Zion Wheeler

2022-06-26

The problem is this:
The impulse response of a system is the output from this system when excited by an input signal $\delta \left(k\right)$ that is zero everywhere, except at $k=0$, where it is equal to 1. Using this definition and the general form of the solution of a difference equation, write the output of a linear system described by:
$y\left(k\right)–3y\left(k–1\right)–4y\left(k–2\right)=\delta \left(k\right)+2\delta \left(k–1\right)$
The initial conditions are: $y\left(–2\right)=y\left(–1\right)=0$.
My question is: How can the particular solution be found using the method of undetermined coefficients if the non-homogeneous equation is also a difference equation?

Do you have a similar question?

Daniel Valdez

Expert

OK. So, I suppose you want to solve your difference equation for $k\ge 0$. If you set $k=0$ we easily get $y\left(0\right)=1$ and you could find $y\left(k\right)$ recursively for $k>0$, right? that's the point. To obtain the impulse response, you can see it as a response of an homogeneous equation with initial conditions different from zero. Then, check it out that the impulse response $h\left(k\right)$ to your problem is the same of this one:
$y\left(k\right)-3y\left(k-1\right)-4y\left(k-2\right)=0,\phantom{\rule{1em}{0ex}}y\left(-2\right)=1/4,y\left(-1\right)=0$
Why? simply because they have the same solution for $k\ge 0$. Now its easy to find $h\left(k\right)$ right? The final solution is then $h\left(k\right)+2h\left(k-1\right)$.

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