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APPENDIX 1
The Kronecker delta,  m,n
It is useful to define a function of two integers n and m in such a way that the function is 1 if m = n and 0 if m  n.
This function is called the Kronecker delta function:
1 if m  n
0 if m  n
 m ,n  
Examples:
 2,3
= 0,
 5,5
= 1.
The orthogonality property of sinusoids.
If n and m are any two integers and T 
2

then:

T
 sin( n  t ) sin( m t )dt   
m,n

T
 m,n
2

T
 m,n
2
0

T
 cos(n t ) cos(m t )dt   
m,n
0
T
 sin( n  t ) cos(m t )dt  0
0
Examples.
T
T
0
0
 sin( 27t ) sin( 15t )dt  0
T
 cos(8t ) cos(8t )dt 
0
T
 sin( 15t ) sin( 15t )dt  2
T
T
2
 sin( 4t ) cos(6t )dt  0
0
(Check these results using DERIVE or Mathcad)
44
The four figures below illustrate how the area is zero under a product of two different frequency sinusoids and not
zero (area = T/2 =  ) when the frequencies are the same.
45
APPENDIX 2
The Fourier transform of a unit pulse.
In the following Mathcad exercise
(i) we describe a unit rectangular pulse of height 1 volt and width a =1 millisecond in both time and
frequency domains
.
(ii) The Fourier Transform of the pulse is taken and is seen to be a sinc function with most of the
frequency content of the pulse lying between +1/a and -1/a Hz.
(iii) An investigation is carried out on the effects of filtering out some of the high frequency components
from the Fourier Transform and reconstructing the pulse in the time domain using only the remaining
components.
(iv) in (iii) we find the reconstructed pulse is distorted and this is related to what happens to a rectangular
pulse that is transmitted over a bandwidth limited channel. The pulse may suffer distortion so that the
pulse that is received is not exactly the same shape as the pulse that was sent.
Unit Pulse
define a unit pulse in the time domain, height Vo = one volt and width
one millisecond
Vo  1
pulse( t)  if ( t  .5 Vo  0)
graph the pulse in the time domain
t  2 1.999 2
2
pulse( t )
1
2
1
0
1
2
t
t (ms)
Fourier Transform
Find an expression for the Fourier transform of a unit height pulse of width a seconds.
a
2

sin  a  f

exp  i 2  f t d t simplify 
 f

 a
2
46
3
a  1 10
In the above expression we set the pulse width a to equal 1 millisecond
Now graph the Fourier transform in the frequency domain
5 10
4
sin( a   f)
(   f)
8000 6000
4000
2000
5 10
0
2000
4000
6000 8000
4
f
f Hz
Find the total area under the transform by integration (total area = 1)





sin  a  f
  f
df  1

In practice we truncate the transform at +/- 6 side frequencies and take the modulus of the truncated
transform. We then determine the total area under the truncated transform and find the frequency interval
required to contain 90% of the truncated area.


trunc( f)  if  f 
5 10
6 sin a  f 

0
a
  f 
4
trunc( f)
8000
6000
4000
2000
5 10
0
4
f
take the modulus
47
2000
4000
6000
8000
0.001
5 10
trunc( f)
8000
6000
4000
2000
4
6
0
2000
4000
6000
8000
f
a

A  
trunc( f) d f


6

Find the truncated area, A
a
A  1.899
truncated area between +/- 6 side frequencies is A
Now find the upper frequency, fo necessary to contain 90% of the truncated area
fo  5
guess value



Given
fo
trunc( f) d f
A  .9
 fo
3
2 3673  7.346  10
3
Find( fo)  3.673  10
Hz
Hence the bandwidth, B, is 2 times fo
Now we can reconstruct the pulse in the time domain using our reduced bandwidth and taking the inverse fourier
transform. Below we plot the reconstructed and the original pulses.

reconstruct ( t)  

3673
trunc( f)  exp  i 2  f t d f
inverse transform
 3673
original( t)  if  t 

x  2 a  2 a 
a
200
a
2
 1 0

 2 a
48
2
volts
1
0.002
0.001
0
0.001
0.002
1
t
The above figure shows the original pulse superimposed on the reconstructed pulse. They are not quite the same
due to the missing high frequencies.
49
APPENDIX 3
Filtering noise from a signal.
50
12. Tutorials on Fourier series
Tutorial 1
In Q1 and Q2 sketch 2 cycles of each waveform and find the Fourier series from tables:
Q1
A sawtooth wave of frequency f = 100 Hz, and peak value V = 5 V.
Q2
A rectangular pulse train of frequency f = 1 kHz, peak value 10 V and duty cycle k = 1/2.
Q3
For the Fourier series
v(t )  2.5  4.6sin(200t  0.4)  5.5sin(400t  0.5)  1.8sin(600t  0.3)
(a) State the period (b) Write the series in sine/cosine form
Q4
For the Fourier series
v(t )  2.4  2.6sin(100t )  1.5cos(100t )  3.2sin(200t )  4.1cos(200t )  1.5cos(300t )
2.1sin(500t )  3.2 cos(500t )
(a) State the period (b) Write the series in magnitude/phase form
Q5
A square wave switches between +10 V and – 10V Find the magnitude of the third harmonic.
Q6
Write the equation for one full cycle of the square wave shown in Table 1 varying between + 5 V and
-5 V with period T.
Q7
Write the equation for one full cycle of a triangular wave varying between + 20 V and -20 V having
period T.
Q8
Q9
A square wave has period T = 20  s . The frequency component at 150 kHz is measured on a spectrum
analyser and found to have magnitude 4.8 volts. Find the peak value of the square wave.
Use the table of Fourier series to find the Fourier series up to and including the third harmonic for a
rectangular pulse train of duty cycle k = 1/4 and peak value 10 V, the period is 1 ms. Evaluate Fourier
coefficients to three significant figures, sketch one cycle of the waveform and state the fundamental
frequency.
51
Q10
The figure below is taken from the Applet by Steve Crutchfield, updated by Hsi
Chen Lee Use the information to find the duty cycle and the pulse height for the rectangular pulse train:
(ignore the reference to ‘phase = 3.141’ for the dc term – phase is not defined for dc)
Q11
In a rectangular pulse train the pulse width is 1 ms and the period is 3 ms. The magnitude of the
fundamental is 5 volts. What is the peak value of the pulse train?
Q12.
A rectangular pulse train has peak value 10 volts and the magnitude of the second harmonic is 3 volts.
Find the pulse width if the period is 10 ms..
52
Answers to Tutorial 1
Q1
Using Table 1
f = 100 Hz, T = 1/f = 1/100 = 0.01s = 10 ms ω = 2πf = 628.3 rad/s
x(t) = 3.18sin ω (t) - 1.59sin(2 ω t) + 1.06sin(3 ω t) - 0.80sin(4 ω t).
Q2
Using Table 1
k = 1/2, V = 10, f = 1000 Hz, T = 1/f = 0.001 s = 1 ms, ω = 2πf = 6283.2 rad/s
to /T = k so to = 1/2* 1 ms = 0.5 ms.
x(t) = 5.00 + 6.37cos(ω t) - 2.12cos(3 ω t) + 1.27cos(5 ω t) -...
Q3.
(a) T = 31.4 ms
(b)
v(t )  2.5  1.791cos(200t )  2.637 cos(400t )  0.5319 cos(600t )
 4.237 sin(200t )  4.827 sin(400t )  1.720sin(600t )
Q4
(a) T = 62.8 ms
(b)
v(t )  2.4  3.002sin(100t  2.618)  5.201sin(200t  0.9081)
 1.5sin(300t  1.571)  3.828sin(500t  2.152)
Q5
4.24 volts
Q6
 5

v (t )   5
-5

Q7
-T / 2 < t  -T/4
T/4 < t  T/4
T/4 < t  T/2
80

20 + T t
v (t )  
 20  80 t

T
-T / 2  t  0
0  t  T /2
Q8
11.3 volts
Q9:
2.50 + 4.50cos(t) + 3.18cos(2t) + 2.12cos(3t), where  = 6283 rad/s,
fundamental = 1 kHz.
Q10
Show that cos( k ) 
55.815
61.664
Solve for k to find k = 0.14 and then find V = 228.04. These values
reproduce the given magnitudes of harmonics 1,2,3 and 4. Perhaps there is an error in the value given for
the dc term as it should equal 31.9 not 41.078? Any better suggestions?
Q11
9.07 volts
Q12
1.96 ms
53
Tutorial 2
Q1.
Decide if each of the following function products has any symmetry and, if so, state
whether it is odd or even:
(i)
t 2 cos(3t )
(ii)
(iv) (3t  2)sin(3t )
Q2.
The odd function
(v)
t 2 sin(4t )
(iii)
sin(t ) cos(3t )
(3t 2  2) cos(3t ) (vi) e2t cos(400t )
y  t 3 and T is any number:
T /2
Verify that
 ydt  0
T / 2
Interpret this result graphically.
Q3.
Table 1 shows the Fourier series for a sawtooth wave of amplitude V and period T.
Use the Table to write the Fourier series for a sawtooth with V = 10 V and T = 1 ms. Show terms up to the
third harmonic.
Use integration to calculate the values of a4 and b4 .(Note Integration by parts:
dv
du
 u dt dt  uv   v dt dt )
Q4.
The Fourier coefficients for the first three harmonics of a periodic waveform with period T = 1 ms are
given below.
a1 = 2.6 b1 = 1.8 a2 = -1.4 b2 = -1.0 a3 = -0.7 b3 = 0.9
Write the Fourier series in magnitude/phase form.and draw a diagram to show the corresponding
magnitude/phase spectra.
Q5.
In a certain Fourier series a3 = -6.1 and b3 = -2.4. Find R3 and φ3
Q6
In a certain Fourier series R3 = 5.5 and φ3 = -0.5 rad. Find a3 and b3
Q7
Use integration and any symmetry to derive all the Fourier series given in Table 1. If you are unable to
do some of the integrals manually you can use Mathcad to help you..
54
Answers to Tutorial 2
Q1. (i) even
(ii) odd
(iii) odd (iv) no symmetry (v) even
(vi) no symmetry
Q3. (a) T = 0.001 s, f = 1 kHz, ω = 6283 rad/s
x(t) = 6.37sin(ωt) - 3.18sin(2 ωt) + 2.12sin(3 ωt)
(b) by integration a4 = 0, b4 = -1.59.
Q4. R1 = 3.16 Φ1 = 55.3o ,
R2 = 1.72 Φ2 = -125.5o
R3 = 1.14 Φ3 = -37.9o
x(t) = 3.16sin(ω t+55.3o )+1.72sin(2 ω t-125.5o )+1.14sin(3 ω t-37.9o)
magnitude
3.16
phase
55.3o
1.72
1.14
1
2
3
f (kHz)
-37.9o
-125.5o
Q5
R3 = 6.56 , φ3 = -1.95 rad
Q6
a3 = -2.64 and b3 = 4.83
Q7
Answers in Table 1
55
Tutorial 3
Q1. An a.c. circuit contains a 1000 Ω resistor in series with a 1.0 H inductor. If the supply voltage is
v = 200sin(500t) + 100sin(1500t) + 50sin(2500t) volts
deduce an expression for the current in the circuit and calculate the power dissipated in the circuit.
Q2
A voltage waveform has Fourier series v(t) = 2.5+3.0sin(t)+2.0sin(3t)+1.5sin(5t) volts where
 = 1000 rad/s. The voltage is applied to a 10 ohm resistor in series with a 5 mH inductor. Find the Fourier
series for the voltage across the resistor. Calculate also the power dissipated in the circuit by the fundamental
component alone.
Q3. The amplitude and phase spectra for a periodic signal are shown in Fig. Q2. Determine for this wave
(a) the amplitude/phase form of the Fourier series
(b) the sine cosine form of the Fourier series.
(c) The mean value of the wave
magnitude spectrum
phase spectrum
30
20
4
3
0
0
50
100
150
50
100
150
freq (Hz)
freq (Hz)
-110
-300
-45o
Figure Q2 (phase with respect to sine)
Q4.
(a)
The voltage v(t) is given by its Fourier series as follows:
v(t )  65  100sin(1000 t )  45sin(2000 t ) ,
The waveform v(t) is applied to a coil of inductance L = 10 mH and resistance r = 10 . Show your
working to calculate the corresponding terms in the Fourier series for the current flowing in the circuit.
(b)
The instantaneous magnetic field energy, E, stored by a coil of inductance L carrying a current i,
is given in joules by E 
1 2
Li . What is the average field energy stored by the coil in Q4(a)?
2
56
Answers to Tutorial 3
Q1.
i = 0.179sin(500t-0.464) +0.055sin(1500t-0.983) + 0.0186sin(2500t - 1.19) A.
` Power = 17.71 W
Q2
2.5 + 2.68sin(t-0.4636)+1.109sin(3t-0.9828)+0.557sin(5t-1.1903)
Power in fundamental = 0.36 W.
(a) f(t) = 20 + 30sin(ωt-11o ) + 4sin(2ωt-30o )+3sin(3ωt-45o )
ao =20, a1 = -5.72, b1 =29.45, a2 =-2, b2 =3.46, a3 =-2.12, b3 =2.12.
(You should write the series in full yourself)
(c) mean value = 20.
Q3.
Q4. (a) Impedance seen by nth harmonic is Z (n)  10  jn L  10  j10n .
n
Z(n)
0
1
2
10
10+j10
10+j20
i
Z (n) ohms
10
14.14
22.36
arg(Z(n)) rad
0
0.785
1.107
65 100
45

sin(1000t  0.785) 
sin(2000t  1.107)
10 14.14
22.36
i  6.50  7.07 sin(1000t  0.785)  2.01sin(2000t  1.107)
(b)
Instantaneous energy stored is mean energy stored is E 
1 2
Li The mean energy stored is
2
1 2
Li where i 2 is the mean value of i 2 . But by definition.
2
1
1
2
i 2  I rms
 6.52  7.072  2.012  8.32 Hence the mean energy stored is
2
2
1
102  8.322  346 mJ
2
E
57
Tutorial 4
Q1 and Q2 are taken from "Fourier series and Harmonic Analysis" by K.A. Stroud (QA 404 STR)
Q1. Determine the rms value of the periodic voltage defined by
v=
5.04 - 0.679 cos(ωt) - 0.432 cos(2ωt) - 0.005 cos(3ωt)
+ 3.852 sin(ωt) - 1.186 sin(2ωt) + 0.480 sin(3ωt)
V
Answer: 5.83 V
Q2.
An alternating voltage v(t) volts applied to a circuit produces a current i(t) amps where
v(t) = 4.70 -1.49 cos(ωt) - 0.50 cos(2ωt) - 0.16 cos(3ωt)
+ 0.53 sin(ωt)-0.52 in(2ωt) -0.23 sin(3ωt)
and
i(t) = 2.04 -0.08 cos(ωt)-0.05 cos(2ωt) + 0.01 cos(3ωt)
+1.75 sin(ωt) -0.36 sin(2ωt) +0.15 sin(3ωt)
Determine the power dissipated.
Answer 10.2 W
Q3. From "Electric Circuits" by Morris and Senior (TK 454.2 MOR)
The current in a circuit is
i = 10sin(ωt-0.7) +sin(3ωt-0.15) +0.25sin(5ωt+1.3) A
where ω = 350 rad/s.
The voltage applied to the circuit is
v = 282.8sin(ωt+0.085) + 63.25sin(3ωt + 1.1) + 25.5sin(5ωt + 2.674) V
Calculate
(a) the rms voltage
(b) the rms current
(c) the power supplied to the circuit
(d) the overall power factor.
Answers (a) 205.7 V (b) 7.11 A (c) 1010.6 W (d) 0.69
Q4.
(From “Fourier Series by Bolton QA404 BOL)
The voltage v  30  40sin(1000t )  25sin(2000t )  15sin(4000t ) V is applied to a
20  resistor connected in series with a 20  F capacitor.
overall power factor.
Find the total power dissipated and the
Would you expect more power or less power to be dissipated if the capacitor were removed and the same
voltage applied across just the resistor? Why?
Answer P = 15.7 W, OPF = 0.43.
58
Q5
a rectangular pulse train of duty cycle k = ½ and peak value 10 V, the period is 20 ms is applied to a
10 ohm resistor. Calculate the total power dissipated in the resistor and the percentage of this power due
to the fundamental.
Answer total power = 5W, % due to fundamental = 40.5%.
Q6
In a certain passive circuit
v = 1.5 + 2.4sin(300t+0.5) + 1.5sin(600t+0.8) V
i = 1.2 + 2.5sin(300t+0.9) + 0.5sin(600t)+1.5cos(600t) A
Calculate
(i)
(ii)
(iii)
Answer
the rms value of v
the total power dissipated
the overall power factor.
2.5 V, 5.63 W, 0.93
59
Engineering Mathematics C
Time Allowed = 1 hour
Past Example of Phase Test 1
Answer all questions, write your answers in the spaces provided.
A formula sheet and a table of Fourier series are provided.
Q1(a).
v(t)
volts
25
-40
0
-20
20
40
t ms
-25
Fig. Q1 (a)
For the periodic voltage waveform in Fig.Q1(a) state, giving appropriate units:
(i)
The period, T
__________________________________________
(ii)
The fundamental frequency
(iii)
The fundamental angular frequency ______________________________
(iv)
The frequency of the third harmonic
(v)
The symmetry (if any) __________________________________________
____________________________________
______________________________
Ans: (i) 40 ms (ii) 25 Hz (iii) 157 rad/s (iv) 75 Hz (v) even and half-wave
Q1(b)
[5 marks]
Use the tables provided to obtain the Fourier series for the waveform in
Fig. Q1(a) up to and including the third harmonic term (evaluate the Fourier coefficients to 3 significant
figures).
[5 marks]
Ans: 20.3cos( t )  2.55cos(3 t )  0.811cos(5 t )
Q1(c)
x(t)
50
t ms
-40
-20
0
Fig Q1(c)
20
40
Use your answer to Q1 (b) (or use the tables provided) to obtain the Fourier series in sine/cosine form up to and
including the third harmonic term for the waveform, x(t), in Fig.Q1(c). (Integration is NOT required)
[5 marks]
Ans: x(t )  25  20.3cos( t )  2.25cos(3 t )  0.811cos(5 t )
60
Q2
A d.c.-shifted sawtooth voltage waveform is low-pass filtered to remove most of the higher harmonics
without attenuating the lower frequencies. After filtering it is given by the first three terms of its Fourier
series as follows:
v(t) = 10 .0 + 9.55sin(1000t) + 4.77sin(2000t + ) volts.
(i) Calculate the rms value of the voltage v(t). (Ans 12.5 V)
[5 marks]
(ii) The voltage, v(t), is applied to a certain passive component and the current, i(t), flowing in the
component is found to be
i(t) = 2 .0 + 4.62sin(1000t-0.60) + 2.15sin(2000t +2.24) amps
Calculate the total power dissipated. (Ans 41.4 W)
[5 marks]
(iii) What was the maximum voltage of the sawtooth wave before filtering? (Ans 25 V)
[5 marks]
Q3.
The voltage, v(t), in Q2, is applied to the terminals of a coil having inductance L = 10 mH and resistance
r = 5 . Show your working to find the Fourier series for the current in the coil.
[10 marks]
Ans: i  2.00  0.854sin(1000t  1.107)  0.231sin(2000t  1.816)
Q4
v(t)
volts
20
-30
-20
-10
t ms
0
10
20
30
Fig. Q4
(i) Taking into account the symmetry, use integration to find the magnitude of the fundamental component of the
periodic rectangular pulse train shown in Fig. Q4. (Ans 12.7 V)
(ii) In Fig. Q4 the period and pulse height are held constant while the pulse width is reduced until the rms voltage
of the waveform reaches 5 volts. Use tables to determine the new magnitude of the fundamental component. (Ans
2.48 V)
[10 marks]
61
13. Miscellaneous examples with answers.
18.1 Find the effective voltage, effective current and average power supplied to a passive network if the applied
voltage is
v = 200 + 100cos(500t+30o ) + 75cos(1500t+60o ) V
and the resulting current is
i = 3.53cos(500t+75o ) +3.55cos(1500t+78.45o ) A.
Answer: 218.5 V, 3.54 A, 250.8 W (Edminster p212, Q12.39)
18.2 A voltage
v = 50 + 25sin(500t)+10sin(1500t)+5sin(2500t) V
is applied to the teminals of a passive network and the resulting current is
i = 5+2.23sin(500t-26.6o )+0.556sin(1500t-56.3o ) +0.186sin(2500t-68.2o )
Find the effective voltage, effective current and the average power.
Answer: 53.6 V, 5.25 A, 276.5 W (Edminster p212 Q12.40)
18.3 A three-element series circuit with R = 5Ω, L = 5mH and C = 50 μF, has an applied voltage
v = 150sin(1000t)+100sin(2000t)+75sin(3000t) V.
Find the effective current and average power for the circuit. Sketch the line spectrum of the voltage and current
and note the effect of series resonance.
Answer: 16.58 A, 1374 W (Edminster p212, Q12.41)
18.4 A two-element series circuit with R=10 Ω and L = 20 mH, has current
i = 5sin(100t)+3sin(300t) + 2sin(500t) A.
Find the effective voltage and the average power.
Answer: 48 V, 190 W (Edminster p213, Q12.42)
18.4 Derive all the Fourier series shown in Table 1 page 7
18.5 Write the Fourier series in amplitude/phase for a certain 20 Hz periodic waveform given that:
ao = 8.00 a1 = -6.63 a2 = 0.26 a3 = -0.97 a4 =0.26
a5 = -0.52 a6 = 0.26
b1 =10.00 b2 = -4.99 b3 = 3.33 b4 = -2.49 b5 =1.99 b6 = -1.65
(Schaum '3000 solved problems in Electric Circuits', p469)
Answer:
v = 8 + 12sin(40πt-33.54o ) + 5.0sin(80πt+177.02o )
+ 3.47sin(120πt-16.24o ) +2.5sin(160πt+174.04o )
+2.06sin(200πt-14.64o ) +1.67sin(240πt+171.05o )
18.6 Determine the rms value of the waveform in 18.5 above.
Answer: 12.69
62
18.7
A series RL circuit in which R=5Ω, L=20 mH has an applied voltage
v = 100 + 50sin(ωt) + 25sin(3ωt) V, with ω=500 rad/s.
Find the instantaneous current.
(Schaum '3000 solved problems in Electric Circuits', p475)
Answer: i = 20 + 4.48sin(wt-63.4 ) + 0.823sin(3wt-80.54 ) A.
18.8 Determine the power dissipated in the resistor of 18.7 above using P  I rms R .
2
Answer: P = (410.6).(5) = 2053 W
18.9
Calculate the power contributed by each harmonic in the circuit of 18.7.
powers is the same as the power obtained in 18.8.
Verify that the sum of these
Answer: P=2000 + 50.1 + 1.69 = 2052 W (round-off error)
18.10 Determine the instantaneous and effective voltages across the resistor in 18.7.
dissipated in the resistor.
Answer:
vR  iR = 100+22.4sin(ωt-63.40 )+4.11sin(3ωt-80.54o )
VR (rms) = 101.3 V
P  v R2 (rms) / R = 2052 W
18.11 A voltage waveform is defined by
V 0  t  T / 4

v   V T / 4  t  3T / 4
V 3T / 4  t  T

Determine the Fourier coefficients.
(Schaum '3000 solved problems in Electric Circuits', p482)
4V
Answer: ao = 0 (by inspection) a1 
n
4V / n n  1,5,9...

a n   4V / n n  3,7,11...
0 n even

63
Hence obtain the power