1. Calculate the DFT of order m of the following sequences. m 1. Furthermore, the only nonzero entry in the DFT of x k occurs when j = 0 thus, m 1

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1 APPM 5600: uerical Analysis Hoework 4 Calculate the DFT of order of the following sequences (a x k, 0 k Solution: ˆxk xe πik/ e πik/ {, k 0 0, k 0 Furtherore, the only nonzero entry in the DFT of x k occurs when 0 thus, (b x k ( k, 0 k, even ˆxk {, 0, 0, 0, 0, } Solution: let ω e πi/ We can now write the DFT as ˆxk ( ω k ( ω k, a geoetric series hence, ˆxk ( ω k ( ω k ( ω k ( ω k 0 k except k / For the case of k /, we have ˆx ( ω k ( ( 0 Furtherore, (c x k k, 0 k ˆxk {, k 0, k Solution: let ω e πi/ We can write the DFT as ˆxk ω k By eleentary theory of geoetric series, we know r r r d r d r dr dr r r r r ( r r

2 hence, r r r r ( r r r r r( r ( r r r ow, ˆxk 0 (ω k ω k ( ω k ωk ( ω k ω k ω k, and, Furtherore, ˆxk { ˆx0 ω k, k 0, k 0 ˆxk ( { i sin( πk ( ( πk, k 0, k 0 What happens to the vector, x, of data if you apply the DFT to it twice? Solution: The Discrete Fourier Transfor is given by: ˆxk xe πik/ xω k, k 0,,,,, ω e πi/ This can be expressed in atrix for as follows: ω ω ω ω ω 4 ω ( ω ω ( ω ( x x x 3 x ˆx ˆx ˆx 3 ˆx Hence, perforing the DFT twice on the vector, x, consists of ultiplying x by the above atrix twice ote that any entry in the DFT atrix can be expressed as A i ωi, thus, if A and B are two atrices representing the DFT, their product is given by AB ik A i B k ω i ω k ω (i+k (ω α, where α i + k ote, this is a finite geoetric series hence, for α 0, we have (ω α (ω α 0, α 0 ω α ωα

3 Additionally, we have (ω α ( (, α 0, ±, ±, ow, α 0, ±, ±, i + k 0, ±, ±, {(i, k} {(0, 0, (,, (,,, (,, (, } Thus, the resulting atrix of applying the DFT twice is a perutation, ie Since this is a perutation atrix, the entries of the original vector are siply rearranged, no scaling or skewing occurs in the output vector: x x x x x 3 x x x ote: Depending on how the DFT is defined deterines whether or not the appears in front In the definition given in class, the noralization is invoked during the FT hence why it appears above If the noralization factor is on both the FT and IFT instead, this ter will vanish fro above and there will only be a perutation atrix left x 3 x 3 Given the following periodic data, x {0,,, 3, 4}, y {,, 3,, } plot the lowest degree trigonoetric interpolant Solution: xodes 0 3 4; yodes 3 ; 3 length(xodes; 4 y hat fft(yodes/; 5 6 nuzeropad 95; 7 y hatpad y hat(:3 zeros(,nuzeropad + y hat(end :end; 8 new length(y hatpad; 9 y Pad ifft(y hatpad*new; 0 x linspace(0,5,length(y Pad+; x(end ; plot(x,y Pad,'b',xodes,yodes,'ro' 3 xlabel('x' 4 ylabel('f(x' 5 legend('data','trigonoetric Interpolant' 3

4 This code generates the following plot: 4 Verify analytically that the Discrete Sine Transfor (DST is identical to its inverse transfor (discounting that it ust be noralized Solution: Consider the Discrete Sine Transfor, â k a k sin ( πk,,,, a k, â R This transfor can be expressed in atrix for as follows: sin π sin π sin 3π sin ( π sin π sin 4π sin 6π sin ( π sin 3π sin 6π sin 9π sin 3( π sin ( π sin ( π sin 3( π sin ( π a a a 3 a â â â 3 â By definition, if a atrix is identical to its inverse, ultiplying the atrix by itself results in the identity atrix In this scenario, since the DST is identical to its inverse (neglecting noralization, ultiplying the above atrix by itself will result in a scalar ultiple of the identity atrix ow, let A sin π sin π sin 3π sin ( π sin π sin 4π sin 6π sin ( π sin 3π sin 6π sin 9π sin 3( π sin ( π sin ( π sin 3( π sin ( π 4

5 ow, any entry in i th -row and th -colun of the DST atrix can be expressed as ( πi A i sin and siilarly, if B is another DST atrix, B k sin ( πk By defining the DST atrix entries as above, we can express the DST of the DST as a ultiplication of the two atricies, naely (AB ik sin ( πi sin ( πk ( π(i k ( π(i + k where the last expression arises fro the given identity, sin α sin β (α + β + (α β ow, let α i + k and β i k We have, ( πβ This can be rewritten as ( πβ ( πβ ( πα (βπ ( ( πβ ( πα + ( α ( β, ( πα ( πα (απ ow, consider α i + k and β i k First, it can be shown that α and β are either both even or odd hence, ( α ( β 0 Additionally, we know k and i hence, 0 < i + k < < i k < Thus, by given result (ii, we have ( πα {, i k 0, i k Finally, our resultant atrix, AB, can be expressed as: AB