I have just started to study Oppenheim's "Signals & Systems, Second Edition" and there is this easy-looking problem that has evaded me its solution for the past 48 hours:
consider systems $S1$, $S2$, and $S3$, where
$ x[n] \xrightarrow{S1} y_{(1)}[n] = \begin{cases} x[n/2] & \text{if n is even;}\\ 0 & \text{if n is odd.}\\ \end{cases}$
$ x[n] \xrightarrow{S2} y_{(2)}[n] = x[n] + (1/2)x[n-1] + (1/4)x[n-2] $
$ x[n] \xrightarrow{S3} y_{(3)}[n] = x[2n]$
To work out the result of $x[n] \xrightarrow{S1,S2,S3} y_{(1,2,3)}[n]$, I first tried to work out the result of $x[n] \xrightarrow{S1,S2} y_{(1,2)}[n]$ as follows:
$x[n] \xrightarrow{S1,S2} y_{(1,2)}[n] = \\ y_{(1)}[n] + (1/2)y_{(1)}[n-1] + (1/4)y_{(1)}[n-2] = \\ \begin{cases} x[n/2] + (1/4)x[(n-2)/2] & \text{if n is even;}\\ (1/2)x[(n-1)/2] & \text {if n is odd.}\\ \end{cases}$
after that, feeding $y_{(1,2)}[n]$ to $S3$, we have:
$y_{(1,2)}[n] \xrightarrow{S3} y_{(1,2,3)}[n] = \\ y_{(1,2)}[2n] = \\ \begin{cases} x[2n/2] + (1/4)x[(2n-2)/2] & \text{if 2n is even;}\\ (1/2)x[(2n-1)/2] & \text {if 2n is odd.}\\ \end{cases}$
which simplifies to $y_{(1,2,3)}[n] = x[n] + (1/4)x[n-1].$
but the solution manual says the output is in fact $x[n] \xrightarrow{S1,S2,S3} y_{(1,2,3)} = x[n] + (1/2)x[n-1] + (1/4)x[n-2].$ :(
What I want to know is, who is actually wrong: me or the manual? Thanks a lot in advance!