The document discusses the Fourier transform of the Heaviside unit step function, noting its non-existence due to properties that make it neither absolutely nor square summable. Various approaches are presented for deriving the Fourier transform, including the differential method, the signum function approach, and the z-transform approach, all yielding different results that incorporate impulse signals. Conclusions highlight that the results from different methods hold valid within their respective domains of continuous and discrete time functions.

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A NOTE ON THE FOURIER TRANSFORM OF HEAVISIDE UNIT STEP
FUNCTION
S Anand Krishnamoorthy
Project Associate, ADI DSP Learning Centre, IIT Madras
I. INTRODUCTION
The Heaviside unit step function is defined as follows –
Table .I
Continuous time Discrete time
𝒖(𝒕) = {
𝟏 ; 𝒕 ≥ 𝟎
𝟎 ; 𝒕 < 𝟎
} 𝒕 ∈ ℝ … (𝑰. 𝟏) 𝑢[𝑛] = {
1 ; 𝑛 ≥ 0
0 ; 𝑛 < 0
} 𝑛 ∈ ℤ … (𝐼. 2)
Fig.1 Continuous and Discrete time Unit step signal
The Unit step function can also be defined as the derivative of the standard ramp
function-
𝑢(𝑡) =
𝑑
𝑑𝑥
max{𝑡, 0} … (𝐼. 3)
Or as the integral of the Dirac delta function (impulse function)
𝑢(𝑡) = ∫ 𝛿(𝑠)𝑑𝑠
𝑡
−∞
… (𝐼. 4)
The unit step function is causal, bounded in its amplitude but not stable,
∑ 𝑢[𝑛]
∞
−∞
= ∑ 1
∞
0
→ ∞ … (𝐼. 5)

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The z transform of the unit step function is
𝐻[𝑧] =
1
1 − 𝑧−1
… (𝐼. 6)
It is clear that the pole at 𝑧 = 1 lies on the unit circle and consequently, the Region of
Convergence (ROC) is the region outside the unit circle.
Fig.2 Z plane plot of Unit step signal
The evaluation of the Fourier transform unit step function has been a confounding
procedure. This is specifically due to its property that it is neither absolutely
summable nor square summable. In the strict sense, the Fourier transform of the
Heaviside unit step function does not exist.
To incorporate the unit step function and other functions with similar properties
in the Fourier domain, the Fourier transform must contain impulses at finite number
of frequencies corresponding to the poles which lie on the unit circle. Thus, by
allowing impulses in the frequency domain, it is possible to extend the Fourier
transform representation to signals which are neither absolutely summable nor
square summable.
II. DERIVATION OF FOURIER TRANSFORM OF UNIT STEP FUNCTION
II.1 Differential Approach
The first approach for finding the Fourier transform of the unit step function is
from the previous statement that the unit step function is the integral of the delta
function. So, the derivation is as follows –

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We know that the Fourier transform of the impulse function is
ℱ{𝛿(𝑡)} = 1 … (𝐼𝐼. 1)
And
𝛿(𝑡) =
𝑑
𝑑𝑡
𝑢(𝑡) = 𝑢′(𝑡) … (𝐼𝐼. 2)
And from the property of Fourier transform,
ℱ{𝑋′(𝑡)} = 𝑗𝜔ℱ{𝑋(𝑡)} … (𝐼𝐼. 3)
Therefore,
ℱ{𝑢(𝑡)} = 𝑈(𝑗𝜔} =
1
𝑗𝜔
ℱ{𝛿(𝑡)} =
1
𝑗𝜔
… (𝐼𝐼. 4)
But the above solution is not completely correct. Note that differentiation methods
does not include the integral constants. Thus, the result must be strictly,
ℱ{𝑢(𝑡)} =
1
𝑗𝜔
+ ℱ{𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡} … (𝐼𝐼. 5)
Fourier transform of a function is unique and has a one-to-one mapping. So, to find
the constant term, the inverse Fourier transform of
1
𝑗𝜔
can be used-
ℱ−1
{
1
𝑗𝜔
} =
1
𝑗2𝜋
∫
𝑒 𝑗𝜔𝑡
𝜔
𝑑𝜔
∞
−∞
=
1
𝑗2𝜋
[∫
𝑒 𝑗𝜔𝑡
𝜔
𝑑𝜔
0
−∞
+ ∫
𝑒 𝑗𝜔𝑡
𝜔
𝑑𝜔
∞
0
] … (𝐼𝐼. 6)
Combining into a single integral, we get
ℱ−1
{
1
𝑗𝜔
} =
1
𝑗2𝜋
[− ∫
𝑒−𝑗𝜔𝑡
𝜔
𝑑𝜔
∞
0
+ ∫
𝑒 𝑗𝜔𝑡
𝜔
𝑑𝜔
∞
0
] =
1
𝑗2𝜋
∫
𝑒 𝑗𝜔𝑡
− 𝑒−𝑗𝜔𝑡
𝜔
𝑑𝜔
∞
0
… (𝐼𝐼. 7)
Which can be simplified into
ℱ−1
{
1
𝑗𝜔
} =
1
𝜋
∫
sin 𝜔𝑡
𝜔
𝑑𝜔
∞
0
=
1
2
[∵ ∫
sin 𝜔𝑡
𝜔
𝑑𝜔
∞
0
=
𝜋
2
] … (𝐼𝐼. 8)
We can obtain −
1
2
if we evaluate in terms of the negative part of the integral. Thus,
the inverse Fourier transform yields the function
𝑢1(𝑡) = {
1
2
; 𝑡 ≥ 0
−
1
2
; 𝑡 < 0
} … (𝐼𝐼. 9)
But, from the equations I.1 and II.9,
𝑢(𝑡) = 𝑢1(𝑡) +
1
2
… (𝐼𝐼. 10)
Thus, this is the term which was missing from the Fourier transform. Now the
complete transform is given by-
ℱ{𝑢(𝑡)} = 𝑈(𝑗𝜔} =
1
𝑗𝜔
+ ℱ {
1
2
} =
1
𝑗𝜔
+ 𝜋𝛿(𝜔) … (𝐼𝐼. 11)

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II.2 Signum function Approach
Another approach followed by Oppenheim and others [2] is to derive the Fourier
transform from another similar function – the signum function. This methods is very
similar to the first method and yields the identical result.
Consider the signum function 𝜎(𝑡) defined as
𝜎(𝑡) = {
1 ; 𝑡 ≥ 0
−1 ; 𝑡 < 0
} … (𝐼𝐼. 12)
Fig.3 Continuous time Signum Function
The Fourier transform of the signum function is given by
ℱ{𝜎(𝑡)} = ∫ 𝜎(𝑡)𝑒−𝑗𝜔𝑡
𝑑𝑡
∞
−∞
… (𝐼𝐼. 13)
But we know that differential of the signum function is also the delta impulse
function. Thus,
𝑑
𝑑𝑡
{
𝜎(𝑡)
2
} = 𝛿(𝑡) … (𝐼𝐼. 14)
𝑗𝜔 ℱ{𝜎(𝑡)} = 2ℱ{𝛿(𝑡)} … (𝐼𝐼. 15)
⇒ ℱ{𝜎(𝑡)} =
2
𝑗𝜔
[∵ ℱ{𝛿(𝑡)} = 1] … (𝐼𝐼. 16)
The relationship between the unit step function and the signum function is
given by
𝜎(𝑡) = 2𝑢(𝑡) − 1 … (𝐼𝐼. 17)

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Note the similarity of II.10 and II.17.
⇒ ℱ{𝜎(𝑡)} = 2ℱ{𝑢(𝑡)} − ℱ{1} … (𝐼𝐼. 18)
⇒ ℱ{𝑢(𝑡)} =
1
2
(ℱ{𝜎(𝑡)} + ℱ{1}) … (𝐼𝐼. 19)
⇒ ℱ{𝑢(𝑡)} =
1
2
(
2
𝑗𝜔
+ 2𝜋𝛿(𝜔)) … (𝐼𝐼. 20)
ℱ{𝑢(𝑡)} = 𝑈(𝑗𝜔) =
1
𝑗𝜔
+ 𝜋𝛿(𝜔) … (𝐼𝐼. 21)
II.3 Z Transform Approach
It is possible to derive the Fourier Transform of the unit step function is many
ways. The second approach for finding the Fourier transform of the unit step function
is as follows
The Z transform of the unit step function is given by-
𝐻[𝑧] = ∑ 𝑢[𝑛]𝑧−𝑛
∞
−∞
= ∑ 𝑧−𝑛
∞
0
=
1
1 − 𝑧−1
=
𝑧
𝑧 − 1
… (𝐼𝐼. 22)
Substituting 𝑧 = 𝑒 𝑗𝜔
, except at 𝑧 = 1,
𝐻(𝑗𝜔) =
𝑒 𝑗𝜔
𝑒 𝑗𝜔 − 1
=
𝑒
𝑗𝜔
2
𝑒
𝑗𝜔
2 − 𝑒−
𝑗𝜔
2
=
𝑒
𝑗𝜔
2
2𝑗 sin
𝜔
2
… (𝐼𝐼. 23)
This formula yields finite results at all frequencies except at 𝜔 = 2𝜋𝑘, 𝑘 ∈ {0,1,2, … }
since 𝐻(𝑗𝜔) becomes indeterminate at those frequencies. This is due to the fact that
the unit step signal is not piecewise continuous at the origin where there is an abrupt
“jump”. Note that this approach disregards the impulse at the poles.
II.4 Even Odd function Approach
This is a very interesting approach, yet uses elementary concepts. This derivation,
used in the Proakis et.al book [3] is as follows.
Firstly, the discrete unit step function is written as a linear combination of an even
and odd sequence.
𝑢[𝑛] = 𝑢 𝑒[𝑛] + 𝑢 𝑜[𝑛] … (𝐼𝐼. 24)

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Fig.4 Even and Odd parts of the Unit step signal
Where the even function is
𝑢 𝑒[𝑛] =
1
2
[𝑢[𝑛] + 𝑢[−𝑛]] … (𝐼𝐼. 25)
𝑢 𝑒[𝑛] =
1
2
+
1
2
𝛿[𝑛] … (𝐼𝐼. 26)
Taking Fourier transform of the even part,
𝑈𝑒(𝑗𝜔) = 𝜋𝛿(𝜔) +
1
2
… (𝐼𝐼. 27)
And the odd part is given by
𝑢 𝑜[𝑛] =
1
2
[𝑢[𝑛] − 𝑢[−𝑛]] … (𝐼𝐼. 28)
𝑢 𝑜[𝑛] = 𝑢[𝑛] −
1
2
−
1
2
𝛿[𝑛] … (𝐼𝐼. 29)
Now, shift the odd signal to the right by one sample,
𝑢 𝑜[𝑛 − 1] = 𝑢[𝑛 − 1] −
1
2
−
1
2
𝛿[𝑛 − 1] … (𝐼𝐼. 30)

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Subtracting II.31 from II.30,
𝑢 𝑜[𝑛] − 𝑢 𝑜[𝑛 − 1] = 𝑢[𝑛] − 𝑢[𝑛 − 1] −
1
2
[𝛿[𝑛] − 𝛿[𝑛 − 1]] … (𝐼𝐼. 31)
⇒ 𝑢 𝑜[𝑛] − 𝑢 𝑜[𝑛 − 1] = 𝛿[𝑛] −
1
2
[𝛿[𝑛] − 𝛿[𝑛 − 1]] =
1
2
[𝛿[𝑛] + 𝛿[𝑛 − 1]] … (𝐼𝐼. 32)
Taking Fourier transform of the odd part and using the shifting property of the
Fourier transform,
𝑈 𝑜(𝑗𝜔) [1 − 𝑒−𝑗𝜔
] =
1
2
(1 + 𝑒−𝑗𝜔
) … (𝐼𝐼. 33)
𝑈 𝑜(𝑗𝜔) =
(1 + 𝑒−𝑗𝜔
)
2(1 − 𝑒−𝑗𝜔)
… (𝐼𝐼. 34)
Now, we can combine the two individual Fourier transform to yield the final Fourier
transform of the unit step function.
ℱ{𝑢[𝑛]} =
1
2
(ℱ{𝑢 𝑒[𝑛]} + ℱ{𝑢0[𝑛]}) … (𝐼𝐼. 35)
ℱ{𝑢[𝑛]} = 𝑈(𝑗𝜔) = 𝜋𝛿(𝜔) +
1
2
(1 +
(1 + 𝑒−𝑗𝜔
)
(1 − 𝑒−𝑗𝜔)
) … (𝐼𝐼. 36)
⇒ 𝑈(𝑗𝜔) = 𝜋𝛿(𝜔) + (
1
(1 − 𝑒−𝑗𝜔)
) … (𝐼𝐼. 37)
III.CONCLUSION
Now, there are three results for the Fourier transform of the Heaviside unit step
signal, obtained from different methods. These results are tabulated in Table.II.
Table. II Fourier transforms of Unit step signal
Approach Fourier Transform 𝑼(𝒋𝝎)
Differential approach & Oppenheim
approach
𝜋𝛿(𝜔) +
1
𝑗𝜔
Z Transform approach 𝑒
𝑗𝜔
2
2𝑗 𝑠𝑖𝑛
𝜔
2
𝜔 ≠ 2𝜋𝑘, 𝑘 ∈ {0,1,2, … }
Even Odd function approach (Proakis)
𝜋𝛿(𝜔) + (
1
(1 − 𝑒−𝑗𝜔)
)

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The result from the Z transform approach is not well-behaved at 𝜔 = 0. The other
two results incorporate the impulse signal at the pole, thus can be considered valid.
Note that the Oppenheim approach is for the continuous time unit step signal and
the Proakis approach is for the discrete time unit step signal. This can be observed from
their derivations. Therefore, both are valid in their own domain.
Moreover, since the Z transform approach also deals in the discrete domain, it is
equivalent to the even-odd approach except for the impulse at the pole.
Observe that,
1
(1 − 𝑒−𝑗𝜔)
=
1
𝑒−𝑗𝜔/2(𝑒 𝑗𝜔/2 − 𝑒−𝑗𝜔)
=
𝑒−𝑗𝜔/2
2𝑗 𝑠𝑖𝑛
𝜔
2
, 𝜔 ≠ 2𝜋𝑘, 𝑘 ∈ {0,1,2, … } … (𝐼𝐼𝐼. 1)
Therefore,
Table.III
Domain Fourier Transform 𝑼(𝒋𝝎)
Continuous Unit step function
𝜋𝛿(𝜔) +
1
𝑗𝜔
Discrete Unit step function
𝜋𝛿(𝜔) + (
1
(1 − 𝑒−𝑗𝜔)
)
IV.REFERENCES
[1] The Fourier Transform of the Heaviside function: A tragedy, Mathematical Physics Vol I,
2005
[2] A. Oppenheim, Shafer, Buck, Discrete time Signal Processing 2nd edition Prentice Hall
[3] Proakis and Manolakis, Digital Signal Processing Algorithms and applications 3rd
edition,
chapter 8 pp 617-619; chapter 4 pp 267-269
[4] Digital Signal Processing Lecture series by Prof. Dutta Roy, IITK Lecture 11, NPTEL