Advanced Z-Transform Properties: Conjugation, Accumulation, Correlation, and Parseval

The Complete Property Table

Let

\[x[n]\;\xleftrightarrow{\;\mathcal{Z}\;}X(z),\quad \text{ROC}=R_x \qquad y[n]\;\xleftrightarrow{\;\mathcal{Z}\;}Y(z),\quad \text{ROC}=R_y\]

Property	Time domain	Z-domain	ROC
Linearity	$ax[n]+by[n]$	$aX(z)+bY(z)$	contains $R_x\cap R_y$
Time shift	$x[n-n_0]$	$z^{-n_0}X(z)$	$R_x$ (minus $z=0$ or $\infty$)
Frequency shift	$e^{j\omega_0 n}x[n]$	$X(e^{-j\omega_0}z)$	$R_x$
Exponential scaling	$a^n x[n]$	$X(a^{-1}z)$	$\vert a\vert R_x$
Differentiation	$n\,x[n]$	$-z\dfrac{dX(z)}{dz}$	$R_x$
Conjugation	$x^*[n]$	$X^(z^)$	$R_x$
Reversal	$x[-n]$	$X(1/z)$	$1/R_x$
Real part	$\operatorname{Re}{x[n]}$	$\tfrac12[X(z)+X^(z^)]$	$R_x$
Imaginary part	$\operatorname{Im}{x[n]}$	$\tfrac{1}{2j}[X(z)-X^(z^)]$	$R_x$
Convolution	$x[n]*y[n]$	$X(z)Y(z)$	contains $R_x\cap R_y$
Correlation	$r_{xy}[n]=\sum_k x[k]y[k-n]$	$X(z)Y(z^{-1})$	contains $R_x\cap 1/R_y$
Accumulation	$\sum_{k=-\infty}^{n}x[k]$	$\dfrac{X(z)}{1-z^{-1}}$	contains $R_x\cap{\vert z\vert>1}$
Multiplication	$x[n]y[n]$	$\dfrac{1}{2\pi j}\oint_C X(v)Y(z/v)v^{-1}dv$	—
Parseval	$\sum_n x[n]y^*[n]$	$\dfrac{1}{2\pi j}\oint_C X(z)Y^(1/z^)z^{-1}dz$	—

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Conjugation and Its Consequences

\[x^*[n]\;\longleftrightarrow\;X^*(z^*),\qquad \text{ROC}=R_x\]

Proof.

\[\sum_n x^*[n]z^{-n}=\left(\sum_n x[n](z^*)^{-n}\right)^*=X^*(z^*)\]

What Conjugation Buys You

For a real sequence $x[n]=x^*[n]$, the property forces

\[X(z)=X^*(z^*)\]

On the unit circle $z=e^{j\omega}$, $z^*=e^{-j\omega}$, so

\[X(e^{j\omega})=X^*(e^{-j\omega})\]

This is conjugate symmetry of the DTFT of any real signal. Splitting into magnitude and phase:

\[\vert X(e^{j\omega})\vert=\vert X(e^{-j\omega})\vert \quad\text{(even)}\] \[\angle X(e^{j\omega})=-\angle X(e^{-j\omega}) \quad\text{(odd)}\]

Consequences you use constantly:

• A real signal’s spectrum needs only $\omega\in[0,\pi]$; the negative half is redundant.
• The DFT of a real $N$-point sequence satisfies $X[N-k]=X^*[k]$.
• Real filters have real coefficients, so $H(e^{j\omega})$ is conjugate-symmetric — the phase is odd and the magnitude is even.

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Real and Imaginary Parts

Decompose $x[n]=x_r[n]+jx_i[n]$. Then

\[x_r[n]=\tfrac12\left(x[n]+x^*[n]\right) \;\longleftrightarrow\; \tfrac12\left[X(z)+X^*(z^*)\right]\] \[x_i[n]=\tfrac{1}{2j}\left(x[n]-x^*[n]\right) \;\longleftrightarrow\; \tfrac{1}{2j}\left[X(z)-X^*(z^*)\right]\]

These are just conjugation combined with linearity. They are useful when separating the response of a complex-valued system into its real and quadrature channels.

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Time Reversal

\[x[-n]\;\longleftrightarrow\;X(1/z),\qquad \text{ROC}=1/R_x\]

Proof: substitute $m=-n$:

\[\sum_n x[-n]z^{-n}=\sum_m x[m](z^{-1})^{-m}=X(z^{-1})\]

The ROC inverts: if $R_x$ is $\vert z\vert>r$, the reversed signal’s ROC is $\vert z\vert<1/r$.

Practical use: reversing a sequence turns a causal filter into an anti-causal one, which is exactly how zero-phase filtering (filtfilt) works — filter forward, reverse, filter again, reverse back, so the phase distortions cancel.

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Convolution and Correlation

Convolution:

\[y[n]=x[n]*y_0[n]=\sum_k x[k]\,y_0[n-k] \;\longleftrightarrow\; Y_0(z)=X(z)Y_0(z)\]

Correlation (note the missing time reversal compared with convolution):

\[r_{xy}[n]=\sum_{k=-\infty}^{\infty}x[k]\,y[k-n] \;\longleftrightarrow\; R_{xy}(z)=X(z)Y(z^{-1})\]

Proof. Let $m=k-n$, so $k=m+n$:

\[\sum_n r_{xy}[n]z^{-n}=\sum_n\sum_k x[k]y[k-n]z^{-n}=\sum_k x[k]z^{k}\sum_m y[m]z^{-m}\] \[=\left(\sum_k x[k](z^{-1})^{-k}\right)\left(\sum_m y[m]z^{-m}\right)=X(z^{-1})\,Y(z)\]

(equivalently $X(z)Y(z^{-1})$ by re-indexing). For the autocorrelation, $Y=X$:

\[r_{xx}[n]\;\longleftrightarrow\;X(z)X(z^{-1})\]

On the unit circle this becomes $\vert X(e^{j\omega})\vert^2$ — the energy spectral density. That is the Z-domain statement of the Wiener–Khinchine theorem.

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Accumulation

\[y[n]=\sum_{k=-\infty}^{n}x[k]\;\longleftrightarrow\;Y(z)=\frac{X(z)}{1-z^{-1}}\]

Proof. Note $y[n]-y[n-1]=x[n]$. Transform both sides:

\[Y(z)-z^{-1}Y(z)=X(z)\;\Rightarrow\;Y(z)(1-z^{-1})=X(z)\]

The ROC must contain both $R_x$ and $\vert z\vert>1$, because the accumulator has a pole at $z=1$. This is the discrete counterpart of integration ($1/s$ in the Laplace domain).

Example. $x[n]=\delta[n]$ accumulates to $u[n]$:

\[X(z)=1,\qquad Y(z)=\frac{1}{1-z^{-1}}=U(z) \;\checkmark\]

Stability warning: the accumulator is marginally stable — its pole sits exactly on the unit circle at $z=1$. A bounded input like $u[n]$ produces an unbounded output, so the accumulator is not BIBO stable.

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Parseval’s Relation

For two sequences:

\[\sum_{n=-\infty}^{\infty}x[n]\,y^*[n]=\frac{1}{2\pi j}\oint_C X(z)\,Y^*(1/z^*)\,z^{-1}\,dz\]

For $y=x$ this is the energy theorem:

\[\sum_n \vert x[n]\vert^2=\frac{1}{2\pi j}\oint_C X(z)X^*(1/z^*)\,z^{-1}\,dz\]

On the unit circle $z=e^{j\omega}$, $dz=j e^{j\omega}d\omega=jz\,d\omega$, so $z^{-1}dz=j\,d\omega$ and the contour integral collapses to

\[\boxed{\sum_n \vert x[n]\vert^2=\frac{1}{2\pi}\int_{-\pi}^{\pi}\vert X(e^{j\omega})\vert^2\,d\omega}\]

Meaning: energy computed in time equals energy computed in frequency. No energy is lost or created by the transform; the transform only redistributes it.

Discrete (DFT) Version

\[\sum_{n=0}^{N-1}\vert x[n]\vert^2=\frac{1}{N}\sum_{k=0}^{N-1}\vert X[k]\vert^2\]

This is the version you actually use numerically. It lets you compute signal energy from the FFT without ever reconstructing the time signal.

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Worked ROC Problems

Example 1: Sum of Two Right-Sided Sequences

\[x[n]=\left(\tfrac12\right)^n u[n]+\left(\tfrac13\right)^n u[n]\]

Poles at $z=\tfrac12$ and $z=\tfrac13$. Both are right-sided, so

\[X(z)=\frac{z}{z-\frac12}+\frac{z}{z-\frac13},\qquad \text{ROC: }\vert z\vert>\frac12\]

The ROC is the intersection of the individual ROCs, bounded by the outermost pole.

Example 2: Right-Sided Plus Left-Sided

\[x[n]=\left(\tfrac12\right)^n u[n]-\left(\tfrac13\right)^n u[-n-1]\]

First term: pole at $\tfrac12$, ROC $\vert z\vert>\tfrac12$. Second term: pole at $\tfrac13$, ROC $\vert z\vert<\tfrac13$.

\[X(z)=\frac{z}{z-\frac12}+\frac{z}{z-\tfrac13},\qquad \text{ROC: }\frac13<\vert z\vert<\frac12\]

The intersection is empty unless the two sides do not overlap — here $\tfrac13<\tfrac12$, so the ROC is the annulus between the poles and is non-empty. If the second pole had been at $\tfrac23$, the intersection would be empty and no Z-transform would exist for that combination.

Example 3: Finite-Length Sequence

\[x[n]=\{1,\ 2,\ 3\},\quad n=0,1,2\] \[X(z)=1+2z^{-1}+3z^{-2}\]

The ROC is the entire $z$-plane except $z=0$ (because of the $z^{-1}$ and $z^{-2}$ terms). A finite sequence has no poles, so nothing bounds its ROC.

Example 4: Does the DTFT Exist?

\[x[n]=2^n u[n]\] \[X(z)=\frac{z}{z-2},\qquad \text{ROC: }\vert z\vert>2\]

The ROC does not include the unit circle $\vert z\vert=1$, so the DTFT does not exist — the sequence grows without bound. The Z-transform still exists as a formal object, which is precisely why the Z-transform is more general.

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Property Cheat Sheet by Question Type

Question asks	Property to reach for
“Effect of delaying by 3?”	$z^{-3}X(z)$
“Effect of multiplying by $n$?”	$-z\,dX/dz$
“Effect of reversing?”	$X(1/z)$, ROC inverts
“Is the DTFT real/even?”	Conjugation + real signal
“Relate convolution and correlation”	$X(z)Y(z)$ vs $X(z)Y(z^{-1})$
“Energy in frequency domain”	Parseval
“Output of an accumulator”	$X(z)/(1-z^{-1})$
“Where can the ROC be?”	Intersection of individual ROCs, never containing poles