The Fourier Transform
Let f (t) be a function which maps complex numbers to
complex numbers. So f (t) is a complex function of one
complex variable,
f : C −→ C
Its Fourier Transform is defined to be
Z +∞
f (t)e−jωt dt
F (jω) =
−∞
This amounts to exchanging a function of t for a function
of jω. It turns out (but we do not prove) that the process
is generally reversible. In fact,
Z +∞
1
f (t) =
F (jω)ejωt dω
2π −∞
We say that F (jω) is the Fourier transform of f (t), and
that f (t) is the inverse Fourier transform of F (jω). We
write F (jω) = F{f (t)} and f (t) = F −1 {F (jω)}.
We sat that f (t) lives in the time domain and that F (jω)
lives in the frequency domain.
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Transform 1
Next, we compute the transform of f (t) = e−at u(t) where
a is a constant and u(t) denotes the unit step function.
Using the definition,
Z +∞
f (t)e−jωt dt
F (jω) =
−∞
+∞
Z
e−at u(t)e−jωt dt
=
−∞
+∞
Z
e−(a+jω)t u(t)dt
=
−∞
+∞
Z
e−(a+jω)t dt
=
0
∞
e−(a+jω)t =
−(a + jω) 0
If the real part of a less than or equal to zero, then the
upper limit term does not exist. If it is strictly positive,
Re(a) > 0, the upper limit term is zero, giving
−e−(a+jω)t F (jω) =
−(a + jω) t=0
=
1
a + jω
This proves that
F e−at u(t) =
2
1
a + jω
Property 1 - Linearity
It’s clear that if F (jω) = F{f (t)} and G(jω) = F{g(t)}
then
F{c1 f (t) + c2 g(t)} = c1 F (jω) + c2 G(jω)
where c1 and c2 are any constants.
Property 2 - Differentiation in the Time Domain
Since
Z +∞
1
f (t) =
F (jω)ejωt dω
2π −∞
and because the integration is with respect to the variable ω, it’s generally permissible to differentiate both
sides with respect to t, giving
df
1
=
dt
2π
Z
+∞
F (jω)jωejωt dω
−∞
Now, the above equation says that
df
= F −1 {jωF (jω)}
dt
and then
jωF (jω) = F
df
dt
So differentiation in the time domain corresponds to multiplication by jω in the frequency.
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Transform 2
1 if |t| ≤ T
0 if |t| > T
where T is a real constant. Using the definition,
Z +∞
f (t)e−jωt dt
F (jω) =
Next, we compute the transform of f (t) =
−∞
Z
+T
=
e−jωt dt
−T
+T
e−jωt =
−jω −T
e−jωT
e+jωT
−
−jω
−jω
jωT
− e−jωT
2 e
=
ω
2j
=
=
4
2 sin (ωT )
ω
Property 3 - Time Delay
Suppose that F (jω) = F {f (t)}. Suppose that the graph
of f (t) is shifted right by T . (If f (t) represents a signal,
this corresponds to a time delay of T seconds.) This
gives the function f (t − T ). How it its Fourier transform
related to F (jω)?
Using the definition,
Z
+∞
f (t − T )e−jωt dt
F {f (t − T )} =
−∞
Z
+∞
=
f (t − T )e−jω(t−T ) e−jωT dt
−∞
The integration is with respect to t, not T . So,
Z +∞
−jωT
f (t − T )e−jω(t−T ) dt
F {f (t − T )} = e
−∞
Do a change in variable, letting τ = t − T ,
Z +∞
f (τ )e−jωτ dτ
F {f (τ )} = e−jωT
−∞
= e−jωT F {f (t)} = e−jωT F (jω)
A time delay in the time domain corresponds to multiplication by e−jωT in the frequency domain.
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