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Properties of z-Transforms

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  • Last Updated : 11 May, 2022

Prerequisite: What is Z-transform? 

A z-Transform is important for analyzing discrete signals and systems. In this article, we will see the properties of z-Transforms. These properties are helpful in computing transforms of complex time-domain discrete signals.

1. Linearity: If we have two sequences x1[n] and x2[n], and their individual z-transforms as X1(z) and X2(z), then the linearity property permits us to write:

\begin{aligned}X(z) & = Z { ax_{1}[n]+bx_{2}[n] } \\     &  = aX_{1}(z) + bX_{2}(z)\end{aligned}

This is easily proved. First consider  x[n]=ax_{1}[n]+bx_{2}[n]

Then, from the definition, we see:

\begin{aligned}X(z)&= \sum _{ n=0 }^{ \infty }{ } x[n]z^{-n}\\&= \sum _{ n=0 }^{ \infty }{ }( ax_{1}[n]+bx_{2}[n])z^{-n}\\&= a\sum _{ n=0 }^{ \infty }x_{1}[n]z^{-n}+b\sum _{ n=0 }^{ \infty }x_{2}[n]z^{-n} \\&= aX_{1}(z) + bX_{2}(z)\end{aligned}

2. Time Shifting: If we have a time-shifted sequence such as x[n-k], then its z-transform is given by  Z{ x[n-k]} = z^{-k}X(z).

Let’s take n – k = m, i.e., n = k + m and y[n] = x[n-k]. Now here we are assuming that x[n] starts from n=0, hence x[n-k] starts from n=k, or n-k=0, or from m=0. 

\begin{aligned}Y(z) & = \sum_{m=0}^{\infty }y[n]z^{-n}       \\& = \sum_{m=0}^{\infty }x[n-k]z^{-n}      \\&= \sum_{m=0}^{\infty }x[m]z^{-m-k}     \\& = z^{-k}\sum_{m=0}^{\infty }x[m]z^{-m}      \\&= z^{-k}X(z)\end{aligned}

3. Time reversal: Time reversal property states that Z{ x[-n]} = X(1/z)

We are going to formally prove this statement by taking y[n]=x[-n]. 

\begin{aligned} \sum_{n=0}^{\infty }y[n]z^{-n}   = \sum_{n=0}^{\infty }x[-n]z^{-n}\end{aligned}

Now let’s take -n=m. Then

 \begin{aligned}\sum_{m=0}^{ -\infty}x[m]z^{m} & = \sum_{m=-\infty}^{0 }x[m]z^{m}\\ & = \sum_{-m=\infty}^{0 }x[m](z^{-1})^{-m}\\ & = X(z^{-1})\end{aligned}

4. Scaling in z domain: When we multiply the signal sequence x[n] in the time domain with an exponential factor an, the equivalent z-transform of the new signal is scaled by a factor of a.

Basically, Z{ anx[n]}=X(z/a) .

Proof is elementary and is shown below.

\begin{aligned} \sum _{ n=0 }^{ \infty } a^{n}x[n]z^{-n} & = \sum _{ n=0 }^{ \infty } x[n](a^{-1}z)^{-n}\\ &=\sum _{ n=0 }^{ \infty } x[n](z/a)^{-n} \\ = X(z/a)\end{aligned}

5. Differentiation in z domain: We know:   X(z)=\sum_{n=0}^{\infty }x[n]z^{-n}

Differentiating with respect to z, we get

\begin{aligned} \frac{\mathrm{d} X(z)}{\mathrm{d} z} = -\sum_{n=0}^{\infty }nx[n]z^{-n-1} & \\ = -z^{-1} \sum_{n=0}^{\infty }nx[n]z^{-n}\end{aligned}

\begin{aligned} -z\frac{\mathrm{d} X(z)}{\mathrm{d} z} & = \\ \sum_{n=0}^{\infty }(nx[n])z^{-n}\end{aligned}

Hence, we can deduce that for k differentiations, we get  Z{ n^kx[n]} = (-1)^kz^k\frac{\mathrm{d^{k}}X(z) }{\mathrm{d} z^{k}}

6. Convolution: Convolution of two sequences x[n] and h[n] is defined as  x[n]*h[n]=\sum_{k=0}^{\infty }x[k]h[n-k]

Now z-transforms of x[n] and h[n] are X(z) and H(z) respectively. Using this notation, we have

\begin{aligned}\sum_{n=0}^{\infty }(x[n]*h[n])z^{-n} =\sum_{n=0}^{\infty }\sum_{k=0}^{\infty }x[k]h[n-k]z^{-n}\\=\sum_{n=0}^{\infty }\sum_{k=0}^{\infty }x[k]h[n-k]z^{-(n-k)}z^{-k}\\ =\sum_{n=0}^{\infty }(\sum_{k=0}^{\infty }x[k]z^{-k})h[n-k]z^{-(n-k)}\\ =\sum_{n=0}^{\infty }X(z)h[n-k]z^{-(n-k)}\\ =X(z)\sum_{n-k=0}^{\infty }h[n-k]z^{-(n-k)}\\ =X(z)H(z)\end{aligned}

Hence, convolution in time domain is multiplication in z domain.

7. Initial value theorem: Initial value theorem gives us a tool to compute the initial value of the sequence x[n], that is, x[0] in the z domain by taking a limit of the value of X(z). It states that the following equivalence is feasible. 

x[0] = \lim _{ z\rightarrow \infty }  x(Z)

The proof, as before, relies on the definition of X(z). 

\begin{aligned} X(z) =  \sum_{n=0}^{\infty }x[n]z^{-n} & \\ =  x[0]z^0+x[1]z^{-1}+z[2]z^{-2}+....&\\ \end{aligned}

Clearly, if we want to get x[0], we can make z approach to infinity so that all the other terms die out. What is left behind is precisely the statement of the theorem presented before.

8. Final value theorem: The final value theorem lets us know the final value of x[n], or the value at infinity of x[n], using appropriate limits of X(z). 

It states that x(\infty ) = \lim _{ z\rightarrow 1 }  X(z)(1-z^{-1})

If we take the z transform of x[n]-x[n-1], then we get \begin{aligned} X(z)-z^{-1}X(z) & =\sum_{n=0}^{\infty }x[n]z^{-n}-z^{-1} \sum_{n=0}^{\infty }x[n]z^{-n} \\& = x[0]+x[1]z^{-1}+x[2]z^{-2}+...-z^{-1}(x[0]+x[1]z^{-1}+x[2]z^{-2}+...) \end{aligned}

Now taking the limit z⇢1, we see that we get  x[1]-x[0]+x[2]-x[1]+x[3]-x[2]+...+x[\infty]-x[\infty -1]  in the right hand side, which simplifies to x[\infty] basically. Hence the theorem is proved. 

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