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Given z transformed function is E(z)=1/(z+4)

I know there are several ways to get the inverse z transform of this function.

  1. use partial fraction

E(z)=1/(z+4)

E(z)/z=1/z(z+4) apply partial fraction here,

E(z)/z=(1/4)*((1/z)-(1/(z+4)) so E(z) is,

E(z)=(1/4)*(1-(z/(z+4)) as you know, it is easy to use inverse z transform here.

e(k)=(1/4)*(delta(k)-(-4)^k)

  1. inversion formula method

E(z) has simple pole at z=-4, the residue is evaluated as

{(residue)|z=-4} = {(z+4)*E(z)*z^(k-1)|z=-4

             = {1*z^(k-1)|z=-4}

             = (-4)^(k-1)

             = -(1/4)*(-4)^k

The problem is that inverse z transform of the same function E(z)

has 2 different answers. As you see, there is no delta function when I used

inversion formula method.

At first, I thought it was because a simple pole,z=-4, is outside of the unit

circle, but same things happen when I used a simple which is inside of the unit circle.

Why does this difference happen?

What is the real answer?

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  • \$\begingroup\$ This isn't explicitly an EE question - there are better sites for raising questions like this. \$\endgroup\$ – Andy aka Sep 26 '17 at 10:27
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The equations do produce the same results. \$\small e(k)\$ is assumed unilateral, so the equations in method 2 should be qualified by \$\small k>0\$, thus:

$$\small e(k)=\begin{cases} (-4)^{k-1}= -(1/4)(-4)^{k}, \:\:\:k>0 & \\\\ 0, \:\:\text{otherwise}\end{cases}$$

Also, remember that:

$$\small \delta (k)=\begin{cases} 1, \:\:\:k=0 & \\\\ 0, \:\:\text{otherwise}\end{cases}$$

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