Example of Cascading sampling rate converters:

Example Given x(n) = e^-n/2 u(n), find x(5n/3).

Answer We borrow this from an earlier section. Our objective is to present the earlier solution and then reformulate it in the context of cascading up- and down-samplers. The sequence

 x(n) =   e^-n/2 u(n) = (e^-1/2)n u(n) = (0.606)ⁿ u(n) = aⁿu(n)

where a = e^-1/2= 0.606, is sketched below:

With y(n) = x(5n/3), we evaluate y(.) for several values of n (we have assumed here that x(n) is zero if n is not an integer):

y(0) = x(5 . 0 / 3) = x(0) = e^-0/2 = 1

y(1) = x(5 . 1 / 3) = x(5 / 3) = 0

y(2) = x(5 . 2 / 3) = x(10 / 3) = 0

y(3) = x(5 . 3 / 3) = x(5) = e^-5/2 = a⁵

...

y(6) = x(5 . 6 / 3) = x(10) = e^-10/2 = a¹⁰

...

The general expression for y(n) can be written as

 y(n) = x(5n/3) = e^-(5n/3)/2, n as specified below

The sequence is sketched below:

We shall recast this problem in terms of cascading the up- and down-samplers. In the expression y(n) = x(5n/3) there is a 3-fold up-sampling and a 5-fold down-sampling. Since the numerator 5 is greater than the denominator 3 there is a net down- ampling by a factor of 1.67 (= 5/3). Let us first do a 3-fold up-sampling of x(n) followed by a 5-fold down-sampling of the resulting sequence. That is, given the sequence x(n)

we define y_u(n) = x(n/3), and then y(n) = y_u(5n) = x(5n/3). The sequences yu(n) and y(n) are given below.

Alternatively, we may first do a 5-fold down sampling followed by a 3-fold up-sampling:

The net effect is that between the first two terms (1 and a ) of the final output y(.) we have dropped four original terms and inserted two zeros.

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