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Like continuous time signal Fourier transform, discrete time Fourier Transform can be used to represent a discrete sequence into its equivalent frequency domain representation and LTI discrete time system and develop various computational algorithms.

X (jω) in continuous F.T, is a continuous function of x(n). However, DFT deals with representing x(n) with samples of its spectrum X(ω). Hence, this mathematical tool carries much importance computationally in convenient representation. Both, periodic and non-periodic sequences can be processed through this tool. The periodic sequences need to be sampled by extending the period to infinity.

From the introduction, it is clear that we need to know how to proceed through frequency domain sampling i.e. sampling X(ω). Hence, the relationship between sampled Fourier transform and DFT is established in the following manner.

Similarly, periodic sequences can fit to this tool by extending the period N to infinity.

Let an Non periodic sequence be, $X(n) = \lim_{N \to \infty}x_N(n)$

Defining its Fourier transform,

$X(\omega ) = \sum_{n=-\infty}^\infty x(n)e^{-jwn}X(K\delta \omega)$...eq(1)

Here, X(ω) is sampled periodically, at every δω radian interval.

As X(ω) is periodic in 2π radians, we require samples only in fundamental range. The samples are taken after equidistant intervals in the frequency range 0≤ω≤2π. Spacing between equivalent intervals is $\delta \omega = \frac{2\pi }{N}k$ radian.

Now evaluating, $\omega = \frac{2\pi}{N}k$

$X(\frac{2\pi}{N}k) = \sum_{n = -\infty}^\infty x(n)e^{-j2\pi nk/N},$ ...eq(2)

where k=0,1,……N-1

After subdividing the above, and interchanging the order of summation

$X(\frac{2\pi}{N}k) = \displaystyle\sum\limits_{n = 0}^{N-1}[\displaystyle\sum\limits_{l = -\infty}^\infty x(n-Nl)]e^{-j2\pi nk/N}$ ...eq(3)

$\sum_{l=-\infty}^\infty x(n-Nl) = x_p(n) = a\quad periodic\quad function\quad of\quad period\quad N\quad and\quad its\quad fourier\quad series\quad = \sum_{k = 0}^{N-1}C_ke^{j2\pi nk/N}$

where, n = 0,1,…..,N-1; ‘p’- stands for periodic entity or function

The Fourier coefficients are,

$C_k = \frac{1}{N}\sum_{n = 0}^{N-1}x_p(n)e^{-j2\pi nk/N}$k=0,1,…,N-1...eq(4)

Comparing equations 3 and 4, we get ;

$NC_k = X(\frac{2\pi}{N}k)$ k=0,1,…,N-1...eq(5)

$NC_k = X(\frac{2\pi}{N}k) = X(e^{jw}) = \displaystyle\sum\limits_{n = -\infty}^\infty x_p(n)e^{-j2\pi nk/N}$...eq(6)

From Fourier series expansion,

$x_p(n) = \frac{1}{N}\displaystyle\sum\limits_{k = 0}^{N-1}NC_ke^{j2\pi nk/N} = \frac{1}{N}\sum_{k = 0}^{N-1}X(\frac{2\pi}{N}k)e^{j2\pi nk/N}$...eq(7)

Where n=0,1,…,N-1

Here, we got the periodic signal from X(ω). $x(n)$ can be extracted from $x_p(n)$ only, if there is no aliasing in the time domain. $N\geq L$

N = period of $x_p(n)$ L= period of $x(n)$

$x(n) = \begin{cases}x_p(n), & 0\leq n\leq N-1\\0, & Otherwise\end{cases}$

The mapping is achieved in this manner.

It states that the DFT of a combination of signals is equal to the sum of DFT of individual signals. Let us take two signals x_{1}(n) and x_{2}(n), whose DFT s are X_{1}(ω) and X_{2}(ω) respectively. So, if

$x_1(n)\rightarrow X_1(\omega)$and$x_2(n)\rightarrow X_2(\omega)$

Then $ax_1(n)+bx_2(n)\rightarrow aX_1(\omega)+bX_2(\omega)$

where **a** and **b** are constants.

The symmetry properties of DFT can be derived in a similar way as we derived DTFT symmetry properties. We know that DFT of sequence x(n) is denoted by X(K). Now, if x(n) and X(K) are complex valued sequence, then it can be represented as under

$x(n) = x_R(n)+jx_1(n),0\leq n\leq N-1$

And $X(K) = X_R(K)+jX_1(K),0\leq K\leq N-1$

Let us consider a signal x(n), whose DFT is given as X(K). Let the finite duration sequence be X(N). Then according to duality theorem,

If, $x(n)\longleftrightarrow X(K)$

Then, $X(N)\longleftrightarrow Nx[((-k))_N]$

So, by using this theorem if we know DFT, we can easily find the finite duration sequence.

Suppose, there is a signal x(n), whose DFT is also known to us as X(K). Now, if the complex conjugate of the signal is given as x*(n), then we can easily find the DFT without doing much calculation by using the theorem shown below.

If, $x(n)\longleftrightarrow X(K)$

Then, $x*(n)\longleftrightarrow X*((K))_N = X*(N-K)$

The multiplication of the sequence x(n) with the complex exponential sequence $e^{j2\Pi kn/N}$ is equivalent to the circular shift of the DFT by L units in frequency. This is the dual to the circular time shifting property.

If, $x(n)\longleftrightarrow X(K)$

Then, $x(n)e^{j2\Pi Kn/N}\longleftrightarrow X((K-L))_N$

If there are two signal x_{1}(n) and x_{2}(n) and their respective DFTs are X_{1}(k) and X_{2}(K), then multiplication of signals in time sequence corresponds to circular convolution of their DFTs.

If, $x_1(n)\longleftrightarrow X_1(K)\quad\&\quad x_2(n)\longleftrightarrow X_2(K)$

Then, $x_1(n)\times x_2(n)\longleftrightarrow X_1(K)© X_2(K)$

For complex valued sequences x(n) and y(n), in general

If, $x(n)\longleftrightarrow X(K)\quad \&\quad y(n)\longleftrightarrow Y(K)$

Then, $\sum_{n = 0}^{N-1}x(n)y^*(n) = \frac{1}{N}\sum_{k = 0}^{N-1}X(K)Y^*(K)$

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