finding the output of these three systems when connected in series?

Question

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!

Answer 1

You are right and the manual is wrong.

Given $S_1, S_2, S_3$ and the respective input-output signals as below :

$$\\x[n] \rightarrow \boxed{S_1} \rightarrow v[n] \rightarrow \boxed{S_2} \rightarrow w[n] \rightarrow \boxed{S_3}\rightarrow y[n] \\$$

$\\$

$ x[n] \xrightarrow{S1} v[n] = \begin{cases} x[n/2] & \text{if n is even;}\\ 0 & \text{if n is odd.}\\ \end{cases}$

$ v[n] \xrightarrow{S2} w[n] = v[n] + (1/2)v[n-1] + (1/4)v[n-2] $

$ w[n] \xrightarrow{S3} y[n] = w[2n] \\ \\$

You would compute the output $y[n]$ in three steps as: (you have compined step 1 into 2 actually)

$v[n] \xrightarrow{S2} w[n] $

$ \begin{align} w[n] &= v[n] + (1/2)v[n-1] + (1/4)v[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} \end{align} $

$\\$

and then, feeding $w[n]$ to $S_3$, we have:

$w[n] \xrightarrow{S3} y[n] = w[2n] \\$

$ y[n] = \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}$

Hence (your) the solution is right: $$y[n] = x[n] + (1/4)x[n-1]. $$

solution manual is wrong: $$x[n] \xrightarrow{S1,S2,S3} y[n] = x[n] + (1/2)x[n-1] + (1/4)x[n-2].$$

You can verify this simply by putting some test signals, or even easier by writing a simple MATLAB/Octave/Python script to implement the $S_1,S_2,S_3$ system.