Discrete-time Fourier transform Slecture by Fabian Faes - Rhea

Discrete Time Fourier Transform (DTFT) with example

a Slecture by ECE Student Fabian Faes

Partly based on the ECE438 Fall 2014 lecture material of Prof. Mireille Boutin.

Definition

The discrete time fourier transform (DTFT) of a finite energy aperiodic signal x[n] can be given by the equation listed below. IT is a representation in terms of a complex exponential sequence e^jωn, where ω is a real frequency variable.

$ X(e^{j{\omega}}) = X({\omega}) = {\sum_{n = -{\infty}}^{\infty} x[n]e^{-j{\omega}n}} $

X(ω) can also be represented in terms of its magnitude and phase as shown below:

magnitude

X(ω) = | X(ω) | e^jθω

phase

$ {\theta}({\omega}) = {\angle}X({\omega}) $

Periodicity Property

We note that $ X({\omega}) $ is periodic with period $ 2{\pi} $ since

$ X({\omega} + 2{\pi}) = {\sum_{n = -{\infty}}^{\infty} x[n]e^{-j({\omega} + 2{\pi})n}} $

$ X({\omega} + 2{\pi}) = {\sum_{n = -{\infty}}^{\infty} x[n]e^{-j{\omega}n}e^{-j2{\pi}n}} $

$ X({\omega} + 2{\pi}) = {\sum_{n = -{\infty}}^{\infty} x[n]e^{-j{\omega}n}} $

$ X({\omega} + 2{\pi}) = X({\omega}) $

for clarification $ e^{-j2{\pi}n} = 1 $ since

$ e^{-j2{\pi}n} = cos(2{\pi}) + jsin(2{\pi}n) $

$ e^{-j2{\pi}n} = 1 + j0 $

$ e^{-j2{\pi}n} = 1 $

However it is interesting to note that this does not apply if we are dealing with a continous signal. This is a common mistake made by students.

$ e^{-j2{\pi}n} = (e^{-j2{\pi}})^n = 1^n = 1 $

$ e^{-j2{\pi}t} = (e^{-j2{\pi}})^t = 1^t {\neq} 1 $

This is because t can lead to a $ {\pm} $ solution due to t not being a definite value such as n.

Complex Exponential Example

$ x[n] = e^{j{\omega}_{o}n} {\Longleftrightarrow} X({\omega}) = 2{\pi}rep_{2{\pi}}({\delta}({\omega} - {\omega}_{o})) $

First we shall insert the x[n] into the general DTFT equation given in the definition

$ X({\omega}) = {\sum_{n = -{\infty}}^{\infty} e^{j{\omega}_{o}n}e^{-j{\omega}n}} $

$ X({\omega}) = {\sum_{n = -{\infty}}^{\infty} e^{-j({\omega} -{\omega}_{o})n}} $

now by inspection of the above equation we can see that if we were to have $ {\omega} $ be set to $ {\omega}_{o} $ we get

$ X({\omega}_{o}) = {\sum_{n = -{\infty}}^{\infty} (1) = +{\infty}} $

Note that for all other values of $ {\omega} $ the summation does not converge due to the oscillatory nature of the complex exponential.

Instead of now taking a shortcut and solving the above summation with a piece-wise function (which would be wrong), we try to take an educated guess as to what the input should be to give us the desired output.

we want

$ e^{j{\omega}_{o}n} = {\frac{1}{2{\pi}}}{\int_{0}^{2{\pi}}}{X({\omega})e^{-j{\omega}n}} $

without loss of generality

$ 0 {\leq} {\omega} {\leq} 2{\pi} $

now if

$ X({\omega}) = 2{\pi}{\delta}({\omega} - {\omega}_{o}) $

it works for

$ 0 {\leq} {\omega} {\leq} 2{\pi} $

since

$ X({\omega}) = {\frac{1}{2{\pi}}}{\int_{0}^{2{\pi}}}{2{\pi}{\delta}({\omega} - {\omega}_{o})e^{-j{\omega}n}} $

$ X({\omega}) = {\int_{0}^{2{\pi}}}{{\delta}({\omega} - {\omega}_{o})e^{-j{\omega}n}} $

$ X({\omega}) = {\int_{0}^{2{\pi}}}{e^{-j{\omega}_{o}n}} $

$ X({\omega}) = e^{j{\omega}_{o}n} $

However we are not done yet since we now need to repeat this signal with a period of $ 2{\pi} $ to make it apply for all n

$ X({\omega}) = 2{\pi}rep_{2{\pi}}({\delta}({\omega} - {\omega}_{o})) $

Conclusion