- How to get the value of $y_p(n)$ using substitution method?
- What exactly do we need to substitute to get the value?
1
$\begingroup$
$\endgroup$
1
$\begingroup$
$\endgroup$
Substitution is not a method to obtain the particular solution $y_p(n)$, but the author simply claims that $y_p(n)=e^{-\beta n}u(n)$ is a particular solution of the recursion (1.70), and encourages you to check his claim by substituting $y_p(n)$ for $y(n)$ in Eq. (1.70) and see if it's correct or not:
$$e^{-\beta n}u(n)=e^{-\beta}e^{-\beta (n-1)}u(n-1)+\delta(n)=e^{-\beta n}u(n-1)+\delta(n)\tag{1}$$
Now check if (1) is satisfied. Distinguish 3 cases: $n<0$, $n=0$, and $n>0$. As a result you should be able to see that (1) (and, therefore, Eq. (1.70)) is satisfied for $y(n)=y_p(n)$, showing that the suggested sequence $y_p(n)$ indeed is a particular solution of (1.70).
-
$\begingroup$ thanks for explaining it to the bits. You made my day. I have one major doubt "how do we deduce the claim like author that Yp(n)=(e^−βn) * u(n) is a solution." $\endgroup$ – vCillusion Nov 9 '14 at 17:34
-
$\begingroup$ @LusionNectar: You can simply compute the first few elements of $y(n)$ according to the equation. Use the initial condition $y(-1)=0$, then you'll get $y(0)=1$, $y(1)=a$, $y(2)=a^2$, etc. Do you see a pattern? $\endgroup$ – Matt L. Nov 9 '14 at 19:56
