K-NotesKrash+Sample+(2018).pdf · 1 K 11 f f o 1 G s H s k magnitude criterion k S 11 G s H s 2i 1 K0t o odd multiples of 180 angle critq erion SG s H s 2i 11 d K 0 even multiples

EE Krash (Sample)

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Control Systems

Root Locus Technique

Root – Loci is important to study trajectories of poles and zeroes as the poles & zeroes

determine transient response & stability of the system.

Characteristic equation

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

Assume G(s)H(s) =

Condition of Root locus

,

,

The difference between the sum of the angles of the vectors drawn from the zeroes and

those from the poles of G(s) H(s) to s1 is an odd multiple of 180° if K > 0.

The difference between the sum of the angles of the vectors drawn from the zeroes and

those from the poles of G(s) H(s) to s1 is an even multiple of 180° if K < 0. This is called as

Complimentary Root Locus.

Properties of Roots loci of

1. K = 0 points: These points are poles of G(s)H(s), including those at s = .

2. K = points: The K = points are the zeroes of G(s)H(s) including those at s = .

3. Total numbers of Root loci is equal to order of equation.

4. The root loci are symmetrical about the axis of symmetry of the pole- zero configuration

G(s) H(s).

5. For large values of s, the RL (K > 0) are asymptotes with angles given by:

For CRL(complementary root loci) (K < 0)

where i = 0, 1, 2, ……….,

1 1KG s H s

1 11 KG s H s 0

1 11G s H s

K

1 1

1G s H s k magnitude criterion

k

1 1G s H s 2i 1 odd multiples of 180 angle critK 0 erion

1 1G s H s 2i even multiples oK f 1800

1 11 KG s H s 0

1 11 KG s H s 0

i

2i 1180

n m

i

2i180

n m

n m 1

K-Notes

EE Krash (Sample)


n = number of finite poles of G(s) H(s)

m = number of finite zeroes of G(s) H(s)

6. The intersection of asymptotes lies on the real axis in s-plane. The point of intersection is

called centroid ( ) =

7. Root locus is found in a section of the real axis only if total number of poles and zeros to

the right side of section is odd if K > 0. For CRL (K < 0), if the number of real poles & zeroes

to right of given section is even, then that section lies on root locus.

8. The angle of departure or arrival of roots loci at a pole or zero of G(s) H(s) say s1 is found

by removing term (s – s1) from the transfer function and replacing ‘s’ by ‘s1’ in the remaining

transfer function to calculate

It can also be calculated as

Angle of Departure (only applicable for poles) = 1800 +

Angle of Arrival (only applicable for zeroes) = 1800 -

For Complimentary Root Locus,

Angle of Departure (only applicable for poles) = 00 +

Angle of Arrival (only applicable for zeroes) = 00 –

9. The crossing point of root-loci on imaginary axis can be found by equating coefficient of

s1 in Routh table to zero & calculating K.

Then roots of auxiliary polynomial give intersection of root locus with imaginary axis.

10. Break-away & Break-in points

These points are determined by finding roots of . On substituting, if K>0, only then it

is a valid breakaway/break-in point.

For breakaway points :

For break in points :

A breakaway point lies between consecutive poles and break in point lies between two

consecutive zeroes.

11. Value of K on Root locus is

The value of gain at any point on root locus can be determined geometrically as,

1

real parts of poles G(s)H(s) real parts of zeroes G(s)H(s)

n m

1 1

G s H s

1 1Angle of s from zeroes Angle of s from poles

1 1

G s H s

1 1

G s H s

1 1

G s H s

1 1

G s H s

dK

0ds

2

2

d K0

ds

2

2

d K0

ds

1 11 1

1K

G s H s

Pr oduct of vector length of polesK

Pr oduct of vector length of zeroes

EE Krash (Sample)


Addition of poles & zeroes to G(s) H(s)

Addition of a pole to G(s) H(s) has the effect of pushing of root loci toward right half

plane & hence stability decreases.

Addition of left half plane zeroes to the function G(s) H(s) generally has effect of moving

& bending the root loci toward the left half s-plane & hence stability increases.

Signal Flow Graphs

Node: It represents a system variable which is equal to sum of all incoming signals at it.

Outgoing signals do not affect value of node.

Output node: Node having only incoming branches

Input node: Node having only outgoing branches

Loop: It is a path which originates & terminates at the same node.

Branch: A signal travels along a branch from one node to another in the direction

indicated by the branch arrow & in the process gets multiplied by gain or transmittance

of branch

Forward Path: Path from input node to output node.

Non-Touching loops: Loops that do not have any common node.

Mason’s Gain Formula

Ratio of output to input variable of a signal flow graph is called net gain.

= path gain of kth forward path

= determinant of graph = 1 – (sum of gain of individual loops)

+ (sum of gain product of 2 non touching loops)

– (sum of gain product of 3 non touching loops) +………

= gain product of all ‘r’ non touching loops.

= the value of obtained by removing all the loops touching kth forward path.

T = overall gain

Example :

Forward path ,

k kK

1T P

kP

m1 m2 m3K

1 P P P ............

mrP

K

P G G G G1 1 2 3 4 P G G

2 1 5

EE Krash (Sample)


To find :

Individual loops , ,

, ,

Non – Touching loops (2) ,

Since, both forward paths touch all loops,

Gain =

Tip: For objective type questions, always prefer Mason’s gain formula instead of block

diagram reduction for finding transfer function.

Effect of Feedback

Due to negative feedback the gain reduces and due to positive feedback the gain

increases.

Feedback can improve stability or be harmful to stability if not applied properly.

Eg. Gain = & GH = –1, output is infinite for all inputs.

Sensitivity

Sensitivity is the ratio of relative change in output to relative change in input

For open loop system

T=G

For closed loop system

(Sensitivity decreases)

Thus, by the use of feedback sensitivity of Transfer Function w.r.t. forward gain reduces.

Sensitivity w.r.t. feedback path is,

L G H1 2 2 L G H

2 4 4

L H3 3 L G G G H

4 2 3 4 1 L G H

5 5 1

L' L L G H G H1 1 2 2 2 4 4

L' L L G H H2 1 3 2 2 3

1 G H G H H G G H G G H G H G H G H H2 2 4 4 3 2 3 1 4 5 1 2 2 4 4 2 2 3

11 2

Y P P5 1 1 2 2

Y1

G G G G G G1 2 3 4 1 5

G

1 GH

TG

TTS

GG

TG

G T GS 1 1

T G G

GT

1 GH

TG 2

G 1 GHG T 1 1S 1

T G G 1 GH1 GH

T T H TTSH H H T H

2T G

2H 1 GH

EE Krash (Sample)


If GH ≫ 1

This implies that effect of variation in H is not reduced even in a closed loop system so

change in feedback parameter is undesirable. Also, the closed loop control system is more

sensitive to changes in feedback elements than forward path elements.

Effect of disturbance (noise) signals

: Response due to noise input

: Response due to applied input

To minimize , we need to increase so effect of noise can be reduce by increasing the

gain of path (forward) earlier to noise.

Due to feedback the gain of noise decreases.

Note: Transfer function can only be written for single input & single output system. Hence,

for finding CR , assume N=0 & for finding CN , assume R=0.

Type and Order of a System

The number of poles of an Open Loop Transfer Function at origin is termed as Type of the

System.

The highest power of ‘s’ in Characteristic Equation or the denominator of Closed Loop

Transfer Function is called as Order of the system.

Characteristic Equation is given as, 1+G(s)H(s) = 0

2 H 1 GHGTSH 2 G1 GH

GH

1 GH

TS 1H

C C CN R

CN

CR

G G1 2C R

R 1 G G H1 2

G2C N

N 1 G G H1 2

CN

G1

EE Krash (Sample)


Signal & Systems

Fourier Transfoem

Fourier series exists only for periodic signals, Fourier series converges to Fourier Transform

which is continuous as compared to Fourier series which is discrete.

Continuous Time Fourier Transform

Analysis equation : j tX j x t e dt

Synthesis equation : j t1x t X j e d

2

Properties of Continuous Time Fourier Transform

Linearity 1 2 1 2

F.T.ax t bx t aX bX

Time shifting 0j t

0

F.T.x t t e X

Frequency shifting 0j t

0 0

F.T.x t e X or X f f

Time scaling F.T. 1

x at Xaa

Duality F.T.X t 2 x or x f

Time reversal F.T.x t X or X f

Differentiation in time domain F.T.d

x t j X or j2 fX fdt

Differentiation in frequency

domain F.T. d

tx t j Xd

F.T. dj2 tx t X f

df

Integration in time domain

t

F.T.X

x d X 0 if X 0 0j

Conjugate * * *F.T.x t X or X f

Convolution in time domain 1 2 1 2

F.T.x t * x t X X

Convolution in frequency domain 1 2 1 2

F.T. 1x t .x t X * X

2

Parseval’s power theorem

21x t dt X d

2

Area under signal x t X 0 x t dt

and 1

x X 0 X d2

EE Krash (Sample)


Some common Fourier Transform Pairs

Signal Fourier Transform

0jk t

kK

a e

0kk

2 a k

0jk te

02

cos 0t

0 0

sin 0t

0 0j

1 2

n

t nT

K

2 2 k

T T

1

1

1, t Tx t

0, t T

1

2sin T

sint

t

1, wx

0, w

t 1

u(t)

1

j

0t t 0j t

e

ate u t ,Re a 0 1

a j

Signum function 2

j

Relation between nature of signal in time domain and frequency domain

Energy Spectral Density

ESD specifies the distribution of energy among the different frequency components in the

signal.

EE Krash (Sample)


ESD, 2

( ) X( )

E = 2 21

x(t) dt X( ) d2

=

2

X(f) df

Power spectral density (PSD)

PSD defines distribution of power in the signal.

2

T

X fS lim

T

Fourier Transform of periodic signal

0jn t

nn

f t C e

0n

nF 2 C n

0n

nF 2 C n

So Fourier transform spectrum of periodic signals is always discrete.

Discrete Time Fourier Transform (For ECE)

The discrete-time Fourier transform (DTFT) or the Fourier transform of a discrete–time

aperiodic sequence x[n] is a representation of the sequence in terms of the complex

exponential sequence ejωn

DTFT

Discrete Continuous

j j x n XX e or X e

As in Fourier transform, jX e is also called spectrum and is a continuous function of the

frequency parameter ω and is a periodic function of frequency with period 2 .

j j n

n

X e x n e

Analysis equation

The infinite series in equation 1 may be converges or may not. x[n] is absolutely summable.

n

When x n

An absolutely summable sequence has always a finite energy but a finite-energy sequence is

not necessarily to be absolutely summable.

Inverse Discrete Time Fourier Transform

j j n

2

1x n X e e d

2

Synthesis equation

Properties of DTFT

(i) Linearity

If DTFT j

1 1x n X e and DTFT j

2 2x n X e

EE Krash (Sample)


(ii) Time shifting

If DTFT jx n X e

0j nDTFT j

0Then x n n e X e

(iii) Frequency shifting

If DTFT jx n X e

0 0j n j( )DTFTThen x n e X e

(iv) Time Scaling

If DTFT jx n X e

DTFT j knThen x X e

k

(v) Time Scaling

If DTFT jx n X e

DTFT j knThen x X e

k

(vi) Conjugate

If DTFT jx n X e

Then * DTFT * jx n X e

(vii) Frequency differentiation

If DTFT jx n X e

Then DTFT jdnx n j X e

d

(viii) Convolution in n-domain

If DTFT j DTFT j

1 1 2 2x n X e and x n X e

Then DTFT j j

1 2 1 2x n x n X e X e

(ix) Windowing or modulation property

If DTFT j DTFT j

1 1 2 2x n X e and x n X e

Then j pDTFT j j jp

1 2 1 2 1 2

1 1x n x n X e * X e or X e X e dp

2 2

EE Krash (Sample)


(x) Periodicity

If DTFT jx n X e

j 2 k jThen x e x e

(xi) First difference:

If DTFT jx n X e

DTFT j jThen x n x n 1 1 e x e

(xii) Parseval’s theorem:

If DTFT jx n X e

2

2j

n 2

1Then x n X e d

2

Distortion less transmission through LTI system

For distortion less transmission, output response must be exact replica of input response.

0y t kx t t

0j tY e kX

0j tY

keX

0

H K

H n t

So the conditions for Distortion less Transmission through a LTI System are:

Magnitude response must be constant.

Phase response must be linear.

Hilbert transform

Hilbert transform of a signal x(t) is defined as the transform in which phase angle of all

components of the signal is shifted by ±900. Hilbert transform of x(t) is represented by x t .

Correlation

Correlation is a measure of similarity between two signals. It is used in RADAR or SONAR to

measure similarity between the transmitted signal and the signal received after reflection.

Types of correlation

1. Cross correlation function (CCF)

2. Auto correlation function (ACF)

Cross correlation function

Cross correlation is the measure of similarity between two different signals.

* *

XYR x t .y t dt x t y t dt

EE Krash (Sample)


Auto correlation function

Correlation is said to be cross correlation if similarity between a signal and its shifted version.

For Energy signal

* *

XXR x t .x t dt x t x t dt

For Power signal

T

*

xT

T

1R Lt x t .x t dt

2

T

TT

1Lt x t x * t dt

2

Properties of ACF

(i) ACF is an even function of “ ” i.e. XX XXR R

(ii) ACF has its maximum magnitude at origin i.e. XX XXR R 0

(iii) 2

XXR 0 x t dt

Energy in x(t)

T

2

TT

1Lt x t dt

2T

Average power in x(t)

(iv) 2 21

x t dt X d2

2

x f df

Where 2

X Power spectral density (PSD)

(v) 2

XX XX XX

F.T.R S S X PSD

EE Krash (Sample)


Engineering Maths

Calculus

Important Series Expansion

a.

b.

c.

d.

e.

f. tan x =

g. ln (1 + x) =

h.

i.

j.

Important Limits

a.

b.

c.

d.

e.

f.

g.

h.

n

n n

r 0

rr1 x C x

1 21 x 1 x x ............

2 3

2 3x x xa 1 x log a xloga xloga ................

2! 3!

3 5x xsinx x .................3! 5!

2 4x xcosx 1 + ......................

2! 4!3

5x 2x x

3 15

2 3x xx + + ............, x < 12 3

2 3x x

ln 1 x x ............ x 12 3

2 3x x x x

e 11! 2! 3!

2 3x x x x

e 11! 2! 3!

lt sinx 1

x 0 x

lt tanx 1

x 0 x

1 nx

lt 1 nx e

x 0

lt cos x 1

x 0

1

xlt

1 x ex 0

xlt 1 1 exx

2

2

1 cos mxlt m

2x 0 x

lt sinx0

xx

EE Krash (Sample)


L – Hospital’s Rule

If f (x) and g(x) are two functions such that

and

Then,

If f’(x) and g’(x) are also zero as , then we can take successive derivatives till this

condition is violated.

For continuity,

For differentiability, exists and is equal to

If a function is differentiable at some point then it is continuous at that point but converse

may not be true.

Mean Value Theorems

Rolle’s Theorem

If there is a function f(x) such that f(x) is continuous in closed interval a ≤ x ≤ b and f’(x) is

existing at every point in open interval a < x < b and f(a) = f(b). Then, there exists a point

‘c’ such that f’(c) = 0 and a < c < b.

Lagrange’s Mean value Theorem

If there is a function f(x) such that, f(x) is continuous in closed interval a ≤ x ≤ b; and f(x)

is differentiable in open interval (a, b) i.e., a < x < b,

Then there exists a point ‘c’, such that

Differentiation

Properties: (f + g)’ = f’ + g’ ; (f – g)’ = f’ – g’ ; (f g)’ = f’ g + f g’

Important derivatives

a. → n

b.

c.

d.

e.

f. sin x → cos x

g. cos x → –sin x

h. tan x →

lt

f x 0x a

lt

g x 0x a

lt ltf x f' x

x a x ag x g' x

x a

lim

f x =f ax a

00f x h f xlim

h 0 h

0f ' x

f b f af ' c

b a

nx n 1x

1nxx

a a1log x (log e)

x

x xe ex x

ea a log a

2sec x

EE Krash (Sample)


i. sec x → sec x tan x

j. cosec x → – cosec x cot x

k. cot x → – cosec2 x

l. sin h x → cos h x

m. cos h x → sin h x

n.

o.

p.

q.

r.

s.

Increasing & Decreasing Functions

, then f is increasing in [a, b]

, then f is strictly increasing in [a, b]

, then f is decreasing in [a, b]

, then f is strictly decreasing in [a, b]

Maxima & Minima

Local maxima or minima

There is a maximum of f(x) at x = a if f’(a) = 0 and f”(a) is negative.

There is a minimum of f (x) at x = a, if f’(a) = 0 and f” (a) is positive.

To calculate maximum or minima, we find the point ‘a’ such that f’(a) = 0 and then decide if it

is maximum or minima by judging the sign of f”(a).

Global maxima & minima

We first find local maxima & minima & then calculate the value of ‘f’ at boundary points of

interval given eg. [a, b], we find f(a) & f(b) & compare it with the values of local maxima &

minima. The absolute maxima & minima can be decided then.

Partial Derivative

If a derivative of a function of several independent variables be found with respect to any

one of them, keeping the others as constant, it is said to be a partial derivative.

Homogenous Function

is a homogenous function of x & y, of degree ‘n’

=

1

2

1sin x

1 - x

2

1 -1cos x

1 x

2

1 1tan x

1 x

2

1 -1cosec x

x x 1

2

1 1sec x

x x 1

12

-1cot x

1 x

f ' x 0 V x a, b

f ' x 0 V x a, b

f ' x 0 V x a, b

f ' x 0 V x a, b

n 2 2n n 1 nn0 1 2

a x a x y a x y ............. a y

2 n

n0 1 2n y y y

x a a a .................... ax x x

EE Krash (Sample)


Euler’s Theorem

If u is a homogenous function of x & y of degree n, then

Maxima & minima of multi-variable function

; ;

Maxima

rt – s2 >0 ; r < 0

Minima

rt – s2>0 ; r > 0

Saddle point

rt – s2< 0

Integration

Indefinite integrals are just opposite of derivatives and hence important derivatives must

always be remembered.

Some standard integral formula

u ux y nu

x y

2

2

x a

y b

flet r

x

2

x a

y b

fs

x y

2

2

x a

y b

ft

y

n 1n x

x dx cn 1

xxa loga dx a c

xxe dx a c

uv dx u v u v c

sinx dx cosx c cosx dx sinx c

tanx dx ln secx c cosecx dx ln cosecx cotx c

secx dx ln secx tanx c cotx dx ln sinx c

1

2

1 dx sin x c

1 x

1

2

1 dx cos x c

1 x

1

2

1dx tan x c

1 x

1

2

1dx cot x c

1 x

EE Krash (Sample)


Properties of definite integral

a.

b.

c.

d.

e.

f.

g.

1

2

1dx cosec x c

x x 1

1

2

1dx sec x c

x x 1

1

2 2

1 xdx sin c

aa x

1

2 2

1 1 xdx tan c

a aa x

2 2

1 1 x adx ln c

2a x aa x

2 2

1 1 x adx ln c

2a x ax a

2 2

2 2

1dx ln x x a c

x a

2 2

2 2

1dx ln x x a c

x a

b b

a a

f x dx f t dt

b a

a b

f x dx f x dx

b c b

a a c

f x dx f x dx f x dx

b b

a a

f x dx f a b x dx

a

2a

0

0

2 f(x)dx if f(2a-x)=f(x)f(x)dx

0 if f(2a-x) f(x)

a

a

0

a

2 f(x)dx if f(x)=f(-x)f(x)dx

0 if f(x) f(x)

t

t

df x dx f t ' t f t ' t

dt

EE Krash (Sample)


Analog Electronics

BJT Amplifier & MOSFET Amplifier

Early effect or Base – width Modulation

Mode JE JC Property Application

1 Cut off RB RB Very high internal

resistance

OFF Switch

2 Saturation FB FB Very low internal

resistance

ON Switch

3 Forward

Active

FB FB Excellent transistor

action

Amplifier

4 Reverse

active

RB FB Poor transistor action Attenuator, logic

circuits

B Base width

W Width of depletion region of Base – collector junction.

VPB Potential barrier due to depletion region.

As VCB [reserve bias voltage] varies, width of depletion region [w] varies, hence width of base

[B] varies. This is known as early effect or base width Modulation

Due to early effect,

IB : AS VCB , W , B . Therefore, time for which electron remains in base decreases. This

decreases recombination in base. Hence IB decreases.

IC : Since number of electrons recombining in base decreases, number of electrons

reaching to collector increases. Hence IC .

IE : AS VCB , W , VPB across JBC, number of e-denting from B to C , no. of e- in B ,

Concentration gradient across JBE, deficiency of e- in Emitter , e-Supply by negative

terminal of VFB IE .

EE Krash (Sample)


Current equations for different configurations.

For common Base configuration

IC = IE + ICBO [ICBO = ICO]

For common Emitter configuration

IC = IB + ICEO

For common Collector configuration

IE = IB + [I + ] ICBO

= 1

, =

1

1 = 1 +

ICFO = (1 + ) ICBO = (1 + ) ICO

Where:

Common base forward current gain

Common Emitter forward current gain

Common Collector forward current gain

ICBO Conventional leakage current from collector to base with emitter open.

[Common base configuration]

ICEO Conventional leakage current from collector to emitter with base open.

[Common emitter configuration]

EE Krash (Sample)


Comparison of Different BJT Configuration.

Parameters CB

CE CC

Circuit

diagram

Common

terminal

Base is common to

input & output for

AC

Emitter is common to

input & output for AC

Collector is common to

input & output for AC

AC input

resistance

low, 20 to 100 Moderate, 1k to 3k high, in mega ohm‘s

AC output

resistance

High, 1 to 3 M Moderate, 40 to 80k very low in ohm‘s

Current

gain

, Very low, < 1 , moderate, 20 to 300 , large, = 1 +

Voltage

gain

> 1, but power gain

is low

>1, power gain is high 1, unity gain

Phase Input & output in

Phase

input & output out of

phase

input & output in phase

Impedance

matching

Cascading is not

efficient

efficient cascading is

possible with common

collector

works as impedance

matching amplifier

Leakage

current

very small

[ICBO] 5µA for Ge

[ICBO]1µA for Si

very large [ICEO]

500 µA for Ge

20 µA for Si

same as that of CE [ICFO]

Applications Constant current

source, high band

width amplifier

Voltage amplifier Buffer or Impedance

matching amplifier

Selection of Q Point [VCFQ, ICQ]

While designing CE amplifier, values of different resistors & voltages [DC] must be selected in

such a way that,

Q point should be in active region

Q point should be on DC load line

As for as possible, Q point should be selected in middle of load line to avoid clipping of

signal.

The Q Point Shifts on DC load line, because of following reason

Variation in device parameter.

EE Krash (Sample)


Thermal instability Temperature variation.

(a) Decrease in VBE with increase in temperature.

(b) Increase in β with increase in temperature.

(c) Change in ICO with temperature.

Stability factor [s] = C

CO

I

I

=

B

C

1+β

I1-β

I

Stability factor is Parameter with decides to what extent a biasing method is successful to

keep Q point stable.

1. Fixed bias 1. Use of diode for VBE compensation

2. Fixed bias with self-bias or Emitter f/b bias 2. Use of diode for ICO compensation

3. Collector to base bias 3. Use of thermistor

4. Potential divider biasing with for compensation emitter bias or self-bias

Type of biasing Fixed bias Fixed bias with

self-bias or

Emitter f/b bias

Collector to

base bias

Potential divider

biasing with

emitter bias or self-

bias

Circuit diagram

Stability factor

[S]

S = 1 + S =

E

B E

1

R1

R R

S =

S = E

E B

1

R1

R R

C

C B

1

R1

R R

EE Krash (Sample)


Different models of BJT

re model for common base configuration. [PNP]

0r Common base output resistance

re =

re model for common base configuration [NPN]

0r Common base output resistance

re model for common Emitter configuration [NPN]

0r Common emitter output resistance

High frequency model – Hybrid model or Via collector model [ECE only]

EQ

26 m

I

EE Krash (Sample)


Relation between different Parameters.

(1) hie = r = re (2) gm = e

r

e

1

r

(3) ro = oe

1

h (4) = hfe

AC Equivalent Circuit of MOSFET [ECE Only]

Model:

ro = DQ

1

I = A

DQ

V

I

VA early voltage

IDQ DC drain current

Common Source Amplifier with current minor bias.

m1 0 ng 2I K

EE Krash (Sample)


Common Drain Amplifier or Source Follower Amplifier.

AV = O

S

V

V =

If we choose gm RL >> 1, then AV 1, then Av 1

If rds. [drain–source resistance/ output resistance] considered, then

o

s

v

v =

Ri = R1 II R2 ; Ro = ; gds =

Common Source Amplifier with Source Degeneration

o

s

v

v = M O L

M X

I [R / R ]

1+I R

Ri = R1IIR2

Rout = rds + gm rds Rx + Rx

m L

m L

g R

g R +1

m L ds

m L ds

g R r

1+g R r

m ds

1

g +g ds

1

r

EE Krash (Sample)


Common Gate Amplifier.

o

s

v

v =

Rout = gm Rs rds + rds + RS

D L

S

R R

R

i

m

1R

g

EE Krash (Sample)


Engineering Mechanics

Impulse, Momentum and Collisions

The impulse of a constant force F is defined as the product of the force and the time t for

which it acts.

Impulse = F.t

The effect of the impulse on a body can be found using below equations, where a is

acceleration, u and v are initial and final velocities respectively and t is time.

v u at v u at

So

chaI F ngt m e a t m in momentumv u

So we can say that,

Impulse of a constant force = F.t = change in momentum produced.

Impulse is a vector quantity and has the same units as momentum, Ns or kg m/s

The impulse of a variable force can be defined by the integral

Impulse=t

0

F.dt , where t is the time for which F acts.

By Newton's 2nd law

F m.a m dv / dt

So impulse can also be written

Impulsev v

v

uu u

dvm dt mdv mv

dt

which for a constant mass

Impulse =m (v-u)

Impulsive force

Suppose the force F is very large and acts for a very short time. During this time the

distance moved is very small and under normal analysis would be ignored. Under these

conditions, the only effect of the force can be measured is the impulse, or change in

momentum which is called an impulsive force.

In theory this force should be infinitely large and the time of action infinitely small. Some

applications where the conditions are approached are collision of snooker balls, a

hammer hitting a nail or the impact of a bullet on a target.

Conservation of linear momentum

Consider the direct collision of two spheres A and B.

EE Krash (Sample)


When the spheres collide, then by Newton's third law, the force F exerted by A on B is

equal and opposite to the force exerted by B on A.

The time for contact is the same for both. The impulse of A on B is thus equal and

opposite to the impulse of B on A. It then follows that the change in momentum of A is

equal in magnitude to the change in momentum in B, but it is in the opposite direction.

The total change in momentum of the whole system is thus zero.

This means that the total momentum before and after a collision is equal, or that linear

momentum is conserved. This is called the principle of conservation of linear momentum

and in summary this may be stated:

The total momentum of a system, in any direction, remains constant unless an external

force acts on the system in that direction.

Caution: Take proper sign convention while solving problems.

Impact of inelastic bodies

When two inelastic bodies collide they remain together. They show no inclination to

return to their original shape after the collision.

An example of this may be two railway carriages that collide and become coupled on

impact.

Problems of this type may be solved by the principle of conservation of linear

momentum.

Momentum before impact = Momentum after impact

(Take proper sign convention)

Although momentum is conserved, it is important to realize that energy is always lost in

an inelastic collision (it is converted from mechanical energy to some other form such as

heat, light or sound.)

Impact of elastic bodies

In the last section the bodies were assumed to stay together after impact. An elastic body

is one which tends to return to its original shape after impact. When two elastic bodies

collide, they rebound after collision. An example is the collision of two snooker balls.

If the bodies are travelling along the same straight line before impact, then the collision is

called a direct collision. This is the only type of collision considered here.

Consider the two elastic spheres as shown. By the principle of conservation of linear

momentum

Momentum before impact = Momentum after impact

1 1 2 2 1 1 2 2m u m u m v m v

where the i

u initial velocity of body i.

i

v final velocity of body i.

EE Krash (Sample)


When the spheres are inelastic v1 and v2 are equal as we saw in the last section. For elastic

bodies v1 and v2 depend on the elastic properties of the bodies. A measure of the

elasticity is the coefficient of restitution, for direct collision this is defined as

1 2

1 2

v ve

u u

The values of ‘e’ in practice vary between 0 and 1. For completely inelastic collision, e 0

and for completely elastic collision, e=1. In the latter case, no energy is lost in the

collision.

Both the law of restitution & conservation of momentum are applicable along x and y

directions in case of oblique collision.

Rolling, torque and angular momentum

Rolling motion:

Combination of translational motion & rotational motion.

In rolling motion, the centre of the object moves in a line parallel to the surface.

Relation between length and angle of rotation:

When the object rotates through an angle ' ' , a point at a distance R from the rotation axis

moves through a distance of S

S R

The arc length S is the same as the distance that the wheel translates.

The linear (translational) speed COM

V of the wheel is dS

dt;

COMV is the velocity of centre of mass.

The angular speed of the wheel is d

dt

So, dS d

Rdt dt

COMV R

Rolling motion is the combination of pure rotational motion and pure translational motion.

EE Krash (Sample)


Pure rotation + pure translation = Rolling motion

The velocity of a point at the top of the rolling wheel is twice that of the centre of the wheel

top COMV 2R 2 R 2V

Kinetic energy of rolling:

As an object rolls, the point at the very bottom, the contact point with the surface, is

instantaneously stationary.

We will call this point P and we can treat rolling about this point.

K.E= 2

P

1I

2

PI : Rotational inertia about the point P

Parallel axis theorem says 2

P COMI I MR

2

P

1K.E I

2

2 2 2

COM

1 1K.E I MR

2 2

2 2

COM COM

1 1K.E I MV

2 2

Kinetic energy of a rolling object comes from rotational kinetic energy and translational

kinetic energy.

Forces in rolling:

If a wheel rolls smoothly, there is no sliding at the contact point so there is no friction.

However if there is an external force that produces an acceleration, there will be an

angular acceleration . The acceleration will make the wheel want to slide at the contact

point. Then a frictional force will be on the wheel to oppose the tendency to slide.

Direction of static frictional force:

If a wheel moving to the right were to accelerate, the bottom of the wheel would want to

move to the left compared to the surface. Thus static friction force is to the right.

If same wheel was to slow down, the direction of the acceleration and angular acceleration

would switch and the static friction force will now be pointing towards the left.

EE Krash (Sample)


Rolling down a ramp:

The direction of the static friction force is the confusing part here. It points up along the

ramp. If the wheel were to slide down the ramp, the friction opposing the sliding would be

pointing up.

s COMf Mgsin Ma

I

Only force on the wheel that produces torque is the friction

s COMRf I

We will need to make use of COM

a R

(a is down the ramp, negative X-direction but the wheel rolls counter-clockwise, is positive)

COMa

R

So we can solve for COM

s COM 2

af I

R

COM

COM

2

gsina

I1

MR

Yo-Yo

A Yo-Yo behaves similar to the wheel rolling down a ramp.

1) Instead of rolling down a ramp of angle ' ' , Yo-Yo follows an angle of 090 with horizontal.

2) Yo-Yo rolls down a string on a radius R.

3) Instead of friction, the tension shows up in the Yo-Yo.

COM

COM

2

0

ga

I1

MR

EE Krash (Sample)


Electrical Machines

Transformer

MMF and Flux

C

C

C C

C

C

C

Ni

A l

lNi

A

F R

c

c

lR

A

Reluctance of transformer core

Flux in core

F=MMF

cl = Effective length of Core

cA = Cross-Sectional Area of the core

= permeability of the core

Construction

In a Transformer the primary and secondary windings are wound around the core of a

transformer. Core is a magnetic material which allows the flow of magnetic flux lines to link

both primary and secondary windings. Also it provides mechanical support to the

Transformer.

1. Core should have low reluctance and high permeability to the flow of magnetic flux.

2. Core is generally made of Silicon steel

3. Generally, this steel is cold rolled grain oriented CRGO steel, to increase permeability

along the direction of magnetization and reduce core losses.

4. KVA Rating (Core Dimension)4 5. Voltage Rating (Core Dimension)2

6. Current Rating (Core Dimension)2

7. No-Load Current Core Dimension

8. Core Loss Core Volume

Properties of Ideal Transformer

Permeability of transformer is infinite.

No Iron or Copper Loss.

EE Krash (Sample)


Induced EMF in a Transformer

For a practical transformer with finite permeability the induced emf can be calculated as,

The instantaneous values of induced emf in primary and secondary are given by,

1 1

2 2

de N

dt

de N

dt

The rms values of induced emf is given as,

1 1 m

2 2 m

E (rms) 4.44fN

E (rms) 4.44fN

Where E1 and E2 are emf in primary and secondary windings of Transformer respectively.

Φ is the flux in the transformer and Φm is maximum value of flux.

m m n

B A where Bm is the maximum value of magnetic flux density and An is the cross

sectional area of the core.

The polarity of emf is decided on basis of Lenz Law as currents in primary and secondary

should be such that primary and secondary flux should oppose each other.

Also, primary current enters the positive terminal of primary winding as primary absorbs

power and secondary current leaves the positive terminal of secondary winding as

secondary delivers power and this way we can mark emf polarities.

Induced EMF always lags the emf by 900.

If the frequency of operation of a transformer is reduced then KVA rating of transformer

reduces proportionately as induced emf varies linearly with frequency.

For a given KVA rating and for given maximum value of B of core, more the designed

frequency lesser is size and weight of transformers.

Exact equivalent circuit

'

1 0 1I I I

Exact equivalent circuit with respect to primary

EE Krash (Sample)


2 2 2

1 1 1

2 2 2 2 L L

2 2 2

N N NR = R ; X = X ; Z = Z

N N N

;

Approximate Equivalent Circuit

01 1 2R = R R

01 1 2X = X X

In this approximation, we are neglecting the no-load copper losses and we are

overestimating the core losses.

The phasor diagram for the exact equivalent circuit is shown below,

I: Magnetizing Current

wI : Core-loss Component of No-Load Current

0I : No-Load Current

o70 to 75

0cos 0.2lag (No-Load pf)

Here, o is the angle between E1 and I0

Transformer has poor No load PF because w

I I

1 1 2 2N I ' N I

Load component of primary MMF = Secondary MMF

2 1 2

1 2 1

N I ' EK

N I E = Transformation ratio

EE Krash (Sample)


Synchronous Machine

Synchronous Speed

e m

P

2

If we differentiate both sides,

e m

e

m

m

P

2

2 f

P2 f

2

4 frad / s

P

4 f2 N ; N rev / s

P

2fN rev / s

P

2fN 60rev / min

P

s

120fN rpm

P

Induced emf

Phase voltage ph

4.44 N f

phN : number of turns per phase

: flux per pole

f : frequency

This phase voltage is rms value & it is valid for a full pitched concentrated winding only.

Armature Winding

Usually, coil span is 180 (electrical)

If coil span = 180 (electrical), coil is called as full pitch coil.

If coil span = 180 (electrical), coil is called as Chorded coil or short pitched winding.

Pitch Factor,

PK = cos

2

Induced emf ph P 4.44 N f K ……….for short-pitched concentrated winding only

For thn harmonic

Induced emf ph P 4.44 N f K

P

nK = cos

2

To eliminate thn harmonic

n =

2 2

180

= electricaln

EE Krash (Sample)


Distributed Winding

number of slotsm =

number of poles no. of phase

number of slotsCoil Span =

number of poles

180

= electricalcoil span

;

Distribution Factor, d

msin

2K

m sin2

For thn harmonic, is replaced by n

d

mnsin

2K

nm sin

2

For uniform distribution replace n

sin2

by n

2

Winding Factor, w P d

K = K K

Induced emf ph w

= 4.44 N f K …………….for short-pitched distributed winding

Armature Resistance

Generally winding resistance is measured using voltmeter ammeter –method.

For star connection

m

voltmeter readingVR = =

I ammeter reading

mR = 2R

mR

R = 2

For Delta Connection

m

voltmeter readingR =

ammeter reading

m

2R = R

3

m

3R = R

2

This resistance is dc resistance but ac resistance is higher due to skin effect.

a acR = 1.2 to 1.3 R

EE Krash (Sample)


Armature Reaction

Power factor Generator Motor

Unity

Zero pf lagging

Zero pf leading

Lagging pf cos

Leading pf cos

EE Krash (Sample)


Leakage Flux

Leakage flux links only one winding but not both so if it is present in stator, it won’t link to

rotor & vice-versa (the function of leakage flux is just same as that of choke in a coil; to drop

voltage)

Equivalent Circuit

sX = synchronous reactance

ar l X X = sum of armature reaction reactance & leakage

reactance

a a sE V 0 + I (R jX ) , for Synchronous Generator

a a sE V 0 - I (R jX ) , for Synchronous Motor

( and are positive)

Where Φ is power factor angle (leading)

for lagging power factor we replace Φ by “– Φ”

Voltage Regulation

Voltage regulation E V

100%V

2 2

a a a s

2 2

a a a s

E (Vcos I R ) (Vsin I X ) for lagging pf load

E (Vcos I R ) (Vsin I X ) for leading pf load

Zero Voltage Regulation is only possible for leading pf loads.

For zero regulation 2 2E V 0

a sI Z

cos( )2V

Negative sign because ( )>90

For maximum voltage regulation

=

cos = load pf lagging

EE Krash (Sample)


Power Electronics

Chopper

Buck Converter

When CH is ON o t DT

Voltage across inductor L S OV V V

When CH is OFF (DT < t < T)

Voltage across inductor L 0

V V

Applying volt-sec balance across inductor

S O OV V DT V T DT 0

S O OV V D V 1 D 0

O SV DV

D = duty cycle = ON

T

T

Where T = switching period = 1f

f = switching frequency

Average output voltage = S

DV

rms output voltage = S

DV

Average source current = O

DI

Average current of FD = O1 D I

Ripple in output current

When CH is ON 0 t DT

L S O SV V V 1 D V

During this period, since voltage is positive current

increase from minimum value to maximum value.

max mini I I

t DT 0 DT

S

iL 1 D V

DT

SD 1 D V

ifL

EE Krash (Sample)


The formula below approximate value of output ripple current for maximum ripple, D = 0.5

S

max

Vi

4fL

L

max O

II I

2

L

min O

II I

2

Critical Inductance (LC)

Value of inductance at which inductor voltage waveform is just discontinuous.

c

1 D RL

2f

Critical Capacitance (CC)

Value of capacitance at which current through inductor is just discontinuous.

C

1C

8fR

Step-up chopper (Boost converter)

L Swhen CH is ON 0 t DT , V V

L S Owhen CH is OFF DT t T , V V V

Applying volt-sec balance across inductor

S S OV DT V V 1 D T 0

S

O

VV

1 D

Since D < 1, O S

V V

C Owhen CH is ON 0 t DT , I I

C L Owhen CH is OFF DT t T , I I I

O L O

Applying Ampere sec balance across capacitor

I DT I I 1 D T 0

OL

II

1 D

EE Krash (Sample)


Ripple in inductor current

min max

When CH is ON 0 t DT ,

current increase from I to I

S SL

S L

V DT DViL V i

DT L fL

Ripple in output voltage

C Owhen CHis ON , I I

C

O

VC. I

DT

O

O C

I DTV V

C

-ve sign indicates voltage decrease

O

O

I DTV

C

Critical Inductance (Lc)

L

L

II

2

C

D 1 D RL

2f

Critical Capacitance (Cc)

O

O

VV

2

C

DC

2fR

If inductor also has an internal resistance, then

O S 2

1 DV V

r 1 DR

r = internal resistance of inductor; R = load resistance

EE Krash (Sample)


Control Systems

Transient Analysis

For Concept, refer to Control Systems K-Notes, Time Response Analysis

For the system shown in figure, with a damping ratio of .7 and an un-damped natural

frequency of 4 rad/sec, the values of ‘K’ and ‘a’ are

(A)K=4,a=.35

(B)K=8,a=.455

(C)K=16,a=.225

(D)K=64,a=.9

Solution: (C) is correct option

Characteristic equation

01. A controller of the form is designed for a plant with transfer function

such that the un-damped natural frequency and the damping ratio of the closed

loop 2nd order system are 2 rad/sec and 0.5 respectively. When the steady state error to a

unit step input is zero, then the controller parameters are

(A) k = 4, a = 3, b = 2 (B) k = 1, a = 3, b = 2

(C) k = 4, a = 2, b = 3 (D) k = 1, a = 2, b = 3

0

1 G(s)H(s) 0

K(1 as)1 0

(s 2)2

2

S 2s Kas K 0

S s(Ka 2) K 0

Compare with standard equation

2 2n n

2 2n

S 2 s 0

K 4 16 => K 16

nKa 2 2 2(.7)(4)

2(.7)(4) 2a .225

16

c

K(s a)G (s)

(s b)

1G(s)

s(s 3)

Problems

Sample Problem

Kuestion

EE Krash (Sample)


01. Ans: (A)

Solution:

So overall OLTF=

As it is a second order system. So taking a=3

(a=b not in the option, so not taken & s=3 cannot be an open loop pole also)

Now un-damped natural frequency

Damping factor=

Now second order OLTF=

So K=4, b=2, a=3

02. A feedback system has un-damped frequency of 10 rad/sec and the damping ratio is 0.4

the transfer function is?

(A) (B)

(C) (D)

02. Ans: (A)

Solution:

Given

(Standard form)

TF=

03. A system is required to have a peak overshoot of 4.32 % and a natural frequency of 5

rad/sec. The required location of the dominant pole is:

(A) – 3.535 + j 5.535 (B) – 2.535 + j 2.525

(C) – 2.535 + j 3.535 (D) – 3.535 + j 3.535

03. Ans: (D)

Solution:

,

=(Delivered from the formula)

C

K s a 1G s and G s

s b s s 3

K s a

s s b s 3

A

2

0.5

2

A

A

4

s s 2 s s 2 0.5 2 4

S S 2

2

100

(s 8s 100) 2

10

(s 8s 10)

2

100

(s 4s 100) 2

10

(s 8s 100)

n10 rad / sec 0.4

2

n

2 2

n n

TFs 2 s

2

100

s 0.8 10s 100

2

100

s 8s 100

P%M 4.32% 0.0432

n5rad / sec

2

p

22

p

ln M

ln M

21tan

P%M e e

EE Krash (Sample)


So pole location=

04. Match List – I (Location of roots) with List – II (Step response) and select the correct

answer using the codes given below the lists:

P Q R S

(A) 4 1 2 3

(B) 5 1 2 3

(C) 4 2 1 3

(D) 5 2 3 4

0.707

2

n nj 1 3.535 j3.535

EE Krash (Sample)


04. Ans: (B)

Solution:

(P) So CE=

One pole system

(Q) Left hand poles & conjugate poles

Poles under damped system

(R) Marginally stable system

(S) Right hand poles

Unstable system

05. The open loop transfer function of a system is for a unit step input to

the system, the approximate settling time for 2% criterion is

(A) 100 sec (B) 4 sec

(C) 1 sec (D) none

05. Ans: (D)

Solution:

OLTF= so CLTF=

So CE=1+OLTF=(S+1)(S+100)=

Comparing with

We get

For 2% criterion, not in the answer

S 0

100G(s)

(s 1)(s 100)

100

S 1 S 100

OLTF

1 OLTF

2s 101s 2002 2

n nS 2

n101 50.5

2

n

4 4

50.5

EE Krash (Sample)


06. Consider the R-L-C circuit shown in figure, For a step-input ei, the overshoot in the

output eo will be

(A) 0

(B) 5 %

(C) 16 %

(D) 48 %

06. Ans: (C)

Solution:

Here [Converting to standard from]

For RLC circuit

Now peak overshoot= =0.1630=16.3%

07. Consider the block diagram of the control system shown in the figure below. The peak

overshoot of the system is 9.49%. The value of the constant k and the steady state error for a

unit ramp input are

(A) 0.18 and 0.20 (B) 0.36 and 0.36

(C) 0.52 and 0.18 (D) 0.52 and 2.14

07. Ans: (A)

Solution:

1 sk 10G s

s s 2

For unity negative feedback system

2

10 10skT s

s s 2 10k 10

Compare with standard second order transfer function

n10 3.16rad sec

0

i

1R sCTF

1RR sL

sC

2

1

LCR 1

s sL LC

6

3

R C 10 10 10 15 0.5

2 L 2 101 10

8 4

n3 6

1 110 10

LC 10 10 10

21e

EE Krash (Sample)


n2 2 10k

n2 2

k10

Now, 21

PM % 9.49% e

0.6

Thus, k 0.18

For Type one system

Velocity error constant

V s 0

10k limsG s 5

2

ss

V

1e 0.2

k

08. The pole location of one of the two closed loop poles of a second order control system

with unity negative feedback is shown below. The value of peak overshoot PM is ____________

%.

08. Ans: 13.53

Solution:

The standard transfer function of second order system

2

n

2 2

n n

C s

R s s 2

2

n ns 1

Here, n

2

2

n1

2

2

1

21

P%M e 100%

2

e 100%

2e 100% 13.53%

09. The block diagram of a unity feedback control system is shown:

EE Krash (Sample)


The time period of oscillation before reaching the steady state is

(A) 0.6 sec (B) 0.78 sec

(C) 1.56 sec (D) 1.84 sec

09. Ans: (C)

Solution:

20G s

s 1 s 5

For unity negative feedback system

2

G s 20T s

1 G s s 6s 25

Compare it with general second order system 2

1 n2 2

n n

kT s k k

s 2 s

By comparison, 2

n n25 5rad / sec

n

32 6 0.6.

5

2

d n

91 5 1 4 rad / sec.

25

Time period of oscillationd

2.

2 1.56 sec

4 2

EE Krash (Sample)


Bode Plot

For Concept refer to Control System K-Notes, Bode Plots

The asymptotic approximation of the log-magnitude v/s frequency plot of a system containing only real poles and

zeros is shown. Its transfer function is

(A)

(B)

(C)

(D)

Solution: (B) is correct option

Since initial slope of the bode plot is −40 dB/decade, so no. of poles at origin is 2.

Transfer function can be written in following steps:

Slope changes from −40 dB/dec. to −60 dB/dec. at ω1 = 2 rad/sec., so at ω1 there is a pole in the transfer

function.

Slope changes from −60 dB/dec to −40 dB/dec at ω2 = 5 rad/sec., so at this frequency there is a zero lying in

the system function.

The slope changes from −40 dB/dec to −60 dB/dec at ω3 = 25 rad/sec, so there is a pole in the system at this

frequency.

01. The Bode plot of a system is given its transfer function is

(A)

(B)

10(s 5)

s(s 25)(s 2)

2

1000(s 5)

s (s 25)(s 2)

100(s 5)

s(s 25)(s 2)

2

80(s 5)

s (s 25)(s 2)

2

K(s 5)Transfer Function T(s)

s (s 25)(s 2)

at =.1

2

Constant term can be obained as : T(j ) 80

K(5)So, 80 20log => K 1000

(.1) 50

2

1000(s 5)Therefore the Transfer Function is T(s)

s (s 25)(s 2)

(1 0.1s)(1 0.01s)

(1 s)(1 0.001s)

(1 0.1s)(1 s)

(1 0.01s)(1 0.001s)

Problems

Sample Problem

Common Mistake / Point to remember

Remember the x-axis is always in terms of logω and not in terms of ω, this is a common mistake while

calculating slope.

EE Krash (Sample)


(C)

(D)

02. What is the transfer function for given bode plot shown in the figure

(A)

(B)

(C)

(D)

03. The asymptotic Bode plot of the transfer function is given in figure. The error in phase angle

and dB gain at a frequency of ω = 0.5a are respectively

(A) 4.90, 0.97 dB (B) 5.70, 3 dB

(C) 4.90, 3 dB (D) 5.70, 0.97 dB

04. The asymptotic magnitude Bode plot of an open loop system G(s) with K>0 and all poles and zeroes on the s-

plane is shown in the figure. It is completely symmetric about ωc. The minimum absolute phase angle

contribution by G(s) is given by

(A) 78.60

(B) 900

(C) 101.40

(D) 1800

(1 10s)(1 s)

(1 0.1s)(1 0.001s)

(1 0.001s)(1 s)

s(1 10s)

2

3

4(s 0.5)

(s 10)

2

3

16000(s 0.5)

(s 10)

2

3

160(s 0.5)

(s 10)

2

3

1600(s 0.5)

(s 10)

KT(s)

s(1 )

a

EE Krash (Sample)


05. The open loop transfer function of a system is given by in the straight–line

approximation of the Bode plot

(A) (B)

(C) (D)

06. For the bode plot shown in the below figure. The transfer function is __________________.

(A) 1 s

1 s

(B)

2

1

s 1

(C) 1

1 s (D)

2

1

s s 1

07. The asymptotic Bode magnitude plot for open loop transfer function of a unity negative feedback system is

shown in figure. If the input is 2t

u t2

, then the steady state error will be _________.

08. The Bode plot of a transfer function G(s) is shown

The gain 1020log G s

is 18dB and –2dB at 1rad/sec and 10rad/sec respectively. The value of gain at

ω=5rad/sec is _________ dB.

100(s 100)G(s)H(s)

s(s 10)

G(j )H(j ) and G(j ) at =100 rad/sec are

30 dB and rad

4

0 dB and rad

4

320 dB and rad

4

20 dB and rad

4

EE Krash (Sample)


01. Ans: (A)

Solution:

Between 1 and ω1, the change in magnitude is 20dB and since slope is 20dB/dec so between 1 and ω1 there

should be a decade.

Similarly, 1000 and ω2 there should be a decade.

So should be a decade, so &

as there is a decade between 1 &

So s=1 pole

zero

zero

s=1000 pole

So

02. Ans: (B)

Solution:

At slop changes from 0dB to +40dB/dec

So s=0.5 = two zeros

At , slop +40dB -20dB

So s=10=three poles

At , Gain = 36dB

03. Ans: (A)

Solution:

Now at , gain =

So error=0.97dB in gain

2log1000 log

2100

110

1

1

s 10

2

s 100

s s1 1

1 0.1s 1 0.01s10 100TF

s 1 s 1 0.001s1 s 1

1000

0.5

10

2

3

K s 0.5TF

s 10

22

3

K 236 20log

10

k 16000

K KT s T j

s j1 1

a a

0.5a 2 220logK 20log 1 0.5

20logK 0.97dB

EE Krash (Sample)


At , phase=

Now from the plot at

Let phase is

So

So error=31.45-26.56=4.90

04. Ans: (A)

Solution: At it is symmetrical about

So =10

The initial slope of bode plot is -40dB/decade which indicates there are two poles present at origin

The change of slope at 1 is 20dB/decade so there is a zero of order 1 and change of slope at 100 is -

20dB/decade so there is a pole of order 1

So

Now G(s) =

is a lag network

Here minimum phase angle

05. Ans: (A)

Solution:

Initial slope=-20dB/dec

Corner frequencies=100(zero), 10(pole)

Drawing the bode plot

At Gain=DC gain=

So

At , gain=

So 20log1=0 dB

0.5a 1 0tan 0.5 26.56

0.5a0x

00x

45log0.5a log0.1a

0 0x 45 log5 31.45

C

C

2

s 1TF

s s 100

s 1

s 100

1 s 1 100 0.01s 1 aTs0.01 K

1 Ts100 1 0.01s 1 0.01s

1 0a 1sin 78.6

a 1

100 s 100 100 100 1 0.01sG s H s

s s 10 10s 1 0.15

1000 1 0.01s

s 1 0.1s

1

1

G s H s 1000

20log1000 60dB

100

2

100

100 1001

s

1 1 1 3G j tan tan tan 10 2.2564100 10 2 4 2

EE Krash (Sample)


06. Ans: (A)

Solution:

By observing magnitude plot gain = 0 dB

i.e. magnitude = 1

It is possible only in all pass filter.

Hence option (A) is correct

07. Ans: 10

Solution:

Here 1p 20dB,0.1 and 2 1

p 20dB,

2 1

2 1

y y40

x x

10 10 1

20 0

log 0.1 log

10 10 1

1log 0.1 log

2

10

1

1 11 log

2

10

1

1 1log

2

1

2

1

110

1

1

10

The initial slope of bode plot is -40dB/decade which indicates there are two poles present at origin

2 2

K 1G s

s 10s

At 1

1

10

2

1

K20log G s 20log 0dB

1K

10

Error for standard parabolic input ss

a

1e

K

2

a s 0

1K lims G s

10

ss

a

1e 10

K

08. Ans: 4.02

Solution:

Slope

18 220dB / dec

log1 log10

The transfer function is

The initial slope of bode plot is -20dB/decade which indicates there are one pole present at origin

K

G ss

EE Krash (Sample)


K K

G sj

10 10 1020log G s 20log K 20log

At 1

1018 20log K

1820K 10

At 5

18

2010 10 10

20log G s 20log 10 20log 5 10

18 20log 5 4.02dB

EE Krash (Sample)


Signal & Systems

Continuous Time Fourier Transform

For Concept refer to Signals and Systems K-Notes, Fourier Transform

Find the Fourier transform of the signal atx(t) te u(t) : a>0

(A) 1

a j (B)

a

a j

(C) 2(a j )

(D)

2

1

(a j )

Solution: (D) is correct option

j t j t (a j )tat at

0 0

X( ) te u(t)e dt te e dt te dt

(a j )t (a j )t

2

0 0

e eX( ) t

(a j ) (a j )

2 2

1 10

(a j ) (a j )X( )

The Fourier transform of the given signal x(t) is

(A) cos sin

4 j

(B) 2

cos sin2j

Sample Problem

Sample Problem


Remember to consider the impulse function while taking the derivative of the rectangular function at the edges.

EE Krash (Sample)


(C) 2

cos sin4 j

(D)

2

cos sin2j


Method 1

equation of x t t u t 1 u t 1 t 1 u t 1 u t 1 t 1 u t 1 u t 1

Taking Fourier Transfer j j j j

2 2 2

e e e e 1 2X( j ) [ 2jsin ] cos

j j j

2

cos sinX( j ) 2j

Method 2

equation of x t t u t 1 u t 1

Taking Fourier Transform

2

dF.T. d 2sin 2 cos 2sinX( j ) j j ( ) j(

u t 1 u t

d

1)

d

2

cos sinX( j ) 2 j( )

01. The Fourier transform X(ω) of the signal x(t) = sgn(t) is

(A) X( ) 2 / j (B) X( ) 2 / j

(C) (D) X( ) ( ) 1 / j

01. Ans: (A)

Solution:

1 t 0

x t sgn t 0 t 0

1 t 0

can be written as Sgn(t)=u(t)-u(-t)

1u t

j

1u t

j

2X

j

j/2j)(X

Problems

EE Krash (Sample)


02. Let the signal x(t) have the fourier transform X(ω). Consider the signal dd x(t t )

y(t) ,dt

where td is arbitrary delay. The magnitude of the fourier transform of y(t) is?

(A) x( ) (B) x( )

(C) 2 x( ) (D) dj tx( ) e

02. Ans: (A)

Solution: x t X

d

d

j t

d

j t

d

x t t e X

dx t t j e X M Let

dt

M X

03. If a signal x(t)=1+cos(2πft)+cos(6 πft) is fourier transformed , the number of spectral lines

in the Fourier transform will be ?

(A) 3 (B) 4

(C) 5 (D) 6

03. Ans: (C)

Solution: x t 1 cos t cos 3 t

After Fourier transform

0 0 0 0X 2 3 3

Hence total spectral lines =1+2+2=5

04. if jtFTx(t) X( ) the F.T. of y(t)=x(t-1)e is

(A) j( 1)

X( 1)e

(B) j( 1) X( 1)e

(C) j( 1) X( )e

(D) j

X( 1)e

04. Ans: (A)

Solution: x t X

j

j 1jt

x t 1 e X Y

e x t 1 Y 1 e X 1

05. Fourier transform of t

f ba

is

(A) jab1e F(a )

a

(B)

jaba e F(a )

(C) jab

a e F(b )

(D) jab1e F(b )

a

EE Krash (Sample)


05. Ans: (B)

Solution: t

f ba

jb

Let f t F

Y f t b e F .....................(1)

1

Note; If f t F Then f at Faa

jab

From Equation (1)

tf b a e F a

a

06. Determine the time signal corresponding to the magnitude and phase spectra as shown

in the figure with ω0=π

(A) x(t) 4sin 4 t 2cos 3 t8 4

(B) x(t) 4sin 2 t 2cos 3 t8 4

(C) x(t) 4cos 4 t 2cos 3 t8 4

(D) x(t) 4cos 2 t 2sin 3 t8 4

06. Ans: (C)

Solution: Given 0

Thus M can be written as

0

j jj j8 84 4

0 0 0 0

j jj j8 84 4

M M M n

M 1.e 3 2e 4 1.e 3 2e 4

M e 3 e 3 2e 4 2e 4

Taking inverse Fourier transform we will have

EE Krash (Sample)


j jj j

j3 t j3 t j4 t j4 t8 84 4M t e .e e .e 2e .e 2e .e

j 3 t j 3 t j 4 t j 4 t

4 4 8 8M t e e 2 e e

M t 2cos 3 t 4cos 4 t4 8

07. The signal x(t) is shown as

Then the inverse Laplace transform of X2(s) is?

07. Ans: (D)

Solution: Given x(t)

EE Krash (Sample)


Now 1L X s X s x t * x t

1L X s X s

08. For the signal shown below

(A) Only Fourier transform exists

(B) Only Laplace transform exists

(C) Both Laplace and Fourier transforms exists

(D) Neither Laplace nor Fourier transforms exists

08. Ans: (B)

Solution: Given ty t e u t

Fourier transform will not exist as y t dt

Laplace will exist for 1 0 => >1

09. Find the Inverse Fourier transform of X(ω) for the magnitude and phase spectra of X(ω)

(A) 0

A1 cos t

(B) 0

A1 cos t

t

(C) 0

A1 sin t

(D) 0

A1 sin t

t

09. Ans: (B)

Solution: j t1x t X e d

2 j

EE Krash (Sample)


0 0

0 0

0 0j jj t j t j t j t2 2

0 0

1 1x t X e dt X e d Ae e d A e e d

2 j 2 j

0

0 0

0

0j t j t j tj t

j j j j02 2 2 2

e 1 e e 1eA Ax t e . e . e e

2 j jt jt 2 j jt jt

0 0j t j tj j

2 22 2 02jsin 2jsin tA e e e e A 2 2

x t2 j jt 2 j jt

0

Ax t 1 cos t

t

10. Find the inverse Fourier transform of the spectra F(ω) depicted below

(A) 24sinc (t)cos4t

(B) 22

sinc (t)cos4t

(C) 22sinc (t)sin4t

(D) 24

sinc (t)sin4t

10. Ans: (B)

Solution: Since we know that

F.Ttx(t) A

T

AT 2 T

sinc2

Thus using duality principle

2 F.TtTAT sinc 2 x

2

For T=2 and A=1

EE Krash (Sample)


2 F.T2sinc t

2 F.T1sinc t

2 F.T2x(t) sinc t

Now, the given graph of F can be obtained as

4 jt 4 jt

IFTX 4 X 4 x t e x t eF x t cos4t

2 2

22f t sinc t cos4t

11. Determine the Fourier transform for the waveform shown below

(A) 2 jsin(2 f)(3 e )

f

(B) 2 jsin(2 f)(3 e )

f

(C) 2 jsin(2 f)(3 e )

f

(D) None

11. Ans: (A)

Solution:

Note:- A sinc2

EE Krash (Sample)


f t Can be break as shown

t3rect

2

t

rect 22

3 2sinc 6sinc 2 f j21 2e sinc 2 f

j2 j2 j2

j2

sin2 fF 6sinc 2 f 2e sinc 2 f 2sinc 2 f 3 e 2 3 e

2 f

sin2 fF 3 e

f

EE Krash (Sample)


Engineering Mechanics

Rectilinear Motion

The initial velocity of an object is 40 m/s. The acceleration ‘a’ of the object is given by the

following expression.

a = -0.1 v, Where v is the instantaneous velocity of the object. The velocity of the object after

3 seconds will be _____________

Solution: 29.63 m/s

U 40 m / s a 0.1 V V ? T 3 s Using expression for acceleration

dva 0.1 v

dt

Rearranging and integrating both sides

v 3

40 0

dv0.1dt

v

v

40lnv 0.1 [t]

ln v – ln 40 0.1 3.0 0.3

ln v ln 40 0.3 ln v 3.38887

3.38887v e 29.6327 m/ s

01. A stone is thrown upto a slope of inclination at a speed of V and an angle (beta) to

slope. The stone reaches its greatest distance from the slope after a time of

(a) Vsin

gcos

(b)

Vcos

gsin

(c) Vsin

gcos

(d)

Vcos

gsin

01. Ans: (c)

Solution:

Problems

Sample Problem

EE Krash (Sample)


Take the reference planes as shown

This ‘ gcos ’ will affect the vertical component of the projectile on this reference plane.

i xV Vcos

; i YV Vsin

f f fx YV V ; V 0

Along Y-direction

f i YYV V at

0 Vsin gcos t (Taking -ive sign with gcos it is acting in downward direction)

Vsint

gcos

EE Krash (Sample)


Free body diagram

A mass 35 kg is suspended from a weightless bar AB which is supported by a cable CB and a

pin at A as shown in figure. The pin reactions at A on the bar AB are

(a) Rx = 343.4 N, Ry =755.4 N (b) Rx = 343.4 N, Ry = 0

(c) Rx = 755.4 N, Ry = 343.4 N (d) Rx = 755.4 N, Ry = 0

Solution: (d) is correct option

yT sin R mg

xT cos R

Now, 125

tan275

24.44 Taking moment about A

Tsin mg l l 35 9.81

T 829.74sin 24.44

Rx = 755.4

Ry = 0

01. The collars A and B hang on vertical frame composed of two smooth rods. If the mass of

collar A is 10 kg and the mass of collar B is 5 kg, determine the equilibrium angle (in

degree)

Problems

Problems

EE Krash (Sample)


01. Ans: 24-26

Solution:

Construct free body diagram.

For equilibrium of A

o

ATcos30 N cos30 0 .......(1)

o

A ATsin30 N sin30 W 0 .......(2)

For equilibrium of B

o

BTcos N cos45 0 ........(3)

o

B BTsin N sin45 W 0 .......(4)

From (1) and (3), NA = 0.816NB

From (2) and (4) o o

A B A BN sin30 N sin45 W W A B

g M M 147.15

After putting value of NA

o o

B0.816 sin30 sin45 N 147.15

NB = 131.96 N

NA = 107.679 N

From (1) oTsin 107.679 cos30 = 93.2533

From (2) o

A ATsin W N sin30 o10 9.81 107.679 sin30 44.2601

44.2601tan

93.2533

= 25.39o

EE Krash (Sample)


Analog Electronics

DC Analysis and Biasing Circuits of BJT

For Concept, refer to Analog Electronics K-Notes, Transistor Biasing and Transistor Amplifiers

The transistor used in the circuit shown below has a β of 30 and ICBO is negligible. If the

forward voltage drop of diode is 0.7 V, then the current through collector will be

(A) 168 mA (B) 108 mA

(C) 20.54 mA (D) 5.36 mA4

Solution: (D) is correct option

Assume that the transistor operated in active region then apply KVL to base-emitter loop.

5-103IB-0.7-0.7+12=0

IB=15.6 mA

IC=0.468 A

Apply KVL to collector-emitter loop

0-2.2K IC - VCE+12=0

VCE=-2200 IC+12=-1017.6 V

As 0<VCE<VCC Not satisfying this condition. Transistor operating in saturation region.

V+(sat)=0.2 V

Apply KVL, 0-2.2IC-0.2+12=0

IC=5.36 mA

Sample Problem

Common Mistake / Point to remember Verify the region of operation of Transistor before calculating the Transistor parameters. While calculating

IC,VCE,VC,IB,IE, open circuit all capacitors and disable AC sources.

For pnp transistor, the analysis remains same as npn transistor but polarity of voltage & direction of

currents are reversed.

EE Krash (Sample)


01. For the circuit shown in the given figure, the quiescent point is

(A) 12V, 5mA (B) 12V, 2mA

(C) 11V, 2mA (D) 10V, 5mA

01. Ans: (C)

Solution:

Let transistor is active region

Assume VBE = 0.7V

IE = 30 0.7

15

= 1.95mA

IE = IC (If large)

Apply KVL

20-5IC – VCE – 15IC+30=0

VCE= 50-20 1.955 = 10.9V

02. Assertion (A): A self-biased BJT Circuit is more stable as compared to a fixed biased one

Reason(R): A self-biased BJT circuit uses more components as compared to a fixed biased

one

(A) Both A and R are true and R is the correct explanation of A

(B) Both A and R are true but R is NOT the correct explanation of A

(C) A is true but R is false

(D) A is false but R is true

02. Ans: (B)

Solution:

For fixed bias stability (S) = + 1

For self bias stability (S) = E

B E

1

R1

R R

(S) self bias < (S) fixed bias

S 1 always

Problems

EE Krash (Sample)


03. For the transistor circuit shown below, the value of the output voltage Vo is

(A) 0.2 V (B) 2.5 V

(C) 4 V (D) 5 V

03. Ans: (C)

Solution: c EI I

1

c EI I 1mA

Thus,

0 c

0

5 V I 1K

V 5 1 4V

04. For the transistor circuit shown below hfe = = 100, hie = 1 k. The DC voltage between

the collector and emitter terminals is

(A) 2.52 Volts (B) 5 Volts

(C) 6.207 Volts (D) 7.48 Volts

EE Krash (Sample)


04. Ans: (C)

Solution: Given β=100

x

E x

2V 10 2V..............(i)

2 8

V V 0.7 1.3V

Thus, E

E

VI 1.3mA

1K

Since β=100

C EI I 1.3mA

Now, we can write

C CE E

CE c

10 2 I V 1 I

V 10 3I 6.1V

DC equivalent circuit

05. In the transistor circuit given that = 100, VBE = 0.7 The value of Quiescent collector

current & voltage are

(A) (0.589 mA, 18.22 V) (B) (0.917 mA, 11.56 V)

(C) (11.56 V, 0.589 mA) (D) (18.22 V, 0.917 mA)

05. Ans: (B)

Solution: Drawing Thevenin’s equivalent circuit,

th th

10 10 20 5 10V 0V and R 10 || 5 k k

10 10 3

Applying KVL,

th b b

th b

10 0.7 R I 1 I 10K

9.3 R 101 10K I

c b

100 9.3I I 0.917mA

10101 10

3

CE C E E c

1Now,V 30 10I 10I Where I I

CE

101V 30 9.17 9.17 11.56V

100

EE Krash (Sample)


06. In the Series voltage regulator circuit. Given that VZ = 10 V, VBE = 0.7 V, =50. The value

of VCE = ……………….

(A) 20 V (B) 10 V

(C) 9.3 V (D) 10.7 V

06. Ans: (D)

Solution: 20 10

I 10mA1K

BE 0also, V 0.7 10 V 0.7

0V 9.3V

CE 0V 20 V 10.7V

07. In the amplifier circuit shown in the figure, the values of R1 and R2 are such that the

transistor is operating at V 3V and I 1.5mACE C

when it’s is 150. If the of the transistor

is 200, then the operating voltage VCE

(in volts) is

(A) 2 V (B) 3 V

(C) 4 V (D) 5 V

EE Krash (Sample)


07. Ans: (A)

Solution: The given circuit is in fixed bias stabilization method. Apply KVL to outer loop then

V V6 3 3CC CEV I R V 0, R 1000 2k

CC C C CE C I 1.5mA 1.5C

And we know that I IC B

I1.5mAcI 10 A

B 150

Apply KVL to inner loop

V I R V 0. or active region V 0.7CC B B BE BE

V V6 0.7 6CC BER R 0.53 10 530K

B 1 I 10 AB

Then, given is changed from 150 to 200.

So, I I 200 10 A 2mAc B

By applying KVL to outer loop we can get VCEQ

V I R V 0 V 6 2 2 6 4 2VCC C C CEQ CEQ

08. For the transistor in figure, value of voltage VEC

is __________________ (V). 50

(A) 5.025V (B) 6.074V

(C) 4.125V (D) None

08. Ans: (B)

Solution: I 1mA; 50E

50I I 1mA 0.98mAC E1 51

V I R 9 0.98 4.7 9 4.394VC C C

I 1mAEI 19.6 A, V V VB EB E B1 51

EE Krash (Sample)


3V V V 0.7 50 19.6 10 1.68VE EB B

V V V 1.68 4.394 6.074VEC E C

09. In the circuit shown in figure, the transistor has current gain of 100 and VBE = 0.7V. The

current flowing through 5 k load (in mA) is _________________.

09. Ans: 0.7825

Solution:

Step (1):

KVL for B-E loop:

B B BE E E EE0 I R V I R V 0

B B B E EE BEI R 1 I R V V

B

5 0.7I

10k 101 5k

8.3 A

C BI I 0.835 mA

Step (2):

KCL at node ‘C’

C LI I I

0 012 V V

0.835m5k 5k

0

1 12.4m 0.835m V

5k 5k

0

1.56m 5kV

2

3.9125V

0

R

VI

5k 0.7825 mA

EE Krash (Sample)


10. Consider the circuit shown in figure. If the β of the transistor is 30 and CBO

I is 20nA and

the input voltage is +5V, the transistor would be operating in

(A) Saturation region (B) Active region

(C) Breakdown region (D) Cut-off region

10. Ans: (B)

Solution:

Assume BJT in active region and we neglect CBO

I and drawing Thevenin’s equivalent at input

side

Th

12 15k 5 100kV 2.78V

115k 115k

th

15k 100kR 15k || 100k 13k

115k

c bI I .................. (i)

th BE

b

th

V VI

R

But b 3

2.78 0.7I 1.6 A

13 10

Using bI in eq. (i) with 30

cI 4.8mA

But CC CE

c

C

V VI

R

But CE

V 12 2.2k 4.8m 1.44

As CE

V 0.2V

BJT is in active region

EE Krash (Sample)


Electrical Machine

Equivalent Circuit of Transformer

For Concept, refer to Electrical Machines K-Notes, Transformers

A Single-phase transformer has a turns ratio 1:2, and is connected to a purely resistive load

as shown in the figure. The magnetizing current drawn is 1 A, and the secondary current is 1

A. If core losses and leakage reactance’s are neglected, the primary current is

(A) 1.41 A

(B) 2 A

(C) 2.24 A

(D) 3 A

Solution: (C) is correct option

I0 = 1 amp (magnetizing current)

Primary current IP = ?

I2 = 1 A

I2p = secondary current referred to Primary

2 2

p

p

2

2

o p

21 2 amp

1

I i i 1 4 2.24 amp

I

01. Across the HV side of a single phase 200 V / 400 V, 50 Hz transformer, an impedance of

32 + j24Ω is connected, with LV side supplied with rated voltage & frequency. The supply

current and the impedance seen by the supply are respectively:

(A) 20 A & (128 + j96)Ω

(B) 20 A & (8 + j6)Ω

(C) 5 A & (8 + j6)Ω

(D) 20 A & (16 + j12)Ω

Problems

Sample Problem


Sometimes we add primary current to no-load current as scalar though they must be added as phasors.

EE Krash (Sample)


01. Ans: (B)

Solution: 2

Z 32 j24

T

400a 2

200

So load impedance seen by primary side 2

2 2

T

ZZ 8 j6

a

Current at supply side200

8 j6

020 36.86

02. Consider the circuit shown below

If the ideal source supplies 1000W, half of which is delivered to the 100 Ω load, then the

value of b is____

(A) 1.5 (B) 0.89

(C) 0.56 (D) 0.67

02. Ans: (B)

Solution:

Since source is supplying 1000W

So current from voltage source 1

1000I 10A

100

Since 500W is delivered to 100Ω resistor 2

3 3I 100 500 I 5A

Power consumed in 4Ω resistor= 2

10 4 400W

Total power consumed=Power consumed is 4Ω+power consumed in 25Ω+Power consumed

in 100Ω

Power consumed in 25Ω=1000-500-400=100W 2

2I 25 100

2I 2A

EE Krash (Sample)


Since 2

3

I b

I 1

2

3

I 2b 0.89433

I 5

03. A 1200/300, turns transformer when loaded, current on the secondary is 100A at 0.8

power factor lagging and primary current is 50A at 0.707 power factor lagging. Determine

the no-load current of the transformer with respect to the voltage?

(A) 25.5A (B) 26.5A

(C) 27.5A (D) 28.5A

03. Ans: (A)

Solution: T

1200a 4

300

Taking voltages at reference phasor

Then 1

2I 100 cos 0.8 0100 36.86

0

2

2

I 100 36.86I

a 4

025 36.86 A

1

1I 50 cos 0.707

050 45 A

No load current 0 1 2I I I 0 0 050 45 25 36.86 25.5 52.97

EE Krash (Sample)


Salient Pole Synchronous Machines

For Concept, refer to Electrical Machines K-Notes, Synchronous Machines

The direct axis and quadrature axis reactance’s of a salient pole alternator are 1.2 p.u and 1.0

p.u respectively. The armature resistance is negligible. If this alternator is delivering rated kVA

at upf and at rated voltage then its power angle is

(A) 30° (B) 45°

(C) 60° (D) 900


Power angle for salient pole alternator is given by

t a q

t a a

V sin I Xtan

V cos I R

Since the alternator is delivering at rated kVA rated voltage

a tI 1 pu , V 1 pu,

00 ,sin 0, cos 1

0

q d

Since 0 , =

X = 1 pu, X = 1.2 pu

0

1

45

1 0 1 1tan

1 0

01. A 1500KVA, star connected 6.6KV salient pole synchronous motor has Xd = 23.2 and

Xq= 14.5/ph respectively. Armature resistance is negligible. The excitation emf (line value)

when the motor is supplying rated load at 0.8 lead P.F. is.

(A) 6.1KV (B) 10.57KV

(C) 15KV (D) 18.3KV

Problems

Sample Problem


While using EMF – Terminal Voltage Equation direct axis and quadrature axis currents, make sure that you

treat them as phasors and use their phase angles as well.

EE Krash (Sample)


01. Ans: (B)

Solution: S=1500KVA, V=6.6kV, d qX 23.2 & X 14.5

a fr 0,e ? at 0.8 lead p.f

t a

6.6K 1500KV 3810,I 131A

3 6.6K 3

f t d de V cos I X

d aI I sin

t a q

t a a

V sin I Xtan

V cos I r

3810 0.6 131 23.21.747

3810 0.8

060.214

36.87 0

d

23.34 ,

I 131sin60.214 113.69

fe 3810cos 23.34 113.61 23.2 6135V

fL L fe 3e 10.62kV

02. A 50-Hz, 3-phase, 480-V, delta connected salient pole synchronous generator has xd =

0.1 ohm and xq = 0.075 ohm. The generator is supplying 1200 Amp. at 0.8 p.f. lagging. The

load angle is

(A) 41.5o (B) 14.5o

(C) 4.6o (D) 6.4o

02. Ans: (C)

Solution: V=480V, d qX 0.1 ,X 0.075

aI 1200A at 0.8 lag

a/p

1200I 692.82A

3 [delta connection]

t a q

t a a

V sin I Xtan

V cos I r

480 0.6 692.82 0.075.885

480 0.8

041.5 04.6

03. A 10KVA, 380V, 50Hz, 3- star connected salient pole alternator has direct axis and

quadrature axis reactance’s of 12 and 8 respectively. The armature has a resistance of 1

per phase. The generator delivers rated load at 0.8 p.f lagging with the terminal voltage

being maintained at rated value. Find the direct axis component of armature current, if the

load angle is 16.15

(A) 12.14A (B) 9.14A

(C) 14.21A (D) 5.37A

EE Krash (Sample)


03. Ans: (A)

Solution: 10KVA, 380V, d q aX 12 & X 8 ,r 1

Pf=0.8 lag, t

V Constant , 0

t

38016.15 ,V 220V

3

a

10KI 15.193A

380 3

220 .6 15.193 8tan 1.326

220 .8 15.193 1

053

d aI I sin 12.14A

04. In a 3 phase, salient pole synchronous generator with the ‘a’ phase terminal voltage of0

V 0 volts, the phase current delivered by the machine is 30A, lagging the terminal voltage by

300. The induced emf per phase is at angle 150 from the terminal voltage. The currents Id and

Iq are, respectively

(A) 0 015 90 A, 26 0 A (B) 030 30 A, 0A

(C) 0 029 15 A, 7.76 105 A (D) 0 021.21 75 A, 21.21 15 A

04. Ans: (D)

Solution: aI 30A , p.f=0.866lag,

0

q d15 ,I and I ?

0 0 030 15 45

q aI I cos 21.21A

d aI I sin 21.21A

w.r.t V, 0

qI 21.21 15 [aligned with

fe ]

0

dI 21.21 75 [

fto E ]

EE Krash (Sample)


Power Electronics

Buck Converter

For Concept, refer to Power Electronics K-Notes, Chopper

A buck converter, as shown in Figure (a) below, is working in steady state. The output voltage

and the inductor current can be assumed to be ripple free. Figure (b) shows the inductor

voltage during a complete switching interval. Assuming all devices are ideal, the duty

cycle of the buck converter is ___________.

Solution: 0.4

Voltage wave shape across inductor is shown below,

vL

Sample Problem


For approximate ripple in Inductor Current we can directly use the concept of Voltage across the inductor

so no need to memorize the formulas for maximum and minimum current in chopper.

Instead of memorizing expression for different parameters of converter, better understand the method to

derive such parameters.

For discontinuous conduction replace D by D

in every expression.

EE Krash (Sample)


Since average voltage across the inductor should be zero over a cycle,

Duty cycle

Therefore,

01. A step down chopper feeding an inductive load and current is ripple free in the output

circuit. The ratio of average currents through freewheeling diode and chopper is ____ (Take

duty cycle D=0.8)

(A) 0.25 (B) 4

(C) 2 (D) 0.5

01. Ans: (A)

Solution: Chopper current ch 0I DI

FD current F.D 0I 1 D I

F.D.

ch

1 DI 1 0.80.25

I D 0.8

02. Type-A chopper supplies power to a resistive load of 10 from 200 V dc source. The

output voltage is ON for 4 ms duration and OFF for 6 ms duration. The rms value of

fundamental component of output voltage is

(A) 80 V (B) 126.5 V

(C) 90 V (D) 85.6 V

02. Ans: (D)

Solution: on

on off

T 4D 0.4

T T 4 6

For Type A chopper

RMS value of fundamental component of output voltage

dc

01

2V 2 200V sin D sin 0.4 85.625V

2 2

L avgV 0

T

L

0

V dt 0

on offT T

0 0

30dt 20dt 0

on off30T 20T 0

off on

3T T

2

on

total

TD

T on

on off

T

T T

on

on on

T

3T T

2

20.4

5

duty cycle 0.4

Problems

EE Krash (Sample)


03. A step down DC chopper feeding a resistive load of R = 20 and input voltage =220 V.

When the chopper remains ON, it’s voltage drop is 1.5 V and chopping frequency is 10 KHz.

What is the chopper efficiency?

(A) 99.3 % (B) 89%

(C) 99% (D) 98.3%

03. Ans: (A)

Solution: i/p S 0 0

i/p S 0

P loss DV I D 1.5 I 218.599.3%

P DV I 220

Here, source conducts for DT where T is switching time period.

04. A DC chopper circuit is connected to a 120 V DC source supplies to inductive load having

50mH in series with a resistance of 6. A Freewheeling diode is placed across the load. The

load current varies between the limits of 12 A and 16 A. Time ratio (Ton / Toff) of chopper is

(A) 2.122 (B) 1.222

(C) 2.333 (D) 1.732

04. Ans: (C)

Solution: For a Buck converter

max min

0 avg

I I 12 16I I 14A

2 2

0 avg avg

0 S

V V I R 14 6 84

V DV

ON

OFF

84 120 D =>D 0.7

T DT D 0.72.33

T 1 D 1 0.71 D T

05. A simple D.C chopper is operating at a frequency of 4 kHz from a 100V DC source to

supply a load resistance of 6. The load time constant is 6ms, average load voltage is 62 V.

Assume load current variations are linear. Magnitude of ripple current is

(A) 0.127 A (B) 0.164 A

(C) 0.237 A (D) 0.282 A

05. Ans: (B)

Solution: For a Buck converter

0 SV DV => 62 D 100

D 0.62

Load Time constant 3 3L6 10 =>L 36 10

R

Magnitude of ripple S

V 1 D DTI

L

3 3

100 1 0.62 0.62I 0.164A

36 10 4 10

EE Krash (Sample)


06. The chopper shown in figure operates at a frequency of 100 Hz. The average value of

load current is 25A, with a peak to peak ripple of 6A. Average value of source and diode

currents are

(A) 10A, 8A (B) 12.5A,12.5A

(C) 13.7A, 7.4A (D) 9A, 7A

06. Ans: (B)

Solution: Accordingly to given load current

0 oV I R 25 4 100V

For Buck converter

0 SV DV => 100=D200

D 0.5

For D=0.5 both currents are equal

Average source current=average diode current=0 0

DI I 0.5 12.5A

07. A battery with it’s terminal voltage of 200 volts is supplied with power from type-A

chopper circuit. The output voltage of the chopper consists of rectangular pulses of 2 ms

duration in an overall cycle time of 5 ms. The average, rms values of output voltage, Ripple

voltage and ripple factors are respectively -------

(A) 80 V, 126.49 V, 149.66 V, 1.87

(B) 80 V, 126.49 V, 97.91 V, 1.87

(C) 80 V, 126.49 V, 149.66 V, 1.581

(D) 80 V, 126.49 V, 97.91 V, 1.2247

07. Ans: (D)

Solution: 2

D 0.45

0 S

or S

V DV 0.4 200 80V

V DV 0.4 200 126.44V

Ripple voltage= 2 2 2 2

or oV V 126.44 80 97.91V

Ripple factor=1 1

1 1 1.2247D 0.4

EE Krash (Sample)


08. A load is operating in a chopper circuit as shown in the figure. It has a duty cycle of 0.4

and chopping frequency is 150 Hz. (consider continuous operation). The value of E is _______.

(A) 70V (B) 80V

(C) 60V (D) 50V

08. Ans: (B)

Solution: Under steady state condition the average voltage drop across L is zero.

So, in that case

in

s

V E

E DV 0.4 200V 80V

09. Given, s 0

V 20V, V 50V, R 50 , D 0.5

Find average inductor current and ripple in inductor current.

(A) 2.5A, 5A

(B) 2.5A, 4A

(C) 2.5A, 6A

(D) 6A, 2.5A

09. Ans: (C)

Solution:

s

0

V 20V 40V

1 D 1 0.5

, 0

V given 50V 40V (discontinuous)

EE Krash (Sample)


s

0

V 20V 50

D 0.5

30 25

25 5

30 6

Average inductor current

C on C offI T I T 0

0 L 0I DT I I D T 0

0 L L 0 0I DT I T I DT I T I DT 0

L L 0I I D I

0

L

0

0

II

D

V 50I 1 A

R 50

L L

1 5/6I 2.5A I avg

5/6 0.5

L

L avg

1I T

2IT

L

12.5 I

2

L

2.5 2I A 6A

5/6

10. Given, s 0

V 20V, V 50V, R 50 , D 0.5

Find average diode and switch current.

(A) 2A, 2.5A

(B) 1A, 1.5A

(C) 2.5A, 2A

(D) 1.5A, 1A

10. Ans: (B)

Solution:

s

0

V 20V 40V

1 D 1 0.5

0V given 50V 40V , discontinuous mode.

EE Krash (Sample)


s

0

VV

D

2050

0.5

5

6

Applying Ampere-sec balance

C on C offI T I T 0

0

0

V 50I 1A

R 50

0 L 0I DT I I D T 0

0

L

51I 6I 2.5A

5D0.5

6

L

L avg

1I T

2IT

L

12.5 I

2

L

I 6 A

Average diode current

L

1I D T

1 52 6 0.5 1 AT 2 6

Average switch current

L

1I DT

12 6 0.5 1.5 AT 2

11. A boost converter has input voltage of 5V and average output voltage of 15V. The

switching frequency is 20 KHz and load resistance is 15 .

Find ripple in capacitor voltage C 20 F

(A) 1.5V (B) 1.6V

(C) 1.67V (D) 1.95V

11. Ans: (C)

Solution:

s

0

V 5 1 2V 15 1 D D

1 D 1 D 3 3

Load current 15

1 A15

.

Ripple in capacitor voltage

C

C 0

DVdvC I C I

dt DT (discharging)

0

C 6 3

21I D 23V 1.67V

Cf 3 0.420 10 20 10

Documents

K-NotesKrash+Sample+(2018).pdf · 1 K 11 f f o 1 G s H s k magnitude criterion k S 11 G s H s 2i 1 K0t o odd multiples of 180 angle critq erion SG s H s 2i 11 d K 0 even multiples