (d)

Sol: Conditions to be satisfied for parallel operation of transformers:

Necessary conditions for possible parallel operation:

1. Voltage ratings mentioned on the name plate of transformers to be connected in parallel must

be same.

2. The transformers must be connected in parallel with correct polarity.

For this, dotted terminals should be connected to same bus bars and undotted terminals should

be connected to another bus bar.

3. The phase sequence of 3- transformers to be connected in parallel must be same.

To get correct phase sequence, the terminals of both secondaries which attain their peak value

simultaneously must be connected to same bus bar, so that potential difference between the

terminals connected to same bus bars equal to zero and there will not be any circulating current.

4. A part from phase sequence matching the phase displacement between the secondaries of both

transformers must be zero. This condition can be fulfil if the two transformers belongs to same

Phasor group.



Desirable conditions for satisfactory parallel operation:

1. The voltage ratios of transformers to be connected in parallel should be same to avoid

circulating currents.

If voltage ratios are not same, there is a possibility of circulating current given by

BA

B2A2C ZZ

EEI

E2A=Secondary voltage of transformer A

E2B =secondary voltage of transformer B

ZA and ZB are the equivalent leakage impedances in ohms referred to the secondary of

transformer A & B respectively.

2. The two transformers should share the common load proportional to their KVA ratings.

More KVA then more sharing of load.

Less KVA then less sharing of load.

3. TheRx

ratios of transformers to be connected in parallel should be equal to avoid

operation of transformers at different power factors.

B

B

A

A

R

X

R

X

A = tan-1

A

A

R

X, B = tan-1

B

B

R

X

A = B

cosA = cosB = cosL

04. (a)

Sol: Mechanical loss W9003000100

30

Stator core loss = Power input at no-load – Mechanical loss– stator I2R loss at no load

Neglecting stator I2R loss,

Stator core loss = 3000 – 900 = 2100 W

Power input during blocked rotor test

= stator I2R loss + Rotor I2R loss = 4000 W



Stator I2R loss = Rotor I2R loss

= W20002

4000

(i) At rated load, air-gap power,

Pg = Output power + mechanical loss + rotor I2R loss

Pg =60,000 +900+2000 = 62900W

But rotor I2R loss = 2000 W = s.Pg

Slip at rated load, 0318.062900

2000s

(ii) At rated voltage, power input to motor during blocked rotor test

kW444.4430

1004

2

Air-gap power,

Pg=Power input– stator core loss – stator I2R loss

= 44444 – 2100 – 2000

= 40344 Watt

Synchronous speed,

s/rad504

504

P

f4s

Starting torque .Nm84.25650

40344P

s

g

04. (b)

Sol: The samples of a signal are highly correlated with each other. This is because the signal does not

change fast. This means that its value from present sample to next sample does not differ by large

amount. The adjacent samples of the signal carry the same information with a little difference.

When these samples are encoded by a standard PCM system, the resulting encoded signal contains

some redundant information. To overcome this redundancy, DPCM is used.



Differential Pulse Code Modulation (DPCM) Transmitter

Working Principle:

The DPCM works on the principle of prediction, the value of the present sample is predicted from

the past samples. The prediction may not be exact but it is very close to the actual sample value.

The sampled signal is denoted by x(nTs) and the predicted signal is denoted by snTx . The

comparator finds out the difference between the actual sample value x(nTs) and predicted sample

value snTx . This is known as Prediction error and it is denoted by e(nTs). It can be defined as,

e(nTs)= x(nTs) – snTx ….. (1)

The error is the difference between unquantized input sample x(nTs) and prediction of it snTx .

The predicted value is produced by using a prediction filter. The quantizer output signal gap

eq(nTs) and previous prediction is added and given as input to the prediction filter. This signal is

called xq(nTs). The prediction is more close to the actual sampled signal. The quantized error signal

eq(nTs) is very small and can be encoded by using small number of bits. Thus number of bits per

sample are reduced in DPCM.

The quantizer output can be written as,

eq(nTs) = e(nTs)+ q(nTs) ….. (2)

x(nTs)

e(nTs) eq(nTs)

snTx

Predictionfilter

Quantizer Encoder DPCMsignal

xq(nTs)

+

+

+

SampledInput

Fig: A Differential PULSE CODE MODULATION Transmitter.



Here, q(nTs) is the quantization error. The prediction filter input xq(nTs) is obtained by sum snTx

and quantizer output i.e.,

xq(nTs) = snTx + eq(nTs) ….. (3)

substituting the value of eq(nTs) from equation 2 in equation 3, we get,

xq(nTs) = snTx + e(nTs) +q(nTs)

Equation 1 is written as,

e(nTs) = x(nTs) – snTx

e(nTs) + snTx = x(nTs)

xq(nTs) = x(nTs) + q(nTs)

This equation does not depend on the prediction filter characteristics.

Hence, the quantized version of the signal xq(nTs) is the sum of original sample value and

quantization error q(nTs). The quantization error can be positive or negative.

DPCM receiver:

The decoder first reconstructs the quantized error signal from incoming binary signal. The

prediction filter output and quantized error signals are summed up to give the quantized version of

the original signal.

Thus the signal at the receiver differs from actual signal by quantization error q(nTs), which is

introduced permanently in the reconstructed signal.

Predictionfilter

Decoder

DPCMsignal +

+

Output

Fig: DPCM receiver



Table: Comparison between PCM and Differential Pulse Code Modulation

S.No.

Parameter

of

comparison

Pulse Code

Modulation(PCM)

Differential Pulse Code

Modulation (DPCM)

1.Number of bits. It can use 4, 8 or 16 bits

per sample

Bits can be more than one but less

than PCM

2.Levels and step

size

The number of levels depend on

number of bits. Level size is

kept fixed

Here, fixed number of levels are used.

3.Transmission

Band width

Highest band width is required

since number of bits are high

Bandwidth required is lower than

PCM.

04. (c)

Sol: dt

x(t)d

/2–/2 0

2/

t

–2/

2

2

dt

x(t)d

/2–/2 0

2/

t

–4/

2/



2t

2)t(

42

t2

dt)t(xd

2

2

Take Fourier transform on both sides

2j

2j

2 e24

e2

)(X)j(

2

2j

2j

ee22

)(X

2

2

2

4sin

82cos1

4)(X

4

Sinc2

)(X 2

04. (d)

Sol: (i) X(pu) = 2)kV(MVA)(X

HV side 0.12 = 2

3

)6.6(1075)(X

X(Ω) = 3

2

1075)6.6(12.0

= 69.696 Ω

LV side 0.12 = 2

3

)4.0(1075)(X

= 0.256 Ω

(ii) Star-Star connection

(x) Line voltage

HV 6.6 3 = 11.43 kV

LV 400 3 = 692.8 V



Rating = 3 × 75 = 225 kVA

(y) X(pu) = 2)linekV(

)phase3(MVA)(X

=2

3

)36.6(

10225696.69 = 0.12

(z) HV side

X(Ω) = 69.696 Ω/phase

LV side

X(Ω) = 0.256 Ω/phase

(iii) Star-Delta

(x) Line voltages

Star side 6.6 3 = 11.43 kV

Delta side = 400 V

Rating = 3 × 75 = 225 kVA

(y) X(pu), calculated from delta side

X(pu) = 2

3

)14.0(10225)3/256.0(

= 0.12.

X(pu) = 0.12

(z) Star side X = 69.69 Ω/phase

Delta side X = 0.256 Ω/phase



05. (a)

Sol: The short circuit test has been performed on the delta-connected hv side, with the star connected lv

side is short circuited.

Per phase ratings on the hv side:

Given voltage and current ratings should always be taken as line quantities unless otherwise

specified. Similarly given kVA rating should be taken as 3-phase kVA rating unless otherwise

specified. Per phase ratings can then be obtained based on whether the transformer windings are

connected in star or delta. In this problem; the phase ratings are, rated voltage = 6600 V.

kVA/phase = 500.

rated current = 500 103/6600 = 75.6 A

From the short - circuit test data,

phase voltage (applied in the test) = 300 V

phase current (drawn by the transformer) = 75.75 A

power input per phase = 10,000 W.

Hence, 75.752 reqhv = 10,000;

where reqhv = resistance/ph (ref hv)

reqhv = 1.74 .

Percentage resistance

= 100voltagerated

)ph/cetanresis)(currentphaserated(

=6600

74.16.75 100 = 2%

zeqhv =75.75

300 = 3.96 /ph

xeqhv =22 74.196.3 = 3.56 /ph.

Percentage reactance =6600

56.36.75 100

= 4.08%

At full load and 0.8 pf lag, percentage regulation= 1006.056.38.074.16600

6.75

= 4.04%



Total losses = Full load copper loss + Iron loss = 30 kW + 25 kW = 55 kW.

Full load output = 1500 x 0.8 = 1200 kW

100poweroutputlosses

powerputOut%

100551200

1200

= 95.61%

05. (b)

Sol: (i) From (1) & (2) X (s) is of the form

X(s) =)bs)(as(

K

Since x(t) is real & one of the pole of X(s) is –1+j

X(s) =)j1s)(j1s(

K

b = a*

From (5) as X(0) = 8 K = 16

X(s) =2s2s

162

Let R denotes ROC of X(s). From pole location we know there are 2 possible choices of R. It may

be Res< –1 (or) Res> –1

The ROC of Y(s) is R shifted by 2 to the right. Since y(t) is not absolutely integrable, the ROC of

Y(s) should not include j-axis

X(s) =2s2s

162

; Res > –1

(ii)2s

4)s(X

,Y(s)=X(s)H(s)

Y(s) =)2s3s)(2s(

42

=2)2)(1(

4

ss

=)2s(

4

)2s(

4

)1s(

42

y(t) = (4e–t –4te-2t – 4e-2t)u(t)

LTFrom (4) y(t) = e2t x(t) Y(s ) = X(s–2)



05. (c)

Sol: (i) Asynchronous mod-11 counter

T0

TF/F

Q0

CLR

T1 Q1

CLR

TF/F

T2 Q2

CLR

TF/F

T3 Q3

CLR

TF/F

1 1 1Q0 Q1 Q2 Q31

clk

1 2 3 4 5 6 7 8 9 10 11 12

1 1 1 1 10 0 0 0 0 0

0 0 1 1 0 0 0 01 11

0 0 0 0

1 1 1 1

0 0 0 0

0 0 0 0 0 0 0 0

1 1 1

clk

Q0

Q1

Q2

Q3

0

0



(ii) Aim: To design a Mod-11 counter

State table:

Present state Next state

Q3 Q2 Q1 Q03Q

2Q 1Q

0Q T3 T2 T1 T0

0 0 0 0

0 0 0 1

0 0 1 0

0 0 1 1

0 1 0 0

0 1 0 1

0 1 1 0

0 1 1 1

1 0 0 0

1 0 0 1

1 0 1 0

0 0 0 1

0 0 1 0

0 0 1 1

0 1 0 0

0 1 0 1

0 1 1 0

0 1 1 1

1 0 0 0

1 0 0 1

1 0 1 0

0 0 0 0

0 0 0 1

0 0 1 1

0 0 0 1

0 1 1 1

0 0 0 1

0 0 1 1

0 0 0 1

1 1 1 1

0 0 0 1

0 0 1 1

1 0 1 0

States 11,12,13,14,15 are don’t cares in this case

T3 = m(7,10)+ d(11,12,13,14,15)

T2 = m(3,7)+ d(11,12,13,14,15)

T1 = m(1,3,5,7,9,10)+ d(11,12,13,14,15)

T0 = m(0,1,2,3,4,5,6,7,8,9)+ d(11,12,13,14,15)

0

Q3Q2

Q1Q0

00 01 11 10

00

01

11

10

0 1 3 2

4 5 7 6

12 13 15 14

8 9 1011

1

1

T3

T3 = Q2Q1Q0+ Q3 2Q Q1

Q3Q2

Q1Q0

00 01 11 10

00

01

11

10

1

T2

T2 = Q1Q0

1



To check the self starting counter or not

First take the unused states then check its output

Present sate

Q3 Q2 Q1 Q0

Input

T3 = Q1(Q2Q0+Q3 2Q ) T2 = Q1Q0 T1 = Q0+Q3Q1 T0 = 13 QQ

Output status

3Q

2Q 1Q

0Q

1 0 1 1

1 1 0 0

1 1 0 1

1 1 1 0

1 1 1 1

1 1 1 0

0 0 0 10 0 1 1

0 0 1 1

1 1 1 0

0 1 0 1

1 1 0 1

1 1 1 0

1 1 0 1

0 0 0 1

Q3Q2

Q1Q0

00 01 11 10

00

01

11

10

1

T1

T1 = m(1,3,5,7,9,10)+ d(11,12,13,14,15)

T1 = Q0+Q3Q1

1

1

1

1

1

Q3Q2

Q1Q0

00 01 11 10

00

01

11

10

1

T0 = m(0,1,2,3,4,5,6,7,8,9)+ d(11,12,13,14,15)

130 QQT

1

1

1

1

1

1

11

1

T0 T F/F(LSB)

T1 T F/F

Q0

Q3

Q1

Q0Q1

T2 T F/F

Q0

Q1

T3 T F/F(MSB)

Q2

Q0

Q3

2Q Q1

Q2Q3

3Q

1Q



From the above state table the unused state 1101 is going to another unused state 1110 and

from there again it goes to 1101 state so, it is locked between these two unused states, Hence it is

not a self starting counter.

06. (a)

Sol: (i)

(x) X(z) = 11

1

212

11

21

ZZ

Z

1Z

2

11

1)z(X

If x(n) is absolutely summalbe Z >2

1 (ROC includes unit circle)

x(n) = )n(u2

1n

(y)

2

21

1

1

2111

Z2

1

Z2

1Z

Z

Z2

11

Z2

1Z1(Z

2

11)Z

2

11

X(Z) = 1–Z-1 + 32 Z4

1Z

2

1 + -----

x(n) = (n) (n1) + 34

12

2

1nn

(z)21

1

z8

1z

4

11

z3)z(X



X(Z) =

11

1

Z4

11Z

2

11

Z3

=11 Z

41

1

4

Z21

1

4

If x(n) is absolutely summable Z >2

1 (ROC includes unit circle)

x(n) = 4

)(2

1nu

n

)(4

14 nu

n

(ii) Y(Z)+ 1.5[Z-1Y(Z) + y(–1)] + 0.5[Z–2Y(Z) + Z-1y(–1) + y(–2)] =1Z5.01

1

Y(Z)[1+1.5Z-1+0.5Z-2]+3+Z-1–2=1Z5.01

1

)Z5.0Z5.11)(Z5.01(

1

)Z5.0Z5.11(

)Z1()Z(Y

21121

1

=)Z1()Z5.01(

1

Z5.01

11211

Solve for Partial fraction expansion Inverse Z-transform is

y(n) = [4(–1)n – 4(–0.5)n +0.5n(–0.5)n–1]u(n)



06. (b) (i)

Sol : No.of chips required =capacityAvailabe

designedbetoMemory

Available ROM chips = 20488 = 210248 = 22108 = 2k8

Memory to be designed = 8kB

No.of chips required = 4kB2

kB8

So for 4 chips address lines assigned A0 to A10

So decoder required is 2:4

Total no.of address lines of Micro processor = n =16

No.of address lines required for memory to be interfaced = x [8kB = 2x = 213 x =13]

Lines for decoding logic = y

n–x =y 16–13 = 3 (decoding logic)

2×4

A11

A12 y0y1y2

y3

2kB

1CS

RD WR

2kB

2CS

RD WR

2kB

3CS

RD WR

2kBA0

A10

C000H

C7FFH

C800H

CFFFH

D000H

D7FFH

D800H

DFFFH

A14

13A

E

Decoder

A150CS

RD WR



06. (b) (ii)

Sol:

T-state

MVI B, XX 7 without loop

LOOP: DCR B 4 with loop

JNZ LOOPTrue (10)/ False (7) with loop

HLT 5 Without loop

Let no.of count value = n

Given Texe= 500s, fop = 5106 Hz

Texe = TwoL +TwL = 500s

valuecountstatesTof.Nof

1T

clkexe

valuecountT105

1T woLstate6woL

s4.2112105

1175

105

1T

66woL

TwL = Texe–TwoL = 500 –2.4 = 497.6 s

497.610–65106 = 14n–14+11

497.65 = 14n–3

H10 2B17892.17714

356.497n

XXB

TwL = T(condition satisfied)+ T(condition not satisfied)

(True) (False)

)False()True(

111)1n(14105

1106.497

66

Tn–1

F1

n

T=4+10=14F= 4+7 = 11

loop



06. (c)

Sol: Total induction motor load

= 4 25 = 100 kW

Total load on transformers is kVA

=)866.0()95.0(

100

= 121.55 kVA

The open delta must be able to supply 121.55 kVA to the load.

(i) 866.0CapacityinstalledtotalrsTransforme

kVAinCapacityAvialble

kVA rating of two transformers

= kVA34.140866.0

55.121

Rating of each transformer =2

34.140

= 70.17 kVA

(ii) Secondary line current

= 1000400

17.70 = 174.43 A

Primary line current (H.V side)

=11000

40043.174

= 6.34 A

Load power-factor angle

= cos–1 (0.866) = 30

One of the transformer power factor

= cos (30 – 30)

= 1

Second transformer power factor

= cos (30 + 30)

= 0.5



(iii) Real power supplied by one

= 70.17 1 = 70.17 kW

and another transformer

= 70.17 0.5 = 35.085 kW

Total power supplied open delta

= 70.17 + 35.085

= 105.26 kW

(iv) When third transformer is installed the available capacity is 70.17 3

= 210.51 kVA

06. (d)

Sol: 5 kW, 200V dc shunt motor

Back emf Eb = VIa Ra= 200 4 1 = 196V

Power developed (P) = EbIa= 1964 = 784W

This developed power is used in over coming the rotational Loss of the machine.

Rotational Loss = 784W

P = 78460

NT2

No load Torque(T) =10002

60784

=7.48N-m

8A

200V

+

2A

Rsh=100

6A

Ra= 1



07. (a)

Sol: Three-phase problems can be solved in per unit by treating them as single-phase problems.

Here Xs = 0.8pu, ra = 0, Ef = 1.2 pu,

Vt = 1.00 pu. For an input kVA of 100% at Vt = 1.00, VtIa = 1.00 and therefore

Ia = 1.00 p.u.

As Ef = 1.2pu is more than Vt = 1.00, synchronous motor is working at a leading pf.

2sat2

aat2f XIsinVrIcosVE

222 )8.0(sincos)2.1(

sin6.164.0sincos44.1 22

Or 125.06.1

2.0sin

Power factor, cos = 0.9922 leading.

Mechanical power developed by the motor

.pu9922.09922.011cosIV at

When excitation emf is reduced to 1.00 pu for the same load, then



9922.0sinX

VE

s

tf

o54.52or9922.0sin8.0

11

with ra = 0,

cosVE2VEXI tf2t

2f

2sa

o222a 54.52cos112)1()1(I8.0

or Ia = 1.1065 p.u.

Input kVA 1065.11IV at

.kVA%65.110or.u.p1065.1

07. (b)

Sol: (i)

1120 R

K211

R

K112

R

K10V

1R

K10

2

2R

K10

2R

K1010 R2 = 1k, R1 can be any value.

+

–R2

R2

10k

10k

11I

12I

20 R

K21V

R

K11V

R

k10V

–

–

+

+

VI1 = 1V

VI2 = 2V

R1

R1

2k

1k

11I R

k21V

12I R

k11V



(ii)

159K2

K10V0

= 5[–6]

= –30V

Op-amp saturates

V0 = –15V

07. (c)

Sol: x(n) = 0.5, 0.5, 0.5, 0.5, 0, 0, 0, 0

+

–2k

2k10k

10k

V0

–

–

+

+

5V

4.5V

R1

R1

2k

1k

151

215

9)11(5.4

1W08

1W08

1W08

1W08

1W08

1W08

1W08

jW 28

jW 28

707.0j707.0

W18

707.0j707.0

W 38

jW 28

X(0)

X(1)

X(2)

X(3)

X(4)

X(5)

X(6)

X(7)

x(0)

x(4)

x(2)

x(6)

x(1)

x(5)

x(3)

x(7)

–1

–1

–1

–1–1–1

–1

–1

–1

–1

–1

–1



Input Stage-1 outputs Stage-2 outputs Stage-3 outputs

0.5 0.5 1 2 = X(0)

0 0.5 0.5 – j0.5 0.5 – j1.207 = X(1)

0.5 0.5 0 0 = X(2)

0 0.5 0.5 + j0.5 0.5 – j0.207 = X(3)

0.5 0.5 1 0 = X(4)

0 0.5 0.5 – j0.5 0.5 + j0.207 = X(5)

0.5 0.5 0 0 = X(6)

0 0.5 0.5 + j0.5 0.5 + j1.207 = X(7)

07. (d)

Sol: Number of poles, 4P

Total lamp load, W500010050p

Terminal voltage Volts200V

Field resistance, 50R sh

Armature resistance 2.0R a

Voltage drop/brush = 1V

(i) Armature current, Ia:

Load current,Voltageterminal

ConsumedPowerI

25A200

5000

V

P

100W Lamps(50 No.s)

Ia

V=

200V

Ish

Rsh=50

+

Ra=0.2

I



Shunt field current,sh

sh R

VI

A450

200

Armature current,sha

III

.A29425

(ii) Current per path:

Current per path4

29

a

Ia = 7.25 A.

[ ,4pa generator being lap wound]

(iii) General e.m.f., Eg:

dropbrushRIVE aag

V8.207122.029200

Hence, generated e.m.f. = 207.8 V.

(iv) Power output of D.C. armature:

Power output of D.C. armature

kW026.6kW1000

298.207

1000

IE ag

: 3 : TEST-10 (Solutions) · 2017. 4. 16. · ACE Engineering Academy...

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Transcript of : 3 : TEST-10 (Solutions) · 2017. 4. 16. · ACE Engineering Academy...