aeroplan damages

here u can see when anything strucks in the motor of aeroplane.

DSP

DIGITAL SIGNAL PROCESSING(EC 1361)
1. Define signal?
Any physical quantity that carries information varies with other independent or
dependent variables.
2. What are the main types of signals with respect to time as independent variable?
Continous time (analog) signals &discrete time (discrete) signals
3. What is analog signal?
The analog signal is a continous function of independent variabls.The analog
signal is defined for every instant of independent variable and so magnitude of
independent variable is continous in the specified range.here both the independent
variable and magnitude are continous.
4. What is discrete signal?
The discrete signal is a function of discreted independent variabls.The
indindependent variable is divided into uniform interval and is represented by an
integer,The discrete signal is is defined for every integer value of independent
variable.here both the values of signal and independent variable are discrete.
5. What is digital signal?
The digital signal is same as discrete signal except that the magnitude of signal is
quantised.
6.What are the different types of signal representations?
a. Graphical representation
b. Functional representations
c. Tabular representation
d. Sequence representation
7. Define periodic and non periodic discrete timesignals?
If the discrete time signal repeated after equal samples of time then it is called
periodic signal.When the discrete time signal x[n] satis fies the condition
x[n+N]=x(n), then it is called periodic signal with fundamental period N samples.if
x(n) x(n+N) then it is called nonperiodic signals.
0 __'HILQH_XQLW_VDPSOH_VHTXHQFH"WKH_XQLW_VDPSOH_VHTXHQFH_ _Q__DQG_LV_GHILQHG_DV__________________________________________ _Q_ _____IRU_Q _
0 for n_ __ _Q_
•1
n
9. .Define unit step sequence
A unit step sequence is denoted as u(n)=1 for n._
u(n) 0 other wise
1
…….. n
10.Define unit ramp sequence?
A unit ramp sequence is defined as r(n)= n for n._
0 other wise
r(t)
t
11..Define a system?
A system is a physical device or algorithm that performs an operation on the
signal
12.What is digital signal processing?
The dsp refers processing of signal by digital system.
13. What are the steps involved in digital signal processing?
a. Converting the analog signal to digital signal ,which is performed by A/D
converter
b. Processing the digital signal by digital systems.
c. Converting the digital output signal from the digital system to analog
signal by D/A converter.
14. What are the advantages of DSP?
a. The programme can be modified easily for better Performance.
b. Better accuracy can be achieved by using adaptive algorithm.
c. The digital signal can be easily stored and transported.
d. Digital systems are cheaper than analog equallent.
15. Give some applications of DSP?
a. Speech processing
b. Communications
c. Biomedical
16. Write the difference equation governing the Nth order LTI system.
N M
Y(n)=ak y(n-k) +bk x(n-k)
k=1 k=0
a. N is the order of the system
b. ak & bk are constant coefficients
c. y(n)&x(n) are output and input to the system
17. List the various methods of classifying discrete time systems?
a. Static and dynamic systems.
b. Time invariant and time variant
c. Linear and nonlinear
d. Causal and noncausal
e. Stable and unstable
f. FIR and IIR systems
g. Recursive and non recursive systems
18. What are static and dynamic systems? Give examples?
A discrete time system is called static(memory less)if it’s output at any instant
n dependent on the input sample at the same time (but does not depend on past or
future samples).If the response depends on past or future samples, then the system
is called dynamic system.
Eg.y(n)=ax(n) static system
Y(n)=ax(n)+bx(n-1)
19. Define time invariant system?
A system is said to be time invariant if it’s input output characteristics does
not change with time. Let H be a system and H{X(n)}=Y(n).now if H{X(nk)}=
Y(n-k) then the system H is called time invariant.
20. What is linear and nonlinear systems?
If a system satisfies superposition and homogeneity principles then the system
is called linear otherwise it is called nonlinear
If H is a system,X1(n) and X2(n) are inputs a and b are constants then
H{aX1(n)+BX2(n)}=aH{X1(n)}+bH{X2(n)} then His linear.
21. What is a causal system give an example?
A system is said to be causal, if the output of the system at any time n depends
on present and past inputs ,but does not depend on future inputs.
Eg.y(n)=x(n)+x(n-1)
22. Define a stable system?
Any relaxed system is said to be bounded input bounded output stable if and
only if every bounded input yields a bounded output.

h(n)< where h(n)is impulse response of the system
n=-
23. What is LTI system?
A linear time invariant system is defined that a system obeys both linearity
and time invariant properties.
If a system satisfies superposition and homogeneity principles then the system
is called linear
A system is said to be time invariant if it,s input output characteristics does
not change with time.
24. What are FIR and IIR systems?
FIR (finite impulse response):this type of system has an impulse response
which is zero outside the finite time interval eg. h(n)=0 for n<0 and n>N
IIR (Infinite Impulse Response):An IIR system exhibits an impulse response
of infinite duration.
25. State sampling theorem.
A band limited continous time signal ,with higher frequency fc Hz can be
uniquely recovered from it’ s samples provided that the sampling rate F>2fc
samples per second.
26. Show whether the system is linear?
Y(n)=n x(n)
H{aX1(n)+BX2(n)}=a H{X1(n)}+b H{X2(n)} then H is linear.
a H{X1(n)}+b H{X2(n)}=anx1(n)+bnx2(n) ------------(1)
H{aX1(n)+BX2(n)}= anx1(n)+bnx2(n) --------------(2)
(1)=(2) So the system is linear.
27. Show whether the system is linear?
Y(n)=nx2(n)
Since x2(n) term is present in the system which implies non linearity in to the
system. Therefore the system is nonlinear.
28. Determine if the following system is time invariant or time variant?
Y(n)=x(n)+x(n-1)
If the input is delayed by k units in time we have y(n,k)=H{x(n-k)}=x(nk)+
x(n-k-1)
If we delay the output by k units then y(n-k)= x(n-k)+x(n-k-1)
So the system is time invariant.
29. Determine if the system described by the following equation is causal or not?
Y(n)=x(n2)
For n = -1
Y(-1)=x(1)
For n = 2 Y(2) = x(4)
Therefore the output of the system depends on future input and hence the
system is non causal.
30. Define unit sample response of a system and what is it’ s significance?
The response of a system denoted as h(n),obtained from a discrete time
system when the input signal is a unit sample sequence is known as unit sample
response.
31. Define z transform?
The Z transform of a discrete time signal x(n) is defined as
.
X(z) = ._[_Q_]-n
n= -.
where z is a complex variable. In polar form z=re-jw
32. What is meant by ROC?
The region of convergence (ROC) is defined as the set of all values of z for
which x(z) converges.
33. Explain about the roc of causal and anti-causal infinite sequences?
For causal system the roc is exterior to the circle of radius r.
For anti causal system it is interior to the circle of radius r.
34. Explain about the roc of causal and anti causal finite sequences
For causal system the roc is entire z plane except z=0.
For anti causal system it is entire z plane except z=._
35. What are the properties of roc?
a. The roc is a ring or disk in the z plane centered at the origin.
b. The roc cannot contain any pole.
c. The roc must be a connected region
d. The roc of an LTI stable system contains the unit circle.
36. Explain the linearity property of the z transform
If z{x1(n)}=x1(z) and z{x2(n)}=x2(z) then z{ax1(n)+bx2(n)}=ax1(z)+bx2(z)
a&b are constants.
37. State the time shifting property of the z transform
If z{x(n)}=x(z) then z{x(n-k)}=z-kx(z)
38. State the scaling property of the z transform
If z{x(n)}=x(z) then z{anx(n)}=x(a-1z)
39. State the time reversal property of the z transform
If z{x(n)}=x(z) then z{x(-n)}=x(z-1)
40. Explain convolution property of the z transform
If z{x(n)}=x(z) & z{h(n)}=h(z) then z {x(n)*h(n)}=x(z)h(z)
41. Explain the multiplication property of z transform
If z{x(n)}=x(z) & z{h(n)}=h(z) then ]_^[_Q__K_Q_` ____ M_.c_[_ _K_]_ __ -1G_ _
42. State Parseval’ s relation in z transform
If x1(n) and x2(n) are complex valued sequences then . .[1(n)x2_Q_ ____ M_.c x1_ _[2*(1/ __ -1G_ _
n=-.
43. State and proveinitial value theorem of z transform
If x(n) is causal then x(0)= lt x(z)
z .
proof: .
X(z)= .[_Q_]-n ----------------------(1)
n=-.
in(1) put n=0 [_Q_ [_]_ .
hence proved
44. State final value theorem of z tramsform
If x(n) is causal z{x(n)}=x(z), where the roc of x(z) includes, but it is not
necessary to confined to _]_ !__DQG__]-1)x(z)has no pole on or outside the unit
circle then
x(.__ __OW__]-1) x(z)
z _
45. Define system function?
The ratio between z transform of out put signal y(z) to z transform of input
signal x(z) is called system function of the particular system
Y(z)
H(z)= ---------
X(Z)
46. What are the conditions of stability of a causal system ?
All the poles of the system are with in the unit circle.
The sum of impulse response for all values of n is bounded
.
._K_Q____.
n = -.
47. Determine z transform and roc of the signal {1,2,3,4}
-.
X(z) = ._[_Q_]-n
n =-.____
3
= .[_Q_]-n =x(0)z-0+x(1)z-1+x(2)z-2+x(3)z-3
n=0
= 1z-0+2z-1+3z-2+4z-3
roc is entire z plane except z = 0
48. Determine z transform and roc of the signal {1,2,3,4}
.
X(z) = ._[_Q_]-n
n=-.
0
X(z)= ._[_Q_]-n = x(-3)z3+x(-2)z2+x(-1)z1+x(0)
n=-3
= 4+3z1+2z2+1z3
ROC is entire z plane except z=.
49. Determine z transform and roc of the signal {1,2,3,4}
.
X(z)= ._[_Q__]-n
n=-.
2
X(z)= .[_Q_]-n = x(-1)z1 + x(0)z0 + x(1)z-1 + x(2)z -2
n=-1
= 1z1+2+3z-1+4z-2
ROC is entire z plane except z=.__
50. Find the z transform and roc of anu(n)
.
X(z) = ._[_Q_]-n
n=-. .
X(z) = ._Dn z-n =1/(1-az-1) roc ] !D_
n=0
51. Find the z transform and roc of -anu(-n-1)
.
X(z)= ._[_Q_]-n
n=-.
-1
X(z)= - .__Dn z-n
n= - . .
= -._D-1z)n = 1/(1-az-1) roc ] _D_
n=1
52. The z-transform of a sequence x(n) is x(z),what is the z transform of nx(n)
If z{x(n)}=x(z) then z{nx(n)}=-zd(x(z))/dz
53. Find the z-transform of (a) A digital impulse (b) A digital step.
(a)Since x(n) is zero except for n = 0, where x(n) is 1, we find x(z) = 1.
(b) Since x(n) is zero except for n.___ZKHUH_[_Q__LV____ZH_ILQG_
.________
x(z) = ._=-n =
n=0 1 – z-1
54. What is the relationship between z-transform and DTFT?
The z-transform of x(n) is given by
.________
x(z) = ._[_Q__=-n ; where z = re􀀀 … … … … … … .. (1)
n=-._____
Substituting z in x(z) we get,
.________
x(z) = ._[_Q__U-ne-􀀀 _ … … … … … … . (2)
n=-._____
The Fourier transform of x(n) is given by
.
x(e􀀀 ) = ._[_Q__H-􀀀 _ … … … … … … ..(3)
n=-.__
Equation (2) and (3) are identical, when r = 1.
In the z-plane this corresponds to the locus of points on the unit circle ] ____
Hence X(e􀀀 ) is equal to H(z) evaluated along the unit circle, or X(e􀀀 ) = x(z) z = e􀀀
For X(e􀀀 ) to exist, the ROC of x(z) must include the unit circle.
55. What are the different methods of evaluating inverse z-transform?
It can be evaluated using several methods.
i. Long division method
ii. Partial fraction expansion method
iii. Residue method
iv. Convolution method
56. Define DFT of a discrete time sequence.
The dft is used to convert a finite discrete time sequence x(n) to an N point
frequency domain sequencex(k).The Npoint DFTof a finite sequence x(n) of
length L,(LN-1
x(k)= ._[_Q_H-􀀀__ ______ K=0,1,2,3,… N-1
n=0
57. Define IDTFT
The IDTFT of the sequence of length N is defined as
N-1
X(n)=(1/N ) .[_N_H􀀀__ ______ n=0,1,2,3,… N-1
k=0
58. Define DTFT and IDTFT of a sequence?
The DTFT (Discrete Time Fourier Transform) of a sequence x(n) is
defined as
.
X(w) = .[_Q_H-jwn
n = -. __________________
The IDTFT is defined as _[_Q_ _____ _.[_Z__Hjwn dw
-
59. What is the drawback in DTFT?
The drawback in discrete time fourier transform is that it is continuous
function of w and cannot be processed by digital systems.
60. What is the relation between DFT and DTFT?
Let x(n) be a sequence. DTFT{x(n)}=x(w) and DFT{X(n)}=x(k).x(k)
is a N point sequence which is obtained by sampling one period of x(w) at N
equal intervals.
;_ ___ = X(K)
___________________________ _ N_1
61. Calculate DFT of the sequence x(n)={1,1,2,2}
N-1
x(k)= ._[_Q_H-􀀀__ ______ K=0,1,2,3,… N-1
n=0
3
x(k)= .[_Q_H-􀀀__ ______ K=0,1,2,3
n=0
N=4
= x(0)+x(1)e-􀀀__ _ _ +x(2)e-􀀀__ +x(3)e-􀀀 __ _ _
= 1+ e-􀀀__ _ _ -2e-􀀀__ -2e-􀀀 __ _ _ K=0,1,2,3
62. List any four properties of DFT
a. Periodicity
b. Linearity
c. Time reversal
d. Circular time shift
63. State periodicity property with respect to DFT.
If x(k) is N-point DFT of a finite duration sequence x(n), then
x(n+N) = x(n) for all n.
x(k+N) = x(k) for all k.
64. State periodicity property with respect to DFT.
If x1(k) and x2(k) are N-point DFTs of finite duration sequences x1(n) and
x2(n), then DFT [a x1(n) + b x2(n)] = a x1(k) + b x2(k), a, b are constants.
65. State time reversal property with respect to DFT.
If DFT[x(n) =x(k), then
DFT[x((-n))N] = DFT[x(N-n)] = x((-k))N = x(N-k)
66. State circular time shifting property with respect to DFT.
If DFT[x(n)] = x(k), then DFT [x((n-l))] = x(k) e-j2 ___ _
67. Assume two finite duration sequences x1(n) and x2(n) are linearly combined. Let
x3(n) = a x1(n) + b x2(n). What is the DFT of x3(n)?
Given x3(n) = a x1(n) + b x2(n).
Let DFT[ x1(n)] = x1(k) and DFT[ x2(n)] = x2(k), then
DFT[ x3(n)] = DFT [a x1(n) + b x2(n) ]
= a DFT[ x1(n)] +b DFT[ x2(n)]
= a x1(k) + b x2(k)
68. &RPSXWH_WKH_')7_RI_[_Q__ _ _Q_– k1)
*LYHQ_[_Q__ _ _Q_– k1) = 1, when n = k1
0, otherwise
N-1
x(k)= .[_Q_H-􀀀__ ______ K=0,1,2,3,… N-1
n=0
N-1
x(k)= ._ _Q_– k1)e-􀀀__ ______ K=0,1,2,3,… N-1
n=0
= e-􀀀__ _
1
k/N
69. What are the two methods used for sectional convolution?
(a) Overlap and add method
(b) Overlap and save method
70. Define circular convolution.
Let x1(n) and x2(n)are finite duration sequences both of length n with
DFTs x1(k) and x2(k). If x3(k) = x1(k) x2(k), then the sequence x3(k) can be
obtained by circular convolution, defined as
N-1
x(k) = ._[1(m) x2((n)) N
n=0
71. Why FFT is needed?
FFT is needed to compute DFT with reduced number of calculations.
DFT is required for spectrum analysis and filtering operations on the signals using
digital computers.
72. Calculate the number of multiplications needed in the calculation of DFT and FFT
with 64 point sequence.
The number of complex multiplications required using direct computation
is N2 = 642 = 4096.
The number of complex multiplications required using FFT is
N log2 N = 64 log264 = 192
2 2
73. What is the main advantage of FFT?
FFT reduces the computation time required to compute discrete fourier
transform.
74. Calculate the number of multiplications needed in the calculation of DFT using
FFT with 32 point sequence.
The number of complex multiplications required using FFT is
N log2N = 32 log232 = 80
2 2
75. What is FFT?
FFT is a method for computing the DFT with reduced number of
calculations using symmetry and periodicity properties of twiddle factor Wk
N .
The computational efficiency is achieved by decomposing of an N-point DFT into
successively smaller DFTs to increase the speed of computation.
76. How many multiplications and additions are required to compute N-point DFT
using radix-2 FFT?
N log2N multiplications and N log2N additions
2
77. What is meant by radix-2 FFT?
If the number of output points N can be expressed as a power of 2, i.e.,
N = 2M Where M is an integer then this algorithm is known as radix-2 algorithm.
78. What is DIT radix2 algorithm.
The radix 2 DIT FFT is an efficient algorithm for computing DFT.The
idea is to break N point sequence in to two sequences ,the DFT of which can be
combined to give DFT of the original N-point sequence. Initially the N point
sequence is divided in to two N/2 point sequences ,on the basis of odd and even
and the DFTs of them are evaluated and combined to give N-point sequence.
Similarly the N/2 DFT s are divided and expressed in to the combination of N/4
point DFTs. This process is continued until we left with 2-point DFTs
79. What is DIF radix2 algorithm.
The radix 2 DIFFFT is an efficient algorithm for computing DFT in this
the out put sequence x(k) is divided in to smaller and smaller. The idea is to break
N point sequence in to two sequences ,x1(n) and x2(n) consisting of the first N/2
points of x(n)and last N/2 points of x(n) respectively. Then we find N/2 point
sequences f(n) and g(n).f(n)=x1(n)+x2(n)and g(n)= (x1(n)+x2(n))WN
n .Similarly
the N/2 DFT s are divided and expressed in to the combination of N/4 point
DFTs. This process is continued until we left with 2-point DFTs
80. What are the differences between DIT and DIF algorithms?
For DIT the input is bitreversed and the output is in natural order ,and in
DIF the input is in natural order and output is bitreversed.In butterfly the phase
factor is multiplied before the add and subtract operation but in DIF it is
multiplied after add-subtract operation
81. What is the basic operation of DIT algorithm?
The basic operation DIT algorithm is called butterfly in which two inputs
G(n) and H(n)are combined to give x1(k) and x2(k)
x1(k)= G(n)+WN
kH(n)
x2(k)= G(n)-WN
kH(n)
WN
k is the twiddle factor
82. What is the basic operation of DIF algorithm?
The basic operation DIF algorithm is called butterfly in which two inputs
G(n) and H(n)are combined to give x1(k) and x2(k)
x1(k)= G(n)+ H(n)
x2(k)={G(n)- H(n)} WN
k
WN
k is the twiddle factor
83. Draw the flow-graph of a two-point DFT for a decimation in time decomposition
The flow-graph of a two-point DFT for decimation in time algorithm is
G(n) x1(k)= G(n)+ H(n)
W2
0
x2(k)= G(n)- H(n)
H(n)
84. Draw the flow-graph of a two-point DFT for a decimation in frequency
decomposition
The flowgraph of a twopoint DFT for decimation in frequency algorithm
is
G(n) W2
0 x1(k)= G(n)+ H(n)
x2(k)= G(n)- H(n)
H(n)
85. Draw the basic butterfly diagram for decimation in time algorithm
The flowgraph of a twopoint DFT for a decimation in time algorithm is
G(n) x1(k)= G(n)+WN
kH(n)
WN
k
x2(k)= G(n)- WN
kH(n)
H(n)
G(n)andH(n) are inputs and x1(k) ,x2(k) are outputs WN
k is phase factor
86. Draw the basic butterfly for a decimation in frequency decomposition
The butterflyof a twopoint DFT for a decimation in frequency algorithm is
G(n) WN
k x1(k)= G(n)+ H(n)
H(n)
x2(k)= {G(n)- H(n)} WN
k
G(n)andH(n) are inputs and x1(k) ,x2(k) are outputs WN
k is phase factor
87. Arrange the 8 point sequence x(n)={1,2,3,4,-1,-2,-3,-4} inn bit reversed order.
Normal order x(n)={1,2,3,4,-1,-2,-3,-4}
Bit reversal order x(n)={1,-1,3,-3, 2,-2,4,-4}
88. How we can calculate IDFT using FFT algorithm?
-The inverse DFT of N point sequence x(k) is defined as
N-1
X(n)=(1/N ) .>[_N_:N
nk ] * n=0,1,2,3,… N-1
k=0
a. Take conjugate of x(k)
b. Compute N point DFT of x*(k) using radix 2 FFT.
c. Take conjugate of output sequence.
d. Divide the output sequence by N.
89. What are the applications of FFT?
1. linear filtering
2. correlation
3. spectrum analysis
90. What are the twiddle factors involved in the first stage of computation in 8 point
DIT radix-2, FFT algorithm?
W8
0, W8
1, W8
2, W8
3
91.What is filter?
Filter is a frequency selective device ,which amplify particular range of
frequencies and attenuate particular range of frequencies.
92.What are the types of digital filter according to their impulse response?
IIR(Infinite impulse response )filter
FIR(Finite Impulse Response)filter.
93. How phase distortion and delay distortion are introduced?
The phase distortion is introduced when the phase characteristics of a filter is
nonlinear with in the desired frequency band.
The delay distortion is introduced when the delay is not constant with in the desired
frequency band.
94.whar are FIR filters?
The filter designed by selecting finite number of samples of impulse response (h(n)
obtained from inverse fourier transform of desired frequency response H(w)) are called
FIR filters
95. Write the steps involved in FIR filter design
Choose the desired frequency response Hd(w)
Take the inverse fourier transform and obtain Hd(n)
Convert the infinite duration sequence Hd(n) to h(n)
Take Z transform of h(n) to get H(Z)
96. What are advantages of FIR filter?
Linear phase FIR filter can be easily designed .
Efficient realization of FIR filter exists as both recurrisive and non recursive structures.
FIR filter realized non recursively are stable.
The round off noise can be made small in non recursive realization of FIR filter
97. what are the disadvantages of FIR FILTER
The duration of impulse response should be large to realize sharp cutoff filters.
The non integral delay can lead to problems in some signal processing applications.
98.what is the necessary and sufficient condition for the linear phase characteristic of a
FIR filter?
The phase function should be a linear function of w, which inturn requires
constant group delay and phase delay.
99. List the well known design technique for linear phase FIR filter design?
Fourier series method and window method
Frequency sampling method.
Optimal filter design method.
100.Define IIR filter?
The filter designed by considering all the infinite samples of impulse response are
called IIR filter.

solid state drives

B.E./B.Tech. DEGREE EXAMINATION, MAY/JUNE 2007.
Sixth Semester
(Regulation 2004)
Electrical and Electronics Engineering
EE 1351 - SOLID STATE DRIVES
(Common to B.E. (Part-Time) Fifth Semester Regulation - 2005)

Time : Three hours Maximum : 100 marks
Answer ALL questions
PART A - [10 X 2 = 20 MARKS]

1. What are all the components of load torque?

2. What are all the conditions to be satisfied for the regenerative braking operation to take place?

3. When is discontinuous conduction mode expected with the operation of converter fed dc drives?

4. Explain whether discontinuous conduction will occur in the operation of chopper fed dc drives?

5.State the advantages of PI controller used in closed loop control of induction motor drives.

6. Compare voltage source and current source inverter fed drives.

7. Mention any two advantages of self control of synchronous motor.

8. Write down the Torque equation of synchronous motor.

9. List out the factors concerned with selection of converters.

10. What are the advantages of closed loop speed control?


PART B - [5 X 16 = 80 MARKS]

11 (a). (i) Derive the mathematical condition for steady state stability of equilibrium point. [MARKS 8]

(ii) Based on the mathematical condition, examine the stability of equilibrium points A, B, C &
D given in figures (1) & (2). [MARKS 8]










FIGURE 1
FIGURE 2


OR

(b) (i)Explain in detail the multi quadrant dynamics in the speed-torque plane. [MARKS 8]

(ii) A motor drives two loads. one has rotational motion. It is coupled to the motor through a
reduction gear with a = 0.1 and efficiency of 90 %. The load has a moment of inertia of 10
Kg-m and a torque of 10 N-m2 . The other load has translational motion and consists of
1000 Kg weight to be lifted up at a uniform speed of 1.5 m/sec. coupling between this load
and the motor has an efficiency of 85%. Motor has an inertia of 0.2 Kg-m and runs at a
constant Speed of 1420 r.p.m. Determine the equivalent inertia referred to the motor shaft and
power delivered by the motor. [MARKS 8]

12. (a) Explain in detail the operation and steady state analysis of 1 phase fully controlled converter fed dc
drive with neat waveforms in continuous and discontinuous conduction mode. [MARKS 16]

OR

(b) (i) Explain the operation of four quadrant chopper control in dc motor drives. [MARKS 8]

(ii) A 250 V separately excited dc motor has an armature resistance of 2.5O. When driving a load at
600 r.p.m. with constant torque, the armature takes 20 A. This motor is controlled by a chopper
circuit with a frequency of 400 Hz and an input voltage of 250 V.
(1) What should be the value of the duty ratio if one desires to reduce the speed from 600 to 540
r.p.m with the load torque maintained constant.
(2) Find out the value of duty ratio for which the per unit ripple current will be maximum.
[MARKS 8]

13. (a) Explain in detail with suitable diagrams and waveforms the (v/f) control applied to induction motor
drives. [MARKS 16]

OR

(b) (i) Explain with neat diagram and equations the static Scherbius system of slip power recovery
scheme. [MARKS 8]

(ii) A 3 phase, star connected, 60Hz, 4 pole induction motor has the following parameters for its
equivalent circuit. Rs = Rr = 0.024O and Xs = Xr = 0.12O. The motor is controlled by the
variable frequency control with a constant (v/f) ratio. For an operating frequency of 12 Hz,
calculate:

(1) The breakdown torque as a ratio of it’s value at the rated frequency for both motoring and
braking.

(2) The starting torque and rotor current in terms of their values at the rated frequency.
[MARKS 8]

14. (a) (i) Explain open loop speed control of synchronous motor with constant (v/f) ratio. [MARKS 8]

(ii) Explain power factor control of synchronous motor with relevant vector diagram. [MARKS 8]

OR
(b) (i) Explain self control of synchronous motor in detail. [MARKS 8]

(ii) Write short notes on permanent magnet synchronous motor. [MARKS 8]


15. (a) (i) Derive the transfer function of dc motor-load system. [MARKS 8]

(ii) Give the design procedure of current controller. [MARKS 8]

OR
(b) Explain the armature voltage control of dc motor with constant field and field weakening modes.
[MARKS 16]
Solid State Drives (EE1351)
S6 EEE
Two Marks Questions and Answers

Two Marks Questions and Answers
Solid State Drives (EE1351) S6 EEE
1. What is meant by electrical drives?
Systems employed for motion control are called drives and they employ any of the prime
movers such as diesel or petrol engines, gas or steam turbines, hydraulic motors and electric
motors for supplying mathematical energy for motion control. Drives employing electric motion
are called electric drives.
2. Draw the electric drive system.
Electric Drive
Fig. An electric-drive system
Command
Fig. Modern Electric Drive system using power electronic converter
3. Specify the functions of power modulator.
Power modulator performs one or more of the following four functions.
a. Modulates flow of power form the source to the motor in such a manner that motor is
imparted speed-torque characteristics required by the load.
b. During transient operations, such as starting, braking and speed reversal, it restricts
source and motor currents within permissible values; excessive current drawn from source
may overload it or may cause a voltage dip.
4. Mention the different types of drives.
1)Group drive
2)Individual drive
3)Multimotor drive
5. List the different types of electrical drives.
1)dc drives
2)ac drives
6. What are the advantages of electric drives?
Main
Power
Source
Power
Controller Working
Machine
Motor
Working
Machine
Main
Power
Source
Power
Controller
Motor
Rotor position or
speed sensor
1) They have flexible control characteristics. the steady state and dynamic characteristics
of electrical drives can be shaped to satisfy load requirements.
2) Drives can be provided with automatic fault detection systems, programmable logic
controllers and computers can be employed to automatically ctrl the drive operations in a
desired sequence.
3) They are available in which range of torque, speed and power.
4) It can operate in all the four quadrants of speed-torque plane. Electric braking gives
smooth deceleration and increases life of the equipment compared to other forms of
braking.
5) Control gear required for speed control, starting and braking is usually simple and easy
to operate.
7. What are the functions performed by electric drives?
Various functions performed by electric drives include the following.
a. Driving fans, ventilators, compressors and pumps etc.
b. Lifting goods by hoists and cranes
c. Imparting motion to conveyors in factories, mines and warehouses and
d. Running excavators and escalators, electric locomotives, trains, cars, trolley buses, lifts
and drums winders etc.
8. What are the disadvantages of electric drives?
The disadvantages of electric drives.
a. Electric drives system is tied only up to the electrified area.
b. The condition arising under the short circuits, leakage from conductors and breakdown of
overhead conductor may lead to fatal accidents.
c. Failure in supply for a few minutes may paralyses the whole system.
9. What are the advantages of group drive over individual drive?
The advantages of group drive over individual drive are
a. Initial cost : Initial cost of group drive is less as compared to that of the individual drive.
b. Sequence of operation : Group drive system is useful because all the operations are
stopped simultaneously.
c. Space requirement : Less space is required in group drive as compared to individual
drive.
d. Low maintenance cost : It requires little maintenance as compared to individual drive.
10. What the group drive is not used extensively.
Although the initial cost of group drive is less but yet this system is not used extensively because
of following disadvantages.
a. Power factor : Group drive has low power factor
b. Efficiency : Group drive system when used and if all the machines are not working
together the main motor shall work at very much reduced load.
c. Reliability : In group drive if the main motor fails whole industry will come to stand still.
d. Flexibility : Such arrangement is not possible in group drive i.e., this arrangement is not
suitable for the place where flexibility is the prime factor.
e. Speed : Group drive does not provide constant speed.
f. Types of machines : Group drive is not suitable fro driving heavy machines such as
cranes, lifts and hoists etc.
11. Write short notes on individual electric drives.
In individual drive, each individual machine is driven by a soparate motor. This motor also
imparts motion to various other parts of the machine. Examples of such machines are single
spindle drilling machines (Universal motor is used) and lathes. In a lathe, the motor rotates the
spindle, moves the feed and also with the help of gears, transmits motion to lubricating and cooling
pumps. A three phase squirrel cage induction motor is used as the drive. In many such
applications the electric motor forms an integral part of the machine.
12. Mention the different factors for the selection of electric drives?
1) Steady state operation requirements.
2) Transient operation requirements.
3) Requirements related to the source.
4) Capital and running cost, maintenance needs, life.
5) Space and weight restriction.
6) Environment and location.
7) Reliability.
13. Mention the parts of electrical drives.
1) Electrical motors and load.
2) Power modulator
3) Sources
4) Control unit
5) Sensing unit
14. Mention the applications of electrical drives
Paper mills
Electric traction
Cement mills
Steel mills
15. Mention the types of enclosures
Screen projected type
Drip proof type
Totally enclosed type
Flame proof type
16. Mention the different types of classes of duty
Continuous duty
Discontinuous duty
Short time duty
Intermittent duty
17. What is meant by regenerative braking?
Regenerative braking occurs when the motor speed exceeds the synchronous speed. In
this case the IM runs as the induction m\c is converting the mechanical power into electrical power
which is delivered back to the electrical system. This method of braking is known as regenerative
braking.
18. What is meant by dynamic braking?
Dynamic braking of electric motors occurs when the energy stored in the rotating mass is
dissipated in an electrical resistance. This requires a motor to operate as a gen. to convert the
stored energy into electrical.
19. What is meant by plugging?
It is one method of braking of IM. When phase sequence of supply of the motor running
at the speed is reversed by interchanging connections of any two phases of stator with respect to
supply terminals, operation shifts from motoring to plugging region.
20. What is critical speed?
It is the speed that separates continuous conduction from discontinuous conduction mode.
21. Which braking is suitable for reversing the motor?
Plugging is suitable for reversing the motor.
22. Define equivalent current method
The motor selected should have a current rating more than or equal to the current. It is also
necessary to check the overload of the motor. This method of determining the power rating of the
motor is known as equivalent current method.
23. Define cooling time constant
It is defined as the ratio between C and A. Coolng time constant is denoted as Tau
Tau = C/A
Where C=amount of heat required to raise the temp of the motor body by 1 degree Celsius
A=amount of heat dissipated by the motor per unit time per degree Celsius.
24. What are the methods of operation of electric drives?
Steady state
acceleration including starting
deceleration including starting
25. Define four quadrant operation.
The motor operates in two mode: motoring and braking. In motoring, it converts electrical
energy into mechanical energy which supports its motion. In braking, it works as a generator,
converting mathematical energy into electrical energy and thus opposes the motion. Motor can
provide motoring and braking operations for both forward and reverse directions.
26. What is meant by mechanical characteristics?
The curve is drawn between speed and torque. This characteristic is called mechanical
characteristics.
27. Mention the types of braking
Regenerative braking
Dynamic braking
Plugging
28. What are the advantage and disadvantages of D.C. drives?
The advantages of D.C. drives are,
a. Adjustable speed
b. Good speed regulation
c. Frequent starting, braking and reversing.
The disadvantage of D.C. drives is the presence of a mechanical commutator which limits the
maximum power rating and the speed.
29. Give some applications of D.C. drives.
The applications of D.C. drives are,
a. Rolling mills b. Paper mills
c. Mine winders d. Hoists
e. Machine tools f. Traction
g. Printing presses h. Excavators
i. Textile mils j. Cranes.
30. Why the variable speed applications are dominated by D.C. drives?
The variable speed applications are dominated by D.C. drives because of lower cost, reliability ad
simple control.
31. What is the use of flywheel? Where it is used?
It is used for load equalization. It is mounted on the motor shaft in compound motor.
32. What are the advantages of series motor?
The advantages of series motors are,
a. High starting torque
b. Heavy torque overloads.
33. How the D.C. motor is affected at the time of starting?
A D.C. motor is started with full supply voltage across its terminals, a very high current will flow,
which may damage the motor due to heavy sparking at commuter and heating of the winding.
Therefore, it is necessary top limit the current to a safe value during starting.
34. Define and mention different types of braking in a dc motor?
In braking the motor works as a generator developing a negative torque which opposes
the motion. Types are regenerative braking, dynamic or rheostat braking and plugging or reverse
voltage braking.
35. List the drawbacks of armature resistance control?
In armature resistance control speed is varied by wasting power in external resistors that
are connected in series with the armature. since it is an inefficient method of speed control it was
used in intermittent load applications where the duration of low speed operations forms only a
small proportion of total running time.
36. What is static Ward-Leonard drive?
Controlled rectifiers are used to get variable d.c. voltage from an a.c. source of fixed
voltage controlled rectifier fed dc drives are also known as static Ward-Leonard drive.
37. What is a line commutated inverter?
Full converter with firing angle delay greater than 90 deg. is called line commutated
inverter. such an operation is used in regenerative braking mode of a dc motor in which case a
back emf is greater than applied voltage.
38. Mention the methods of armature voltage controlled dc motor?
When the supplied voltage is ac,
Ward-Leonard schemes
Transformer with taps and un controlled rectifier bridge
Static Ward-Leonard scheme or controlled rectifiers
When the supply is dc:
Chopper control
39. How is the stator winding changed during constant torque and constant horsepower operations?
For constant torque operation, the change of stator winding is made form series – star to
parallel – star, while for constant horsepower operation the change is made from series-delta to
parallel-star. Regenerative braking takes place during changeover from higher to lower speeds.
40. Define positive and negative motor torque.
Positive motor torque is defined as the torque which produces acceleration or the positive rate of
change of speed in forward direction. Positive load torque is negative if it produces deceleration.
41. Write the expression for average o/p voltage of full converter fed dc drives?
Vm=(2Vm/pi)cospi.................continuous conduction
Vm=[Vm(cos alpha-cos beta)+(pi+alpha+beta)]/pi]........discontinuous conduction
42. What are the disadvantages of conventional Ward-Leonard schemes?
Higher initial cost due to use of two additional m\cs.
Heavy weight and size.
Needs more floor space and proper foundation.
Required frequent maintenance.
Higher noise and higher loss.
43. Mention the drawbacks of rectifier fed dc drives?
Distortion of supply.
Low power factor.
Ripple in motor current
44. What are the advantages in operating choppers at high frequency?
The operation at a high frequency improves motor performance by reducing current ripple
and eliminating discontinuous conduction.
45. Why self commutated devices are preferred over thyristors for chopper circuits?
self commutated devices such as power MOSFETs power transistors, IGBTs, GTOs and
IGCTs are preferred over thyristors for building choppers because they can be commutated by a
low power control signal and don’t need commutation circuit.
46. State the advantages of dc chopper drives?
Dc chopper device has the advantages of high efficiency, flexibility in control, light
weight, small size, quick response and regeneration down to very low speed.
47. What are the advantages of closed loop c of dc drives?
Closed loop control system has the adv. of improved accuracy, fast dynamic response and
reduced effects of disturbance and system non-linearities.
48. What are the types of control strategies in dc chopper?
Time ratio control.
Current limit control.
49. What are the adv. of using PI controller in closed loop ctrl. of dc drive?
Stabilize the drive
Adjust the damping ratio at the desired value
Makes the steady state speed error close to zero by integral action and filters out noise
again due to the integral action.
50. What are the different methods of braking applied to the induction motor?
Regenerative braking
Plugging
Dynamic braking.
51. What are the different methods of speed control of IM?
Stator voltage control
Supply freq. control
Rotor resistance control
Slip power recovery control.
52. What is meant by stator voltage control.?
The speed of the IM can be changed by changing the stator voltage. Because the torque is
proportional to the square of the voltage.
53. Mention the application of stator voltage control.
This method is suitable for applications where torque demand reduced with speed, which
points towards its suitability for fan and pump drives.
54. Mention the applications of ac drives.
AC drives are used in a no. of applications such as fans, blowers, mill run-out tables,
cranes, conveyors, traction etc.
55. What are the three regions in the speed-torque characteristics in the IM?
Motoring region (0<=s<=1)
Generating region(s<0)
Plugging region (1<=s<=2) where s is the slip.
56. What are the adv. of stator voltage control method?
The ctrl circuitry is simple
Compact size
Quick response time
There is considerable savings in energy and thus it is economical method as compared to
other methods of speed ctrl.
57. What is meant by soft start?
The ac voltage controllers show a stepless control of supply voltage from zero to rated
volt. they are used for soft start for motors.
58. List the adv of squirrel cage IM?
Cheaper
light in weight
Rugged in construction
More efficient
Require less maintenance
It can be operated in dirty and explosive environment
59. Define slip
The difference between the synchronous speed (Ns)and actual speed(N)of the rotor is
known as slip speed. the % of slip is gn by,
%slip s=[(Ns-N)/Ns]x 100
60. Define base speed.
The synchronous speed corresponding to the rated freq is called the base speed.
61. What is meant by frequency control of IM?
The speed of IM can be controlled by changing the supply freq because the speed is
directly proportional to supply frequency. This method of speed ctrl is called freq control.
62. What is meant by V/F control l?
When the freq is reduced the i/p voltage must be reduced proportionally so as to maintain
constant flux otherwise the core will get saturated resulting in excessive iron loss and magnetizing
current. This type of IM behavior is similar to the working of dc series motor.
63. What are the advantages of V/F control?
Smooth speed ctrl
Small i/p current and improved power factor at low freq. start
Higher starting torque for low case resistance
64. What is meant by stator current control?
The 3 phase IM speed can be controlled by stator current control. The stator current can
be varied by using current source inverter.
65. What are the 3 modes of region in the adjustable-freq IM drives characteristics?
Constant torque region
Constant power region
High speed series motoring region
66. What are the two modes of operation in the motor?
The two modes of operation in the motor are, motoring and braking. In motoring, it
converts electrical energy to mechanical energy, which supports its motion. In braking, it works as
a generator converting mechanical energy to electrical energy and thus opposes the motion.
67. How will you select the motor rating for a specific application?
When operating for a specific application motor rating should be carefully chosen that the
insulation temperature never exceed the prescribed limit. Otherwise either it will lead to its
immediate thermal breakdown causing short circuit and damage to winding, or it will lead to
deterioration of its quality resulting into thermal breakdown in near future.
68. What is braking ? Mention its types.
The motor works as a generator developing a negative torque which opposes the motion
is called barking.
It is of three types. They are,
a. Regenerative braking.
b. B. Dynamic or rheostat braking.
c. Plugging or reverse voltage braking.
69. What are the three types of speed control?
The three types of speed control as,
a. Armature voltage control
b. Field flux control
c. Armature resistance control.
70. What are the advantages of armature voltage control?
The advantages of armature voltage control are,
a. High efficiency
b. Good transient response
c. Good speed regulation.
71. What are the methods involved in armature voltage control?
When the supply in A.C.
a. Ward-Leonard schemes
b. Transformer with taps and an uncontrolled rectifier bridge.
c. Static ward Leonard scheme or controlled rectifiers when the supply in D.C.
d. Chopper control.
72. Give some drawbacks and uses of Ward-Leonard drive
The drawbacks of Ward . Leonard drive are.
a. High initial cost
b. Low efficiency
The Ward-Leonard drive is used in rolling mills , mine winders, paper mills, elevators,
machine tools etc.
73. Give some advantages of Ward-Leonard drive.
The advantages of Ward-Leonard drive are,
a. Inherent regenerative barking capability
b. Power factor improvement.
74. What is the use of controlled rectifiers?
Controlled rectifiers are used to get variable D.C. Voltage form an A.C. Source of fixed
voltage.
75. What is known as half-controlled rectifier and fully controlled rectifier?
The rectifiers provide control of D.C. voltage in either direction and therefore, allow
motor control in quadrants I and IV. They are known as fully-controlled rectifiers.
The rectifiers allow D.C. Voltage control only in one direction and motor control in
quadrant I only. They are known as half-controlled rectifiers.
76. What is called continuous and discontinuous conduction?
A D.C. motor is fed from a phase controlled converter the current in the armature may flow in
discrete pulses in called continuous conduction.
A D.C. motor is fed from a phase controlled converter the current in the armature may flow
continuously with an average value superimposed on by a ripple is called discontinuous
conduction.
77. What are the three intervals present in discontinuous conduction mode of single phase half and
fully controlled rectifier?
The three intervals present in half controlled rectifier are,
a. Duty interval
b. Free, wheeling interval
c. Zero current interval.
The two intervals present in fully controlled rectifier are
a. Duty interval
b. Zero current interval.
78. What is called inversion?
Rectifier takes power from D.C. terminals and transfers it to A.C. mains is called inversion.
79. What are the limitations of series motor? Why series motor is not used in traction
applications now a days?
1. The field of series cannot be easily controlled. If field control is not employed, the
series motor must be designed with its base speed equal to the highest desired speed of the drive.
2. Further, there are a number of problems with regenerative braking of a series motor.
Because of the limitations of series motors, separately excited motors are now preferred even
for traction applications.
80. What are the advantages of induction motors over D.C. motors?
The main drawback of D.C. motors is the presence of commutate and brushes, which
require frequent maintenance and make them unsuitable for explosive and dirty environments. On
the other hand, induction motors, particularly squirrel-cage are rugged, cheaper, lighter, smaller,
more efficient, require lower maintenance and can operate in dirty and explosive environments.
81. Give the applications of induction motors drives.
Although variable speed induction motor drives are generally expensive than D.C. drives,
they are used in a number of applications such as fans, blowers, mill run-out tables, cranes,
conveyors, traction etc., because of the advantages of induction motors. Other applications
involved are underground and underwater installations, and explosive and dirty environments.
82. How is the speed controlled in induction motor?
The induction motor speed can be controlled by supplying the stator a variable voltage, variable
frequency supply using static frequency converters. Speed control is also possible by feeding the slip
power to the supply system using converters in the rotor circuit, Basically one distinguishes two
different methods of speed control.
a. Speed control by varying the slip frequency when the stator is fed from a constant voltage, constant
frequency mains.
b. Speed control of the motor using a variable frequency variable voltage motor operating at constant
rotor frequency.
83. How is the speed control by variation of slip frequency obtained?
Speed control by variation of slip frequency is obtained by the following ways.
a. Stator voltage control using a three-phase voltage controller.
b. Rotor resistance control using a chopper controlled resistance in the rotor circuit.
c. Using a converter cascade in the rotor circuit to recover slip energy.
d. Using a cyclconverter in the rotor circuit.
84. Mention the effects of variable voltage supply in a cage induction motor.
When a cage induction motor is fed from a variable voltage for speed control the following observations
may be made.
a. The torque curve beyond the maximum torque point has a negative shape. A stable operating point
in this region is not possible for constant torque load.
b. The voltage controlled must be capable of withstanding high starting currents. The range of speed
control is rather limited.
c. The motor power factor is poor.
85. Classify the type of loads driven by the motor.
The type of load driven by the motor influences the current drawn and losses of the motor as the slip
various. The normally occurring loads are
a. Constant torque loads.
b. Torque varying proportional to speed.
c. Torque varying preoperational to the square of the speed.
86. What are the disadvantages of constant torque loads?
The constant torque loads are not favored due to increase in the losses linearly with slip and becoming
maximum at s= 1.0. This is obvious form the variation of flux as the voltage is varied for speed control.
To maintain constant torque the motor draws heavy current resulting in poor torque/ampere, poor
efficiency ad poor power factor at low speeds.
87. In which cases, torque versus speed method is suitable.
Torque versus speed method is suitable only for the following cases.
a. For short time operations where the duration of speed control ids defined.
b. For speed control of blowers or pumps having parabolic or cubic variations of torque with speed.
This is not suitable for constant torque loads due to increases and heating.
c. For speed control of motor having poor efficiencies under normal operation.
88. How is the speed of a squirrel cage induction motor controlled?
The speed of a squirrel cage induction motor can be controlled very effectively by varying the stator
frequency. Further the operation of the motor is economical and efficient, if it operates at very small
slips. The speed of the motor is therefore, varied by varying the supply frequency and maintaining the
rotor frequency at the rated value or a value corresponding to the required torque on the linear portion of
the torque-speed curve.
89. Why the control of a three-phase indication motor is more difficult than D.C. motors.
The control of a three-phase induction motor, particularly when the dynamic performance involved
is more difficult than D.C. motors. This is due to
a. Relatively large internal resistance of the converter causes voltage fluctuations following load
fluctuations because the capacitor cannot be ideally large.
b. In a D.C. motor there is a decoupling between the flux producing magnetizing current and torque
producing armature current. They can be independently controlled. This is not the case with
induction motors.
c. An induction motor is very poorly damped compared to a D.C. motor.
90. Where is the V/f control used?
The V/f control would be sufficient in some applications requiring variable torque, such as centrifugal
pumps, compressors and fans. In these, the torque varies as the square of the speed. Therefore at small
speeds the required torque is also small and V/f control would be sufficient to drive these leads with no
compensation required for resistance drop. This is true also for the case of the liquid being pumped with
minimal solids.
91. What are the components of the applied voltage to the induction motor?
The applied voltage to the induction motor has two components at low frequencies. They are
a. Proportional to stator frequency.
b. To compensate for the resistance drop in the stator.
The second component deepens on the load on the motor and hence on rotor frequency.
92. What is indirect flux control?
The method of maintaining the flux constant by providing a voltage boost proportional to slip
frequency is a kind of indirect flux control . This method of flux control is not desirable if very good
dynamic behaviour is required.
93. What is voltage source inverter?
Voltage source inverter is a kind of D.C. link converter, which is a two stage conversion device.
94. What is the purpose of inductance and capacitance in the D.C. link circuit?
The inductance in the D.C. link circuit provides smoothing whereas the capacitance maintains the
constancy of link voltage. The link voltage is a controlled quality.
95. What are the disadvantages of square wave inverter in induction motor drive?
Square wave inverters have commutation problems at very low frequencies, as the D.C. link
voltage available at these frequencies cannot charge the commutating capacitors sufficiently
enough to commutate the thrusters. Those puts a limit on the lower frequency of operation. To
extend the frequency towards zero, special charging circuits must be used.
96. What is slip controlled drive?
When the slip is used as a controlled quantity to maintain the flux constant in the motor the drive is
called slip enrolled drive. By making the slip negative (i.e., decreasing the output frequency of the
inverter) The machine may be made to operate as a generator and the energy of the rotating parts fed back
to the mains by an additional line side converter or dissipated in a resistance for dynamic barking. By
keeping the slip frequency constant, braking at constant torque and current can be achieved. Thus
braking is also fast.
97. What are the effects of harmonics in VSI fed induction motor drive?
The motor receives square wave voltages. These voltage has harmonic components. The
harmonics of the stator current cause additional losses and heating. These harmonics are also responsible
for torque pulsations. The reaction of the fifth and seventh harmonics with the fundamental gives rise to
the seventh harmonic pulsations in the torque developed. For a given induction motor fed from a square
wave inverter the harmonic content in the current tends to remain constant independent of input
frequency, with the rang of operating frequencies of the inverter,
98. What is a current source inverter?
In a D.C. link converter, if the D.C. link current is controlled, the inverter is called a current source
inverter, The current in the D.C. link is kept constant by a high inductance and he capacitance of the
filter is dispensed with . A current source inverter is suitable for loads which present a low impedance to
harmonic currents and have unity p.f.
99. Explain about the commutation of the current source inverter.
The commutation of the inverter is load dependent. The load parameters form a part of the commutation
circuit. A matching is therefore required between the inverter and the motor. Multimotor operation is not
possible. The inverter must necessarily be a force commutated one as the induction motor cannot provide
the reactive power for the inverter. The motor voltage is almost sinusoidal with superimposed spikes.
100. Give the features from which a slip controlled drive is developed.
The stator current of an induction motor operating on a variable frequency, variable voltage supply
is independent of stator frequency if the air gap flux is maintained constant. However, it is a
function of the rotor frequency. The torque developed is also a function of rotor frequency. The
torque developed is also a function of rotor frequency only. Using these features a slip controlled
drive can be developed employing a current source inverter to feed an induction motor.
101. How is the braking action produced in plugging?
In plugging, the barking torque is produced by interchange any two supply terminals, so that the direction
of rotation of the rotating magnetic field is reversed with respect to the rotation of the motor. The
electromagnetic torque developed provides the braking action and brings the rotor to a quick stop.
102. Where is rotor resistance control used?
Where the motors drive loads with intermittent type duty, such as cranes, ore or coal unloaders,
skip hoists, mine hoists, lifts, etc. slip-ring induction motors with speed control by variation of
resistance in the rotor circuit are frequently used. This method of speed control is employed for a
motor generator set with a flywheel (Ilgner set) used as an automatic slip regulator under shock loading
conditions.
103. What are the advantages and disadvantages of rotor resistance control?
Advantage of rotor resistance control is that motor torque capability remains unaltered even at low
speeds. Only other method which has this advantage is variable frequency control. However, cost of
rotor resistance control is very low compared to variable frequency control.
Major disadvantage is low efficiency due to additional losses in resistors connected in the rotor
circuit.
104. Where is rotor resistance control used?
Where the motors drive loads with intermittent type duty, such as cranes, ore or coal unloaders,
skip hoists, mine hoists, lifts, etc. slip-ring induction motors with speed control by variation of
resistance in the rotor circuit are frequently used. This method of speed control is employed for a
motor generator set with a flywheel (Ilgner set) used as an automatic slip regulator under shock
loading conditions.
105. What are the advantages and disadvantages of rotor resistance control?
Advantage of rotor resistance control is that motor torque capability remains unaltered
even at low speeds. Only other method which has this advantage is variable frequency
control. However, cost of rotor resistance control is very low compared to variable
frequency control.
Major disadvantage is low efficiency due to additional losses in resistors connected in the
rotor circuit.
106. Give the equation of slip of the motor
S’ = S R’
2
+R’
ex
R’
2
Where , R’
2 = Rotor resistance
R’
ex = Resistance included
The external resistance can be added very conveniently to the phases of a slip
ring rotor.
107. How is the resistance in the output terminals of a chopper varied?
The resistance connected across the output terminals of a chopper can be varied form O to
R by varying the time ratio of the chopper. When the chopper is always OFF, the supply is always
connected to the resistance R. The time ratio in this case is zero and the effective resistance
connected in R. Similarly when the chopper is always ON, the resistance is short circuited. The
time ratio in the case is unity and the effective resistance connected is 0. Hence by varying the
time ratio from 0 to 1, the value of resistance can be varied from R to O.
108. What is the function of inductance L and resistance R in the chopper resistance circuit?
A smoothing inductance L is used in the circuit to maintain the current at a constant value.
Any short circuit in the chopper does not become effective due to L.
The value of R connected across the chopper is effective for all phases and its value can
be related to the resistance to be connected in each phase if the conventional method has been
used. The speed control range is limited by the resistance.
109. What are the disadvantages and advantages of chopper controlled resistance in the rotor circuit
method?
The method is very inefficient because of losses in the resistance. It is suitable for
intermittent loads such as elevators. At low speeds, in particular the motor has very poor
efficiency. The rotor current is non-sinusoidal. They harmonics of the rotor current produce
torque pulsations. These have a frequency which is six times the slip frequency.
Because of the increased rotor resistance, the power factor is better.
110. How is the range of speed control increased?
The range of speed control can be increased if a combination of stator voltage control and
rotor resistance control is employed. Instead of using a high resistance rotor, a slip ring rotor with
external rotor resistance can be used when stator voltage control is used for controlling the speed.
111. Why the static scherbius drive has a poor power factor?
Drive input power is difference between motor input power and the power fed back. Reactive
input power is the sum of motor and inverter reactive power. Therefore, drive has a poor power
factor throughout the range of its options.
112. How is super synchronous speed achieved?
Super synchronous speed can be achieved if the power is fed to the rotor from A.C. mains.
This can be made possible by replacing the converter cascade by a cycloconverter. A
cycloconverter allows power flow in either direction making the static sherbets drive operate at
both sub and supper synchronous speeds.
113. Give the features of static scherbius drive
The torque pulsations and other reactions are minimal. The performance of the drive
improves with respect to additional losses and torque pulsations. A smooth transition is possible
from sub to super synchronous speeds without any commutation problems. Speed reversal is not
possible. A step up transformer may be interposed between the lines and the converter, to reduce
the voltage rating of the converter.
114.Where is Kramer electrical drive system used?
Some continuous rolling mills, large air blowers, mine ventilators, centrifugal pumps and
any other mechanisms including pumps drives of hydraulic dredgers require speed adjustment in
the range from 15 to 30% below or above normal . If the induction motor is of comparatively big
size (100 to 200 KW) it becomes uneconomical to adjust speed by mean’ s pf external resistances
due to copper losses as slip power is wasted as heat in the retort circuit resistance. In these case ,
the Kramer electrical drive system is used , where slip power recovery takes places.
115. What is the use of sub synchronous converter cascades?
Sub synchronous converter cascades have been used, till now, in applications requiring one
quadrant operation. These can be employed for drives where at least one electrical barking is
required. A four quadrant operation can also be made possible in these cascades, using suitable
switching.
116. How is the speed control obtained in static Kramer drive?
For speed control below synchronous speed, the slip power is pumped back to the supply, where as
for the case of speed above synchronous speed, additional slip power is injected into the rotor
circuit.
117.What is static Kramer drive?
Instead of wasting the slip power in the rotor circuit resistance, it can be converted to 60
Hz A.C. and pumped back to the line. The slip power controlled drive that permits only a sub
synchronous range of speed control through a converter cascade is know as static Kramer drive.
118. What is the use and functions of step down transformer is static Kramer drive?
For a restricted speed range closer to synchronous speed, the system power factor can be
further improved by using a step –down transformer.
The step-down transformer has essentially two functions: besides improving the line
power factor, it also helps to reduce the converter power ratings.
119. What are the advantages of static Kramer drive?
The static Kramer drive has been very popular in large power pump and fan-type drives,
where the range of speed control is limited near, but below the synchronous speed. The drive
system is very efficient and the converted power rating is low because t has to handle only the slip
power, In fact, the power rating becomes lower with a more restricted range of speed control. The
additional advantages are that the drive system has D.C. machine like characteristics and the
control is very simple.
120. What are the causes of harmonic currents in static Kramer drive?
The rectification of slip power causes harmonic currents in the rotor, and these harmonics
are reflected to the stator by the transformer action of the machine. The harmonic currents are also
injected into the A.C. line by the inverter. As a result, the machine losses are increased and some
amount of harmonic torque is produced. Each harmonic current in the rotor will create a reading
magnetic filed and its direction of rotation will depend on the order pf the harmonic.
121. Give the four modes of operation of a Scherbius drive
The four modes of operation of static Scherbius drive are,
Sub synchronous motoring.
Sub synchronous regeneration
Super synchronous motoring
Super synchronous regeneration
122. How is the static Scherbius drive operated in super synchronous motoring mode?
In super synchronous motoring mode, the shaft speed increases beyond the synchronous speed, the
slip becomes negative and the slip power is absorbed by the rotor. The slip power supplements the
air gap power for the total mechanical power output. The line therefore supplies slip power in
addition to stator input power. At this condition, the phase sequence of slip frequency is reversed so
that the slip current – induced rotating magnetic filed is opposite to that of the stator.
123. Give the use of synchronous motors.
Synchronous motors were mainly used in constant speed applications. The development
of semiconductor variable frequency sources, such as inverters and cycloconverters, has allowed
their use in draft fane, main line traction, servo drives, etc.
124. How are the stator and rotor of the synchronous motor supplied?
The stator of the synchronous motor is supplied from a thyristor power converter capable of
providing a variable frequency supply. The rotor, depending upon the situation, may be
constructed with slip rings, where it conforms to a conventional rotor. It is supplied with D.C.
through slip rings. Sometimes rotor may also be free from sliding contacts (slip rings), in which
case the rotor is fed from a rectifier rotating with rotor.
125. What is the difference between an induction motor and synchronous motor?
An induction motor operates at lagging power factor and hence the converter supplying
the same must invariable is a force commutated one. A synchronous motor, on the other hand, can
be operated at any power factor by controlling the field current.
126. List out the commonly used synchronous motors.
Commonly used synchronous motors are,
b. Wound field synchronous motors.
c. Permanent magnet synchronous motors
d. Synchronous reluctance synchronous motors.
e. Hysterias motors.
127. Mention the main difference between the wound field and permanent magnet motors.
When a wound filed motor is started as an induction motor, D.C. field is kept off. In case
of a permanent magnet motor, the field cannot be ‘turned off’ .
128. Give the advantages and applications of PMSM.
The advantages of PMSM are,
b. High efficiency
c. High power factor
d. Low sensitivity to supply voltage variations.
The application of PMSM is that it is preferred of industrial applications with large duty cycle
such as pumps, fans and compressors.
129. Give the uses of a hysteresis synchronous motor.
Small hysteresis motors are extensively used in tape recorders, office equipment and fans.
Because of the low starting current, it finds application in high inertia application such as
gyrocompasses and small centrifuges.
130. Mention the two modes employed in variable frequency control
Variable frequency control may employ and of the two modes.
a. True synchronous mode
b. Self-controlled mode
131. Which machine is said to be self controlled?
A machine is said to be self controlled if it gets its variable frequency from an inverter
whose thrusters are freed in a sequence, using the information of rotor position or stator
voltages. In the former a rotor position sensor is employed which measures the rotor position
with respect to the stator and sends pulses to the thyristors. Thus frequency of the inverter
output is decided by the rotor speed.
132. What is Commutator Less Motor ( CLM)?
The self controlled motor has properties of a D.C. Motors both under steady state and
dynamic conditions and therefore is called commutator less motor(CLM). These machines have
better stability behaviors. They do not fall out of step and do not have oscillatory behaviors, as
in normal synchronous motors.
133. Give the application of self controlled synchronous motor.
A self controlled synchronous motor is a substitute for a D.C. motor drive and finds
application where a D.C. motor is objectionable due to its mechanical commutator, which limits
the speed range and power output.
134. Define load commutation
Commutation of thyristors by induced voltages pf load is known as load commutation,
135. List out the advantages of load commutation over forced commutation.
Load commutation has a number of advantages over forced commutation
It does not require commutation circuits
Frequency of operation can be higher
It can operate at power levels beyond the capability of forced commutation.
136. Give some application of load commutated inverter fed synchronous motor drive.
Some prominent applications of load commutated inverter fed synchronous motor drive
are high speed and high power drives for compressors, blowers, conveyers, steel rolling mills,
main-line traction and aircraft test facilities.
137. How the machine operation is performed in self-controlled mode?
For machine operation in the self-controlled mode, rotating filed speed should be the
same as rotor speed. This condition is relaised by making frequency of voltage induced in the
armature. Firning pulses are therefore generated either by comparison of motor terminal
voltages or by rotor position sensors.
138. What is meant by margin angle of commutation?
The difference between the lead angle of firing and the overlap angle is called the margin angle
of commutation. If this angle of the thyristor, commutation failure occurs. Safe commutation is
assured if this angle has a minimum value equal to the turn off angle f the thyristor.
139. What are the disadvantages of VSI fed synchronous motor drive?
VSI synchronous motor drives might impose fewer problems both on machine as well as
on the system design. A normal VSI with 180o conduction of thyristors required forced
commutation and load commutation is not possible.
140. How is PNM inverter supplied in VSI fed synchronous motor?
When a PWM inverter is used, two cases may arise the inverter may be fed from a constant D.C.
source in which case regeneration is straight forward. The D.C. supply to the inverter may be
obtained form a diode rectifier. In this case an additional phase controlled converter is required on
the line side.
141. What is D.C. link converter and cycloconverter?
D.C. link converter is a two stage conversion device which provides a variable voltage,
variable frequency supply.
Cycloconverter is a single stage conversion device which provides a Variable voltage,
variable frequency supply.
142. What are the disadvantages of cycloconverter?
A cycloconverter requires large number of thyristors and ts control circuitry is complex.
Converter grade thyristors are sufficient but the cost of the converter is high.
143. What are the applications of cycloconverter?
A cycloconverter drive is attractive for law speed operation and is frequently employed in
large, low speed reversing mils requiring rapid acceleration and deceleration. Typical
applications are large gearless drives, e.g. drives for reversing mills, mine heists, etc.
144. Give the application of CSI fed synchronous motor.
Application of this type of drive is in gas turbine starting pumped hydroturbine starting,
pump and blower drives, etc.
145. What are the disadvantages of machine commutation?
The disadvantages of machine commutation are,
a. Limitation on the speed range.
b. The machine size is large
c. Due to overexciting it is underutilized.
146. What is the use of an auxiliary motor?
Sometimes when the power is small an auxiliary motor can be used to run up the
synchronous motor to the desired speed.
147. What are the advantages of brushless D.C. motor?
The brushless D.C. motor is in fact an inverter-fed self controlled permanent synchronous
motor drive. The advantages of brushless D.C. motor are low cost, simplicity reliability and
good performance.
BIG QUESTIONS AND ANSWER KEY
EE1351- Solid State Drives
1. Give a brief explanation on the selection of drives.[8M]
Ans: 1.Selection of Electric Motors
2.Kinds of Motors
3.Kinds of enclosures
Refer: ‘Solid state drives’ by V.Thiyagarajan pg.no. 2.41
‘Fundamentals of Electric drives’ by G.K Dubey pg.no.51
2. Give a brief explanation on the rating of drives.[8M]
Ans: 1.Requirement of a drive motor
2.Power losses &heating of motors
3.Heating &cooling of a Electric motor
Refer: ‘Solid state drives’ by V.Thiyagarajan pg.no. 2.37
‘Fundamentals of Electric drives’ by G.K Dubey pg.no.51
3.Explain the four quadrant operation of electric drives.[8M]
Ans: First Quadrant-Forward Motoring
Second Quadrant-Forward Braking
Third Quadrant-Reverse Motoring
Fourth Quadrant- Reverse Braking
T-Wm characteristics
Refer: ‘Solid state drives’ by V.Thiyagarajan pg.no. 2.33
‘Fundamentals of Electric drives’ by G.K Dubey pg.no.14
4.Explain the constant torque and constant HP operations.
Ans:
•V/f operation,
•constant torque and constant power operation
•V,T,Pm,Is, Wsl with respect to ‘a’ per unit frequency
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no.233
5. Describe the operation of 1fully controlled rectifier control of DC separately excited motor and
obtain the expression of motor speed for continuous and discontinuous modes of operations.
Ans: Circuit diagram of 1fully controlled rectifier
Mode of operation under Motoring – Continuous conduction mode
Discontinuous conduction mode
Mode of operation under braking – Continuous conduction mode
Discontinuous conduction mode
Va = 2Vmcos

wm = 2Vmcos- Ra Ta
k k2
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no.72
6. Explain the motoring operation and braking operation of three phase fully controlled rectifier control
of dc separately excited motor with aid of diagrams and waveforms. Also obtain the expression for
motor terminal voltage and speed.
Ans: Circuit diagram of 3fully controlled rectifier
Mode of operation under Motoring – Continuous conduction mode
Discontinuous conduction mode
Mode of operation under braking – Continuous conduction mode
Discontinuous conduction mode
Va = 3Vmcos

wm = 3Vmcos- Ra Ta
k k2
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no.103
7. Explain the operation of chopper for forward motoring and braking control of separately excited dc
motor with aid of diagrams, waveforms and speed-torque curves.
Ans: i)Circuit diagram of chopper drive for forward motoring
Waveforms of motor terminal & current – Continuous conduction
Discontinuous conduction
a)Duty interval b)Freewheeling interval equivalent circuits
Va=V, m = V - Ra Ta
k k2
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no.151
ii) Circuit diagram of chopper drive for Regenerative braking of DC motor
Waveforms of motor terminal & current – Continuous conduction
Discontinuousconduction
a)Energy storage interval b)Energy transfer interval equivalent circuits
Va= (1-)V
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no.159
8. Explain the operation of 1half controlled rectifier control of dc separately excited motor for
continuous and discontinuous modes of operations with aid of diagrams and waveforms. Also obtain
the expression for motor terminal voltage and speed.
Ans: Circuit diagram of 1half controlled rectifier
Mode of operation under Motoring – Continuous conduction mode
Discontinuous conduction mode
Va = Vm(1+cos)

wm = Vm (1+ cos)- Ra Ta
k k2
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no.101
9. Explain the operation of chopper controlled dc series motor for motoring and braking with circuit
diagrams and waveforms.
Ans: i)Circuit diagram of chopper drive for forward motoring
Waveforms of motor terminal & current – Continuous conduction
Discontinuous conduction
a)Duty interval b)Freewheeling interval equivalent circuits
Va=V, m = V - Ra Ta
k k2
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no.164
ii) Circuit diagram of chopper drive for Regenerative braking of DC motor
Waveforms of motor terminal & current – Continuous conduction
Discontinuousconduction
a)Energy storage interval b)Energy transfer interval equivalent circuits
Va= (1-)V
m - T curves, m - Prg curves
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no.164
10. Explain the closed loop speed control of DC drives.
Ans: Block diagram, operation – i)drive with current limit control
ii)drive with inner current loop control
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no.184
11. Explain the Four Quadrant operation by using Class-E chopper with aid of diagrams and waveforms.
Ans: Circuit diagram & operation – Mode I,II,III,IV
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no.175
12. Explain the two Quadrant operation by using Class-D and Class-C chopper with aid of diagrams and
waveforms.
Ans: Class-C Circuit diagram & operation – Mode I, II
Class-D Circuit diagram & operation – Mode I, IV
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no.169, 172
13. A 200 V, 875 rpm, 150 A separately excited motor has an armature resistance of 0.06. It is fed from
a 1fully controlled rectifier with an ac source voltage of 220 V, 50 Hz. Assuming continuous
conduction, Calculate
i)firing angle for rated motor torque and 750 rpm
ii)firing angle for rated motor torque and (–500)rpm.
iii)Motor speed for =1600 and rated torque.
Solution: E at rated operation E=V-IaRa=191 v
i) =29.30 ii) =1200 iii)N=-893.2 rpm
Refer: ‘Fundamentals of Electric drives’ by G.K Dubey pg.no.117
14. A 230 V, 960 rpm and 200 A separately excited dc motor has an armature resistance of 0.02. The
motor is fed from a chopper which provides both motoring and braking operations. The source has a
voltage of 230 V. Assuming continuous conduction, Calculate
i) duty cycle ratio of chopper for motoring operations at 350 rpm.
ii) duty cycle ratio of chopper for braking operation at rated torque and
350 rpm.
iii) If maximum duty ratio of chopper is limited to 0.95 and maximium
permissible motor current is twice the rated. Calculate the maximum permissible
motor speed and power fed to the source.
Solution: E at rated operation E=V-IaRa=226 v
i)Va = E+IaRa, = Va/ V ; =0.376
ii)Va = E-IaRa, 1-= Va/ V ; =0.66
iii)N=962 rpm, power= VaIa= 87.4 KW
Refer: ‘Fundamentals of Electric drives’ by G.K Dubey pg.no.139
15. A 220 V, 1500 rpm 50 A separately excited motor with armature resistance of 0.5, is fed from a 3
fully controlled rectifier. Available ac source has a line voltage of 440 V,50 Hz. A star-delta
connected transformer is used to feed the armature so that motor terminal voltage equals rated voltage
when converter firing angle is zero.
d. Calculate the transformer turns ratio.
e. Determine the value of firing angle when
a) motor is running at 1200 rpm & rated torque
b) When motor is running at (-800) rpm & twice the rated torque.
Assume continuous conduction.
Solution:
i) Va = 3Vm cos; t/f turns ratio=1.559

ii) a) =34.650 b) = 104.200
Refer: ‘Fundamentals of Electric drives’ by G.K Dubey pg.no.129
16. Discuss in brief about the control of an induction motor by stator voltage variation
using 3 phase voltage controller.
Ans:
•Ac voltage controllers circuit- star and delta
•Four quadrant ac voltage controllers circuit & operation
•Speed-torque curves
•Closed loop speed control
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no.274
17. Explain the of operation constant V/f control of induction motor and draw the
waveforms.
Ans:
•V/f operation,
•constant torque and constant power operation
•V,T,Pm,Is, Wsl with respect to ‘a’ per unit frequency
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no.233
18. Describe briefly the PWM inverter fed induction motor drive.
Ans: PWM inverter circuits
Control signal generation- sinusoidal PWM
Operation
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no.298
19. Explain the operation of induction motor fed by current source inverter.
Ans: Current source inverter circuit
CSI variable frequency drives – Block diagram, operation
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no.325, 335
20. Starting from the approximate equivalent circuit, derive an expression for the
torque-speed characteristics, based on this expression. How does this
characteristics change,
i) when stator voltage is varied (keeping frequency constant)
ii) when the rotor resistance is varied?
Ans :
•performance of 3IM & equivalent circuits
•speed-torque curves
•expression for torque T
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no.206, 226
21. How is dynamic/ regenerative braking achieved in a variable frequency V.S.I/C.S.I fed induction
motor drives?
Ans: VSI fed IM drive Block diagram, operation – dynamic braking
- Regenerative braking
CSI fed IM drive Block diagram, operation – dynamic braking
- Regenerative braking
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no.316, 336
22. In a pump drive, the fluid flow is to be varied from full down to 50 percent. Stator voltage controlled
3induction motor is used for driving the pump. If full load slip of the motor is 0.15. Evaluate
i)the maximum motor current to rated motor current ratio.
Also derive the expressions used.
Ans: Imax = 2 1.169
Irated 33 (1-s rated)s rated
Expression for Imax/ Irated
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no.279
23. Discuss the operation of an open-loop variable frequency voltage source inverter
fed induction motor drive.
Ans: Block diagram
Operation
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no.315
24. A 3, star connected, 50 Hz, 4 pole induction motor has the following
parameters in ohms per phase referred to the stator.
Rs =Rr’ = 0.034 & Xs =Xr’ = 0.18 
The motor is connected by the variable frequency control with a constant (V/f).
Determine the following for an operating frequency of 15 Hz.
i) The breakdown torque as a ratio of its value at the rated frequency for motoring
and braking.
ii) The starting torque and rotor current in terms of their values at the rated
frequency.
Solution:
i ) Motoring Tmax (a=0.2)
= 0.68
Tmax (a= 1)
Braking Tmax (a=0.2)
= 1.46
Tmax (a= 1)
ii) Ts (a=0.2) = 2.6 ; Irs’ (a=0.2) = 0.72
Ts (a=1) Irs’ (a= 1)
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no.235
25. Explain the closed loop scheme of
i) speed control of poly phase induction motor by rotor resistance control.
ii) Compare this scheme of control with the slip power recovery scheme.
Ans: i) Block diagram, operation
ii) Block diagram, operation
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no.362, 380
26. Derive an expression for the torque T for the rotor resistance control using the
chopper scheme.
Ans:
•Circuit diagram
•Operation, equivalent circuits
•Derivation for torque
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no. 356
27. State the merits and demerits of rotor resistance control of wound rotor induction
motor. What are its applications. (8M)
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no. 356
28. Explain schematic diagram two methods of super synchronous speed control of
slip ring induction motor under slip power recovery scheme. What are the
advantages and disadvantages of slip power recovery scheme?
Ans:
•Super synchronous speed control operation
•Modified Scherbius drive – diagram, operation
•Modified Kramer drive – diagram, operation
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no. 383
29.With block diagram explain the closed loop operation of slip power recovery
scheme of induction motor. Discuss on the power factor and locus diagram of
supply current for constant torque operation of slip power recovery scheme.
Ans:
•Closed speed control operation , block diagram
•Power factor and locus diagram
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no. 380, 368
30. Explain with schematic diagram of sub-synchronous speed control of slip-ring
Induction Motor under slip power recovery scheme. Derive the expression for
the torque. (16)
Ans:
•Sub synchronous speed control operation
•static Scherbius drive – diagram, operation
•Expression for torque
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no. 366
31. A 3, 400 V, 50 Hz, 960 rpm, 6 pole , star connected, wound rotor induction motor
has the following parameters per phase referred to the stator.
Rs = 0.3, Rr’ =0.5, X1= X2= 1.6, Xm= 35
Stator to rotor turns ratio is 2. The motor speed is controlled by the static rotor
resistance. The filter resistance is 0.01. The value of the external resistance is
chosen such that =0, the break-down torque is obtained at standstill. Dtermine the
value of external resistance.
Solution: R=1.37 
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no. 365
32. A 440 V, 50 Hz, 960rpm, star- connected wound rotor Induction Motor has:
Rs=0.15, Xr’ =0.6and Xm=20. The stator to rotor turns ratio is 2. This motor is
controlled by a rotor-chopper scheme. The filter inductor has a resistance of 0.01.
The external resistance is 4. For a duty cycle of 0.7 and a speed of 600 rpm,
evaluate the torque developed.
Solution: T= 258.98 Nm
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no. 365
33. Explain the operation of a ‘torque- angle control’ based self –controlled
synchronous motor drive.
Ans: Block diagram
Operation
Refer: ‘Power semiconductor controlled drives’ by Murphy and Turnbull
34. Write short notes on Brushless Excitation system. (8M)
Ans: diagram (4M)
Operation (4M)
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no. 407
35. Explain the working of a self controlled synchronous mode fed from a three phase
inverter. Why a self controlled synchronous motor is free from hunting oscillations?
Ans: Rotor position encoder operation -diagram
Brushless AC & DC motor –diagram, Operation
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no. 418, 423
36. i)Describe the self control of synchronous motor fed from VSI. Discuss about
separately controlled synchronous motor fed from VSI.
ii)Compare the above two schemes.
Ans: Self controlled mode - diagram, Operation
True synchronous mode - diagram, Operation
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no. 423, 412
37. Explain the closed loop control scheme of adjustable speed synchronous motor
drive.
Ans: VSI and CSI fed Synchronous motor drives
Block diagram (8M)
Operation (8M)
Refer: ‘Power semiconductor controlled drives’ by G.K Dubey pg.no. 421
38. Explain the operation of a ‘power factor control’ based self –controlled
synchronous motor drive.
Ans: Block diagram
Operation
Refer: ‘Power semiconductor controlled drives’ by Murphy and Turnbull

Human resource development

Human resource development
In terms of recruitment and selection it is important to consider carrying out a thorough job analysis to determine the level of skills/technical abilities, competencies, flexibility of the employee required etc. At this point it is important to consider both the internal and external factors that can have an effect on the recruitment of employees. The external factors are those out-with the powers of the organization and include issues such as current and future trends of the labor market e.g. skills, education level, government investment into industries etc. On the other hand internal influences are easier to control, predict and monitor, for example management
styles or even the organizational culture.
In order to know the business environment in which any organization operates, three major trends should be considered:
• Demographics – the characteristics of a population/workforce, for example, age, gender or social class. This type of trend may have an effect in relation to pension offerings, insurance packages etc.
• Diversity – the variation within the population/workplace. Changes in society now mean that a larger proportion of organizations are made up of "baby-boomers" or older employees in comparison to thirty years ago. Traditional advocates of "workplace diversity" simply advocate an employee base that is a mirror reflection of the make-up of society insofar as race, gender, sexual orientation, etc.
• Skills and qualifications – as industries move from manual to a more managerial professions so does the need for more highly skilled graduates. If the market is "tight" (i.e. not enough staff for the jobs), employers will have to compete for employees by offering financial rewards, community investment, etc.
In regard to how individuals respond to the changes in a labour market the following should be understood:
• Geographical spread – how far is the job from the individual? The distance to travel to work should be in line with the pay offered by the organization and the transportation and infrastructure of the area will also be an influencing factor in deciding who will apply for a post.
• Occupational structure – the norms and values of the different careers within an organization. Mahoney 1989 developed 3 different types of occupational structure namely craft (loyalty to the profession), organization career (promotion through the firm) and unstructured (lower/unskilled workers who work when needed).
• Generational difference –different age categories of employees have certain characteristics, for example their behavior and their expectations of the organization.
While recruitment methods are wide and varied, it is important that the job is described correctly and that any personal specifications are stated. Job recruitment methods can be through job centres, employment agencies/consultants, headhunting, and local/national newspapers. It is important that the correct media is chosen to ensure an appropriate response to the advertised post.
Human Resources Development is a framework for the expansion of human capital within an organization. Human Resources Development is a combination of Training and Education that ensures the continual improvement and growth of both the individual and the organisation. Adam Smith states,“The capacities of individuals depended on their access to education”.Kelly D, 2001[1]Human Resources Development is the medium that drives the process between training and learning. Human Resources Development is not a defined object, but a series of organised processes, “with a specific learning objective” (Nadler,1984)[2] Human Resources Development is the structure that allows for individual development, potentially satisfying the organisation’s goals. The development of the individual will benefit both the individual and the organisation. The Human Resources Development framework views employees, as an asset to the enterprise whose value will be enhanced by development, “Its primary focus is on growth and employee development…it emphasises developing individual potential and skills” (Elwood, olton and Trott 1996)[3] Human Resources Development can be in-room group training, tertiary or vocational courses or mentoring and coaching by senior employees with the aim for a desired outcome that will develop the individual’s performance. A successful Human Resources Development program will prepare the individual to undertake a higher level of work, “organised learning over a given period of time, to provide the possibility of performance change” (Nadler 1984). Human Resources Development is the framework that focuses on the organisations competencies at the first stage, training, and then developing the employee, through education, to satisfy the organisations long-term needs and the individuals’ career goals and employee value to their present and future employers. Human Resources Development can be defined simply as developing the most important section of any business its human resource by, “attaining or upgrading the skills and attitudes of employees at all levels in order to maximise the effectiveness of the enterprise” (Kelly 2001)[4]. The people within an organization are its human resource. Human Resources Development from a business perspective is not entirely focused on the individual’s growth and development, “development occurs to enhance the organization's value, not solely for individual improvement. Individual education and development is a tool and a means to an end, not the end goal itself”. (Elwood F. Holton II, James W. Trott Jr)[5].