5.1
Periodic Signals:
A signal f(t) is periodic iff for some T0 > 0,
f (t ) = f (t + T )
∀ t
The smallest T0 value that satisfies the above conditions is called the period of f(t).
i
0
Consider a signal examined over 0 to 5 seconds shown below. What values of T0
satisfy the condition in the definition of periodic? What is the period of this signal?
s i n(2 *p i *t) + s i n(2 *p i *2 *t + .3 )
2
1 .5
Amplitude
1
0 .5
0
-0 .5
-1
-1 .5
-2
0
1
2
3
S e c o nd s
4
5
5.2
Properties of Periodic Signals:
The integration over one full period has the same result, independent of where the
integration limits begin or end.
a + T0
b + T0
a
b
∫ g ( t )dt = ∫ g (t )dt = ∫ g ( t )dt
T0
The Sinusoid:
⎛ 2π
⎞
g (t ) = sin(ω t + θ ) = sin( 2π ft + θ ) = sin⎜ t + θ ⎟
⎝T
⎠
Phase - θ, Frequency (radians per second) - ω, Frequency (Hz) - f, Period - T0
0
Harmonics:
Consider a signal consisting of a sum of sinusoids whose frequencies are integer
multiples of a fundamental frequency f0:
g ( t ) = C + C sin(2π f t + θ ) + C sin( 2π 2 f t + θ ) + C sin(2π 3 f t + θ ) +L
0
1
0
1
2
0
2
3
0
3
where f0 is the fundamental frequency and the sinusoid with frequency kf0 is called
the k-th harmonic in the series and C0 is the zero-th harmonic (or DC).
Examples: Find the periods and fundamental frequencies of the signals below.
5.3
a) g(t) = 6sin(20πt) -13cos(30πt)
(Hint: For f0 find the greatest common factor between the frequencies, the least
common factor will also work)
(Hint: For T0 find the least common multiple between the periods)
b) g(t) = 4 + 3sin(120πt) + 10sin(150πt) + 2sin(240πt)
c) g(t) = 4cos(8πt) - 11cos(20πt) + cos(10πt)
d) g(t) = 6sin(3.2πt) -13cos(7.2πt)
e) g(t) = 6sin(20πt) - 13sin(20t)
(Hint: Must the sum of periodic sinusoidal signals be periodic?)
Fourier Series:
5.4
A periodic signal, f(t), of almost any form can be expressed as a series of harmonic
sinusoidals:
Trig-forms
f ( t ) = C + ∑ C cos( nω t + θ ) = a + ∑ a cos( nω t ) + b sin( nω t )
∞
0
∞
n=1
n
n
0
0
n=1
n
where C = a , a = C cos(θ ), b = C sin(θ ) , C =
0
0
0
n
n
n
n
n
n
0
n
0
⎛ −b ⎞
a + b , θ = tan ⎜ ⎟
⎝ a ⎠
2
2
n
n
−1
n
n
n
Exponential-forms
∞
f ( t ) = ∑ D exp( jnω t )
n =−∞
n
0 n
where Dn belongs to the set of complex numbers and Dn = D*-n
5.5
Example: Show by plotting the harmonic components of a Fourier series that the
periodic extension of r(t+1)-2r(t) on the interval [-1, 1], has Fourier coefficients:
1
2 (1 − cos(π n ))
, b =0
a = , a =
2
(π n )
0
2
n
n
>>t = [-4:.001:4];
% Set up time axis to show 4 periods
>>s = .5*ones(size(t)); % Initialize summing array with a0
>>for k=1:10,
% Loop to sum up 10 harmonics
>>s = s + 2*((1-cos(k*pi))/(pi*k)^2)*cos(k*pi*t);
>>end
>>plot(t,s,'w')
first 5 harmonics
Up to 11-th harmonic
first harmonic
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0.5
0
0
-4
-3
-2
-1
0
time
1
2
3
0
-4
4
-3
-2
Up to third harmonic
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
-3
-2
-1
0
time
1
0
time
1
2
3
4
Up to 51-st harmonic
1
0
-4
-1
2
3
4
0
-4
-3
-2
-1
0
time
1
2
3
4
-0.5
-4
-3
-2
-1
0
time
5.6
Computing Fourier Series Coefficients (trig-form):
Given the Fourier Series representation of a periodic signal f(t):
f (t ) = a + ∑ a cos( nω t ) + b sin( nω t )
∞
0
n =1
n
0
n
0
it can be shown that the coefficients for f(t) can be computed with:
1
2
2
a = ∫ f (t ) dt , a = ∫ f (t ) cos( nω t ) dt , b = ∫ f (t ) sin( nω t ) dt
T
T
T
0
n
T0
0
T0
0
n
0
0
T0
0
Derive the Fourier Series coefficients in trig form for the waveform in the previous
example.
Derive the Fourier Series coefficients in trig form for the periodic extension of p (t )
on the interval [0, T0], where 0<Δ<T0
Δ
show that: a =
0
1
,
T
0
a =
n
sin(ω n Δ )
,
( nπ Δ )
0
b =
n
(1 − cos(ω π n ))
( nπ Δ )
0
5.7
Computing Fourier Series Coefficients (exponential-form):
Given the Fourier Series representation of a periodic signal f(t):
f ( t ) = ∑ D exp( jnω t )
∞
n =−∞
n
0 n
it can be shown that the coefficients for f(t) can be computed with:
1
D = ∫ f ( t ) exp( − jnω t )dt
T
Note the differences between the Trig and Exponential form:
1. Coefficients for the trig for are real for real f(t), while for the exponential form
they are complex.
2. The index for summation is from 0 to ∞ for the trig form and from -∞ to +∞ for
the exponential form
3. The formulas for the exponential form are more compact and easier to work with
than those for the trig form.
n
0
T0
0
The coefficients for compact trig representation can be determine from the
exponential form coefficients: C = D , and C = 2 D , θ = ∠D for n>0
0
Examples:
0
n
n
n
n
5.8
Derive the Fourier Series coefficients in exponential form for the periodic extension
of p (t ) on the interval [0, T0], where 0<Δ<T0. Plot the magnitudes and phases of
the coefficients (for T0 = 10 ms and Δ = 1 ms) as a function of frequency (f) and
discuss their meaning.
⎛
nΔπ ⎞ ⎛ nπ Δ ⎞
exp⎜ − j
⎟ sin⎜
⎟
T ⎠ ⎝ T ⎠
⎝
1
Show: D =
for n≠0
D =
T
nπ Δ
Δ
0
n
0
0
0
>>t0 = 10e-3
% Period
>>del = 1e-3
% Duration of pulse
>>n = [-60:-1, eps, 1:60]; % Harmonic index (eps is close to zero)
>>f0 = 1/t0;
% Fundamental Frequency
>>d = exp(-j*pi*del*f0*n).*sin(f0*n*pi*del)./(pi*del*n); % Fourier Coefficients
>>figure(1)
>>stem(f0*n,abs(d),'-w')
% Stem plot of magnitudes
>>title('Magnitude of Fourier Coefficients')
>>xlabel('Hz')
>>figure(2)
>>stem(f0*n,angle(d),'-w')
% Stem plot of phases
>>title('Phase of Fourier Coefficients')
>>xlabel('Hz')
5.9
M a g n i tu d e o f F o u r i e r C o e ffi c i e n ts
100
80
60
40
20
0
-6 0 0 0
-4 0 0 0
-2 0 0 0
0
H z
2000
4000
P ha s e o f F o urie r C o e ffic ie nts
4
3
2
1
0
-1
-2
-3
-4
-6 0 0 0
-4 0 0 0
-2 0 0 0
0
Hz
2000
4000
5.10
Examples:
Derive the Fourier Series coefficients in exponential form for the periodic extension
of r(t) on the interval [0, 0.5]. Plot the magnitudes and phases of the coefficients as a
function of frequency (f) and discuss their meaning.
j
1
for n≠0
Show: D =
D =
4 nπ
4
n
0
>>t0 = .5
% Period
>>n = [-30:-1, eps, 1:30]; % Harmonic index (eps is close to zero)
>>f0 = 1/t0;
% Fundamental Frequency
>>d = j./(4*pi*n); % Fourier Coefficients.
>>d(31) = 1/4;
% Must substitute DC term in directly, since expression was
>>
% simplified beyond the point of using L'Hopital's rule
>>figure(1)
>>stem(f0*n,abs(d),'-w')
% Stem plot of magnitudes
>>title('Magnitude of Fourier Coefficients')
>>xlabel('Hz')
>>figure(2)
>>stem(f0*n,angle(d),'-w')
% Stem plot of phases
>>title('Phase of Fourier Coefficients')
>>xlabel('Hz')
5.11
M a g n i tu d e o f F o u ri e r C o e ffi c i e n ts
0 .2 5
0 .2
0 .1 5
0 .1
0 .0 5
0
-6 0
-4 0
-2 0
0
20
Hz
P h a s e o f F o u ri e r C o e ffi c i e n ts
40
6
-4 0
-2 0
40
60
2
1 .5
1
0 .5
0
-0 .5
-1
-1 .5
-2
-6 0
0
Hz
20