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# Inverse DFT using the FFT algorithm Assignment Help

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Fast Fourier transform - Inverse DFT using the FFT algorithm
Inverse DFT using the FFT algorithm

The inverse DFT of the N-point sequence {X(k), k = 1, 2, ... , (N-1)} can be defined as

,

here . Take complex conjugate of x(n) and multiply by N to obtain

The right hand side of above equation is simply DFT of the sequence X*

(k) and can be calculated by using any FFT algorithm. The output sequence which is desired is then found by taking the conjugate of result and dividing by N

Example Given DFT sequence X(k) = {0, (-1-j), j, (2+j), 0, (2-j), -j, (-1+j)} obtain the IDFT x(n) by using the DIF FFT algorithm.

Solution

This is an 8 point IDFT. The 8 point twiddle factors are, as computed earlier,

The elementary computation (Butterfly) is shown as follows:

The signal flow graph id drawn as follows:

The output at the stage 3 gives us values in bit-reversed order:

The IDFT can be given by arranging the data in normal order, taking complex conjugate of the

sequence and dividing it by 8:

Because of the conjugate symmetry of {X(k)}, we should expect sequence {x(n)} to be

real-valued.

The MATLAB program:

Example 3.4.2 Given DFT sequence X(k) = {0, (1-j), j, (2+j), 0, (2-j), (-1+j), -j}

obtain IDFT x(n) using the DIF FFT algorithm.

Solution

There exists no conjugate symmetry in {X(k)}. By using MATLAB

The IDFT is

x(n) = {0.5, (-0.44 + 0.037i), (0.375 - 0.125i), (0.088 + 0.14i), (-0.75 + 0.5i),

(0.44 + 0.21i), (-0.125 - 0.375i), (-0.088 - 0.39i)}

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Example Given DFT sequence X(k) = {0, (-1-j), j, (2+j), 0, (2-j), -j, (-1+j)} obtain the IDFT x(n) by using the DIF FFT algorithm.

The elementary computation (Butterfly) is shown as follows:

The signal flow graph id drawn as follows:

The output at the stage 3 gives us values in bit-reversed order:

The MATLAB program:

Example 3.4.2 Given DFT sequence X(k) = {0, (1-j), j, (2+j), 0, (2-j), (-1+j), -j}

There exists no conjugate symmetry in {X(k)}. By using MATLAB

The IDFT is

x(n) = {0.5, (-0.44 + 0.037i), (0.375 - 0.125i), (0.088 + 0.14i), (-0.75 + 0.5i),