EGR 261 – Signals and Systems
Lecture #18
Read: Ch. 16 in Electric Circuits, 9th Edition by Nilsson
Fourier Series
Recall that:

 a
f(t)  a v 
n 1
n
cosnw 0 t   b n sin nw 0 t 
(Fourier Series
representation for f(t))
where
av 
1
T
to  T
2
ak 
T
 f(t)dt
to
to  T
 f(t)cos(kw
to
o
t)dt
2
bk 
T
to  T
 f(t)sin(kw
o
t)dt
to
1
EGR 261 – Signals and Systems
Lecture #18
Example: A) Find the Fourier Series for v(t) shown below.
v(t)
T
Vm
-1
Result : v(t)  a v 
0

 a
k 1
k
1
2
3
t
coskw 0 t   b k sin kw 0 t 
Vm
-V
, a k  0 , and b k  m cos(k ) - 1
2
k
 2Vm
 k  for k  odd

or b k  
0
for k  even


where a v 
2
Lecture #18 EGR 261 – Signals and Systems
b)
Write out the first several terms in the Fourier Series.
c)
Discuss how some of the results above could be determined by
symmetry. The symmetry in v(t) above is clearer if the average value is
removed as shown below.
v(t)
T
0.5Vm
1
0
1
2
3
t
-0.5Vm
We will soon investigate the effects of symmetry on a Fourier Series. If we
had already done so, we could have predicted that ak = 0 for all k and
that
bk = 0 for k = even. This is because the waveform above has:
Half-wave symmetry – thus it has no even harmonics
Odd symmetry – thus it has no cosine terms (i.e., ak = 0)
3
Lecture #18 EGR 261 – Signals and Systems
The effect of symmetry in a Fourier Series
4 types of symmetry will be considered:
• Even-function symmetry
• Odd-function symmetry
• Half-wave symmetry
• Quarter-wave symmetry
4
Lecture #18 EGR 261 – Signals and Systems
Even-function symmetry
A function is even if f(t) = f(-t).
This type of symmetry depends on where t = 0.
The y-axis is the axis of symmetry (the waveform is mirrored about the y-axis).
Examples:
f(t)
0
f(t)
t
f(t) = Kcos(wt)
t
0
f(t)
t
t
0
0
5
Lecture #18 EGR 261 – Signals and Systems
Even-function symmetry
For all functions :
T
1
a v   f(t)dt
T0
T
2
a k   f(t)cos(kw o t)dt
T0
T
2
b k   f(t)sin(kw o t)dt
T0
For functions with even symmetry :
T/2
2
av 
f(t)dt
T 0
T/2
4
ak 
f(t)cos(kw o t)dt
T 0
bk  0
(no sines since sines are odd)
6
Lecture #18 EGR 261 – Signals and Systems
Even-function symmetry - Proof
Show that the results above apply to any general function f(t), such as the one shown
below, which has even symmetry. Determine expressions for bk and ak in general to
verify the results above. (Integrate from –T/2 to 0 and 0 to T/2 and then use a
substitution on the first integral: t = -x so dt = -dx and when t = T/2 then x = -T/2.)
f(t)
-T
2
+T
2
t
0
7
Lecture #18 EGR 261 – Signals and Systems
Odd-function symmetry
A function is odd if f(t) = -f(-t).
This type of symmetry depends on where t = 0.
It is often helpful to remove the average (DC) value to see the symmetry.
This symmetry is present if the waveform is mirrored and flipped about the y-axis.
Examples: (If Av0, sketch the waveform with the DC level removed.)
f(t)
t
f(t) = Ksin(wt)
t
f(t)
t
8
Lecture #18 EGR 261 – Signals and Systems
Odd-function symmetry
For all functions :
T
1
a v   f(t)dt
T0
T
2
a k   f(t)cos(kw o t)dt
T0
T
2
b k   f(t)sin(kw o t)dt
T0
For functions with odd symmetry :
T/2
2
av 
f(t)dt
T 0
ak  0
(no cosines since cosines are even)
T/2
4
bk 
f(t)sin(kw o t)dt

T 0
9
Lecture #18 EGR 261 – Signals and Systems
Half-wave symmetry
A function has half-wave symmetry if f(t) = -f(t – T/2). In other words, if you shift the
function one-half period and invert it, it is identical to the original function.
This type of symmetry does not depend on where t = 0.
This symmetry is present if one half of a period is a mirrored image of the other half.
Example: Note that the waveform below is neither even nor odd, but has half-wave
symmetry. If the waveform were shifted, it could also have either even or odd
symmetry.
f(t)
axis of
symmetry
T
2
0
T
2
t
10
Lecture #18 EGR 261 – Signals and Systems
Half-wave symmetry
For all functions :
T
1
a v   f(t)dt
T0
T
ak 
2
f(t)cos(kw o t)dt
T 0
T
2
b k   f(t)sin(kw o t)dt
T0
For functions with half - wave symmetry :
av  0

0
for even k

ak  
 4 T/2
  f(t)cos(kw o t)dt for odd k
 T 0

0

bk  
 4 T/2
  f(t)sin(kw o t)dt
 T 0
for even k
for odd k
So, a periodic function with half-wave symmetry has zero average
and contains only odd harmonics.
11
Lecture #18 EGR 261 – Signals and Systems
Quarter-wave symmetry
To have quarter-wave symmetry, there must also be half-wave symmetry.
If a waveform with half-wave symmetry also is symmetric about the center of
each half-cycle, then it also has quarter-wave symmetry.
Example: The waveform below has both half-wave and quarter-wave
symmetry. Does it have any other type of symmetry?
v(t)
axis of half
symmetry
Vm
Vm
2
0
T
4
T
2
3T
4
axes of quarter
symmetry
T
t
12
Lecture #18 EGR 261 – Signals and Systems
Quarter-wave symmetry
For all functions :
T
1
a v   f(t)dt
T0
T
ak 
2
f(t)cos(kw o t)dt
T 0
T
2
b k   f(t)sin(kw o t)dt
T0
For functions with quarter - wave symmetry :
av  0

0
for even k

ak  
 8 T/4
  f(t)cos(kw o t)dt for odd k
 T 0

0

bk  
 8 T/4
  f(t)sin(kw o t)dt
 T 0
for even k
for odd k
Additional ly, note that :
1) if the function is even, then b k  0 for all k
2) if the function is odd, then a k  0 for all k
13
Lecture #18 EGR 261 – Signals and Systems
Example: Find the Fourier Series representation for v(t) below if Vm = 10V
v(t)
and T = 2ms.
Vm
Vm
2
0
T
4
T
2
3T
4
T
t
14
Lecture #18 EGR 261 – Signals and Systems
Example: Find the Fourier Series representation for v(t) below.
v(t)
10V
5V
0
2
4
6
10 12
t
15