(a) Find the d.c.t., F[k], of the sequence f[n] = 2, 4, 6.
(b) Apply the inverse d.c.t. to F[k] and show that the original sequence, f[n], is obtained.

Question

Accepted Answer

(a) We use the formula

[latex]F[k]=\frac{1}{\sqrt{N}} \sum_{n=0}^{N-1} f[n] \cos \left[\frac{\pi}{N}\left(n+\frac{1}{2} ight) k ight] \quad ext { for } k=0,1,2, \ldots, N-1[/latex]

Here the number of terms, N, is three.
When k = 0

[latex]\begin{aligned} F[0] & =\frac{1}{\sqrt{3}} \sum_{n=0}^2 f[n] \cos \left[\frac{\pi}{3}\left(n+\frac{1}{2} ight) imes 0 ight] \ & =\frac{1}{\sqrt{3}}[2 \cos 0+4 \cos 0+6 \cos 0]=\frac{12}{\sqrt{3}} \end{aligned}[/latex]

When k = 1

[latex]\begin{aligned} F[1] & =\frac{1}{\sqrt{3}} \sum_{n=0}^2 f[n] \cos \left[\frac{\pi}{3}\left(n+\frac{1}{2} ight) imes 1 ight] \ & =\frac{1}{\sqrt{3}}\left[2 \cos \frac{\pi}{6}+4 \cos \frac{\pi}{2}+6 \cos \frac{5 \pi}{6} ight] \ & =\frac{1}{\sqrt{3}}\left[2 imes \frac{\sqrt{3}}{2}+0+6 imes\left(-\frac{\sqrt{3}}{2} ight) ight]=\frac{1}{\sqrt{3}}[\sqrt{3}-3 \sqrt{3}]=-2 \end{aligned}[/latex]

When k = 2

[latex]\begin{aligned} F[2] & =\frac{1}{\sqrt{3}} \sum_{n=0}^2 f[n] \cos \left[\frac{\pi}{3}\left(n+\frac{1}{2} ight) imes 2 ight] \ & =\frac{1}{\sqrt{3}}\left[2 \cos \frac{\pi}{3}+4 \cos \pi+6 \cos \frac{5 \pi}{3} ight] \ & =\frac{1}{\sqrt{3}}\left[2 imes \frac{1}{2}+4 imes(-1)+6 imes \frac{1}{2} ight]=1-4+3=0 \end{aligned}[/latex]

So the d.c.t. of the sequence 2, 4, 6 is [latex]\frac{12}{\sqrt{3}},-2,0[/latex].

(b) We now apply the inverse d.c.t. to F[k].
The inverse d.c.t. is

[latex]\begin{gathered} f[n]=\frac{1}{\sqrt{N}}\left\{F[0]+2 \sum_{k=1}^{N-1} F[k] \cos \left[\frac{\pi}{N} k\left(n+\frac{1}{2} ight) ight] ight\} \ ext { for } n=0,1,2, \ldots, N-1 \end{gathered}[/latex]

When n = 0

[latex]\begin{aligned} f[0] & =\frac{1}{\sqrt{3}}\left\{\frac{12}{\sqrt{3}}+2 \sum_{k=1}^2 F[k] \cos \left[\frac{\pi}{3} k\left(0+\frac{1}{2} ight) ight] ight\} \ & =\frac{1}{\sqrt{3}}\left[\frac{12}{\sqrt{3}}+2 imes(-2) imes \cos \left(\frac{\pi}{6} ight)+2 imes 0 ight] \ & =\frac{1}{\sqrt{3}}\left[\frac{12}{\sqrt{3}}-4 imes \frac{\sqrt{3}}{2}+0 ight]=4-2=2 \end{aligned}[/latex]

When n = 1

[latex]\begin{aligned} f[1] & =\frac{1}{\sqrt{3}}\left\{\frac{12}{\sqrt{3}}+2 \sum_{k=1}^2 F[k] \cos \left[\frac{\pi}{3} k\left(1+\frac{1}{2} ight) ight] ight\} \ & =\frac{1}{\sqrt{3}}\left[\frac{12}{\sqrt{3}}+2 imes(-2) imes \cos \left(\frac{\pi}{2} ight)+2 imes 0 ight] \ & =\frac{1}{\sqrt{3}}\left[\frac{12}{\sqrt{3}}-4 imes 0+0 ight]=4 \end{aligned}[/latex]

When n = 2

[latex]\begin{aligned} f[2] & =\frac{1}{\sqrt{3}}\left\{\frac{12}{\sqrt{3}}+2 \sum_{k=1}^2 F[k] \cos \left[\frac{\pi}{3} k\left(2+\frac{1}{2} ight) ight] ight\} \ & =\frac{1}{\sqrt{3}}\left[\frac{12}{\sqrt{3}}+2 imes(-2) imes \cos \left(\frac{5 \pi}{6} ight)+2 imes 0 ight] \ & =\frac{1}{\sqrt{3}}\left[\frac{12}{\sqrt{3}}-4 imes\left(-\frac{\sqrt{3}}{2} ight)+0 ight]=4+2=6 \end{aligned}[/latex]

So f[n] = 2, 4, 6, which was the original sequence as expected.

Question 24.26: (a) Find the d.c.t., F[k], of the sequence f[n] = 2, 4, 6. (......

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