Can any one help with the $y[n]$ and $x[n]$ relationship in this block diagram, I just keep have a $t[n]$ in my answer that I can't get rid off.

On my best try I got to $y[n] = 2t[n]-x[n-1]-y[n-2]+x[n]$

If you could also tell me how you got there it would be great.

Thanks anyone for help.enter image description here

  • $\begingroup$ Are you allowed to use the $\mathcal{Z}$-transform? $\endgroup$
    – Matt L.
    Mar 30, 2015 at 9:40
  • $\begingroup$ that unit-delay element in the middle is not legit. the two outputs are the same and it should be drawn that way explicitly. $\endgroup$ Mar 30, 2015 at 18:01

2 Answers 2


I prefer to solve such problems using the $\mathcal{Z}$-transform. First, write down the time domain equations:

$$\begin{align}\tag{1} y[n]&=r[n]+x[n]\\ r[n]&=r[n-1]+t[n]+t[n-1]\\ t[n]&=t[n-1]+y[n-1]+y[n-2] \end{align}$$

Taking the $\mathcal{Z}$-transform of these equations gives

$$\begin{align}\tag{2} Y(z)&=R(z)+X(z)\\ R(z)&=z^{-1}R(z)+T(z)+z^{-1}T(z)\\ T(z)&=z^{-1}T(z)+z^{-1}Y(z)+z^{-2}Y(z) \end{align}$$

After simplification you get

$$\begin{align} R(z)&=T(z)\frac{1+z^{-1}}{1-z^{-1}}\\ T(z)&=Y(z)z^{-1}\frac{1+z^{-1}}{1-z^{-1}} \end{align}$$

from which


follows. Plugging (3) into the first equation of (2) gives


from which you finally obtain afters some algebra


In the time domain, Eq. (5) is equivalent to



Two general suggestions: 1-Systematically write the equations at each node. 2-Time index can be changed to get new equations. For example in the top left node we have: $$r[n]=y[n]-x[n]$$, and thus: $$r[n-1]=y[n-1]-x[n-1]$$.

Answer: Write the equations at the two other nodes: $$t[n]-t[n-1]=y[n-1]+y[n-2](*)$$ $$t[n]+t[n-1]=r[n]-r[n-1]=y[n]-x[n]-y[n-1]+x[n-1] $$ Add two equations together: $$t[n]=\frac{y[n]+y[n-2]-x[n]+x[n-1]}{2}$$ change n to n-1 $$t[n-1]=\frac{y[n-1]+y[n-3]-x[n-1]+x[n-2]}{2}$$ replace $t[n]$ and $t[n-1]$ in (*)

$$y[n-1]+y[n-2]=-\frac{y[n-1]+y[n-3]-x[n-1]+x[n-2]}{2}+\frac{y[n]+y[n-2]-x[n]+x[n-1]}{2}$$ simplify....

Or you can use Z transform as above which gives a more compact solution.

  • $\begingroup$ Your last equation is wrong. The easiest way to get the correct result is to plug the expressions for $t[n]$ and $t[n-1]$ into the equation above the starred one. If you do this you obtain the same result as I do. Basically, it's just the plus sign before the last term in your final equation which should be a minus sign. $\endgroup$
    – Matt L.
    Mar 31, 2015 at 19:09

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