The Fourier transform of unit sequence is

The unit step sequence is defined as:

\(u(n) = \begin{cases} 1 &{for~n\ge0}\\ 0&{for~n<0} \end{cases}\)

\(\sum_{-\infty}^{\infty} |u(n)|=\infty \)

Since the unit step sequence is not absolutely integrable, we cannot find the Fourier transform using the standard formula. 

Hence, we will derive the Fourier transform of the unit step sequence starting from the Fourier transform of the signum function. 

The signum function can be defined as follows:

\(\)\(sgn(n) = \begin{cases} 1 &{for~n>0}\\ 0&{for~n=0}\\ -1&{for~n<0} \end{cases}\)

∴ We can see that the signum function is also not absolutely integrable. 

The signum function can be made absolutely integrable by multiplying with the exponential function.

\(sgn(n)= \mathop {\lim }\limits_{a \to 0 }~[e^{-an}u(n) - e^{an}u(-n)]\)

Taking Fourier transform we get:

\(F[sgn(n)]=\mathop {\lim }\limits_{a \to 0 }~[\frac{1}{1-e^{-a-j\omega}}-\frac{1}{1-e^{a+j\omega}}]\)

\(F[sgn(t)]=\mathop {\lim }\limits_{a \to 0 }~[\frac{-2j\omega}{a^2+\omega^2}]\)

\(F[sgn(t)]=\frac{2}{j\omega}\)

Now, the unit step signal can be represented in terms of signum function as follows:

\(u(t)=\frac{1+sgn(t)}{2}\)

Taking Fourier transform we get:

\(F[u(t)]=F[\frac{1}{2}]+\frac{1}{2}F[sgn(t)]\)

We know, the Fourier transform of a DC signal 'A' is given as:

\(A \mathop \leftrightarrow \limits^{\;F.T\;} 2\pi A ~\delta(\omega)\)

∴ \(F[u(t)]=2\pi \times \frac{1}{2}\times\delta(\omega)+\frac{1}{2}\times \frac{2}{j\omega}\)

 \(F[u(t)]=\pi~ \delta(\omega)+ \frac{1}{j\omega}\)

y(n) = u(n)

\(u\left( n \right) = \begin{array}{*{20}{c}} 1&{n = 0,1,2, \ldots }\\ 0&{else} \end{array}\)

Y(e^jω) = ?

If \(y\left( n \right) = \mathop \sum \limits_{m = - \infty }^\infty x\left( m \right)\)

Using Accumlation property of DTFT

\(X\left( {{e^{J\omega }}} \right) \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\leftharpoonup\over {\smash{\rightharpoondown}}$}} x\left( n \right)\)

⇒\(DT{\rm{ }}F.T.\left[ {y\left( n \right)} \right] = \frac{1}{{1 - {e^{ - J\omega }}}}X\left( {eJ\omega } \right) + \pi \times \left( {{e^{J0}}} \right)\mathop \sum \limits_{K = - \infty }^\infty \delta \left( {\omega - 2\pi K} \right)\)

y(n) = u(n)

x(n) = δ(n)

\(u\left( n \right) = \mathop \sum \limits_{m = - \infty }^n \delta \left( m \right)\)

X(e^Jω) = DTFT of δ(n) = 1

By Accumlation property

\(y\left( {{e^{J\omega }}} \right) = \frac{1}{{1 - {e^{ - J\omega }}}} + \pi \mathop \sum \limits_{K = - \infty }^\infty \delta \left( {\omega - 2\pi K} \right)\)

For ω ∈ [-π, π]

\(y\left( {{e^{J\omega }}} \right) = \frac{1}{{1 - {e^{ - J\omega }}}} + \pi \delta \left( \omega \right)\)