Transcript EENGR 3810 Chapter 2 - UNT College of Engineering

EENGR 3810 Chapter 2
Introduction to Signals
1
Chapter 2 Homework
2.1-1 c & d, 2.1-8b & f, 2.3-2, 2.5.-2
Extra credit !!
Prove that:
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Chapter 2 Homework
Continued
Write the equation for the signal shown below:
f(t)
10
0
2
4
6
8
t(s)
-10
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Signals
• Set of information or data
• Function of time (t) in this course
• Processed by systems
– Processes the signal input
– Outputs the processed signal
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Signal Energy Capability
(Based on a load of 1 ohm resistance)
• Measure of signal energy capability for real signal
Eg =
-
g2(t)dt
Where:
Eg = signal energy
g (t) = area under the signal (always positive)
• Measure of signal energy capability for complex signal
Eg =
|g2|(t)dt
Where:
Eg = signal energy
|g|(t) = area under the signal (always positive)
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Signal Energy Capability
• If g(t) = sin t for t = 0 2, find the energy of the signal.
2
Eg =
sin2
t dt = 1/2
0
Eg =
½ t – ¼ sin 2t
0
2
=
2
2
dt – 1/2
cos 2t dt
0
[(2/2) -0 ] - [(sin 4 - sin 0)/4] =  - 0 – 0 = 
0
Eg =  J
(J = Joules of energy)
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Signal Energy Capability
• If g(t) = - sin t, how does this effect the energy?
– Sign change does not affect the signal energy
• If g(t) is shifted in time (2  4), how does this effect the
energy?
– Time shift does not affect the signal energy
• If g(t) is doubled, how does this effect the energy?
– Doubling the signal quadruples its energy
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Signal Energy Capability
• Signal energy must a be finite value for a meaningful measure
of the signal size.
– Signal amplitude must approach  0 as |t|  
– Otherwise Eg = g2(t)dtwill not converge.
– If a signal amplitude does not approach  0 as |t|  ,
the signal energy is infinite.
– The signal shown below has finite energy.
g(t)
t
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Signal Power Capability
• The amplitude of the signal shown below does not approach
 0 as |t|  . Therefore, the signal energy is infinite.
• A more meaningful measure of the signal size in such a case is
the time average of the energy, which is the average power
(Pg) .
g(t)
t
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Signal’s Average Power (Pg)
(Based on a load of 1 ohm resistance)
• The average power for a real signal is defined by
• Average power (Pg)
– time average (mean) of the signal amplitude squared.
– mean-squared value of g(t)
• The root-mean square (rms) value of g(t) is (Pg)1/2.
• A Complex Signal’s Average Power (Pg)
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Ramp Signal g (t) =t
• A ramp signal is shown below
– g(t) =t increases indefinitely as |t|  
– Energy does not exist for this signal
– Power does not exist for this signal
g (t) =t
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Actual Signal Energy Dissipation
• The actual signal energy dissipation depends not only on the
signal, but on the load.
• g (t) = voltage in volts
• Current I = g (t) / R in amperes
• R = resistance in ohms
I
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What is the suitable measure for this signal?
• This signal approaches  0 as |t|  , therefore use the
energy equation.
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What is the suitable measure for this signal?
• This signal does not approach  0 as |t|   and it is a
periodic wave, therefore use the power equation where g2 is
replaced with t2.
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What is the suitable measure for this signal?
g (t) = C cos (0t + )
• This signal is a sinusoid signal.
• Therefore, use the power equation.
• The power of a sinusoid of amplitude C is:
Pg = C2/2
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What is the suitable measure for this signal?
10 cos (100t + /2)
Power of a sinusoid of amplitude of C is:
Pg = C2/2
Therefore,
P = (10)2 /2 = 100 / 2 = 50 Watts
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What is the suitable measure for this signal?
g (t) = C1 cos (1t + 1) + C2 cos (2t + 2)
1 ≠ 2
• This signal is the sum of two sinusoid signals.
• Therefore, use the power equation.
• Therefore, P = (C12 / 2) + (C22 / 2)
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What is the suitable measure for this signal?
2 cos (200t + /4) + 4 sin (10t + /3)
• P = (22 / 2) + (42 / 2) = (4 / 2) + (16 / 2) = 10
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What is the suitable measure for this signal?
g (t) = Dejt
• The signal is complex and periodic. Therefore, use the power
equation averaged over T0.
• |ejt| = 1 so that |Dejt|2 = |D|2 and
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What is the suitable measure for this signal?
ejtsint
• sint = (1/2j)(ejt - e-jt)
• ejtsint = (1/2j)(ej( + )t - e-j ( + )t)
• Therefore, use power equation
P = (1/2j)2 - (1/2j)2 = (-1/4) - (-1/4) = 0
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Classification Of Signals
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Continuous-time Signal
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Discrete-time Signal
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Analog Signals
Continuous-time
Discrete-time
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Digital Signals
Continuous-time
Discrete-time
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Periodic Signal
• A periodic signal starts at -  and continues for ever.
• A signal is said to be periodic if for some positive constant
T0:
g(t) = g(t+T0) for all t
Periodic Signal of period T0
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Comments on Energy and Power Signals
• Every signal generated in the lab is an energy signal.
• Every signal observed in real life is an energy signal.
• A power signal must have an infinite duration. It is impossible
to generate a true power signal in practice because such a
signal has infinite duration and infinite energy.
• Periodic signals are power signals when the area |g(t)|2 over
one period is finite because of periodic repetition. (However,
not all power signals are periodic.)
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Deterministic and Random Signals
• Deterministic Signal
– A signal whose physical description is known
completely, in either a mathematical form or a
graphical form.
• Random signal (Covered in Chapter 11)
– A signal whose physical description is not known
completely, in either a mathematical form or a
graphical form.
– A signal known only in terms of probabilistic
description, such as mean value, mean squared value,
and so on.
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Time Shifting of a Signal
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Time Scaling of a Signal
Compressing a signal by n
(t) = g(nt)
Expanding a signal by n
(t) = g(t/n)
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Time Inversion of a Signal
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For the signal g(t) shown below, sketch g(-t)
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For the signal g(t) shown below, sketch g(t-4)
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For the signal g(t) shown below, sketch g(t/1.5)
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For the signal g(t) shown below, sketch g(2t-4)
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For the signal g(t) shown below, sketch g(2-t)
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Signal Functions
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Unit Step Function
u(t)
1
t
0
u(t) = 0 for t < 0 and 1 for t > 0
Example:
v = 20 u(t) Sketch the waveform
20 u(t)
20
0
t
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Delayed Step Function
u(t-a)
1
0
t
a
U(t-a) = 0 for t  a and 1 for t > a
Example:
V = 20 u(t-3)
Sketch the waveform
20 u(t-3)
20
0
3
t
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Ramp Function
(1st integral of a step function)
F(t)
Ktu(t)
K = a/b = slope
a
t
0
f(t) = k
b
u(t)dt = Ktu(t) = 0 for t  0 and Kt for t > 0
Example:
v(t) = 60 tu(t)
Sketch the waveform
v(t)
60 tu(t)
300 mV
0
K = 300/5 = 60
t
5 ms
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Parabolic Function
(1st integral of a ramp function
or 2nd integral of a step function )
f(t) =( Kt2 / 2)u(t)
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Impulse Function
(1st derivative of a step function)
[du(t)/dt] = (t)
f(t)
(t)
t
[du(t-a)/dt] =(t-a)
f(t)
(t-a)
a
t
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How to use signal functions to
write equation of signals as a
function of time.
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The figure shown below is the voltage across a 5 H inductor. Write
the voltage equation as a function of time “t”.
v (V)
+100
L=5H
1
2
3
4
5
6
t (s)
-100
v (V)
+100
1
-100
2
3
4
-100tu(t)
5
6
t (s)
Note: K is determined here
100V / 1s.
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Write the voltage equation continued.
v (V)
+100
+200(t-1)u(t-1)
1
2
3
4
5
6
t (s)
-100
v (V)
+100
-100(t-3)u(t-3)
1
2
3
4
5
6
t (s)
-100
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Write the voltage equation continued
v (V)
+100
-100(t-5)u(t-5)
1
2
3
4
5
6
t (s)
-100
v (V)
+100
+100(t-6)u(t-6)
1
2
3
4
5
6
t (s)
-100
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Write the voltage equation continued.
v (V)
+100
L=5H
1
2
3
4
5
6
t (s)
-100
v (t) = -100tu(t) + 200(t-1)u(t-1) - 100(t-3)u(t-3) - 100(t-5)u(t-5) + 100(t-6)u(t-6)
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Calculate the inductive current (For 0  t  1 )
v (V)
L=5H
+100
1
2
3
4
5
t (s)
6
-100
v (t) = -100tu(t) + 200(t-1)u(t-1) - 100(t-3)u(t-3) - 100(t-5)u(t-5) + 100(t-6)u(t-6)
For 0  t  1:
i(t) = (1/5)
v(t) = -100t
1
-100tdt =
0
(0.2)(-100/2)t2
1
= - 10t2 A
0
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Calculate the inductive current (For 1  t  3)
v (V)
+100
L=5H
1
2
3
4
5
6
t (s)
-100
v (t) = -100tu(t) + 200(t-1)u(t-1) - 100(t-3)u(t-3) - 100(t-5)u(t-5) + 100(t-6)u(t-6)
For 1  t  3: v(3) = -100t + 200t - 200 = 100t – 200
i(t=1) = -10t2 = -10A
i(3) = (1/5)
3
(100t -200) dt -10A = (0.2)[(-100/2)t2 -200t]
1
= 10t2 – 40t
3
- 10 = (10t2 – 40t) –(10 -40) - 10
-10
3
1
1
= 10t2 – 40t + 20 A
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Calculate the inductive current (For 3  t  5: )
v (V)
+100
L=5H
1
2
3
4
5
t (s)
6
-100
v (t) = -100tu(t) + 200(t-1)u(t-1) - 100(t-3)u(t-3) - 100(t-5)u(t-5) + 100(t-6)u(t-6)
For 3  t  5: v(t) = 100 V from the waveform. or
v = -100t +200t – 200 – 100t + 300 = 100 V
i(t=3) = 10t2 – 40t + 20 A = 90 – 120 + 20 = - 10 A
= 20t
5
3
-10
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Calculate the inductive current (For 5  t  6)
v (V)
+100
L=5H
1
2
3
4
5
6
t (s)
-100
v (t) = -100tu(t) + 200(t-1)u(t-1) - 100(t-3)u(t-3) - 100(t-5)u(t-5)+ 100(t-6)u(t-6)
For 5  t  6: i(5) = 20t – 70 = 100 – 70 = 30 A
v(6) = 100 – 100t + 500 = -100t + 600
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Calculate the inductive current (For 6  t  )
v (V)
+100
L=5H
1
2
3
4
5
6
t (s)
-100
v (t) = -100tu(t) + 200(t-1)u(t-1) - 100(t-3)u(t-3) - 100(t-5)u(t-5)+ 100(t-6)u(t-6)
For 6  t  : i(6) = -10t2 + 120t – 320 A = -360 + 720 – 320 = 40 A
v(6) = -100t + 600 + 100t -600 = 0
i(t) = 40 A
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Calculate the inductive current (For 6  t  )
t(s)
1
2
3
5
6
6  t 
A
-10
-20
-10
30
40
40
i(2)
= 40 – 80 + 20 = - 20
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Fourier Series
1. Sine-Cosine Form
2. Trigonometric Form
3. Exponential Form
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Fourier Series Representation of a Periodic Function
(Sine-Cosine Form)
f(t) = av +
an cos n0t + bn sin n0t
Where,
n = 1,2,3 ….
av, an, and bn = Fourier Coefficients
0 = the fundamental frequency
n0 = harmonic frequencies
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Analysis Steps for the Fourier Series
1.
2.
Determine f(t).
Calculate Fourier coefficients:
av = (1/T)
f(t) dt
ak = (2/T)
f(t) cos k0t dt
bk = (2/T)
f(t) sin k0t dt
3.
Resolve the periodic source in to a DC (av) plus a sum of sinusoidal
sources (an and bn).
4.
Use the superposition principle to find the steady-state response:
(a) Calculate the response to each source generated by
the Fourier series representation of f(t).
(b) Add individual responses to obtain the total response.
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Example 1
Vm = 9 V
Vm
Vm /3
t
T/3
2T/3
T
4T/3
5T/3
2T
T
2T/3
= (Vm/T)t
+ (Vm/3T)t
0
2T/3
= 2Vm/3 + Vm/3 – 2Vm/9 = Vm -2Vm/9 = 7Vm/9 = 7(9)/9 = 7 V
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Example 1Continued
2T/3
0
(2T/3) - 0
= 2(9) / k(T)(2/T) sin k(2/T)(2T/3) = (9 / k) sin 4k/3
T
3
2T/3
= [2(9) / 3k(T)(2/T) sin k(2/T)(T)] - [ 2(9) / 3k(T)(2/T) sin k(2/T)(2T/3)]
0
= - (3 / k) sin 4k/3)
= [(3 / k) sin k2] - [ (3 / k) sin 4k/3]
= (9 / k) sin 4k/3 - (3 / k) sin 4k/3)
= (6 / k) sin 4k/3
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Example 1Continued
2T/3
0
-
[
(2T/3) - 1] =
2(9) / k(T)(2/T) [1 - cos k(2/T)(2T/3)
= 9 / k [1 - cos 4k/3]
T
3
2T/3
= - [2(9) / 3k(T)(2/T) cos k(2/T)(T)]+ [ 2(9) / 3k(T)(2/T) cos k(2/T)(2T/3)]
= (3/ k) [-cos k2 + cos 4k/3] = - 3/k [ 1 – cos 4k/3]
bk = 9 / k [1 - cos 4k/3] - 3/k [ 1 – cos 4k/3] = 6 / k [1 - cos 4k/3]
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Example 1Continued
What is the average value of the periodic?
b. Compute the numerical values of a1 through
a5 and
b1 through b5 .
c. If T = 125.66 ms, what is the fundamental
frequency in radians per seconds?
d. Write the Fourier series up to and including
the fifth harmonic.
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Example 1 continued
[a] the average value of the periodic voltage is :
[b] ak = (6 / k) sin 4k/3
a1 = 6 sin 4.19 = -5.2 V
a2 = 3 sin 8.38 = 2.6 V
a3 = 2 sin 12.57 = 0 V
a4 = 1.5 sin 16.76 =-1.3 V
a5 = 1.2 sin 20.95 = 1.04 V
bk = 6 / k [1 - cos 4k/3]
b1 = 6 (1-cos 4.19) = 9 V
b2 = 3 (1 - cos 8.38) = 4.5 V
b3 = 2 (1- cos 12.57) = 0 V
b4 = 1.5 (1- cos 16.76) = 2.24V
b5 = 1.2 (1- cos 20.95) = 1.8 v
61
Example 1 Continued
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Determining Symmetry of Waveforms
(No symmetry)
Vg(t)
Vm
t
-5T/3 -2T/3
-T/2
-T/3
-T/6 0
T/6 T/3
T/2
2T/3
5T/3
T
-Vm
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Determining Symmetry of Waveforms
(Odd symmetry)
Vg(t)
Vm
t
-T
-5T/3 -2T/3 -T/2
-T/3
-T/6
0
T/6
T/3 T/2
2T/3
5T/3
T
-Vm
f(t) = - f(-t)
Note how the waveform crosses y-axes at zero.
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Determining Symmetry of Waveforms
(Even symmetry)
Vg(t)
Vm
t
-T -5T/3
-2T/3 -T/2
-T/3 -T/6 0 T/6 T/3
T/2
2T/3
5T/3
T
-Vm
f(t) = f(-t)
Note how the waveform crosses y-axes at maximum value.
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Determining Symmetry of Waveforms
(Half-wave odd symmetry)
Vg(t)
Vm
t
-T
-5T/3 -2T/3 -T/2
-T/3
-T/6
0
T/6
T/3 T/2
2T/3
5T/3
T
-Vm
The above waveform has half-wave odd symmetry because when this
odd symmetry waveform is shifted by T/2 and inverted it would be
identical to its original waveform.
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Determining Symmetry of Waveforms
(Half-wave even symmetry)
Vg(t)
Vm
t
-T -5T/3
-2T/3 -T/2
-T/3 -T/6 0 T/6 T/3
T/2
2T/3
5T/3
T
-Vm
The above waveform has half-wave even symmetry because when
this even symmetry waveform is shifted by T/2 and inverted it would
be identical to its original waveform.
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Determining Symmetry of Waveforms
Quarter-wave symmetry
Quarter-wave symmetry means the periodic function has halfwave symmetry and ,in addition, symmetry about the midpoint of
the positive and negative half-cycles.
f(t)
A
t
0
T/4
T/2
3T/4
T
-A
Quarter-wave can be made even or odd by the choice of the point
Where t = 0. The above is odd because symmetry is about T/4.
Move the symmetry about t = 0, the function will be even.
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Determining Symmetry of Waveforms
(Odd function with half-wave and quarter-wave symmetry)
i (A)
8
2
t (ms)
4
12
16
20
28
32
36
44
48
-2
-8
69
Effects of symmetry on Fourier Coefficients
Even-function symmetry
av = (2/T)
f(t) dt
ak = (4/T)
f(t) cos k0 t dt
bk = 0 for all k
Odd-function symmetry
av = 0 for all k
ak = 0 for all k
bk = (4/T)
f(t) sin k0 t dt
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Effects of symmetry on Fourier Coefficients
Half-wave symmetry
av = 0
ak = 0 for k even
ak = (4/T)
f(t) cos k0 t dt for k odd
bk = 0 for k even
bk = (4/T)
f(t) sin k0 t dt for k odd
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Effects of symmetry on Fourier Coefficients
Quarter-wave symmetry: even function
av = 0, because of the half-wave symmetry
ak = 0, for k even, because of the half-wave symmetry
ak = (8/T)
f(t) cos k0 t dt for k odd
bk = 0, for all k even, because the function is even
Quarter-wave symmetry: odd function
av = 0, because the function is odd
ak = 0, for all k, because the function is odd
bk = 0, for k even, because of the half-wave symmetry
bk = (8/T)
f(t) sin k0 t dt, for k odd
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Example 2
Determine the Fourier series for this quarter-wave periodic voltage:
Vg(t)
Vm
t
0
T/6
T/3 T/2
2T/3
5T/3
T
-Vm
73
Example 2 Continued
= 48 Vm/T2
T/6
= 48 Vm/T2 {[ (1/k0)2 (sin k0t)] - [(t/k0) cos k0t]}
0
= 48 Vm/T2 { [1 / (k242/T2)][sin (k2/T)(T/6)] – [(T/6) / (k2/T)] [cos (k2/T)(T/6)]}
- 48 Vm/T2 { [1 / (k242/T2)][sin (k2/T)(0)]} – [(0) / (4k2/T2)] [cos (k2/T)(0)]}
= 48 Vm/T2 { [1 / (k242/T2)][sin (k/3)]} – [(T2/12k)] [cos (k/3)]} - 0
= [12 Vm/(k22)][sin (k/3)] – [4Vm/k][cos (k/3)]
74
Example 2 Continued
= 8 Vm/T
T/4
= (8 Vm/T)/ (k0)(- cosk0t)
T/6
= (8 Vm/T)/ (k2/T) [-cos (k2/T)(T/4) + cos (k2/T)(T/6)]
= (4Vm)/(k)[- cos (k/2) + cos (k3)]
= -(4Vm)/(k)[cos (k/2)] + (4Vm)/(k)[cos (k/3)]
0
= [12 Vm/(k22)][sin (k/3)] - [4Vm/k][cos (k/3)] - (4Vm)/(k)[cos (k/2)]
+ (4Vm)/(k)[cos (k/3)]
bk = [12 Vm/(k22)][sin (k/3)]
75
Example 2 Continued
f(t) = av +
an cosn0t + bn sin n0t
av = 0, because the function is even
ak = 0, for all k, because the function is even
bn = bk = [12 Vm/(n22)][sin (n/3)]
vg(t) = 0 +
vg(t) =
0 + bn sin n0t
bn sin n0t
76
Fourier Series Alternative Trigonometric Form
(Amplitude-Phase Form)
f(t) = av +
An cos (n0t - n)
an – jbn = An -n
Where,
N = 1,2,3 ….
av and an = Fourier Coefficients
0 = the fundamental frequency
n0 = harmonic frequencies
An = the “n” magnitude
n = Phase angle
an – jbn = Rectangular Form
An -n = Polar Form
77
Example 3
Vm = 9 V
Vm
Vm /3
t
T/3
2T/3
T
4T/3
5T/3
2T
a. Compute A1 through A5 and 1 through 5 for the same
function above.
b. Write the Fourier series for v(t) up to and including the fifth
harmonic.
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Example 3
From Example 1:
Vm = 9 V
Vm
a1 = -5.2 V
a2 = 2.6 V
a3 = 0 V
a4 = -1.3 V
a5 = 1.04 V
b1 =
b2 =
b3 =
b4 =
b5 =
9V
4.5 V
0V
2.24V
1.8 v
Vm /3
t
T/3
2T/3
T
4T/3
5T/3
2T
79
Example 4
Vm = 281.252 mV
T = 200
Vi
Vm
R
+
t
0
T/4
T/2
3/4T
+
100K
C
Vi
100nF
T
_
Vo
_
-Vm
Filter
Quarter-wave symmetry-odd
av = 0, because the function is odd
ak = 0, for all k, because the function is odd
bk = 0, for k even, because of the half-wave symmetry
bk = (8/T)
f(t) sin k0 t dt, for k odd
80
Example 4 Continued
bk = (8/T)
f(t) sin k0 t dt
(4Vm/T)
= (32Vm/T2)
T/4
= 32 Vm/T2 {[ (1/k0)2 (sin k0t)] - [(t/k0) cos k0t]}
0
= 32 Vm/T2 { [1 / (k242/T2)][sin (k2/T)(T/4)] – [(T/4) / (k2/T)] [cos (k2/T)(T/4)]}
- 32 Vm/T2 { [1 / (k242/T2)][sin (k2/T)(0)]} – [(0) / (4k2/T2)] [cos (k2/T)(0)]}
0
2
2
2
2
2
= 32 Vm/T { [1 / (k 4 /T )][sin (k/2)]} – [(T /8k)] [cos (k/2)]} - 0
b= [8Vm/(k22)][sin (k/2)]
81
Example 4 Continued
f(t) = av +
An cos (n0t - n)
av = 0, because the function is odd
An = [8Vm/(k22)][sin (k/2)]
f(t) = av +
An cos (n0t - n)
82
Example 4 Continued
83
Example 4 Continued
84
84
Example 4 Continued
85
Average Power of Periodic Functions
(Using trigonometric form of Fourier series.)
v(t) = VDC +
i(t) = iDC +
P(t) = VDCIDC +
vn cos n0t - vn
in cos n0t - in
vnIn/2 cos (vn - in)
86
Example 5
Given: 12 Vm = 296.09 V and T = 2094.4 ms
30
Vg(t)
Vm
t
0
T/6
T/3 T/2
2T/3
5T/3
T
-Vm
87
Example 5 Continued
88
Example 5 Continued
(n = 1, k = 1 and 0= 3)
30
(
)(
)
89
Example 5 Continued
(n = 3, k = 3 and 0= 3)
30
= sin  = 0
90
90
Example 5 Continued
(n = 5, k = 5 and 0= 3)
30
P = P1 + P3 + P5 = 60.75 + 0 + .00476 = 60.75476 W
91
Exponential Form of Fourier Series
Cnejn t
f(t) =
0
Where
Cn = 1/T
f(t)e-jn t dt = (an – jbn) / 2 = An/2-n
0
e
jn0t
= cos 0t + jsin t
e -jn t = cos 0t - jsin t
0
e jn t + e -jn t
cos 0t =
2
e
e jn t - e -jn t
sin 0t =
2j
0
0
0
0
92
Example 6
i (A)
8
2
Determine Cn
t (ms)
4
12
16
20
28
32
36
44
48
-2
-8
93
Example 6 Continued
T/8
16
0
T/4
64
-
T/8
1
T/8
16
8
(k/4)
-
k
T/4
-
k
T/8
+
8
[1 -
(k/4)]
k
0
(k/2)
32
-
[k2/T](0)
T(k2)/T
8
+
k
64
16
[k2/T](T/8) +
T(k2)/T
0
-
16
-
+
32
(k/4)
k
32
k
(k/4)
94
Example 6 Continued
-
8
[1 -
32
(k/4)] +
k
-
8
k
(k/4)
k
(k/4)
+
8
+
k
32
(k/4)
k
Cn = (an – jbn) / 2 = -jbn/2
95
Example 7a
i (A)
8
2
t (ms)
4
12
16
20
28
32
36
44
48
-2
-8
Irms = ((I2)(t) / T)1/2
Irms = {[(2)2(4ms)+(8)2(8ms)+(2)2( 4ms)+(-2)2(4ms)+(-8)2(8ms)+(-2)2( 4ms)] / 32ms} 1/2
Irms = [(16+512+16+16+512+16) / 32]1/2 = (34)1/2 = 5.831 A
96
Example 7b
97
Example 7c
98
Amplitude and Phase Spectra
Based on :
f(t) = av +
An cos (n0t - n)
Given:
Vm = 40 V, av = Vm / 4, ak = (Vm / k) sin (k/2),
bk = (Vm / k) [1 - cos (k/2)], and
Therefore:
99
Amplitude and Phase Spectra Continued
100
Amplitude and Phase Spectra Continued
Based on :
f(t) =
Cnejn t
0
Given:
Vm = 40 V, av = Vm / 4, ak = (Vm / k) sin (k/2),
bk = (Vm / k) [1 - cos (k/2)],
and
101
Amplitude and Phase Spectra Continued
102