Signals and Systems - Electrical Engineering

5.2 From the Fourier Series to the Fourier Transform 301

AsT 0 →∞,Xnwill tend to zero. To avoid this we defineX(n)=T 0 Xnwhere{n=n 0 }are the
harmonic frequencies.

Letting1= 2 π/T 0 = 0 be the frequency interval between harmonics, we can then write the above
equations as

̃x(t)=

∑∞

n=−∞

X(n)

T 0

ejnt=

∑

n

X(n)ejnt

1

2 π

X(n)=

T∫ 0 / 2

−T 0 / 2

x ̃(t)e−jntdt

AsT 0 →∞, then1→d, the line spectrum becomes denser—that is, the lines in the line spectrum
get closer, the sum becomes an integral, andn=n 0 =n1→, so that in the limit we obtain

x(t)=

1

2 π

∫∞

−∞

X()ejtd

X()=

∫∞

−∞

x(t)e−jtdt

which are the inverse and the direct Fourier transforms, respectively. The first equation transforms
a function in the frequency domainX()into a signal in the time domainx(t), while the other
equation does the opposite.

The Fourier transform measures the frequency content of a signal. As we will see, time and frequency
are complementary, thus the characterization in one domain provides information that is not clearly
available in the other.

Remarks

n Although we have obtained the Fourier transform from the Fourier series, the Fourier transform of a periodic
signal cannot be obtained from the above integral. Consider x(t)=cos( 0 t),−∞<t<∞, which is peri-
odic of period 2 π/ 0. If you attempt to compute its Fourier transform using the integral you do not have a
well-defined problem (try to obtain the integral to convince yourself ). But it is known from the line spectrum
that the power of this signal is concentrated at the frequencies± 0 , so somehow we should be able to find
its Fourier transform. Sinusoids are basic functions.
n On the other hand, if you consider a decaying exponential x(t)=e−|a|tsignal, which has finite energy
and is absolutely integrable and has a Laplace transform that is valid on the jaxis (i.e., the region
of convergence X(s)includes this axis), then its Fourier transform is simply the Laplace transform X(s)
computed at s=j, as we will see. There is no need for the integral formula in this case, although if you
apply it your result coincides with the one from the Laplace transform.
n Finally, consider finding the Fourier transform of a sinc function (which is the impulse response of a
low-pass filter as we see later). Neither the integral nor the Laplace transform can be used to find it. For
this signal, we need to exploit the duality that exists between the direct and the inverse Fourier transforms
(Notice the duality in Equations (5.1) and (5.2)).