We know that that the difference equation formula for computing an output sample at time $ n$ is based upon its past/present input samples and past output samples in time domain:

$$ \begin{aligned} y(n) &= b_0 \,x(n) + b_1 \,x(n - 1) + \cdots + b_M \,x(n - M) \\ &\qquad- a_1 \,y(n - 1) - \cdots - a_N \,y(n - N) \\ &=\sum_{i=0}^M b_i \,x(n-i) - \sum_{j=1}^N a_j \,y(n-j) \end{aligned} $$

My question is why $j$ begins from 1 rather than from value 0 like for $i$ and also how can we write the below output equation in difference equation form:

$$ \begin{aligned} y(n) &= 0.01\ x(n - 5) + 0.002\ x(n - 1) + 0.99\ y(n - 1)\\ y(n) &= 0.01\ x(n) + 0.002\ x(n - 1) \\ y(5) &= 0.01\ x(3) + x(1) \\ \end{aligned} $$

Secondly why $y(n)$ sequence values are calculated as a sum of a series of input / output, cause I have been studying that $$y(n) = \{2,4,6,8\}$$ when $$x(n) = \{1,2,3,4\}$$ and $$n =\{1,2,3,4\}$$ (example per say) i.e. $y(n) = 2\cdot x(n)$.

There was no mention that for $y(n)$ ($y(2)$ say) value we need to add values of $x(n)$ and $y(n)$ i.e. (take values of $x(1)$ or $y(1)$) or $y(n-1)$)?

  • $\begingroup$ If $j$ began at 0 there could exist a term $a_0y(n)$ on the r.h.s of the difference equation. Bringing this term to the l.h.s yields $(1-a_0)y(n) = \ldots$ and you could rewrite this difference equation by dividing both sides by $1-a_0$. This would just mean that all coefficients $b_i$ and $a_j$ are changed. So without a loss of generality, $j$ starts at 1. $\endgroup$ – Deve Aug 21 '14 at 8:26
  • $\begingroup$ Thanks @Deve for your comment but sorry I could not understand as you mentioned - "... This would just mean that all coefficients bi and aj are changed. So without a loss of generality, j starts at 1." $\endgroup$ – Programmer Aug 21 '14 at 8:54
  • $\begingroup$ Any difference equation with coefficents $b_0$ to $b_M$ and $a_0$ to $a_N$ can be rearranged yielding a new difference equation with coefficients $b'_0$ to $b'_M$ and $a'_1$ to $a'_N$. Both equations are equivalent and describe the same system. Thus, in the general definition $j$ starts at 1. $\endgroup$ – Deve Aug 21 '14 at 9:09

The answer to your first question (why $j$ starts at $1$) is contained in your first sentence: the output is computed from the current input value ($i=0$), from past input values ($i\ge 1$), and from past output values (i.e. $j\ge 1$). Imagine that you had the following difference equation with $j$ starting at $0$:


Then you could rewrite (1) as


and consequently


Now you can divide (3) by $1+a_0$ to arrive at an equation like the one given in your question (with $j$ starting at $1$) (and with all coefficients scaled by the constant $1+a_0$). So you can always use the difference equation with $j$ starting at $1$ without loss of generality.

Concerning the 3 examples of difference equations you gave I'll just discuss the first one, the others are totally analogous. Here you have $M=5$ and $N=1$ with coefficients $b_1=0.002$, $b_5=0.01$, and $b_0=b_2=b_3=b_4=0$, and $a_1=-0.99$.

Finally, concerning your last question about the system $y(n)=2x(n)$. That's again a special case of your general difference equation with all $a_i=0$ (i.e. no recursion) and with $b_0=1$ (and all other $b_i=0$, $i\ge 1$). So all your examples are special cases of the general difference equation.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.