Why there is Difference between shapes of ROC of z domain and s domain?How does the $mathcal Z$-transform's “region of convergence” work?Inverse Laplace transform Using Inversion FormulaHow does the ROC (Region of Convergence) related to a real world application?How to identify causality, stability and ROC from the pole-zero plot?Inverse $mathcal Z$-transform when region of convergence goes outwards from the inner pole?Design a filter which passes all frequencies except $omega=pmfracpi2$ and plot its pole-zero diagramZ transform stabilityCan a Fourier Transform exist even if the j$omega$ axis is not in the Region of Convergence in it's Laplace TransformROC of inverse system functionPotential issues arising from too stable discretization
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Why there is Difference between shapes of ROC of z domain and s domain?
How does the $mathcal Z$-transform's “region of convergence” work?Inverse Laplace transform Using Inversion FormulaHow does the ROC (Region of Convergence) related to a real world application?How to identify causality, stability and ROC from the pole-zero plot?Inverse $mathcal Z$-transform when region of convergence goes outwards from the inner pole?Design a filter which passes all frequencies except $omega=pmfracpi2$ and plot its pole-zero diagramZ transform stabilityCan a Fourier Transform exist even if the j$omega$ axis is not in the Region of Convergence in it's Laplace TransformROC of inverse system functionPotential issues arising from too stable discretization
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$begingroup$
ROC(region of convergence) of Z domain is shown by a circular region while ROC in S domain is shown by a rectangular(approximately looking like rectangle) region
What is the reason of this difference in shapes of ROC regions?
z-transform laplace-transform
asked Jul 9 at 11:55
studentstudent
445 bronze badges
$endgroup$
add a comment |
$begingroup$
ROC(region of convergence) of Z domain is shown by a circular region while ROC in S domain is shown by a rectangular(approximately looking like rectangle) region
What is the reason of this difference in shapes of ROC regions?
z-transform laplace-transform
asked Jul 9 at 11:55
studentstudent
445 bronze badges
$endgroup$
add a comment |
$begingroup$
ROC(region of convergence) of Z domain is shown by a circular region while ROC in S domain is shown by a rectangular(approximately looking like rectangle) region
What is the reason of this difference in shapes of ROC regions?
z-transform laplace-transform
asked Jul 9 at 11:55
studentstudent
445 bronze badges
$endgroup$
ROC(region of convergence) of Z domain is shown by a circular region while ROC in S domain is shown by a rectangular(approximately looking like rectangle) region
What is the reason of this difference in shapes of ROC regions?
z-transform laplace-transform
z-transform laplace-transform
asked Jul 9 at 11:55
studentstudent
445 bronze badges
asked Jul 9 at 11:55
studentstudent
445 bronze badges
asked Jul 9 at 11:55
studentstudent
445 bronze badges
asked Jul 9 at 11:55
studentstudent
445 bronze badges
asked Jul 9 at 11:55
studentstudent
445 bronze badges
445 bronze badges
add a comment |
add a comment |
1 Answer
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$begingroup$
Because, the region of convergence in the Laplace transform
$$
X(s) = int_-infty^infty x(t) e^-st dt
$$
is related to the weighting provided by the real part of the complex $s = sigma + j omega$; as this will yield the weight $|e^-st| = e^-sigma t$ applied on the input signal $x(t)$, and is a function of $sigma$ alone and is a rectangular (planar) region on the s-plane.
But the region of convergence in the Z-transform
$$
X(z) = sum_n=-infty^infty x[n] z^-n
$$
is related to the weighting provided by the magnitude of the complex $z = sigma + j omega$ as given by $|z|^-n = |sigma + jomega|^-n = |z|^-n |e^-j n anglez| = |z|^-n$, which is a circular region on th z-plane, due to the magnitude of $z$ being involved.
answered Jul 9 at 12:27
Fat32Fat32
16.8k3 gold badges12 silver badges33 bronze badges
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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oldest
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$begingroup$
Because, the region of convergence in the Laplace transform
$$
X(s) = int_-infty^infty x(t) e^-st dt
$$
is related to the weighting provided by the real part of the complex $s = sigma + j omega$; as this will yield the weight $|e^-st| = e^-sigma t$ applied on the input signal $x(t)$, and is a function of $sigma$ alone and is a rectangular (planar) region on the s-plane.
But the region of convergence in the Z-transform
$$
X(z) = sum_n=-infty^infty x[n] z^-n
$$
is related to the weighting provided by the magnitude of the complex $z = sigma + j omega$ as given by $|z|^-n = |sigma + jomega|^-n = |z|^-n |e^-j n anglez| = |z|^-n$, which is a circular region on th z-plane, due to the magnitude of $z$ being involved.
answered Jul 9 at 12:27
Fat32Fat32
16.8k3 gold badges12 silver badges33 bronze badges
$endgroup$
add a comment |
$begingroup$
Because, the region of convergence in the Laplace transform
$$
X(s) = int_-infty^infty x(t) e^-st dt
$$
is related to the weighting provided by the real part of the complex $s = sigma + j omega$; as this will yield the weight $|e^-st| = e^-sigma t$ applied on the input signal $x(t)$, and is a function of $sigma$ alone and is a rectangular (planar) region on the s-plane.
But the region of convergence in the Z-transform
$$
X(z) = sum_n=-infty^infty x[n] z^-n
$$
is related to the weighting provided by the magnitude of the complex $z = sigma + j omega$ as given by $|z|^-n = |sigma + jomega|^-n = |z|^-n |e^-j n anglez| = |z|^-n$, which is a circular region on th z-plane, due to the magnitude of $z$ being involved.
answered Jul 9 at 12:27
Fat32Fat32
16.8k3 gold badges12 silver badges33 bronze badges
$endgroup$
add a comment |
$begingroup$
Because, the region of convergence in the Laplace transform
$$
X(s) = int_-infty^infty x(t) e^-st dt
$$
is related to the weighting provided by the real part of the complex $s = sigma + j omega$; as this will yield the weight $|e^-st| = e^-sigma t$ applied on the input signal $x(t)$, and is a function of $sigma$ alone and is a rectangular (planar) region on the s-plane.
But the region of convergence in the Z-transform
$$
X(z) = sum_n=-infty^infty x[n] z^-n
$$
is related to the weighting provided by the magnitude of the complex $z = sigma + j omega$ as given by $|z|^-n = |sigma + jomega|^-n = |z|^-n |e^-j n anglez| = |z|^-n$, which is a circular region on th z-plane, due to the magnitude of $z$ being involved.
answered Jul 9 at 12:27
Fat32Fat32
16.8k3 gold badges12 silver badges33 bronze badges
$endgroup$
Because, the region of convergence in the Laplace transform
$$
X(s) = int_-infty^infty x(t) e^-st dt
$$
is related to the weighting provided by the real part of the complex $s = sigma + j omega$; as this will yield the weight $|e^-st| = e^-sigma t$ applied on the input signal $x(t)$, and is a function of $sigma$ alone and is a rectangular (planar) region on the s-plane.
But the region of convergence in the Z-transform
$$
X(z) = sum_n=-infty^infty x[n] z^-n
$$
is related to the weighting provided by the magnitude of the complex $z = sigma + j omega$ as given by $|z|^-n = |sigma + jomega|^-n = |z|^-n |e^-j n anglez| = |z|^-n$, which is a circular region on th z-plane, due to the magnitude of $z$ being involved.
answered Jul 9 at 12:27
Fat32Fat32
16.8k3 gold badges12 silver badges33 bronze badges
answered Jul 9 at 12:27
Fat32Fat32
16.8k3 gold badges12 silver badges33 bronze badges
answered Jul 9 at 12:27
Fat32Fat32
16.8k3 gold badges12 silver badges33 bronze badges
answered Jul 9 at 12:27
Fat32Fat32
16.8k3 gold badges12 silver badges33 bronze badges
16.8k3 gold badges12 silver badges33 bronze badges
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