17-02-06
EECE 2602 -- Signals and Systems in Continuous Time
Week 5 – Signal Representations Using
Fourier Series
(Textbook: Ch. 4.2 – 4.4)
1
1
EECE 2602 -- Signals and Systems in Continuous Time
LTIC systems with sinusoidal input signals
• In lecture 3 (slides 27-29), we investigated the output response of an LTIC
system (i.e. a RC circuit) with a sinusoidal input signal
R
x(t ) = sin (t ) C y(t)
_
C = 0.1F and R = 40Ω, and y(0) = 10V
174 −4t 4 1
y(t ) = e − cos(t ) + sin (t )
17 17 17
2
2
1
17-02-06
EECE 2602 -- Signals and Systems in Continuous Time
LTIC systems with sinusoidal input signals
Output response when x(t) = sin(t)
10
6
total response
2
4 1
ysteady (t ) = − cos(t ) + sin (t )
0
17 17
1
-2 ysteady (t ) = sin (t − 75.96°)
0 5 10 15 20
t [sec]
25 30 35 40
17
We observed that the output response of an RC circuit to a sinusoidal
function was another sinusoidal function of the same frequency except for
changes in amplitude and phase
3
3
EECE 2602 -- Signals and Systems in Continuous Time
LTIC systems with sinusoidal input signals
• The output response of an LTIC system, characterized by a real-valued impulse
response h(t), to a sinusoidal input is another sinusoidal function with the same
frequency , except for possible changes in its amplitude and phase.
x(t) LTIC y(t)
system
k1 sin(ω1t ) → A1k1 sin(ω1t + φ1 )
and
k1 cos(ω1t ) → A1k1 cos(ω1t + φ1 )
where A1, k1, are constant values
4
4
2
17-02-06
EECE 2602 -- Signals and Systems in Continuous Time
LTIC systems with complex exponential signal
• If a complex exponential function is applied to an LTIC system with real-valued
impulse response function, the output response of the system is identical to the
complex exponential function except for changes in amplitude and phase:
k1e jω1t → A1k1e j (ω1t +φ1 )
where A1 and ϕ1 are constants
h(t)
x(t) LTIC y(t)
system
5
5
EECE 2602 -- Signals and Systems in Continuous Time
LTIC systems with complex exponential signal
• Since:
{ }
cos(ω1t ) = Re e jω1t and sin (ω1t ) = Im e jω1t { }
• Similar to the complex exponential case, with a real-valued impulse response
function, then:
{ } {
Re k1e jωt → Re A1k1e j (ωt +φ1 ) }
{ } {
Im k1e jωt → Im A1k1e j (ωt +φ1 ) }
The output response to a sinusoidal function is just another sinusoidal function
of the same frequency except for changes in amplitude and phase
6
6
3
17-02-06
EECE 2602 -- Signals and Systems in Continuous Time
LTIC systems with complex exponential signal
Proof:
∵ x (t ) = k1e jω1t , Impulse response of LTCI is h(t)
∞ ∞
∴ y (t ) = ∫ h (τ ) x (t − τ ) dτ =k1e jω1t ∫ h (τ ) e jω1τ dτ
−∞ −∞
∞
Defining H (ω1 ) = ∫ h(τ )e − jωτ dτ
−∞
y (t ) = A1k1e jω1t H (ω1 )
7
7
EECE 2602 -- Signals and Systems in Continuous Time
Activity 1
Calculate the output response if the input signal x(t) = 2sin(5t)
is applied to an LTIC system with impulse response h(t) =
2e-4tu(t)?
8
8
4
17-02-06
EECE 2602 -- Signals and Systems in Continuous Time
Fourier series (FS) basis
• From Ch. 1, we know that a signal x(t) that is a linear combination of sine and
cosine functions is also periodic if the ratio of their frequencies is a rational
number.
x(t ) = sin(t ) + sin(3t )
x(t ) = sin(t ) + sin(3t ) + sin(5t ) These are all periodic signals
x(t ) = sin(t ) + sin(3t ) + sin(5t ) + sin(7t )
9
9
EECE 2602 -- Signals and Systems in Continuous Time
Trigonometric CT Fourier Series
• Definition: An arbitrary periodic function x(t) with fundamental period T0 can be
expressed as follows: ∞
x(t ) = a0 + ∑ (an cos(nω0t ) + bn sin (nω0t ))
n =1
where ω0 is the fundamental frequency of x(t) and coefficients a0, an, bn are
referred to as the trigonometric Fourier series (FS) coefficients.
1
a0 =
T0 ∫ x(t )dt
T0
Mean value of x(t) or the DC component
2
an =
T0 ∫ x(t )cos(nω t )dt
T0
0
2
bn =
T0 ∫ x(t )sin(nω t )dt
T0
0
10
10
5
17-02-06
EECE 2602 -- Signals and Systems in Continuous Time
Trigonometric CTFS
• In x(t): ∞
x(t ) = a0 + ∑ (an cos(nω0t ) + bn sin (nω0t ))
n =1
• The sine or cosine term with the fundamental frequency ω0 is called the
fundamental component of x(t), that is, when n = 1.
• The sine or cosine term with the fundamental frequency nω0 is called the
harmonic component of x(t) , that is, when n ≠ 1.
11
11
EECE 2602 -- Signals and Systems in Continuous Time
Activity 2
Calculate the trigonometric CT Fourier series of the periodic signal x(t) defined over
one period T0=3 as follows:
⎧⎪ t +1 −1 ≤ t ≤ 1
x (t ) = ⎨
⎪⎩ 0 1≤ t ≤ 2
x(t)
2
t
−8 −6 −4 −2 0 2 4 6 8 10
12
12
6
17-02-06
EECE 2602 -- Signals and Systems in Continuous Time
CTFS coefficients for symmetrical signals
1. If x(t) is zero-mean, then a0 = 0.
2. If x(t) is an even function, then bn = 0. In other words, an
even signal is represented by its DC component and a
linear combination of a cosine function of frequency ω0 and
its higher-order harmonics.
3. If x(t) is an odd function, then a0 = an = 0 for all n. In other
words, an odd signal can be represented by a linear
combination of a sine function of frequency ω0 and its
higher-order harmonics.
4. If x(t) is a real function, then the CTFS coefficients a0 , an
and bn are also real-valued for all n.
13
13
EECE 2602 -- Signals and Systems in Continuous Time
Activity 3
Consider the function w(t)=Ev[x(t)-a0] shown in figure below. Express w(t) as a
trigonometric CTFS.
Ev{x (t) − a0}
1/3
t
−9 −6 −4 0 3 6 9
2/3
14
14
