Correct Answer - Option 1 : Retaining sequence values of X
p[n] other than zeros
Concept of time scaling in discrete signal:-
Let we have a discrete signal f[n] is we apply scaling then new discrete signal will be f[an]
\(f\left( n \right)\;\mathop \to \limits^{Scaling} \;f\left[ {an} \right]\)
If a > 1 → decimation
a < 1 → interpolation
\(\begin{array}{*{20}{c}} {f\left( n \right)}\\ {} \end{array}\;\begin{array}{*{20}{c}} \Rightarrow \\ {} \end{array}\;\left\{ {\begin{array}{*{20}{c}} { - 5,\;2,\;3,\; - 2,\;7}\\ \uparrow \end{array}} \right\}\)
CASE-1
f [2n] → here a = 2 since a > 1, it is case of decimation
let f’[n] = f[2n] f’(-1) = f(-2), f’(1) = f[2], f’[2] = f[4]
\(\begin{array}{*{20}{c}} {f\left( {2n} \right)}\\ {} \end{array}\;\begin{array}{*{20}{c}} \Rightarrow \\ {} \end{array}\;\left\{ {\begin{array}{*{20}{c}} { - 5,\;3,\;7}\\ {\;\;\; \uparrow } \end{array}} \right\}\)
so here we can see if we put n = 2, f[4] which is gero so we removed it. Here we Retaing only that which are not zero
CASE-2
Now \(f''\left( n \right) = f\left( {\frac{n}{2}} \right)\)
Here we can see \(a = \frac{1}{2}\) since a < 1, it is a case of interpolation
Let \(a = \frac{1}{b} - b = \frac{1}{a}\)
Here we add (b – 1) zeros between every two samples on this example \(a = \frac{1}{2}\), b = 2, so we will add 1 zero between every two sample
\(f\left[ {\frac{1}{2}n} \right] = \left\{ {\begin{array}{*{20}{c}} { - 5,\;0,\;2,\;0,\;3,\;0,\; - 2,\;0,\;7}\\ \uparrow \end{array}} \right\}\)
