0
1.0kviews
IDFT of 4 pt using matrix method
1 Answer
0
23views

TOFT (Inverse Discrete Fourice Transform)

IDFT $x(n)=\frac{1}{N} \sum_{k=0}^{N-k} \times \left(k) e^{\frac{j 2 \pi k}{N}} n\right.$

DFT $x(k)=\sum_{n=0}^{N-1} x(n) e^{-j \frac{2 \pi k}{N}} n$

DFT

$\left[\begin{array}{c}{x(0)} \\ {x(1)} \\ {x(2)} \\ {x(3)}\end{array}\right]\left[\begin{array}{cccc}{1} & {1} & {1} & {1} \\ {1} & {-j} & {x} & {j} \\ {1} & {-1} & {1} & {-1} \\ {1} & {j} & {-1} & {-j}\end{array}\right]\left[\begin{array}{c}{x(0)} \\ {x(1)} \\ {x(2)} \\ {x(3)}\end{array}\right]$

IDFT

$\left[\begin{array}{c}{x(0)} \\ {x(t)} \\ {x(2)} \\ {x(3)}\end{array}\right]=\left[\begin{array}{ccc}{1} & {0} & {1} \\ {1} & {j} & {2} \\ {1} & {-1} & {1} & {-1} \\ {-1} & {-1} & {-1}\end{array}\right]-\frac{1}{4}\left[\begin{array}{c}{x(0)} \\ {x(1)} \\ {x(2)} \\ {x(3)}\end{array}\right]$

(1) $I F \times(K)=6,1-j, 0,1+j$ find $x(n)$ ?

$\left[\begin{array}{c}{x(0)} \\ {x(1)} \\ {x(2)} \\ {x(3)}\end{array}\right]=\frac{1}{4}\left[\begin{array}{cccc}{1} & {1} & {1} & {1} \\ {1} & {j} & {-1} & {-1} \\ {1} & {-1} & {1} & {-1} \\ {1} & {-j} & {-1} & {j}\end{array}\right]\left[\begin{array}{c}{6} \\ {1-j} \\ {0} \\ {1+j}\end{array}\right]$

$=\frac{1}{4}\left[\begin{array}{c}{6+1-j+0+1+j=8} \\ {6+j+1+0-j+1=8} \\ {6-1+j+0-1-j=4} \\ {6-j-1+0+j-1=4}\end{array}\right]=\left[\begin{array}{l}{2} \\ {2} \\ {1} \\ {1}\end{array}\right]$

Please log in to add an answer.

Continue reading...

The best way to discover useful content is by searching it.

Search