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Ali Alyoussef
 
Ali Alyoussef
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Properties of ROC
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The existence of Laplace transform $X(s)$ of a given $x(t)$ depends on whether the transform integral converges
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\begin{displaymath}X(s)=\int_{-\infty}^\infty x(t)e^{-st} dt =\int_{-\infty}^\infty x(t)e^{-\sigma t} e^{-j\omega t} dt < \infty \end{displaymath}
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which in turn depends on the duration and magnitude of $x(t)$ as well as the real part of $s$ $Re[s]=\sigma$ (the imaginary part of $s$ $Im[s]=j\omega$ determines the frequency of a sinusoid which is bounded and has no effect on the convergence of the integral).
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Right sided signals: $x(t)=x(t)u(t-t_0)$ may have infinite duration for $t>0$, and a positive $\sigma>0$ tends to attenuate $x(t)e^{-\sigma t}$ as $t \rightarrow \infty$.
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Left sided signals: $x(t)=x(t)u(t_0-t)$ may have infinite duration for $t<0$, and a negative $\sigma<0$ tends to attenuate $x(t)e^{-\sigma t}$ as $t \rightarrow -\infty$.
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Based on these observations, we can get the following properties for the ROC:
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    * If $x(t)$ is absolutely integrable and of finite duration, then the ROC is the entire s-plane (the Laplace transform integral is finite, i.e., $X(s)$ exists, for any $s$).
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    * The ROC of $X(s)$ consists of strips parallel to the $j\omega$-axis in the s-plane.
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    * If $x(t)$ is right sided and $Re[s]=\sigma_0$ is in the ROC, then any $s$ to the right of $\sigma_0$ (i.e., $Re[s]>\sigma_0$) is also in the ROC, i.e., ROC is a right sided half plane.
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    * If $x(t)$ is left sided and $Re[s]=\sigma_0$ is in the ROC, then any $s$ to the left of $\sigma_0$ (i.e., $Re[s]<\sigma_0$) is also in the ROC, i.e., ROC is a left sided half plane.
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    * If $x(t)$ is two-sided, then the ROC is the intersection of the two one-sided ROCs corresponding to the two one-sided components of $x(t)$. This intersection can be either a vertical strip or an empty set.
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    * If $X(s)$ is rational, then its ROC does not contain any poles (by definition $X(s)\vert _{s=s_p}=\infty$ dose not exist). The ROC is bounded by the poles or extends to infinity.
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    * If $X(s)$ is a rational Laplace transform of a right sided function $x(t)$, then the ROC is the half plane to the right of the rightmost pole; if $X(s)$ is a rational Laplace transform of a left sided function $x(t)$, then the ROC is the half plane to the left of the leftmost pole.
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    * A signal $x(t)$ is absolutely integrable, i.e., its Fourier transform $X(j\omega)$ exists (first Dirichlet condition, assuming the other two are satisfied), if and only if the ROC of the corresponding Laplace transform $X(s)$ contains the imaginary axis $Re[s]=0$ or $s=j\omega$.

Latest revision as of 04:22, 30 July 2009

Ali Alyoussef

Properties of ROC

The existence of Laplace transform $X(s)$ of a given $x(t)$ depends on whether the transform integral converges

\begin{displaymath}X(s)=\int_{-\infty}^\infty x(t)e^{-st} dt =\int_{-\infty}^\infty x(t)e^{-\sigma t} e^{-j\omega t} dt < \infty \end{displaymath}

which in turn depends on the duration and magnitude of $x(t)$ as well as the real part of $s$ $Re[s]=\sigma$ (the imaginary part of $s$ $Im[s]=j\omega$ determines the frequency of a sinusoid which is bounded and has no effect on the convergence of the integral).

Right sided signals: $x(t)=x(t)u(t-t_0)$ may have infinite duration for $t>0$, and a positive $\sigma>0$ tends to attenuate $x(t)e^{-\sigma t}$ as $t \rightarrow \infty$.

Left sided signals: $x(t)=x(t)u(t_0-t)$ may have infinite duration for $t<0$, and a negative $\sigma<0$ tends to attenuate $x(t)e^{-\sigma t}$ as $t \rightarrow -\infty$.

Based on these observations, we can get the following properties for the ROC:

   * If $x(t)$ is absolutely integrable and of finite duration, then the ROC is the entire s-plane (the Laplace transform integral is finite, i.e., $X(s)$ exists, for any $s$).
   * The ROC of $X(s)$ consists of strips parallel to the $j\omega$-axis in the s-plane.
   * If $x(t)$ is right sided and $Re[s]=\sigma_0$ is in the ROC, then any $s$ to the right of $\sigma_0$ (i.e., $Re[s]>\sigma_0$) is also in the ROC, i.e., ROC is a right sided half plane.
   * If $x(t)$ is left sided and $Re[s]=\sigma_0$ is in the ROC, then any $s$ to the left of $\sigma_0$ (i.e., $Re[s]<\sigma_0$) is also in the ROC, i.e., ROC is a left sided half plane.
   * If $x(t)$ is two-sided, then the ROC is the intersection of the two one-sided ROCs corresponding to the two one-sided components of $x(t)$. This intersection can be either a vertical strip or an empty set.
   * If $X(s)$ is rational, then its ROC does not contain any poles (by definition $X(s)\vert _{s=s_p}=\infty$ dose not exist). The ROC is bounded by the poles or extends to infinity.
   * If $X(s)$ is a rational Laplace transform of a right sided function $x(t)$, then the ROC is the half plane to the right of the rightmost pole; if $X(s)$ is a rational Laplace transform of a left sided function $x(t)$, then the ROC is the half plane to the left of the leftmost pole.
   * A signal $x(t)$ is absolutely integrable, i.e., its Fourier transform $X(j\omega)$ exists (first Dirichlet condition, assuming the other two are satisfied), if and only if the ROC of the corresponding Laplace transform $X(s)$ contains the imaginary axis $Re[s]=0$ or $s=j\omega$.

Alumni Liaison

Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin