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In my mind, the steps are

  • 1: Find the homogeneous solution using the characteristic equation
  • 2: Use "undetermined coefficients" method of finding the particular solution.

However, what is the form of the particular solution for these kinds of equations?

I am trying to find the general form of $x[n]$: $$ x[n] + x[n-1] = {\delta}[n] $$

Is there no way to solve this using the Z-transform?

Benjamin Smus
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    We find that $x[n] = (-1)^n \cdot u[n]$ (where $u[n]$ denotes the unit step function) is a particular solution. It is not clear what the correct "guess" corresponding to $\delta[n]$ should have been. – Ben Grossmann Mar 01 '20 at 20:35
  • In general, using $\delta[n]$ as our forcing function and imposing the condition $x[n] = 0$ for $n < 0$ leads to the unit impulse response of the difference operator. Because the unit impulse of operators is so diverse, I'm inclined to think that there is no good Ansatz for the response (i.e. no way to set up the method of undetermined coefficients). – Ben Grossmann Mar 01 '20 at 20:41
  • Note that solving this system with $x[n] = 0$ for $n<0$ is equivalent to solving the homogeneous problem $x[n] + x[n-1] = \delta[n]$ with the initial condition $x[0] = \delta[0] = 1$. – Ben Grossmann Mar 01 '20 at 20:49

1 Answers1

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You can do something like that, 1.) $x[n]=(-1)^n$ is a homogeneous solution to your example problem, to 2.) find via variation of coefficients $$ x[n]=(-1)\cdot y[n]\implies y[n]-y[n-1]=\delta[n], $$ using that the delta sequence is zero almost everywhere, so that the alternating sign change does not influence it.

Now if you set $y[n]=0$ for $n<0$, then if follows that $y[n]=1$ for $n\ge 0$, as the recursion tells you that it is the previous value except at $n=0$, where $1$ is added.

The method of undetermined coefficients is not applicable here, as it requires that the right side is composed of terms $r[n]=p(n)q^n$ for some $q\ne0$ and some polynomial $p(n)$.

Lutz Lehmann
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