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SEMMA13 Signals / Transforms
This covers Fourier Series and Transforms, Laplace Transforms, Z
transforms and Correlation and Convolution
In effect we are looking at representing signals in different ways.
A signal may vary with time, f(t)
That signal, however, is made up of signals of different frequencies,
so we can see how f varies with frequency.
We can transform between the time and frequency domains
The Laplace domain/operator can be used for modelling systems
For sampled systems, the sampled data z operator is used.
Convolution is relevant as the output of a system is in fact the
convolution of its input and the system transfer function.
Correlation is important in processing signals…
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Fourier Series
By way of introduction, consider the signals below, comprising one or
more sinusoids superimposed….
As more
sinusoids are
added, we start
to approximate
a square wave
1
1
0
0
-1
-1
0
4π
1
1
0
0
-1
-1
0
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2π
2π
4π
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0
2π
4π
0
2π
4π
2
Fourier Series
The square wave was approx by the
“Fourier Series”:
1
0
4
sin(3t)
sin(5t)
sin(7t) 
sin(t)
+
+
+

 
3
5
7 
-1
0
2π
4π
Here a sawtooth is approximated
sin(t) -
sin(2t)
sin(3t) sin(4t)
+
2
3
4
In general, series of sin and/or cos functions are needed
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Finding Fourier Series
Any repetitive signal f(t) with period T can be expressed by


a0
f(t) =
  an cos(n2 t/T) +  bn sin(n2 t/T)
2 n=1
n=1
where
an =
T
2
2
 f(t) cos (n2 t/T) dt
T T
2
T
2
2
bn =
 f(t) sin (n2 t/T) dt
T T
2
n is constant integer
NB an and bn are functions of n
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Using Complex Numbers
It is generally easier, however to use complex numbers
If we defined cn = 1 (an  jbn ), so c0 = 1 a0
2
2
T
2
2
cn = 1
f(t) cos (n2 t/T)-j f(t) sin (n2 t/T) dt
2 T T

=
T
2
2
1
 f(t) exp(-j n2 t/T) dt
T T
2
So the Fourier Series is given by

f(t)   cn exp(j n2 t/T)
n  -
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For Squarewave
1
Square wave is 1 for t = 0 to π
and -1 from – π to 0
(or from π to 2 π)
0
-1
0
2π
4π
T
2
2
1 0
1
cn =
 f(t) exp(-jnt) dt =
 -exp(-jnt) dt   exp(-jnt) dt
T T
 
0
2
0

1  exp(-jnt) 
1  exp(-jnt) 
1
= 
+

=
1-exp(jn )-exp(-jn )+1




jn
jn
   
 0  jn
2 1  cos(n ) 
1
=
2  2cos(n ) =
 jn
jn
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Continued
cn =
2 1  cos(n ) 
jn
If n is even, cos(n) = 1, so cn = 0.
4
If n is odd, cos(n) = -1, so cn =
.
jn
d
1-cos(n ) 
sin(n )
dn
c0 = lim
= lim
=0
d
j

n 0
n 0
 jn 
dn


4
f(t) =  cn exp(j nt)=

n  -
n(odd only)  - jn
L’Hopital’s
rule
 cos(nt) + j sin(nt) 
As cos(nt) = cos(-nt), cos(nt)/n + cos(-nt)/-n = 0
So f(t) =
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4
sin(3t) sin(5t) sin(7t)

sin(t)




...

 
3
5
7

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Spectra
If n is even, cn = 0; if n is odd, cn =
4
.
jn
Hence at most frequencies the signal amplitude is 0,
But at odd frequencies it is defined by 4/jnπ
At these frequencies, this has amplitude 4/nπ and phase –π/2 rads
amplitude
4/π
phase
1
3
5
7
1
3
5
7
f
-π/2
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f
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Phase Shifted
1
Square wave is 1 for t = 0 to π/2,
0
-1 from π/2 to 3π/2
and 1 from to 3π/2 to 2π
-1
0
Hence it is the previous square wave shifted by 90O.
2π
4π
As should be expected, the amplitude spectrum is unchanged.
The phase spectrum is 0 at frequencies 1, 3, 5, 7 …
It can be shown that cn =
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2 1  cos(n ) 
n
=
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at n = 1, 3, 5, ...
n
9
FS of Rectangular Pulses
1  /2
A
 /2
cn =
exp  -jn2 t/T  
 A exp  -jn2 t/T  dt = 
 / 2
T  / 2
Tjn2 /T
=
A  exp  jn /T   exp  -jn /T  
Asin(n /T)
=


n
j2
n

A sin(n /T)
A
n 

=
=
sinc  
T n / T
T
T
sin x
NB sinc(x) =
x
This is purely real,
Phase spectrum is 0, amplitude spectrum is modulus of above
If phase shift, amplitude spectrum unchanged…
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Amplitude Spectrum
First note plot of A sinc  x 
Then amplitude spectrum is as follows (note is |sinc|)
|cn|
cn =
f
1/T 2/T 1/
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A
n
sinc  
T
T
The ‘envelope’ is
|sinc| function
2/
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What if vary period
Here period T = 0.4,
0.8 and 2, the pulse
width is  = 0.1, and
the pulse height is 1.
Spectra lines get
closer as T becomes
larger, but amplitude
is reduced
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Fourier Transform
Fourier Series can be used for repetitive signals.
Signals have amplitude and phase at discrete frequencies
Some signals are not repetitive, but it is still valid to consider their
frequency properties
In fact these have amplitude and phase at all frequencies.
So signals can be represented as functions of time or of frequency
A function f(t) can be transformed to one of frequency F(f)
Similarly F(f) can be transformed to f(t)
On the next slide is the derivation of these Fourier Transforms

F(f) =  f(t) exp(-j 2 f t) dt

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
f(t) =  F(f) exp(j 2 ft) df
-
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Fourier Transform
Let’s see what happens to Fourier Series integrals if period T ∞
cn =
T
2
1
 f(t) exp(-j n2 t/T) dt
T T
2
Define

f(t)   cn exp(j n2 t/T)
n  -
1
= f, a small amount of frequency, so f = n f.
T
Frequency Specrum F(f) = lim CnT = lim
T 

 F(f) =  f(t) exp(-j 2 f t) dt

T
2
 f(t) exp(-j n2f t) dt
T   T
2
For f(t) summation of cn terms becomes integral of F(f) …


f(t)   cn exp(j n2 t / T) =  F(f) exp(j 2 ft) df
n  -
-
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Properties of FT
Various properties of FT follow from the definitions
Suppose f1(t) transforms to F1(f) and f2(t) transforms to F2(f)
Basic
f1(t) + f2(t) transforms to F1(f) + F2(f)
K f1(t) transforms to K F1(f), for a constant K.
Time Shift
f1(t-) transforms to F1(f) exp(-j2f)
Freq Shift
F1(f-) transforms to f1(t) exp(j2t)
Differential
df(t)/dt transforms to j 2 f F(f)
Duality
if f(t) transforms to F(f), F(t) transforms to f(-f)
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Fourier Transform of Pulse
A
A
-T T
2T

T
F(f) =  f(t) exp( -j 2 f t) dt =  A exp( -j 2 f t) dt
-
T
A
A
T
=
exp(
-j
2

f
t)
=

T
exp( -j 2 fT)-exp(j 2 fT)
-j 2 f
-j 2 f
= 2AT
sin(2 f T)
= 2AT sinc (2fT)
2 fT
Second Pulse is first shifted by T, so … its FT is
F(f) =2ATsinc(2fT) exp(-j2πfT)
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On Impulse
A unit impulse, denoted
, has area 1 and width 0.
An Impulse of area A can be considered as the first pulse on last
page whose area is 1 but whose width is zero:
2AT = A and T = 0,
So, as T  0,
F(f) = 2AT sinc(2fT)  A sinc(0) = A
By duality principle, if f(t) = A, a constant at all times,
F(f) is a signal at frequency 0 only (the d.c. component)
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Impulse Train
Fourier Series of Train of Pulses is cn =
A
n
1
sinc   at intervals
T
T
T
Pulses become impulses if  0 and A  = A.
As   0, but A = A, cn 
A
1
1
sinc  0  =
at intervals
T
T
T
Hence FT of a train of unit impulses at interval T is a series
of impulses of height 1/T at frequency intervals 1/T
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Exponential
exp(- t)
Exponential f(t) = 
0

t0
t<0

F(f) =  exp(- t)exp( -j 2 f t) dt
0

=  exp(-t   j 2 f ) dt
0
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=
1

exp  -t(  j 2 f )  
0
  j 2 f
=
1
  j 2 f
=
1
  j 2 f
 0  1
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Of a Step …
We find this in stages …
consider signum function
A
sgn(t) = 
-A
t0
if
t0


Not absolutely integrable  | sgn(t) | dt =  A dt = , so cant do its FT


But can do FT of sgnexp and see what happens when
0
A
t0
A exp(- t)
sgn exp(t) = 
if
t0
-A exp( t)
t
1/
-A
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FT of SigExp
0

F(f) =  -Aexp( t)exp( -j 2 f t) dt   Aexp(- t)exp( -j 2 f t) dt
-
0
A
-A
0

=

exp  t(- +j2 f )  
exp  -t( +j2 f )  
  +j2 f
0
 +j2 f
=
A
A
1  0  
  j 2 f
  j 2 f
=
-Aj4 f
A
A

=
  j 2 f   j 2 f
 2  (2 f)2
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 0  1
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FT of Signum and Step
Hence FT of signum is that of above as
-j4 f
-j4 fA
A
lim A
=
=
j f
  0  2  (2 f)2
(2 f)2
0
A unit step is 0 before t = 0, and 1 after.
It is in effect ½ + signum of height ½
Hence FT of a step is
0.5
+ 0.5 
j f
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Convolution
Convolution is useful operation in Signal Theory.
Sometimes called blurring.
Can simplify Fourier Transforms

Convolution of two funcs : c(t) = f1 (t)  f2 (t) =  f1 ( ) f2 (t- ) d
-
f1(t) times f2(t) transforms to F1(f)*F2(f)
f1(t) * f2(t) transforms to F1(f) times F2(f)
Suppose a system has a transfer function TF, and I is its input,
then its output is the convolution of I and TF in time domain.
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Convolving Two Pulses
Convolution is process of sliding second pulse across first pulse, multiplying
and computing area for each amount of sliding.
Where they don’t overlap, multiplication is 0,
Where they partly overlap product rises / falls
Where they totally overlap, product is A1 * A2
Hence convolution of
pulses is trapezium
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Convolving Pulse Trains
If convolve pulse and train of impulses, get series of pulses
centred on impulses
Is in effect a square wave
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FT of Repetitive Signal
Given FT of Pulse is Sinc
Also the FT of a train of
impulses is a train of
impulses of height 1/T at
frequency interval 1/T.
So FT of set of pulses is
product of two, being a
set of impulses whose
envelope is the sinc
This is same as that found
by Fourier Series
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FT of Triangular Pulse


A* 1  t
T


f(t) = A* 1  t
T

0




-T  t  0
0tT
otherwise
T
0

t
F(f) = A   1 
exp(  j2 f t) dt   1  t exp(  j2 f t) dt 
T
T
 -T

0




Do-able, but tricky
Much easier to use convolution
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By convolution
=
*
Want this to be as follows, so B =
A
T
But FT of Rect Pulse height A width 2T is F(f) =ATsinc(2fT)
So the transform of the triangular pulse is then the product
of the transform of each rectangular pulse, namely
A
A
F(f) =
Tsinc(2fT) 
Tsinc(2fT) = ATsinc2 (2fT)
T
T
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Correlation
Signals are also processed by correlating them,.
Suppose have two signals f1(t) and f2(t). Their correlation is

12 (t) =  f1 ( ) f2 (t   ) d
-
12(t) is correlation of f1(t) & f2(t); 21(t) is that of f2(t) & f1(t).
This can be compared with the integral for convolution.

f1 (t)  f2 (t) =  f1 ( ) f2 (t- ) d
-
In both cases, f2 is delayed and slid across f1. For convolution it is
reversed in the time axis, not so with correlation.
If f1 = f2 process is autocorrelation, otherwise cross-correlation
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Autocorrelation
Illustrates stages in autocorrelation
of a pulse
A
Shows pulse and it shifted at a time
A
Blob shows the amount of shifting to
justify the height on the last graph
Last graph is result when shifted
over all t
Result has maximum when shift 0
A
A
f()
T
t

f(-t)

f()*f(-t)
T

11(t)
-t
-T
If f(t) repetitive with period T, its autocorrelation is a
maximum at delay 0, T, 2T etc
Hence autocorrelation is a measure of extent can predict a
future value of a signal
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Cross-Correlation
Process similar, but two functions. Here show two pulses
Bottom graph illustrates general point 12(t) = 21(-t)
Cross-correlation is a measure of similarity between signals
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Application Pattern Recognition
Cross correlation can be used to find
a signal affected by random noise
Graph shows pulse at some time, to
which noise added.
Correlation of pulse and signal shown
in bottom graph
Peak in result indicates time where pulse can be found
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Energy Spectrum
The energy of a signal f(t) can be expressed as the square of f(t)
over all time or of its energy E(f) at all frequencies


2
E =  f (t) dt =  E(f) df
-
-
Suppose f(t) has a Fourier transform F(f)

F(f) =  f(t) exp(-j2 f t) dt
-
We can define its complex conjugate

F (f) =  f(t) exp(j2 f t) dt
-

E =  F(f) F * (f) df
It can be shown that
-

But E =  E(f) df, so E(f) = F(f) F*(f) = |F(f)|2
-
*
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Autocorrelation and ES
Now consider the autocorrelation of a function f(t), whose
spectrum is F(f).

11 ( ) =  f(t) f(t   ) dt
-
It can be shown that this can be rearranged to

11 ( ) =  F(f) F * (f) exp(j2 f ) df
-
But, E(f) = F(f) F*(f), so

11 ( ) =  E(f) exp(j2 f ) df
-
Hence the autocorrelation of a signal f(t) is the Fourier
transform of the energy spectrum of f(t).
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Example to confirm this
Suppose f(t) is a pulse of height A between -T and T.
Previous analysis has shown that F(f) = 2ATsinc(2fT)
The total energy of the signal is, using the time domain

T
2
E =  f (t) dt =  A2dt = 2A2T
-
-T
In freq domain
E(f) = F(f) F * (f) = 2ATsinc(2fT)  * 2ATsinc(2fT)  = 4A2T 2sinc2 (2fT)
Using the spectrum of F(f) (and noting  sinc2(x) = 1 for x = -∞ to ∞)


4A2T 2
*
2
2
2
E =  F(f) F (f) df = 4A T  sinc (2fT) df =
= 2A2T
2T
-
-
Result confirmed
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Continued
When autocorrelate f(t) get triangular pulse,
width -2T to 2T, max value 2A2T
Its FT is
2*2A2T*Tsinc2(2FT) = 4A2T2sinc2(2FT)
Hence, autocorrelation of rect pulse, is FT of its energy spectrum
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FT of Spatial Signals
So far we have considered signals which vary with time and their
corresponding frequency response.
The Fourier Transform works between these domains.
But a signal can vary spatially : the distance between points being
analogous to difference in time.
The points also have a Spatial frequency
Again Fourier Transform can work between spatial distances and
spatial frequency.
If x represents distance and u is 1/x being spatial frequency

The Fourier Transform is F(u) =  f(x) e -j2 ux dx
-
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Two dimensional Images
Often images from a camera are processed
These are typically two dimensional, so can be defined as f(x,y)
They have spatial frequencies in two dimensions F(u,v), v = 1/y
The transforms can be expressed in double integral pairs
 

-j2 (ux  vy)
F(u,v) =    f(x,y) e
dx  dy

- -
 

j2 (ux  vy)
f(x,y) =    F(u,v) e
du  dv

- -
Note, can transform in one dimension then the other. So if

Fy (u,y) =  f(x,y) e-j2 ux dx
-
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
-j2 vy
dy
then F(u,v) =  Fy (u,y) e
-
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38
Example
Consider a two dimensional pulse, of width 2 and height 1, centred
at x=0, y=0. That is, f(x,y) = 1 only if -1 ≤ x ≤ 1 and -1 ≤ y ≤ 1

Fy (u,y) =  f(x,y) e-j2 ux dx
-
1 -j2 ux
=  e
dx
-1
-j2 ux 1
e

= 

  j2 u  1
e-j2 u  e j2 u
=
 j2 u
=
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sin(2 u)
 u

F(u,v) =  Fy (u,y) e-j2 vy dy
-
sin(2 u) 1 -j2 vy
=
dy
 e
 u 1
-j2 vy 1

sin(2 u)  e
=


 u   j2 v 
1
sin(2 u) e-j2 v  e j2 v
=
 u
 j2 v
=
SEMMA13 Eng Maths and Stats
sin(2 u) sin(2 v)
 u
 v
39
Discrete Fourier Transform
When processing images, data are points at discrete positions.
Similarly, when a computer samples a signal at different times, the
sampled signal is known at discrete times, not at all times
The signal being sampled is said to be continuous.
The sampled signal is discrete
Note signals are usually sampled at regular intervals
Simplifies the Maths
The original Fourier Transform assumes signals are continuous
When discrete, the discrete Fourier Transform, DFT, is used.
First we note the effect of sampling on frequency response.
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Sampling and Frequency Response
Sampling at regular intervals is in effect multiplying the continuous
signal by a set of impulses at regular intervals, call the interval T.
Hence the frequency response of the sampled signal is that of the
continuous signal convolved with the frequency response of the
impulses, a set of impulses at frequency interval 1/T.
In effect the Frequency Response is repeated, as illustrated below:
This ok, but problems if F(f) has components > 1/T, spectra overlap
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Discrete Fourier Transfrom
Hence, interested only in freq response of signal, not in the higher
freqs where the spectrum is repeated.
Suppose take N samples of the signal x(t) at intervals T.
Thus x(kT) is the kth value of x, sampled at instance kT
The spectrum produced will itself be discrete.
There will be N values in the frequency spectrum, at interval v
The rth value will be X(rv)
Note, v and T are related, because v * T = 1/N
As signals are now discrete, the Fourier integral becomes a
summation. The rth frequency value of X is defined by a summation.
N-1
X(rv) = T  x(kT) exp (-j 2 r k / N)
k=0
Done for r = 0..N-1. The values X(rv) are complex numbers
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MatLab Function
function x = MyDFT (v);
% x = MyDFT(v)
% v = array of numbers to be processed
% x = array which returns with the DFT of v
% Dr Richard Mitchell, 13.3.00
x = [];
% result set to zero
for r = 1:length(v)
% for all data
ans = 0;
% initialise sum to 0
for k = 1:length(v)
% now perform the sum
ans = ans + v(k)*exp(-2*j*pi*(r-1)*(k-1)/length(v));
end
x = [x ans];
% add sum to result vector
end;
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43
Example
Shown below a square wave, 32 values at 1 and 32 values of 0,
its amplitude and phase spectra (found by abs(x) and angle(x))
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Two Dimensional DFT
This is a sum of a sum: Input x(kT, lT) transformed to X(ru,sv)
N 1 N-1
X(ru, sv) = T   x(kT, lT) exp (-j 2 (r k  s l) / N)
l=0 k=0
eg 2D Rect Pulse
Response is 2D
sinc pulse, found
by sinc(u)*sinc(v)
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45
Fast Fourier Transform FFT
DFT involves (N-1)2 multiplications and N(N-1) additions
FFT has log2(N)*N operations, but N must be power of 2.
To illustrate consider four initial samples x0 transformed to x2
This is done by transforming x0 to x1 and then x1 to x2 as illustrated
by the ‘butterfly’ figure
Value in node at end of dashed line
= value in node at start of dashed line
+ value in node at start of solid line
* e-j2q/T
q is value written in node and T = 2N
Eg
x1(0) = x0(0)+x0(2) e-j20/4 = x0(0)+x0(2)
x2(1) = x1(0) + x1(1) e-j22/4
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It can be shown that
x2(0) = x0(0) + x0(1) + x0(2) + x0(3)
x2(1) = x0(0) - x0(1) + x0(2) - x0(3)
x2(2) = x0(0) + jx0(1) + x0(2) - jx0(3)
x2(3) = x0(0) - jx0(1) + x0(2) + jx0(3)
But applying the DFT formula
X(0) = x0(0) + x0(1) + x0(2) + x0(3)
X(v) = x0(0) + j x0(1) - x0(2) - j x0(3)
X(2v) = x0(0) - x0(1) + x0(2) - x0(3)
X(3v) = x0(0) - j x0(1) -- x0(2) + j x0(3)
Hence X(0) = x2(0), X(v) = x2(2), X(2v) = x2(1) and X(3v) = x2(3)
In general X(pv) = x2(q) where binary p is the binary q reversed
eg 2 = 10, reversed is 01 = 1
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In general
In each operation, ‘pairs’ of nodes in one level are processed from
pairs of nodes in the prev level. Sets of such pairs exist, which
share the same q value
When transforming from xL-1 to xL,
• there are 2L-1 sets of 2T-L consecutive pairs, where a set
contains nodes with the same q value
• In each set, the pairs are separated by 2T-L nodes.
• The q value of the first node in the s'th set is the bit reversal
of 2*s; the q value in the second node in the pair is this number
plus 2T-1.
In final level, need to reorder nodes by bit reversing q values
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48
Laplace Transform
The Laplace Transform is introduced as a means of modelling systems
and solving differential equations for the time response of systems.
Its relationship with Fourier Transforms will be considered later.
First consider the RC circuit below, first introduced in calculus.
RC circuit, At time 0, battery E
volts connected, when V = 0
RC
Instead of writing
dV
d
we write sV, s meaning
dt
dt
Hence RCsV + V = E or V = E *
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dV
+V=E
dt
1
1+sRC
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Continued
V=E 
1
1+sRC
Means Output V =
Input E * System Transfer Function
A transfer function defines how input transferred to output
Strictly, as the transfer function is a function of s, V and E
should be also
1
V(s) = E(s) 
1+sRC
Here we are in the Laplace domain, s being the Laplace operator
Generally V and E are functions of time, and we want to know
how V varies with time,
So we want to transform from Laplace to the time domain.
Note signals are assumed to be 0 when t < 0
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Three Useful Transforms
At this stage we will define the transformation for a few signals.
1 if t  0
u(t) is the unit step function, being 
0 if t < 0
1
U(s) is the Laplace transform of u(t) being
s
Exponential function, u(t) e - t
Laplace transform of this is
1
s+
Impulse,  (t) is pulse of area 1, width 0 and time 0, 0 elsewhere
Laplace transform of this is 1
NB if X(s) transforms to x(t), Y(s) to y(t), then X(s) + Y(s)
transforms to x(t) + y(t). For constant k, k*X(s) transforms to k*x(t)
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Applied to Circuit
V(s) = E(s) 
1
1+sRC
Switch closed at t = 0, so E is in effect 0 before t = 0, and E afterwards
E
s
E
1
So V(s) =

s
1+sRC
We want E(t), but don’t know how to transform back.
Therefore input is a step, so E(s) =
However, can use partial fractions
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Continued
E
1
A
B

=
+
s
1+sRC
s 1+sRC
E
1
A(1+sRC)
Bs


=
+
s
1+sRC
s(1+sRC) s(1+sRC)
V(s) =
Equating numerators : E = A + sRCA + Bs
Equating coefficients of s, E = A and 0 = RCA + B
Hence A = E and B = -RCE
V(s) =
E
ERC
E
E
=
s 1+sRC
s s+ 1
RC
We can transform to functions of time for both of these

V(t) = u(t)  E - Ee

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- t RC 


Same as found by solving ODE,
except here include u(t)
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Formal Transform
Formally there are transforms from t to s and s to t, just as
there are from t to f and f to t in Fourier Transforms

F(s) =  f(t)e stdt
0
1 
st
f(t) =
 F(s)e ds
2 j -
From these integrals the rules given earlier are clear
if X(s) transforms to x(t), Y(s) to y(t),
then X(s) + Y(s) transforms to x(t) + y(t).
For constant k, k*X(s) transforms to k*x(t)
Note, time signals are in effect assumed to be multiplied by u(t) as
the F(s) integral is from time t = 0.
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Applying the integral
Consider exponential f(t) = Ke-at
NB, strictly this is u(t) Ke-at as signal 0 before t = 0



-at st
st
F(s) =  Ke e dt =  f  t  e dt =  Ke-(s  a)tdt
0
0
0

-(s

a)t
K


=  sa e

0
=  sK a  0  1 = sK a
So if f(t) = K e -at then F(s) =
K
sa
Process reversible, if F(s) is
K
, then f(t) is Ke-at
sa
K
If a = 0, f(t) = u(t)K e 0 = u(t) K a step of height K, so F(s) =
s
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Laplace Transform of sin(ωt)
It is tricky to do this, as the integral of e-st * sin(ωt) is awkward.
It is however made easier using complex numbers, and we can find
the transform of cos(ωt) as well as sin(ωt)
K
We know if f(t) = K e -at then F(s) =
; what if a = j ?
s+a
So if f(t) = K e -j t = K(cos( t) - jsin(t))
then F(s) =
K(s-j )
K
Ks
K
=
=
-j
s  j s2  2
s2  2
s2  2
Equating real and
imaginary parts
K cos( t) transforms to
K sin( t) transforms to
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Ks
s2  2
K
s2  2
56
RC Circuit and Sinusoid
For the RC circuit, suppose E is E sin(t). Then we can say that:
V(s) =
1
E
1  sRC s 2 2
Again, this is re-arranged slightly and then put into partial fractions:
1
E
A
Bs  C
RC
RC
=
=

2 2
2 2
1
1
 s s 2 2
s 1
s
s

s
RC
RC
RC
It can be shown that:
V(s) =
V(s) =
So:
E



ERC E  ERCs
ERC ERCs
E

=


2 2
2 2
2 2
s 1
s 1
s


s


s

RC
RC
-t
V(t) = ERC e RC  ERCcos( t)  E sin (t)
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Initial Conditions
In the above, it is assumed that the output is 0 at time 0.
This may not be the case.
Formally, for a function f(t), whose differential is f’(t), whose value
at time 0 is f(0), and whose Laplace Transform is F(s), then
The Laplace Transform of f'(t) = s F(s) - f(0)
For the RC Circuit, where V = V0 at time 0
RC
dV
+V=E
dt
Is transformed to
So
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RC(sV - E0 ) + V = E
RCE0
1
V(s) = E(s)

1+sCR 1+sCR
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If E(s) is a Step of Height E
V(s) =
RCE0
E+sRCE0
E 1
A
B

=
= 
s 1+sCR 1+sCR
s(1+sCR)
s 1+sCR
Getting all over a common denominator and equating numerators
E+sRCE0 = A(1+sCR)  Bs
Equating coefficients of s
E = A and RCE0 = ACR  B
V(s) =
E RCE0 -ECR E E0 -E

= 
s
1+sCR
s s+ 1
CR

-t

V(t) = u(t)  E   E0 -E  e CR 


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Transform Table
f(t)
F(s)
f(t)
F(s)
(t), unit
impulse
1
sin(ωt)

u(t), unit
height step
1
s
1
cos(ωt)
s2 + 2
s
e- tsin(ωt)

n!
e- tcos(ωt)
(s+ )2 + 2
s+
1
s+
1
sinh (t)

t, unit ramp
tn
e-
t e17/07/2015
s2
sn 1
t
t
(s+ )2
cosh (t)
SEMMA13 Eng Maths and Stats
s2 + 2
(s+ )2 + 2
s2 - 2
s
s2 - 2
60
Second Order Systems
The same idea is extended for second order systems
For zero initial conditions
d2O
dt2
becomes s2O
d2O dO
9
2
So 0.01
+
+ 9O = 9 becomes 0.01s O + sO + 9O =
2
dt
s
dt
Hence O(s) =
9
s(0.01s2 + s + 9)
For non zero initial conditions, need initial values of f(t) and f’(t),
then Transform of f''(t) = s2F(s) - s f(0) - f'(0)
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Solving This Example
O(s) =
9
s(0.01s2 + s + 9)
=
9
A
B
C
=
+
+
s(0.1s + 1)(0.1s + 9)
s
0.1s+1
0.1s+9
Getting all over a common denominator and equating numerators
9 = A(0.01s2 +s+9) + B(0.1s2 +9s) + C(0.1s2 +s)
Equating coefficients of s
9 = 9A; 0 = A + 9B + C; 0 = 0.01A + 0.1B + 0.1C
Hence A = 1, B = -0.9/8 and C = 0.1/8
1
0.9/8
0.1/8
1
9/8
1/8
O(s) =
+
=
+
s 0.1s+1
0.1s+9
s s+10
s+90
O(t) = 1 17/07/2015
9 -10t
1
e
+ e-90t
8
8
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Underdamped Example
d2O dO
0.25
+
+ 10O = 10 
2
dt
dt
O(s) =




10
2
0.25s + s + 10 O(s) =
10
s 0.25s2 + s + 10

=
s
A
Bs+C

s 0.25s2 + s + 10

Can show 10 = 0.25s2 + s + 10 A + Bs2 + Cs, so A = 1, C = -1, B = -0.25
O(s) =
1
0.25s+1
1
s+4
1
s+2
2
= = s (0.5s+1)2 + 32
s (s+2)2 + 62
s (s+2)2 + 62 (s+2)2 + 62
O(t) = 1 - e-2t cos(6t) -
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1 -2t
e sin(6t)
3
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Critically Damped Example
1 d2O dO
1 2
9
+
+ 9O = 9  
s + s + 9  O(s) =
36 dt2 dt
s
 36

O(s) =
9
9
=
=
324
=
A
B
C
+
+
s  s + 18   s + 18 2
2
s
s  s +18 2


s  + 3
6

Can show 324 = s2 + 36s + 324 A + Bs2 + 18Bs + Cs
1 2
s 
s +s+9 
 36



So A = 1, B = -1 and C = -18
O(s) =
1
1
18
s s+18  s + 18 2
O(t) = 1 - e-18t - 18te-18t
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Some Theorems
Initial Value : f(0) = lim f(t) = lim  s F(s) 
t 0
s 
Final Value Theorem f() = lim f(t) = lim  s F(s) 
t 
s 0
1
Suppose system has step input
and transfer function G(s).
s
G(s)
Output is
, so steady state value is
s
 G(s) 
lim  s
 = lim  G(s) 
s  s 0
s 0 
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Continued
First Shift : if f(t)  F(s), for constant  , f(t) e- t  F(s+ ),
As sin( t) 

s2 + 2
, therefore e
- t
sin(t) 

(s+ )2 + 2
Second Shift : if f(t)  F(s), for constant  , f(t- )  e-s F(s+ )
1 -2s
e
s
1
Hence a unit pulse of 2 secs is u(t) - u(t-2)  (1-e-2s )
s
So a unit step starting at t = 2, is u(t-2) 
T
If f(t) has period T, F(s) =
*  e-st f(t) dt
1-e-sT 0
1
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Impulse Response / Convolution
For System with input I, output O and transfer function G
O(s) = G(s) x I(s)
If I is a unit impulse, then I(s) = 1, so O(s) = G(s)
Hence, in the time domain o(t) = g(t)
The Transfer function (in time domain) is also Impulse Response
Re Fourier Transforms it was noted that if two functions are
multiplied in the Fourier domain, there time domain functions are
convolved. This is true also of Laplace.
Hence, as O(s) = G(s) x I(s)
So
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t
o(t) = g(t)  i(t) =  g(t-q) i(q) dq
0
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On s, jω, etc
s can be used to represent differentiation wrt t
The inverse of differentiation is integration
So integration can be represented by 1/s
For a capacitor of capacitance C, the voltage across it V, due to
the current through it I is
V(t) =
V(s)
1
=
;
I(s)
sC
1
 I(t) dt
C
or
V(s) =
1
I(s)
sC
1
is the transfer function of the capacitor
sC
This is also the impedance of the capacitor
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Applied to the RC Circuit
Here the resistor has impedance
R, the capacitor impedance 1/sC
The relationship between V and E can be found by the potential
divider rule, being impedance of C divided by that of C + that of R
1
V(s)
1
sC
=
=
1
E(s)
sCR + 1
R+
sC
This confirms the result already quoted
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When signals are sinusoids
1
1
cos(t )
 sin(t )dt = C
C
1
1
But cos(t ) =
sin (t   )
2
C
C
If I(t) = sin(t ), V(t) =
The output is a sinusoid shifted by –π/2 rads scaled by 1/ωC
1
1

But
has modulus
and argument - rads
j C
C
2
So for sinusoid signals
1
j C
is the transfer function of the capacitor
Hence s and jω are interchangeable, so for RC circuit
V(j )
1
=
E(j )
jCR + 1
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On Steady State and Transient
It should be noted that the output is a scaled/shifted sinusoid
under steady state conditions only.
This provides the easiest way of finding the steady state sinusoid
If input is sin( t), system is
1
, output is
jCR + 1


1
1
sin  t  
 =
jCR + 1
jCR + 1 

1
1+2C2R2
sin(t  tan 1 CR)
The transient response is found by modelling the system using s,
and processing that…
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Time Response using FT
Replacing  by 2 f, Transfer function of RC circuit is
Fourier Transform of battery (step input) =
1
j2 fCR + 1
E
E (0)

j2 f
2
 1
 (0) 
1
1
1
E (0)
V(f) = E 

=E


j2

f
2
1

j2

fRC
j2

f
1

j2

fRC
2


As impulse is 1 only at f = 0, when transfer function is 1
V(f) = E
 (0)
2

E
ERC
 (0)
E
E

=E


j2 f 1  j2 fRC
2
j2 f 1  j2 f
-t
Taking Inverse Transform v(t) = E - Ee RC
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Most people use
Laplace though!
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Z Transform
Laplace Transform can be used for continuous signals, but often
computers process signals, in which case they are sampled
For these it is easier to use the z-transform
Differential Equations then become Difference Equations.
Sampling is usually done at regular intervals, 0, T, 2T, etc
The sampled signal at one such time is in effect the value of the
continuous signal multiplied by an impulse at that time
The impulse at time nT is
(t-nT)
If the continuous signal is f(t), the sampled signal at the nth
sampling instant is given by
f[n] = f(nT) *
(t-nT)
A series of samples of a signal are termed a sequence
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Z operator and Transform
The z operator defines relationship between signal at one
instant and the next:
z f[n+1] = f [n]
or f[n+1] = z-1 f[n]
The z transform of a sequence, F(z), is defined as

F(z) =  f[n]z-n
n=0
Unit step signal, f[n] is 1 for all n
0, so its z transform is
 -n
1
F(z) =  z = 1  z 1  z 2  z 3  z 4  .... =
1-z-1
n=0
Sampled exponential signal e-at, at sample n has value e-anT
 -an -n
1
F(z) =  e z = 1  e-az 1  e-a2z 2  e-a3z 3 .... =
1
a
n 0
1- ze
 
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Z Transform Table
f(t)
F(s)
f(t)
[n],
impulse
1
sin(an)
u[n],
step
a-n
n
1
1
1
1-z-1cos(aT)
cos(an)
1-2z-1cos(aT)  z-2
1-z-1
1-z 1e-aT
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z-1sin(aT)
1-2z-1cos(aT)  z-2
e- nsin(bn)
1  z 1 a T
e-
F(s)
z-1e-aT sin(bT)
1-2z-1e-aT cos(aT)  z-2e-2aT
e- ncos(bn)
1-z-1e-aT cos(bT)
1-2z-1e-aT cos(aT)  z-2e-2aT
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On Modelling Systems
In s-domain, a system with input I, output, O and transfer function H:
O(s) = H(s) * I(s)
e.g. For a first order system, H(s) =
K
1+sT
Can form differential equation of O(t) and I(t): T
dO
+O=I
dt
In z-domain, functions are of z (or z -1 ): O(z) = H(z) * I(z)
For a first order system, H(z) =
az-1
1 - bz-1
Can rearrange into difference equation in sampled signals O[n] and I[n]
O[n](1 - bz-1 ) = az-1I[n]
or
O[n] = a I[n-1] + b O[n-1]
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Second Order System
az 2
H(z) =
1  bz 1  cz 2
This can be rearranged as:
(1 - bz-1 - cz-2 )O[n] = a z 2 I[n]
Hence
Or
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O[n] - bO[n-1] - cO[n-2] = a I[n-2]
O[n] = a I[n-2] + bO[n-1] + cO[n-2]
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Properties of Z Transform
Assume f[n] transforms to F(z), g[n] to G(z) and c is a constant.
c f[n] transforms to c F(z)
f[n] + g[n] transforms to F(z) + G(z)
The first shift theorem : for delayed signal:
f[n+k] transforms to zk F(z) – (zk f[0] + zk-1 f[1] + .. + z f[k-1])
For k = 1, this becomes
f[n+1] transforms to z F(z) – z f[0]
For k = 2, this becomes
f[n+2] transforms to z F(z) – z2 f[0] – z f[1]
Similar to the differential properties of the Laplace transform:
Laplace transform of f’(t) = sF(s) - f(0)
Useful for handling the cases where initial conditions are not zero.
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Initial and Final Values
For s domain Final Value is
o() = lim o(t) = lim sO(s)
t 
s 0
For z domain similar, except z tends to 1 not 0
o() = lim o* (t) = lim (1-z-1 )O(z)
t 
z 1
In s-domain to find steady state value to unit step input, set s to 0
in system transfer function.
In z-domain, therefore, set z to 1 in its transfer function
For initial value, providing limit exists:
o(0) = lim o* (t) = lim O(z)
t 
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z 
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First Order Step Response
For the RC circuit, the step response is of the form V = E – E e-t/RC.
The transfer function has steady state gain of 1. Lets do z equivalent
Assume Circuit modelled by H(z) =
(1 - b)z-1
1 - bz 1
As input is step size E, O(z) = H(z) I(z) =
Using Partial Fractions O(z) =
A
1 - bz 1

1 - b  z-1
[NB this is 1 if z = 1]
E
1 - bz 1 1 - z 1
B
1 - z 1
Hence E(1-b)z 1 = A - Az 1 + B - Bbz 1 . Hence A = -E and B = E
Hence O(z) =
E
1 - z 1

E
1 - bz 1
so
O[n] = E - Ebn
If assume b = e-a, then O of form E – Ee-an
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Solving First Order Difference Eqn
For H(z) =
(1 - b)z-1
1 - bz 1
, O[n+1] - bO[n] = (1-b)I[n]
To solve, follow similar procedure to linear differential eqn.
Set RHS to zero, then form auxilliary equation, where O[n+r] = mr.
Each root means transient response has term k * rootn:
So O[n+1] – b O[n] = 0  m1 – b = 0, the root of this is m = b
Hence, the transient solution is O[n] = k bn.
For steady-state, if I[n] is a constant E, so O[n] = C
This must be solution of difference equation
O[n+1] - bO[n] = C - b C = (1-b) E, so C = E
Complete solution is O[n] = E + k bn : Can show k = -E if O[0] = 0
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For Second Order
Again have over-, critically- and under-damped responses
Transient resp depends on roots of auxilliary equation:
If characteristic eqn is (m-a)(m-b) = 0, O[n] = k1 an + k2 bn
O[n] = k1 e-cn + k2 e-dn if a = e-c etc
If characteristic eqn is (m-a)2 = 0, O[n] = (k1 + k2n) an
If characteristic eqn has complex roots are r
O[n] = rn (k1 cos(n ) + k2 sin(n ))
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Relating s and z
Laplace Trans of delayed impulse,
(t-nT), is 1 shifted by nT = e-nTs
At the nth sampling instant, a sampled signal is (t-nT)f(nT)
-nTs
Then f(nT) is constant, so its Laplace transform is: f(nT) e

Overall the sampled signal is  f(nT)  t-nT 
n 0


-nTs
Whose Laplace Transform is  f(nT)e
=  f[n]e-nTs
n 0
n 0

But the z transform of f[n] is F(z) =  f[n]z-n
n=0
These must be equal, so z = eTs
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Other Mappings s to z
Derived by approximating integral of a sampled signal = area:
At an instant, area = prev area + extra are due to next sample
1
1
1
1
f[n] = f[n-1]  extra area or f[n] = f[n]z -1  extra area
s
s
s
s
In Euler Integration, extra area is f[n]*T
1
f[n] (1 - z-1 ) = f[n] *T
s
1-z-1
s=
T
Trapezoidal Integration
1
f[n]+f[n-1]
f[n] (1 - z-1 ) =
*T
s
2
2 1  z-1
s=
T 1  z-1
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Tustin’s Rule
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