Manual for K-Notes
Why K-Notes?Towards the end of preparation, a student has lost the time to revise all the chapters from his / her class notes / standard text books. This is the reason why K-Notes is specifically intended for Quick Revision and should not be considered as comprehensive study material.
What are K-Notes?
A 40 page or less notebook for each subject which contains all concepts covered in GATE Curriculum in a concise manner to aid a student in final stages of his/her preparation. It is highly useful for both the students as well as working professionals who are preparing for GATE as it comes handy while traveling long distances.
When do I start using K-Notes?
It is highly recommended to use K-Notes in the last 2 months before GATE Exam (November end onwards).
How do I use K-Notes?
Once you finish the entire K-Notes for a particular subject, you should practice the respective Subject Test / Mixed Question Bag containing questions from all the Chapters to make best use of it.
BASIC CONCEPTS
In continuous time signals independent variable is continuous and thus these signals are defined for a continuum of values of independent variable.
Discrete time signals are only defined at discrete times and consequently for these signals the independent variable takes discrete set of values.
Representation of continuous time signals
We use symbol ‘t’ to denote independent variable for continuous time signal.
These signals can be represented by a wave form as shown below
If possible, these can also be represented by a mathematical function like
x(t) = sin t
Representation of discrete time signal
We use symbol ‘n’ to denote independent variable for discrete time signal.
These signals can be represented as a series of numbers like x[n] = [5, 4, 5, 7, 9, 2……]
Arrow indicates reference point or x [0]
If possible, we can represent the same by a function like x[n] = sin
n4Energy & Power Signals Interval
,
Energy of continuous time signal
T
2
2 T lim E x t dt x t dt T
Energy of discrete time signal
N
2
2 n n N lim E x n x n T
Power of continuous time signals
2 T T lim 1 P x t dt T 2T
Power of discrete time signals
N
2 n N lim 1 P x n N 2N 1
Signals having non-zero (finite) power and infinite energy are called as Power Signals. ex. x(t) = sint
Signals having finite (non-zero) energy and zero power are called as Energy Signals. ex. x[n] = [1, 2, 3, 4]
The bounded signal radiate finite energy and periodic signal radiate finite average power.
Even & Odd signals
A signal is said to be “even” if it satisfies the condition x(t) = x (–t) or x [n] = x[–n]
A signal is said to be “odd” if it satisfies the condition x(t) = –x(–t) or x [n] = –x[–n]
Any signal (even those which are neither odd nor even) can be broken into odd & even parts
Odd Part x t0
x t x t 2 ;
0 x n x n x n 2 Even Part x te
x t x t 2 ;
e x n x n x n 2 Periodic and Aperiodic Signals
A signal is said to be periodic with period “T” or “N” if x(t + T) = x(t)
x[n + N] = x[n]
Otherwise, the signals are said to be aperiodic. Classification of systems
(i) Linear & Non-Linear Systems For Linearity
if x t1
y t1
x t2
y t2
then, this condition must be true
1 1x t
2 2x t
1 1y t
2 2y t
Example : y(t) = t x (t) is lineary[n] = 2x [n] + 3 is non-linear
(ii) Time Invariant & Time-variant Systems For system to be time-invariant the
following condition must hold true x(t - ) y(t – )
It means that following two realizations must be equivalent
The simplest way to verify this is to check the coefficient of ‘t’ inside x(t) eg. y(t) = tx(t) is time invariant
but y(t) = tx(2t) is time variant as coefficient of ‘t’ in side x(t) is not ‘1’
Otherwise, you need to verify the system equivalence shown above.
(iii) Causal & Non-causal Systems
The output should depend only on present & past values of input.
h t 0 V t 0 For discrete time system
h[n] = 0 V n < 0
(iv) Stable & Unstable Systems
Every Bounded input should produce a bounded output.
K DT : h k
; CT : h
d
(v) LTI systems with or without memory
The output at any time should depend only on value of input at the same time.
For discrete time system h[n] = 0 V n 0 h[n] = k [n]
For continuous time system h(t) = 0 V t 0 h [t] = k [t]
(vi) Invertible Systems
The system is invertible if there exists h1(t) such that Thus h(t) * h1(t) =
tFor discrete time, h[n] * h1[n] =
n Shifting and Scaling operations Shifting Delay
if
shift the waveform right by the amount of delay Advance if
Scaling Compression
if
Replace upper & lower limit by original limit divided by compression factor
Expansion
if
Replace upper & lower limit by original limit multiplied by expansion factor.
Note : If both scaling and shifting are given in the question . Ex. x(3t-2)
1. shift the waveform right by the amount of delay
2. Replace upper & lower limit by original limit divided by compression factor
This method is applicable for both continuous and discrete time signal.
LTI system (Linear Time Invariant Systems)
Any continuous time or discrete time system can be represented in terms of impulses.
x t x t d
k x[n] x k n k LTI systems are characterized on the basis of Impulse Response h(t) or h[n]
The response of a system with impulse as an input is called as impulse response.
Due to time invariance property of LTI system if
n h n
n k
h n k
since
K x n x k n k
k y n x k h n k x n * h n
= convolution sumfor continuous time domain
k
y t x h t x t * h t
The condition for causality of system then becomes h[n] =0 V n < 0 ; h(t) = 0 V t < 0
Calculating convolution sum
Suppose x [n] = u[n] h[n] = [1, 2, 5, 7, 9]
Draw plots of both x[n] & h[n]
Flip either x[n] or h[n] about y-axis Here, we flip x[n]
For calculating y[n], shift x[–k] to right by amount ‘n’ For y[0]
The only overlapping between the two is at k = 0, –1, –2 y [0] = x[0] h [0] + x [1] h [–1] + x [2] h [–2] = 1 x 5 + 1 x 2 + 1 x 1
= 8
y [1] = x [0] h [1] + x [1] h [0] + x [2] h [–1] + x [3] h [–2] = 1 x 7 + 1 x 5 + 1 x 1 x 1 x 2 = 15
Similarly, we can calculate all values of y[n] y[n] = [2, 3, 8, 15, 24, 24……..]
Calculating Convolution Integral Assume x (t) = u (t) h (t) = Step 1 Flip either x(t) or h(t) Here, we flip h(t) Step 2
Shift h() by amount “t” to the right to calculate y(t) by calculating overlapping between
h t & x Overlapping area =
0 1 t 1.1d 1 t
if t < – 1if t > 1
overlapping area = 2
y (t) is shown in adjoining figure:
Properties of Convolution Sum 1) Commutative Property x[n] * h[n] = h[n] * x[n] 2) Distributive Property y1[n] = x[n] * h1[n] y2[n] = x[n] * h2[n] y [n] = y1[n] + y2[n] = x[n] * h1[n] + x2[n]*h2[n] = x[n] * { h1[n] + h2[n] } 3) Associative Property {x[n] * h1[n] }* h2[n] = { x[n]* h2[n] } * h1 [n]
Same properties will apply for continuous time domain for convolution integral.
Parallel & Cascade structure of LTI systems Parallel:
y[n] = x[n] * [h1 [n] + h2 [n]]
Cascade:
y[n] = x[n] * ([h1 [n] + h2 [n]])
Frequency Response
The frequency response of any LTI system is given by its Fourier Transform.
DT:
jw
jwn n H e h n e
CT: H jw
h t e
jwtdt
Group delay & Phase delay
Assuming transfer function of system is H(s) input is x(t)= ejwt Output: H jw e
jwt H jw e
j w ejwt = H jw e
j
wt w
w Arg H jw
Group Delay, g
w d w
dw Phase Delay,
w
w w Continuous – Time Fourier series
Fourier states that any periodic signal can be represented by a set of complex exponential signals provided that it satisfies Drichlet Conditions.
Drichlet conditions
(i) Over any period x(t) is absolutely integrable
i.e., T
0
x t dt
(iii) It should have finite number of discontinuities in the given interval
Note : for distortion less transmission of the of a signal with some finite frequency content through a continuous time LTI system , the frequency response of the system must satisfy these two conditions.
1. The magnitude response H( j ) must be constant for all frequencies of interest ;
that is, we must have
H( j ) C For some constant C
2. For the same frequencies of interest, the phase response arg H( j ) must be linear in
frequency, with slope –to and intercept zero ; that is, we must havearg H( j )
t o Fourier series as generally expressed in 2 forms. Trigonometric
Exponential
Trigonometric Fourier Series Analysis equations 0 T
0 1 a x t dt T
0 T k 0 2 a x t cos k t dt T
where 0 2T
0 T k 0 2 b x t sin k t dt T
Synthesis equations
0
k 0
k 0 k - k k 0 k 0 x t a a cosk t b sink tExponential Fourier Series Analysis equations T
jk t0 k 0 1 C x t e dt T
Synthesis equations
jk t0 K k x t C e where 0 2TRelation between T.F.S. and E.F.S.
co a 0 n n n a jb C 2 n n n a jb C 2
Important facts about Trigonometric Fourier series (i) Any odd signal contains only sine terms in Fourier series. (ii) Any even signal contains only cosine terms in Fourier series. (iii) For half–wave symmetric signal
x t
T2
x t
Only odd harmonics are present i.e., k = 1, 3, 5…….
Properties of complex exponential Fourier Series (i) Linearity If x t
F.S. ak y t
F.S. bk then Ax (t) + By (t) F.S. Aak + Bbk (ii) Time-shifting if x t
F.S. ak
0
0 0 k -jk t F.S. x t t e a where 0 2T (iii) Time-Reversal if x t
F.S.ak x t
F.S.a kFor odd signal For even signal
x(t) = – x(–t) x(t) = x (–t)
ak ak ak ak
(iv) Time – Scaling
if x t
F.S.ak x t
F.S.a kbut 0 is replaced by
0 , though Fourier series coefficients remain same. (v) Multiplication if x t
F.S.ak y t
F.S.bk z t
x t y t
F.S.ck k p k p P C b a
= convolution sum(vi) Parseval’s Relation
Energy in time domain = Energy frequency Domain
2 k 2 k T 1 x t dt a T
where x t
F.S.ak Discrete –Time Fourier seriesFor a discrete-time signal, with period ‘N’ the following equations are used for Fourier series. Analysis equations
N k 2 j KnN C x n e
0 2N
0 N k j Kn C x n e
Synthesis equations
0 N K j Kn x n C e
The properties of Fourier series coefficients are same as continuous time Fourier series with one additional property.
CK N CK
That is, Fourier series coefficients are periodic
IMPORTANT DUALITY
A signal discrete in one domain is periodic in other domain & vice versa.
Example: For continuous Time Fourier Series, x (t) is periodic in time domain & hence Fourier Series exists where coefficients exist for frequency integral multiple of "0" & hence is discrete.
Fourier Transform
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 X jw
x t e
jwtdt
Synthesis equation x t
1 x jw e dw
jwt 2
Properties of Continuous Time Fourier Transform
Signal Fourier Transform
x(t) X(jw) y(t) Y(jw) Ax(t)+By(t) AX(jw)+BY(jw) x(t-t0) x*(t) X*(-w) x(-t) X(-w) x(at) x(t)*y(t) X(jw)Y(jw) jwX(jw) x(t)y(t) tx(t) Ev{x(t)} Re{X(jw)} Od{x(t)} jIm{X(jw)} X(t) 2πx(-w) X(w-w0) Parseval’s Relation x t
2 1 x w dw
2 2
j t0 e X w jw 1 X a a d x(t) dt 1 X(w) * Y(w) 2
t x d 1 X jw
X 0
w jw
d j X jw dw
0 j t e x tSome common Fourier Transform Pairs
Signal Fourier Transform
0 jkw t k K a e
k
0
k 2 a k
0 jkw t e 2
0
cos w t 0
0
0
sin w t 0
0 0 j 1 2
n t nT
K 2 2 k T T
1 1 1, t T x t 0, t T 1 2sin T (sin wt)/πt
1, w x 0, w
t 1 u(t) 1
j
t t0
e j t0
at e u t ,Re a 0 1 a j Discrete Time Fourier Transform Analysis equation
j
j n n X e x n e
Synthesis Equation
2 j j n 1 x n X e e d 2
Properties of Discrete Time Fourier Transform
Signal Fourier Transform x[n] y[n]
X periodic with period 2 Y ax[n] + by [n] aX
bY
0 x n n 0 j n e X x*[n] X *
0 j n e x n X
0
x [–n] X
k x n | k , if n is multiple of k x n 0, is n is not multiple of k
X k x [n] * y [n] X
Y n x [n] dx
j d
Ev x n Re {X()} Od {x [n]} j Im {X()} Parseval’s Relation
2
2 n 2 1 x n X d 2Some common Fourier Transform Pairs
Signal Fourier Transform
k 2 jk N n K N a e
k k 2 k 2 a N
0 j n e
0 2 2
cos0n
0 2 0 2
0 sin n
0 2 0 2 j
x [n] = 1 2
2
1 1 1, n N x n N 0, n N , n 2 and x [n + N] = x [n] k k 2 k 2 a N
k n kN
k 2 2 k N N
1 1 1, n N x n 0, n N
1
1 sin N 2 sin 2 sinWn Wsinc Wn n
1, 0 W x 0, W < 0 n n e j n0Laplace Transform
Laplace Transform is more general than Fourier Transform but can only be computed in Region of Convergence (ROC), so it cannot be computed V s
ROC =
t S jw; such that x t e dt
Laplace transform becomes Fourier transform for 0, if it lies in ROC.
Analysis Equations
for bilateral Laplace Transform H(s) = h t e dt
st
for unilateral Laplace Transform
H(s) =
st 0 h t e dt
Synthesis Equation x(t) = j
st j 1 x s e ds 2 j
Properties of ROC(i) ROC consists of a collection of lines parallel to jw–axis in s–plane.
such that
x t e t dt(ii) If X (s) is rational, then ROC does not contain any poles.
(iii) If x(t) is of finite duration & absolutely integrable, then ROC is entire s-plane.
(iv) If x(t) is right sided signal (i.e., it is zero before some time) and if Re(s) = 0 is in the ROC, then all values of s for which Re(s) > 0 are also in ROC.
(v) If x(t) is left sided, (i.e., if it is zero after some time), and if Re (s) = 0 is in ROC, then all values of s for which Re(s) < 0 are also in ROC.
(vi) If x(t) is two–sided signal and if the line Re (S) = 0 is in ROC, then the ROC consists of a strip in s–plane include the line Re (S) = 0
(vii) If X(s) is rational, and
x(t) is right sided signal, then ROC is right of right most pole. x(t) is left sided signal, then ROC is left of left most pole.
Properties of Laplace Transform
Signal Transform ROC
x(t) X(s) R x1(t) X1(s) R1 x2(t) X2(s) R2 ax1(t) + bx2(t) aX1(s) + bX2(s) At least R1 Ռ R2
0
x t t est0X s
R
0 s te x t X s s
0
Shifted version R [i.e., s is in ROC if
s s 0
is in R]x (at) 1 X
sa a
Scaled ROC i.e., s is ROC if
s a is in R
1 2 x t * x t X s X s1
2
At least R1 Ռ R2
d x t dt sX s
At least R tx(t) dx s
ds R
t x d
1X s
s At least RSome common Laplace Transform Pairs
Signal Transform ROC
t 1 All s u(t) 1 s Re {s} > 0 –u(–t) 1 s Re {s} < 0
n 1 t u t n 1 ! n 1 s Re {s} > 0
n 1 t u t n 1 ! n 1 s Re {s} < 0
at e u t 1 s a Re {s} > –a -e u tat
1 s a Re {s} < –a
n 1 at t e u t n 1 !
n 1 s a Re {s} < –a
n 1 at t e u t n 1 !
n 1 s a Re {s} > –a
t T
esT All s
0 cos t u t 0 2 2 s s Re {s} > 0
0 sin t u t 0 0 2 2 s Re {s} > 0
at 0 e cos t u t
2 02 s a s a Re {s} > –a
at 0 e sin t u t
0 0 2 2 s a Re {s} > –aInitial and Final Value Theorem
lim x 0 sX s initial value s
lim x s X ss 0 Final value, first stability should be ensured, else final value does
not exist.
Analysis of LTI system using Laplace Transform Stability
h t dt
; ROC of H(s) should include 0. Causalityh(t) = 0, t < 0 i.e., right sided signal ROC should be right sided
ROC should include Right half plane. but converse is not true.
Z – Transform
It is generalization of Discrete Time Fourier Transform Analysis Equation
k k H z h k z Synthesis Equation
n 1 1 h[n] H z z dz 2 jIndicates integration around counter clockwise circular contour centered at origin & with radius r.
ROC for Z-Transform
Z – Transform also exists only inside ROC
nn x n r
is the condition for ROC. Mapping from s-plane from z–plane The jw-axis is mapped to unit circle in z–plane.
Right Half plane is mapped to exterior of unit circle.
Left Half plane is mapped to interior of unit circle.
Properties of ROC
(i) The ROC x(z) consists of a ring in the z – plane centered about the origin. (ii) The ROC does not contain any poles.
(iii) If x[n] is of finite duration, then ROC is the entire z – plane except possibility at z = 0 and/or z =
(iv) If x[n] is a right sided sequence and if the circle, | z | = r0 is in the ROC, then all finite values of z, for which | z | > r0 will also be in ROC.
(v) If x[n] is a left sided sequence, and the circle | z | = r0 is in ROC, then all finite value of z, for which 0 < | z | < r0 will be in ROC.
(vi) If x[n] is two sided sequence and if circle | z | = r0 is in the ROC. Then ROC will consist of a ring in z-plane which consist of ring | z | = r0.
(vii) If X (z) is rational and
x[n] is right sided than ROC is outside of outer most pole. x[n] is left sided then ROC is inside of inner most pole. (viii) If x[n] is causal, ROC includes z = provided x[n] = 0, n < 0. If x [n] is anti – causal, ROC includes z = 0 provided x [n] = 0, n > 0.
(ix) A causal LTI system with rational system function is stable if all poles inside the unit circle that is have magnitude, | z | < 1.
Properties of z–Transform
Signal Transform ROC
x[n] X(z) Rx
1 x n X z 1
R 1
2 x n X z 2
R2
1 2 ax n bx n aX z1
bX z 2
At least R1 Ռ R2 0 x n n zn0X z
Rx with addition or deletion of origin
0 j n e x n X e
j 0z
Rx
0 n z x n 0 z X z z R 0 x x[–n] X z
1 1
x z s.t z R
x r , n=rk
w n 0, n rk for some r
k X z 1 k 1k x x R i.e.,z s.t z R
2
1 x n * x n X z X z 1
2 At least R1 Ռ R2 nx[n] zdX z
dz Rx except addition or deletion of zero
n k
x k 1
1 X z 1 z x R z 1Some common Z –Transform pairs
Signal Transform ROC
n 1 All z
u n 1 1 1 z | z | > 1
u n 1 1 1 1 z | z | < 1
n m
zm All z except 0 (if m > 0) or
(if m < 0)
n a u n 1 1 1 az | z | > | a |
n a u n 1 1 1 1 az | z | < | a |
n na u n
1 2 1 az 1 az | z | > | a |
n na u n 1
1 2 1 az 1 az | z | < | a |Initial & Final value Theorem
lim x 0 X z z Initial value
lim 1 x 1 X z z z 1 Final valueIn z – transform also, stability must be verified before using final value theorem.
Sampling
Continuous Discrete Time
Time signal signal
Nyquist Sampling Theorem
signal for the signal to be recovered from its samples.
wS 2wM
Note: For this condition signal spectrum should be centered around y-axis.
Band-pass Sampling Theorem
If the signal spectrum is band-pass which means it has minimum & maximum frequency f = lower frequency ; L fu = upper frequency
u u L f K ,where
f f indicates Greatest Integer function
u S 2f w K xp(t) = x(t) p(t)
n p t t nTT = sampling interval ; x tp
Sampled signal x(t) = continuous time signal
p n x t x t t nT
P 1 X w X w * P w 2
s
k 2 P w w kw T
s
P k 1 X w X w kw T
; ws 2 T The spectrum of sampled signal is just repetition of actual spectrum at integral multiples of ws.
If ws 2wM , adjacent samples of spectrum overlap, called as aliasing.
Discrete Fourier Transform
DFT of n – point sequence is given by: Analysis equation:
N 1
n 0 j2 kn N X k x n .e , k = 0, 1, 2…………., N–1Synthesis equation:
N 1 K 0 j2 kn N 1 x n X k e N , n = 0, 1, 2..………., N – 1Each point of DFT require N complex multiplications and (N – 1) complex additions. Therefore, N point DFT will required N2 complex multiplications and N (N – 1) complex additions. Properties of DFT Sequence Transform x[n] X(k) x1[n] X1[k] x2[n] X2[k] x[n + N] = x[n] X(K+N)= X(k)
2 2
1 1 a x n a x n a x k1 1
a x k2 2
N 1 N 1 2 n 0x n x m n
Where x2
m n
Nx2
N m N
2 1 x k x k
N
x n x N n X(N – k)
N x n
j2 kN X K e
j2 nN X n e X K
N x*(n) X*(N – k)
1 2 x n x n
N 1 2 1X K x k N Circular convolution of 2 DFT sequences Parseval’s Relation N 1
2 N 1
2 n 0 K 0 1 x n x k N
Fast Fourier Transform (FFT) Algorithms
These are the algorithms for computing DFT when the size N is a power of 2 or when it is a power of 4.
Direct computation of DFT is inefficient because it does not exploit the properties of symmetry and periodicity of the phase factor, WN ej2N
Symmetry property : WNK+ 2N WNK Periodicity property : K+N K N N W W DFT can be expressed as : R
N 1 R
I
n 0 2 kn 2 kn X K x n cos x n sin N N
I
N 1 R
I
n 0 2 kn 2 kn X K x n sin x n cos N N
No. of operations required for direct computation of DFT 1) 2N2 evolutions of trigonometric functions.
2) 4N2 Real multiplications. 3) 4N (N – 1) real additions.
Radix – 2 FFT algorithm
There are two types of FFT algorithm: 1) Decimation in time.
2) Decimation in frequency.
Radix – 2 algorithm can be implemented over N – point DFT sequence if N =2 . n
We divide the given time sequence by 2 till we get the prime factor.
We split the N–point data sequence into two N
2 point data sequences f n1
andf n2
corresponding to even – numbered and numbered samples of x(n).
N 1
Nnk n 0 X K x n W
, K = 0, 1,… ,N – 1.
Nnk
Nnk n even
x n W n odd
x n W
N
N N2 1 N2 1 2r 1k 2rk r 0 r 0 X k x 2r W x 2r 1 W =
N N
N N2 1 N2 1 rk rk 2 2 2 r 0 x 2r W W r 0 x 2r 1 W
But N2 N 2 W W . N j2 2 N 2 j N 2 2 N 2 W e e W
N 1 N 1 2 2 N N 2 2 rk rk N r 0 r 0 k X k x 2r W W x 2r 1 W =
N2 1 2 G K W H K f k f k , k = 0, 1, …………, N – 1. Now, computation of X[k] requires 2 N 2 N N2 N
2 2 2 2
complex multiplications.
We can further decimate f n1
and f n2
in time. Thus f n1
would result in two N 4 point sequences. f n decimation by 2 would also result in two N2
4 point sequence.
We can deduce finally that total number of complex multiplication is reduced to
2
Nlog N
FFT computation for 8-point DFT
Order of input is bit reversed, due to decimating twice in time domain.
Number of butterflies in N
2 per stage.
Number of stage islog N2 .
Butterfly computation.
Decimation in frequency
We split the DFT formula into two summations, one of which involves the sum over the first N 2 data points and the second sum involves the last N
Thus,
N 12 N 1 Kn Kn N 2 N N n 0 n X k x n W x n W
N 1 N Nk 2 2 Kn 2 Kn 0 1 N N N n 0 n N X k x n .W W x n W 2 Since
2 Nk Nk jN 2 k j k 2 N W e e 1 Therefore,
N 1 2 Kn k N n 0 N x k x n 1 x n W 2
Now, divide X(k) into even and odd numbered samples.
N 1 2 Kn N 2 n 0 N X 2k x n x n W 2 , k = 0, 1, ………. N 1 2 .
n Kn N 2 N 12 n 0 N N X 2k 1 x n x n W W 2 , k = 0,1, ………, N 12 . Total number of complex multiplications areNlog N2
2 .
Total Number of complex addition are Nlog N. 2 Decimation in frequency for N = 8 FFT algorithm
Note :
1) Order of input is normal while order of output is bit reversed. 2) Number of stages is log N. 2
3) Number of butterflies is N
2 for stage.
Butterfly computation for decimation in frequency.