Transcript Example - personal.rdg.ac.uk

(b) Finite energy signal
The area is simply
And the total energy



f ( t )dt
E


f 2 ( t )dt
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Example: A repetitive triangular waveform as shown in the Figure
below. (i) Find the area, and total energy of one pulse of the waveform.
(ii) Find the mean, mean square and variance of the repetitive waveform.
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(a)


for
t  0

Area 


 t2

 2

A
 2

Energy 
to



f
t  :
f (t ) 
f ( t )dt  2 

0
A
t

A
tdt




  A
0
2
( t )dt  2 

0
2
 A 
t  dt

  

3
2
 A t 
2

 2

A

 
 3 

3

 
0
2
1
f 
T
(b)
f

2
2

T / 2
1

T
 f
T /2

T /2
T / 2
2
2
f ( t )dt 
T
0
2
f 2 ( t )dt 
T
 
 f

2
A
A
tdt 

T



0
2
2
 A 
t  dt 
A2 

3T
  
2
 A 

A2   

3T
 T 
2
A2   2


  3 
T 3
T 
(ii) Frequency domain average
(a) Finite mean power signals
Assume that the filter passes a narrow band of frequencies of bandwidth
δf , with a centre frequency f , and the gain is unity.
0
For a continuous random signal, the plot of meter reading versus
centre frequency f 0 would be similar to
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The power spectrum P(f), sometimes called power spectral density,
is the distribution of mean power per unit bandwidth, and is
obtained by dividing the mean square power voltage by δf.
The total mean power
P


0
2
f (or
P( f )df
P ) is given by
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Example: Repetitive pulses
The zero line is simply the
squares of mean value, i.e.
( A / T )
2
Total mean power is just the
sum of the power Pn in the
lines
P

P
n 0
n
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Fourier series for the function f(t) can be written in the
following trigonometric form:
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%The following Matlab code rectangular.m demonstrates that a rectangular wave can be generated from a series of
sine wave;
clear;
%clear work space;
clf;
%clear figure space;
n=input('how many sine waves?(>2)');
%number of sine wave to be used, try changing n from small to large for better approximation;
i=1:500;
%500 data point;
t=i/5;
%5 points per time step;
a(1,i)=2.5+10/pi*sin(pi*t/4);
% the first sine wave plus a constant term (offset);
for j=2:n;
a(j,i)=10/(2*j-1)/pi*sin(pi*t*(2*j-1)/4);
end;
% wave forms that you generated
wave_rec=sum(a);
%the rectangular wave is form by summing these waves;
plot(t,wave_rec);
xlabel('t');
ylabel('wave forms');
hold on;
plot(t,a(1,:),'r');
%the first sine wave (shown as red);
plot(t,a(2,:),'g');
%the second sine wave (shown as green);
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sin x
x
10t 
f ( t )  2.5 
sin c( nx )

4 n 1
t
x
4
sin c( x ) 
n corresponds to frequency. The amplitude versus frequency is
determined by sinc function. Power spectrum takes square value of
amplitude.
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(b) Finite energy signals
Finite energy signal (discrete) have no power spectrum, but it can be
represented by a continuous energy spectrum.
Total energy
E


0
E( f )df
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A Summary of time and frequency domain average
Type of signal
Time domain
Frequency domain
Finite power continuous
(random)
repetitive
f , f , , etc
Continuous power
spectrum
Discrete, lines at
1/T,T/2, etc
Finite energy
Area, total energy E
2
2
f , f 2 ,2 , etc
Continuous energy
spectrum
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• Time and ensemble averages
(a) Time average, e. g.
1 T/2
f  lim
f ( t )dt
T   T T / 2
1 T/2
 
f ( t )dt if repetitive

T
/
2
T
(b) Ensemble average, e. g.
x


xp( x )dx
If applied to the same signal, both must give the same value.
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Consider Johnson noise produced by a large number of identical resistors
Ensemble average from complete set of resistors at instant τ.
Time average from any one resistor average over a long time.
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Example: Triangle wave as shown below. Find the mean (shown here), mean
square and variance using time average and ensemble averages.
Time average:
2A
T
t
from 0 to
T
2
1 T / 2 2A
2 T / 2 2A
x 
tdt 
tdt
T T / 2 T
T 0
T
2 A 2 T/2
A

t 0

T T
2
x 
 
Ensemble average: p( x )  1
from
A
x 

1

A


x p( x )dx 
 x2

 2

A



0

A
2
0

A
0
to
A
1
x dx
A
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