Fourier Series
To use the Fourier Series, signal 𝑥(𝑡) must be periodic with period 𝑇𝑜 seconds (or seconds/cycle)
and fundamental frequency 𝑓𝑜 Hertz (cycles/seconds), where 𝑓𝑜 = 1/𝑇𝑜
In that case you can write the signal as a sum of cosines and sines, like this
∞
𝑥(𝑡) = ∑ 𝐴𝑘 𝑐𝑜𝑠(2𝜋𝑘𝑓𝑜 𝑡) + 𝐵𝑘 𝑠𝑖𝑛(2𝜋𝑘𝑓𝑜 𝑡)
𝑘=0
Note that k is an integer greater than or equal to zero, 𝐴𝑘 and 𝐵𝑘 are real numbers that can be
positive or negative. The cosine terms will give you the even part of x(t), while the sine terms
will give you the odd part of x(t). You can find the 𝐴𝑘 and 𝐵𝑘 coefficients using the equations:
𝐴𝑘 =
2 𝑇𝑜
∫ 𝑥(𝑡)𝑐𝑜𝑠(2𝜋𝑘𝑓𝑜 𝑡)𝑑𝑡
𝑇𝑜 𝑡=0
2 𝑇𝑜
𝐵𝑘 = ∫ 𝑥(𝑡)𝑠𝑖𝑛(2𝜋𝑘𝑓𝑜 𝑡)𝑑𝑡
𝑇𝑜 𝑡=0
Those equations are only for positive k. At k = 0 use
𝐴0 =
1 𝑇𝑜
∫ 𝑥(𝑡)𝑑𝑡
𝑇𝑜 𝑡=0
𝐵0 = 0
An alternative way to write x(t) is
∞
𝑥(𝑡) = ∑ 𝐶𝑘 𝑐𝑜𝑠(2𝜋𝑘𝑓𝑜 𝑡 + 𝜃𝑘 )
𝑘=0
In this case k is an integer greater or equal to zero, 𝐶𝑘 is a real number greater or equal to zero,
and 𝜃𝑘 is a real number we usually write between – 𝜋 and 𝜋 radians. You can find 𝐶𝑘 and 𝜃𝑘
from 𝐴𝑘 and 𝐵𝑘 like this:
𝐶𝑘 = √𝐴2𝑘 + 𝐵𝑘2
−𝐵𝑘
𝜃𝑘 = 𝑡𝑎𝑛−1 (
)
𝐴𝑘
𝐴𝑘 = 𝐶𝑘 cos(𝜃𝑘 )
𝐵𝑘 = −𝐶𝑘 sin(𝜃𝑘 )
Usually we use the complex exponential form of the Fourier Series, which looks like this
∞
𝑥(𝑡) = ∑ 𝑋𝑘 𝑒 +𝑗2𝜋𝑘𝑓𝑜 𝑡
𝑘=−∞
Note that k is an integer that can be positive, negative or zero, j is the square root of -1, and 𝑋𝑘 is
in general complex. When x(t) is real (which is always is in EE 3400) then the negative
harmonics have a special relationship with the positive harmonics because
∗
𝑋𝑘 = 𝑋−𝑘
where the star operator means complex conjugate (change sign of imaginary part). You can find
𝑋𝑘 using
𝑋𝑘 =
1 𝑇𝑜
∫ 𝑥(𝑡)𝑒 −𝑗2𝜋𝑘𝑓𝑜 𝑡 𝑑𝑡
𝑇𝑜 𝑡=0
You can convert between 𝑋𝑘 and the other Fourier Series coefficients like this:
𝐶𝑘 = 2 |𝑋𝑘 |
𝐴𝑘 = 2 𝑅𝑒𝑎𝑙(𝑋𝑘 )
𝜃𝑘 = 𝑎𝑛𝑔𝑙𝑒(𝑋𝑘 )
𝐵𝑘 = −2 𝐼𝑚𝑎𝑔𝑖𝑛𝑎𝑟𝑦(𝑋𝑘 )
Fourier Transform
If x(t) is not periodic, but it is finite energy, then you cannot use the Fourier Series, but you can
use the Fourier Transform. The energy of a signal is
∞
𝐸𝑛𝑒𝑟𝑔𝑦 = ∫ 𝑥 2 (𝑡)𝑑𝑡
−∞
The Fourier Transform equations are
∞
𝑥(𝑡) = ∫ 𝑋(𝑓)𝑒 +𝑗2𝜋𝑓𝑡 𝑑𝑓
−∞
∞
𝑋(𝑓) = ∫ 𝑥(𝑡)𝑒 −𝑗2𝜋𝑓𝑡 𝑑𝑡
−∞
Here f is a real number, that can be both positive and negative. When x(t) is real (as it always is
in EE 3400) the negative frequencies will always be the complex conjugate of the positive
frequencies
𝑋(𝑓) = 𝑋 ∗ (−𝑓)
Combining Fourier Series and Fourier Transform
If x(t) is the sum of a finite energy (aperiodic) signal, 𝑦(𝑡), and a periodic signal, 𝑧(𝑡), which
has a fundamental frequency 𝑓0 like this
𝑥(𝑡) = 𝑦(𝑡) + 𝑧(𝑡)
And suppose the Fourier Transform of 𝑦(𝑡) is 𝑌(𝑓) and the Fourier Series of 𝑧(𝑡) is 𝑍𝑘 then we
speak of the Fourier Transform of x(t) as
∞
𝑋(𝑓) = 𝑌(𝑓) + ∑ 𝑍𝑘 𝛿(𝑓 − 𝑘𝑓0 )
𝑘=−∞
is not periodic, but it is finite energy, then you cannot use the Fourier Series, but you can use the
Fourier Transform. The energy of a signal is
Fourier Transform Theorems Discussed in EE 3400
∞
Fourier Transform
𝑥(𝑡)
𝑋(𝑓) = ∫𝑡=−∞ 𝑥(𝑡)𝑒 −𝑗2𝜋𝑓𝑡 𝑑𝑡
Linearity
𝑎𝑥(𝑡) + 𝑏𝑦(𝑡)
𝑎𝑋(𝑓) + 𝑏𝑌(𝑓)
Time Delay/Advance
𝑥(𝑡 − 𝜏)
𝑒 −𝑗2𝜋𝑓𝜏 𝑋(𝑓)
Time Expansion/Compression
𝑥(𝑎𝑡)
1
𝑓
𝑋 (𝑎)
|𝑎|
Duality
𝑥(𝑡) → 𝑋(𝑓) → 𝑥(−𝑡) → 𝑋(−𝑓) → 𝑥(𝑡)
Derivative
𝐹𝑇
𝑑
𝑑𝑡
𝐹𝑇
𝐹𝑇
𝑥(𝑡)
𝑡
𝐹𝑇
𝑗2𝜋𝑓𝑋(𝑓)
1
Integral
∫−∞ 𝑥(𝛼)𝑑𝛼
𝑗2𝜋𝑓
𝑋(𝑓)
Convolution
𝑥(𝑡) ∗ 𝑦(𝑡)
𝑋(𝑓)𝑌(𝑓)
Multiplication
𝑥(𝑡)𝑦(𝑡)
𝑋(𝑓) ∗ 𝑌(𝑓)
Even
𝑥(𝑡) = 𝑥(−𝑡)𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡
𝑋(𝑓) 𝑖𝑠 𝑟𝑒𝑎𝑙
Odd
𝑥(𝑡) = −𝑥(−𝑡)𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡
𝑋(𝑓) 𝑖𝑠 𝑖𝑚𝑎𝑔𝑖𝑛𝑎𝑟𝑦
Real
𝑥(𝑡)𝑖𝑠 𝑟𝑒𝑎𝑙 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡
𝑋(𝑓) = 𝑋 ∗ (−𝑓)
Common Fourier Transform Pairs Used in EE 3400
Square Pulse
𝑋(𝑓) = 𝑠𝑖𝑛𝑐(𝑓)
x(t) is one volt high, 1 second wide, centered at t=0
Triangular Pulse
𝑋(𝑓) = 𝑠𝑖𝑛𝑐 2 (𝑓)
x(t) is one volt high, 2 seconds wide, centered at t=0
One-Sided Decaying Exponential
𝑥(𝑡) = 𝑒 −𝑎𝑡 𝑢(𝑡)
1
𝑋(𝑓) = 𝑎+𝑗2𝜋𝑓
Two-Sided Decaying Exponential
𝑥(𝑡) = 𝑒 −𝑎|𝑡|
1
𝑋(𝑓) = 𝑎+𝑗2𝜋𝑓
Bell Shaped Curve
𝑥(𝑡) = 𝑒
−𝑎𝑡 2
𝜋
𝑋(𝑓) = √𝑎 𝑒
−
𝜋2 𝑓2
𝑎
DC
𝑥(𝑡) = 1
𝑋(𝑓) = 𝛿(𝑓)
Impulse
𝑥(𝑡) = 𝛿(𝑡)
𝑋(𝑓) = 1
Cosine
𝑥(𝑡) = 𝑐𝑜𝑠(2𝜋𝑓𝑜 𝑡)
1
1
𝑋(𝑓) = 2 𝛿(𝑓 − 𝑓𝑜 ) + 2 𝛿(𝑓 + 𝑓𝑜 )
Sine
𝑥(𝑡) = 𝑠𝑖𝑛(2𝜋𝑓𝑜 𝑡)
𝑗
𝑗
𝑋(𝑓) = − 2 𝛿(𝑓 − 𝑓𝑜 ) + 2 𝛿(𝑓 + 𝑓𝑜 )
Sinusoid
𝑥(𝑡) = 𝐴𝑐𝑜𝑠(2𝜋𝑓𝑜 𝑡 + 𝜃)
𝑋(𝑓) =
𝐴𝑒 𝑗𝜃
2
𝛿(𝑓 − 𝑓𝑜 ) +
𝐴𝑒 −𝑗𝜃 𝐴
2
𝛿(𝑓 + 𝑓𝑜 )
Sign
𝑥(𝑡) = {
−1 𝑓𝑜𝑟 𝑡 < 0
1 𝑓𝑜𝑟 0 < 𝑡
−𝑗
𝑋(𝑓) = 𝜋𝑓
Unit Step
𝑥(𝑡) = {
0 𝑓𝑜𝑟 𝑡 < 0
1 𝑓𝑜𝑟 0 < 𝑡
−𝑗
1
𝑋(𝑓) = 2𝜋𝑓 + 2 𝛿(𝑓)