4.6 Fourier Coefficients from Laplace 257the Fourier coefficients will be real. Doing this analysis before the computations is important so
we know what to expect.
The Fourier coefficients are obtained directly using their integral formulas or from the Laplace
transform of a period. SinceT 0 =1, the fundamental frequency ofx(t)is 0 = 2 πrad/sec. Using
the integral expression for the Fourier coefficients we have
Xk=1
T 0
∫^3 /^4
− 1 / 4x(t)e−j^0 ktdt=∫^1 /^4
− 1 / 42 e−j^2 πktdt=
2
πk[
ejπk/^2 −e−jπk/^2
2 j]
=
sin(πk/ 2 )
(πk/ 2 )which are real as we predicted. The Fourier series is then
x(t)=∑∞
k=−∞sin(πk/ 2 )
(πk/ 2 )ejk^2 πtTo find the Fourier coefficients with the Laplace transform, let the period bex 1 (t)=x(t)for−0.5≤
t≤0.5. Delaying it by 0.25 we getx 1 (t−0.25)=2[u(t)−u(t−0.5)] with a Laplace transform
e−0.25sX 1 (s)=2
s( 1 −e−0.5s)so thatX 1 (s)=( 2 /s)[e0.25s−e−0.25s], and therefore
Xk=1
T 0
L[x 1 (t)]∣
∣s=jk
0=
2
jk 0 T 02 jsin(k 0 / 4 )and for 0 = 2 π,T 0 =1, we get
Xk=sin(πk/ 2 )
πk/ 2k6= 0Since the above equation gives zero over zero whenk=0 (i.e., it is undefined), the dc value is
found from the integral formula as
X 0 =
∫^1 /^4
− 1 / 42 dt= 1These Fourier coefficients coincide with the ones found before.
The following script is used to find the Fourier coefficients with our functionfourierseriesand to
plot the magnitude and phase line spectra.