1
$\begingroup$

I have a system given by $$y[n] - \frac{1}{4} y[n-1] - \frac{1}{8} y[n-2] =3x[n] $$

I want to solve for $y[n]$ for $x[n]=(\frac{1}{2})^nu[n]$.

The complementary solution evaluates to $[k_1(\frac{1}{2})^n+k_2 (\frac{-1}{4})^n]u[n]$.

But when I evaluate the particular solution consider $y_p[n]=k(\frac{1}{2})^n u[n]$ I get an absurd answer as $0=3$. And boundary conditions are not given either. What initial conditions am I to consider? I want to solve this difference equation without using Z-transform or Fourier transform.

Edit: I took the particular solution of the form $$y_p[n]=kn(\frac{1}{2})^nu[n]$$ since the input is of the same form as the roots of the characteristic equation.

and get k as equal to 2.

and evaluate the coefficients of complementary solution as $$y[n]=\frac{8}{3} (\frac{1}{2})^n u[n]+\frac{1}{3} (\frac{-1}{4})^n u[n]+2n (\frac{1}{2})^n u[n]$$

but the solution given is $$\frac{1}{3} (\frac{1}{4})^n u[n] + 4(n+1)(\frac{1}{2})^{(n+1)} u[n+1] + \frac{2}{3} (\frac{1}{2})^n u[n]$$

I don't where the anomaly in my evaluation is? Or is the given solution wrong? I'm new to the concept of difference equations.

I just want to know if my understanding is right.

I wish to solve this difference equation without using z transform.

Improve this question
$\endgroup$
4
  • $\begingroup$ How did you come up with that particular solution? $\endgroup$
    – Matt L.
    Commented Jan 26, 2021 at 16:34
  • $\begingroup$ Isn't the particular solution suppose to be of the same form as the input. The particular solution for the input $a^n u[n]$ is of the form $ka^n u[n]$....or so was told to me by my professor....or does this take a different form because the input is the same as one of the roots of the characteristic equation...I'm confused.. $\endgroup$
    – Orpheus
    Commented Jan 26, 2021 at 17:05
  • $\begingroup$ Yes, the problem is that the input has the same form as a characteristic mode of the system. $\endgroup$
    – Matt L.
    Commented Jan 26, 2021 at 17:22
  • $\begingroup$ So how do I resolve it?...I can't seem to find anything online $\endgroup$
    – Orpheus
    Commented Jan 26, 2021 at 17:24

1 Answer 1

Reset to default
1
$\begingroup$

The solution you came up with is the correct homogeneous solution (i.e., when $x[n]=0$). Thing is, this is a non-homogeneous difference equation, and its solutions are of the form $$y[n]=y_h[n]+y_p[n]=k_1\Big(\frac{1}{2}\Big)^nu[n]+ k_2\Big(\frac{-1}{4}\Big)^nu[n]+y_p[n]$$

where $y_p[n]$ is the particular solution that solves for the $x[n]$ term, and you are missing the corresponding particular solution in your evaluation. One way to find $y_p[n]$ is to guess the form of $y_p[n]$ with unknown coefficients given the form of $x[n]$, and use the method of undetermined coefficients to find those coefficients.

For $x[n]=r^nu[n]$, you would try $y_p[n]=ar^nu[n]$ at first. But, in this case $x[n]$ corresponds to a mode of the system. Therefore, the particular solution cannot be of the form $a(\tfrac{1}{2})^nu[n]$, but it has to be of the form $an(\tfrac{1}{2})^nu[n]$. Therefore, the solution of the equation is

$$y[n]=k_1\Big(\frac{1}{2}\Big)^nu[n]+ k_2\Big(\frac{-1}{4}\Big)^nu[n]+an\Big(\frac{1}{2}\Big)^nu[n],$$

and you find $a$ by evaluating this solution in the equation.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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