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

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!

## 1 Answer

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.

• This isn't even the first time I have been trapped in a book by Oppenheim; "classic" as his books might be, I really wish there was a better, less atrocious book for those who are new to this field. By the way, I can't thank you enough for saving me from yet another 48 hours of knocking on an empty door! – user45966 Nov 1 '19 at 18:59
• You're welcome! There are known errors in the solution manual. For some of the answers you can use Matlab / Octave to verify the result... – Fat32 Nov 1 '19 at 19:54