Lecture 10
Fourier Transform in
Circuit Analysis (Part 1)
The Derivation of the Fourier
Transform

We begin the derivation of the Fourier
transform, viewed as a limiting case of a
Fourier series, with the exponential form
of the series:
The Derivation of the Fourier
Transform
Allowing the fundamental period T to
increase without limit accomplishes the
transition from a periodic to an aperiodic
function. In other words, if T becomes
infinite, the function never repeats itself
and hence is aperiodic.
 As T increases, the separation between
adjacent harmonic frequencies becomes
smaller and smaller. In particular,

The Derivation of the Fourier
Transform

and as T gets larger and larger, the
incremental separation Δω approaches a
differential separation dω.

As the period increases, the frequency
moves from being a discrete variable to
becoming a continuous variable, or
The Derivation of the Fourier
Transform

As the period increases, the Fourier
coefficients Cn get smaller. In the limit,
Cn→ 0 as T → ∞. This result makes
sense, because we expect the Fourier
coefficients to vanish as the function
loses its periodicity. Note, however, the
limiting value of the product CnT; that is,
The Derivation of the Fourier
Transform

The integral is the Fourier transform of
f(t) and is denoted
(1)

We obtain an explicit expression for the
inverse Fourier transform
The Derivation of the Fourier
Transform
Let's now derive the Fourier transform of
the pulse shown in Figure 1.
 The Fourier transform of v(t) is

The Derivation of the Fourier
Transform
Figure 1

which can be put in the form of (sin x)/x
by multiplying the numerator and
denominator by . Then,
The Convergence of the Fourier
Integral
If f(t) is different from zero over an infinite
interval, the convergence of the Fourier
integral depends on the behavior of f(t)
as t →∞.
 A single-valued function that is nonzero
over an infinite interval has a Fourier
transform if the integral

The Convergence of the Fourier
Integral

exists and if any discontinuities in f(t) are
finite. An example is the decaying
exponential function illustrated in Figure
2. The Fourier transform of f(t) is
Figure 2
The Convergence of the Fourier
Integral
The Convergence of the Fourier
Integral
A third important group of functions have
great practical interest but do not in a
strict sense have a Fourier transform.
 For example, the integral in Equation 1
doesn't converge if f(t) is a constant. The
same can be said if f(t) is a sinusoidal
function, cos ωot, or a step function, Ku(t).
 These functions are of great interest in
circuit analysis, but, to include them in
Fourier analysis, we must resort to some
mathematical subterfuge.

The Convergence of the Fourier
Integral
First, we create a function in the time
domain that has a Fourier transform and
at the same time can be made arbitrarily
close to the function of interest.
 Next, we find the Fourier transform of the
approximating function and then
evaluate the limiting value of F(ω) as this
function approaches f(t).
 Last, we define the limiting value of F(ω)
as the Fourier transform of f(t).

The Convergence of the Fourier
Integral

Let's demonstrate this technique by
finding the Fourier transform of a
constant. We can approximate a
constant with the exponential function

As ε → 0, f(t) → A. Figure 3 shows the
approximation graphically. The Fourier
transform of f(t) is
The Convergence of the Fourier
Integral
Figure 3
The Convergence of the Fourier
Integral
This function generates an impulse
function at ω = 0 as ε → 0.
 The area under F(ω) is the strength of
the impulse and is

The Convergence of the Fourier
Integral
In the limit, f(t) approaches a constant A,
and F(ω) approaches an impulse
function 2πAδ(ω).
 Therefore, the Fourier transform of a
constant A is defined as 2πAδ(ω), or

Using Laplace Transforms to Find
Fourier Transforms

1.
The following rules apply to the use of
Laplace transforms to find the Fourier
transforms of such functions.
If f(t) is zero for t ≤ 0-, we obtain the
Fourier transform of f(t) from the
Laplace transform of f(t) simply by
replacing s by jω. Thus
Using Laplace Transforms to Find
Fourier Transforms
Using Laplace Transforms to Find
Fourier Transforms
2.

Because the range of integration on the
Fourier integral goes from - ∞ to + ∞, the
Fourier transform of a negative-time
function exists. A negative-time function
is nonzero for negative values of time
and zero for positive values of time.
To find the Fourier transform of such a
function, we proceed as follows. First,
we reflect the negative-time function
over to the positive-time domain and
then find its one-sided Laplace
transform.
Using Laplace Transforms to Find
Fourier Transforms

We obtain the Fourier transform of the
original time function by replacing s with
-jω. Therefore, when f(t) = 0 for t ≥ 0+,
Using Laplace Transforms to Find
Fourier Transforms

Both f(t) and its mirror image are plotted
in Figure 4. The Fourier transform of f(t)
is
Using Laplace Transforms to Find
Fourier Transforms
Figure 4
Using Laplace Transforms to Find
Fourier Transforms
3.

Functions that are nonzero over all time
can be resolved into positive and
negative-time functions. We use the
previous two cases to find the Fourier
transform of the positive - and negative
- time functions, respectively.
The Fourier transform of the original
function is the sum of the two
transforms. Thus if we let
Using Laplace Transforms to Find
Fourier Transforms
Using Laplace Transforms to Find
Fourier Transforms

An example involves finding the Fourier
transform of e-a|t| For the original
function, the positive- and negative time
functions are
Using Laplace Transforms to Find
Fourier Transforms
The Fourier Transform of a
Signum Function

The signum function, defined as +1 for t
> 0 and -1 for t < 0.The signum function
is denoted sgn(t) and can be expressed
in terms of unit-step functions, or
Figure 5 shows the function graphically.
 To find the Fourier transform of the
signum function, we first create a
function that approaches the signum
function in the limit:

The Fourier Transform of a
Signum Function
Figure 5

The function inside the brackets, plotted
in Figure 6, has a Fourier transform
because the Fourier integral converges.
The Fourier Transform of a
Signum Function
Figure 6
The Fourier Transform of a
Signum Function
The Fourier Transform of a Unit
Step Function

To find the Fourier transform of a unit
step function, we do so by recognizing
that the unit-step function can be
expressed as
The Fourier Transform of a
Cosine Function
Fourier Transforms of
Elementary Functions