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Can you help me find the z-transform of the impulse response below:

$$h(n) = (\frac{1}{2})^{|n-1|} + (\frac{1}{2})^{|n|} $$

I know the z-transform of $$(\frac{1}{2})^{n}$$ is equal to: $$(\frac{z}{z-b})$$

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You can simplify the expression by segmenting it to two intervals

  • \$ u(n)\$ \$\space\space\space\space\space\$ for \$ n>0 \$
  • \$ u(-n-1)\$ \$\space\space\$ for \$n < 0\$

$$h(n) = (\frac{1}{2})^{n-1}u(n) + (\frac{1}{2})^{n}u(n) +(\frac{1}{2})^{-n+1}u(-n-1) + (\frac{1}{2})^{-n}u(-n-1) $$

According to the superposition theorem, z-transform of two or more functions is the same as taking the z-transform of each one individually, and then adding them. Same Applies here.

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Lastly, you need to determine region of convergence

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