Determine the inverse z-transform of F(z)=(2 z)/(2 z+1) (2 z)/(2 z+1)=(2 z)/(2(z+(1)/(2)))=(z)/(

step 1 So in this session we have to find in the x transform of x. So 2 z upon 2 z plus 1 equal 2 z tak

Determine the inverse z-transform of F(z)=(2 z)/(2 z+1) (2...

Determine the inverse z-transform of $$ F(z)=\frac{2 z}{2 z+1} $$ $$ \frac{2 z}{2 z+1}=\frac{2 z}{2\left(z+\frac{1}{2}\right)}=\frac{z}{z+\frac{1}{2}} $$ From 6 in Table 79.1, $Z\left\{a^{k}\right\}=\frac{z}{z-a}$ hence $\quad Z^{-1}\left\{\frac{z}{z-a}\right\}=a^{k}$ Comparing $\frac{z}{z+\frac{1}{2}}$ with $\frac{z}{z-a}$ shows that $\mathrm{a}=-\frac{1}{2}$ Thus, $Z^{-1}\left\{\frac{2 z}{2 z+1}\right\}=Z^{-1}\left\{\frac{z}{z+\frac{1}{2}}\right\}=\left(-\frac{1}{2}\right)^{k}$

Determine the inverse z-transform of
$$
F(z)=\frac{2 z}{2 z+1}
$$
$$
\frac{2 z}{2 z+1}=\frac{2 z}{2\left(z+\frac{1}{2}\right)}=\frac{z}{z+\frac{1}{2}}
$$
From 6 in Table 79.1, $Z\left\{a^{k}\right\}=\frac{z}{z-a}$ hence $\quad Z^{-1}\left\{\frac{z}{z-a}\right\}=a^{k}$
Comparing $\frac{z}{z+\frac{1}{2}}$ with $\frac{z}{z-a}$ shows that $\mathrm{a}=-\frac{1}{2}$ Thus, $Z^{-1}\left\{\frac{2 z}{2 z+1}\right\}=Z^{-1}\left\{\frac{z}{z+\frac{1}{2}}\right\}=\left(-\frac{1}{2}\right)^{k}$

Key Concepts

Z-transform

The Z-transform is a mathematical tool used to convert discrete-time signals into a complex frequency domain representation. It extends the concept of the Fourier transform to sequences and is extensively used in digital signal processing and control systems to analyze and design linear, time-invariant systems.

Inverse Z-transform

The inverse Z-transform is the process of converting a function from the z-domain back into its original discrete-time sequence. It involves techniques such as partial fraction expansion, power series expansion, or using transform pair tables, and is essential for obtaining time-domain representations from z-domain analyses.

Transform Pair Tables

Transform pair tables provide a list of common functions and their corresponding Z-transforms. They are an efficient reference that can simplify the process of computing inverse Z-transforms by allowing one to match the given z-domain expression to a known transform pair, thereby avoiding more complex computation methods.

Comparison and Substitution Method

The comparison and substitution method involves rewriting the given function in the z-domain into a form that matches a standard transform pair. This requires algebraic manipulation such as factoring constants or shifting terms. Once the expression is properly aligned, the corresponding time-domain sequence can be deduced directly from the known transform pair.

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