Proof of DFT Circular Convolution Property 2
Key property 2 says that multiplication of 2-D DFTs corresponds to the defined circular convolution in the spatial domain, in a manner very similar to that in one dimension. In fact, we can see from (4.2.3) that this operation is separable into a 1-D circular convolution along the rows, followed by a 1-D circular convolution over the columns. The correctness of this property can then be proved by making use of the 1-D proof twice, once for the rows and once for the columns.
Proof of DFT Circular Property 5
Since , it follows that the periodic shift of agrees with the circular shift of x,
for , so the DFS of the left-hand side must equal the DFT of the right-hand side, over the fundamental period in frequency, i.e., . We have thus have the DFT
Figure( 4.4): Example of 2D Circular Convolution
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Let . The diagram in Figure (4.4) shows an example of the 2-D circular convolution of two small arrays x and y. In this figure, the two top plots show the arrays
and
, where the open circles indicate zero values of these 4 4 support signals. The non-zero values are denoted by filled-in-circles in the 3 3 triangle-support x, and various filled-in shapes on the square 2 2 support y. To perform their convolution, over dummy variables
, the middle plots of the figure show the inputs as functions of these dummy variables. The bottom two plots then show y circularly shifted to correspond to the output points
on the left and
on the right. Continuing in this way, we can see that for all
, we get the linear convolution result. This has happened because the circular wrap-around points in y occur only at zero values for x. We see that this due to the large value we took for N1 and N2 , i.e., larger than the necessary value for the DFT. We have seen in this example that the circular or wrap-around does not affect all output point. In fact, by enclosing the small triangular support signal and the small pulse in the large 4 4 square, we have avoided the wrap-around for all . We can generalize this result as follows. Let the support of signals x and y be given as
Then, if the DFT size is taken as or large, and or larger, we get the linear convolution result. Thus to avoid the circular or spatialaliasing result, we simply have to pad the two signals with zeros cout to a DFT size that is large enough to contain the linear convolution result, i.e., the result of ordinary noncircular convolution.
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