Contents
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 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:
magnitude
X(ω) = | X(ω) | ejθω
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})) $
