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.


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 ejωn, 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:


X(ω) = | X(ω) | ejθω


$ {\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} $


$ 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})) $


In this section we have learned the definition of a DTFT and the importance of its periodic property. The two main ideas to be taken away from this is that one should not try to fudge the answer, rather "reverse engineer" the answer by taking an educated guess of what the input should be to achieve the desired output. The second idea is that it is important to note that the simplification in the periodic property with the complex exponential only applies to a discrete case and not a continuous case.


Signals and Systems by K. Raja Rajeswari and B. Visvesvara Rao 2009

Mireille Boutin, "ECE438 Digital Signal Processing with Applications," Purdue University August 26,2009

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