Meaning of double arrow $Leftrightarrow$ symbol in signal processing context












0












$begingroup$


What is the meaning of the double left-right arrow in the following context? This is from a signal processing paper, Alias-Free Digital Synthesis of Classic Analog Waveforms.




For a general discrete-time signal $x(nT_s)$ which has been
uniformly sampled at twice its highest frequency, exact
bandlimited interpolation, which we call sinc interpolation,
is carried out as
$$x(t) = sum_{n=-infty}^infty x(nT_s)h_s(tLeftrightarrow nT_s)$$
$$=frac{sin(pi F_st)}{pi F_s} sum_{n=-infty}^infty x(nT_s)frac{(Leftrightarrow 1)^n}{tLeftrightarrow nT_s}$$
where $$h_s(t) triangleq sinc(F_st) triangleq frac{sin(pi F_st)}{pi F_st}.$$




I have read in Wikipedia's list of mathematical symjbols, and in other questions that it means if and only if. However, this does not make sense to me here.



There is also an answer suggesting "the $pLeftrightarrow q$ symbol is just a strange font substitution for $p−q$", which I would indeed find very strange. Most of the answer is specific to a different context which I don't understand well enough to verify it.



I'm especially confused about its use in the second line: $(Leftrightarrow 1)^n$ where there isn't even a left-hand term.



This must be a specialised usage. What is it?



Edit



If it is being used as iff, could someone walk me through how to interpret that? eg. Could I unpack the first line of the equation as



$$sum_{n=-infty}^infty x(nT_s) begin{cases} h_s(t), & t = nT_s \ [...], & otherwise end{cases} $$



and what would go in place of $[...]$? How to then unpack the expression $(Leftrightarrow 1)^n$?






notation signal-processing







share|cite|improve this question











$endgroup$












  • $begingroup$
    If that is the case, how do I evaluate it when the condition is not satisfied? I've added a follow-up question
    $endgroup$
    – Igid
    Dec 2 '18 at 11:24










  • $begingroup$
    It's a typo for a '$-$' sign. The expression is a convolution.
    $endgroup$
    – Andy Walls
    Dec 2 '18 at 11:45










  • $begingroup$
    @AndyWalls As in minus??
    $endgroup$
    – Igid
    Dec 2 '18 at 11:48
















0












$begingroup$


What is the meaning of the double left-right arrow in the following context? This is from a signal processing paper, Alias-Free Digital Synthesis of Classic Analog Waveforms.




For a general discrete-time signal $x(nT_s)$ which has been
uniformly sampled at twice its highest frequency, exact
bandlimited interpolation, which we call sinc interpolation,
is carried out as
$$x(t) = sum_{n=-infty}^infty x(nT_s)h_s(tLeftrightarrow nT_s)$$
$$=frac{sin(pi F_st)}{pi F_s} sum_{n=-infty}^infty x(nT_s)frac{(Leftrightarrow 1)^n}{tLeftrightarrow nT_s}$$
where $$h_s(t) triangleq sinc(F_st) triangleq frac{sin(pi F_st)}{pi F_st}.$$




I have read in Wikipedia's list of mathematical symjbols, and in other questions that it means if and only if. However, this does not make sense to me here.



There is also an answer suggesting "the $pLeftrightarrow q$ symbol is just a strange font substitution for $p−q$", which I would indeed find very strange. Most of the answer is specific to a different context which I don't understand well enough to verify it.



I'm especially confused about its use in the second line: $(Leftrightarrow 1)^n$ where there isn't even a left-hand term.



This must be a specialised usage. What is it?



Edit



If it is being used as iff, could someone walk me through how to interpret that? eg. Could I unpack the first line of the equation as



$$sum_{n=-infty}^infty x(nT_s) begin{cases} h_s(t), & t = nT_s \ [...], & otherwise end{cases} $$



and what would go in place of $[...]$? How to then unpack the expression $(Leftrightarrow 1)^n$?






notation signal-processing







share|cite|improve this question











$endgroup$












  • $begingroup$
    If that is the case, how do I evaluate it when the condition is not satisfied? I've added a follow-up question
    $endgroup$
    – Igid
    Dec 2 '18 at 11:24










  • $begingroup$
    It's a typo for a '$-$' sign. The expression is a convolution.
    $endgroup$
    – Andy Walls
    Dec 2 '18 at 11:45










  • $begingroup$
    @AndyWalls As in minus??
    $endgroup$
    – Igid
    Dec 2 '18 at 11:48














0












0








0





$begingroup$


What is the meaning of the double left-right arrow in the following context? This is from a signal processing paper, Alias-Free Digital Synthesis of Classic Analog Waveforms.




For a general discrete-time signal $x(nT_s)$ which has been
uniformly sampled at twice its highest frequency, exact
bandlimited interpolation, which we call sinc interpolation,
is carried out as
$$x(t) = sum_{n=-infty}^infty x(nT_s)h_s(tLeftrightarrow nT_s)$$
$$=frac{sin(pi F_st)}{pi F_s} sum_{n=-infty}^infty x(nT_s)frac{(Leftrightarrow 1)^n}{tLeftrightarrow nT_s}$$
where $$h_s(t) triangleq sinc(F_st) triangleq frac{sin(pi F_st)}{pi F_st}.$$




I have read in Wikipedia's list of mathematical symjbols, and in other questions that it means if and only if. However, this does not make sense to me here.



There is also an answer suggesting "the $pLeftrightarrow q$ symbol is just a strange font substitution for $p−q$", which I would indeed find very strange. Most of the answer is specific to a different context which I don't understand well enough to verify it.



I'm especially confused about its use in the second line: $(Leftrightarrow 1)^n$ where there isn't even a left-hand term.



This must be a specialised usage. What is it?



Edit



If it is being used as iff, could someone walk me through how to interpret that? eg. Could I unpack the first line of the equation as



$$sum_{n=-infty}^infty x(nT_s) begin{cases} h_s(t), & t = nT_s \ [...], & otherwise end{cases} $$



and what would go in place of $[...]$? How to then unpack the expression $(Leftrightarrow 1)^n$?






notation signal-processing







share|cite|improve this question











$endgroup$




What is the meaning of the double left-right arrow in the following context? This is from a signal processing paper, Alias-Free Digital Synthesis of Classic Analog Waveforms.




For a general discrete-time signal $x(nT_s)$ which has been
uniformly sampled at twice its highest frequency, exact
bandlimited interpolation, which we call sinc interpolation,
is carried out as
$$x(t) = sum_{n=-infty}^infty x(nT_s)h_s(tLeftrightarrow nT_s)$$
$$=frac{sin(pi F_st)}{pi F_s} sum_{n=-infty}^infty x(nT_s)frac{(Leftrightarrow 1)^n}{tLeftrightarrow nT_s}$$
where $$h_s(t) triangleq sinc(F_st) triangleq frac{sin(pi F_st)}{pi F_st}.$$




I have read in Wikipedia's list of mathematical symjbols, and in other questions that it means if and only if. However, this does not make sense to me here.



There is also an answer suggesting "the $pLeftrightarrow q$ symbol is just a strange font substitution for $p−q$", which I would indeed find very strange. Most of the answer is specific to a different context which I don't understand well enough to verify it.



I'm especially confused about its use in the second line: $(Leftrightarrow 1)^n$ where there isn't even a left-hand term.



This must be a specialised usage. What is it?



Edit



If it is being used as iff, could someone walk me through how to interpret that? eg. Could I unpack the first line of the equation as



$$sum_{n=-infty}^infty x(nT_s) begin{cases} h_s(t), & t = nT_s \ [...], & otherwise end{cases} $$



and what would go in place of $[...]$? How to then unpack the expression $(Leftrightarrow 1)^n$?






notation signal-processing




notation signal-processing






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 11:40







Igid

















asked Dec 2 '18 at 11:00









IgidIgid

1032




1032












  • $begingroup$
    If that is the case, how do I evaluate it when the condition is not satisfied? I've added a follow-up question
    $endgroup$
    – Igid
    Dec 2 '18 at 11:24










  • $begingroup$
    It's a typo for a '$-$' sign. The expression is a convolution.
    $endgroup$
    – Andy Walls
    Dec 2 '18 at 11:45










  • $begingroup$
    @AndyWalls As in minus??
    $endgroup$
    – Igid
    Dec 2 '18 at 11:48


















  • $begingroup$
    If that is the case, how do I evaluate it when the condition is not satisfied? I've added a follow-up question
    $endgroup$
    – Igid
    Dec 2 '18 at 11:24










  • $begingroup$
    It's a typo for a '$-$' sign. The expression is a convolution.
    $endgroup$
    – Andy Walls
    Dec 2 '18 at 11:45










  • $begingroup$
    @AndyWalls As in minus??
    $endgroup$
    – Igid
    Dec 2 '18 at 11:48
















$begingroup$
If that is the case, how do I evaluate it when the condition is not satisfied? I've added a follow-up question
$endgroup$
– Igid
Dec 2 '18 at 11:24




$begingroup$
If that is the case, how do I evaluate it when the condition is not satisfied? I've added a follow-up question
$endgroup$
– Igid
Dec 2 '18 at 11:24












$begingroup$
It's a typo for a '$-$' sign. The expression is a convolution.
$endgroup$
– Andy Walls
Dec 2 '18 at 11:45




$begingroup$
It's a typo for a '$-$' sign. The expression is a convolution.
$endgroup$
– Andy Walls
Dec 2 '18 at 11:45












$begingroup$
@AndyWalls As in minus??
$endgroup$
– Igid
Dec 2 '18 at 11:48




$begingroup$
@AndyWalls As in minus??
$endgroup$
– Igid
Dec 2 '18 at 11:48










1 Answer
1






active

oldest

votes


















0












$begingroup$

It's a '$-$' sign.



$sin(pi F_s [t - nT_s]) =sin(pi F_s t)cos(pi F_s nT_s) - cos(pi F_s t)sin(pi F_s nT_s) = sin(pi F_s t)(-1)^n -cos(pi F_s t)(0)$



since $F_sT_s =1$ by definition.



The overall expression is a convolution.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you, that's a very helpful expansion. But why on earth would someone print that for a minus sign? That's just pure obfuscation.
    $endgroup$
    – Igid
    Dec 2 '18 at 12:21












  • $begingroup$
    Blame computers and software and editors. That's not something a signal processing expert did.
    $endgroup$
    – Andy Walls
    Dec 2 '18 at 12:56










  • $begingroup$
    Fair enough I suppose. Now I’m really curious to know how that confusion could arise. Surely if you’re marking up in LaTeX there’s no proximity between the two.
    $endgroup$
    – Igid
    Dec 2 '18 at 17:11











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1 Answer
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1 Answer
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active

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oldest

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0












$begingroup$

It's a '$-$' sign.



$sin(pi F_s [t - nT_s]) =sin(pi F_s t)cos(pi F_s nT_s) - cos(pi F_s t)sin(pi F_s nT_s) = sin(pi F_s t)(-1)^n -cos(pi F_s t)(0)$



since $F_sT_s =1$ by definition.



The overall expression is a convolution.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you, that's a very helpful expansion. But why on earth would someone print that for a minus sign? That's just pure obfuscation.
    $endgroup$
    – Igid
    Dec 2 '18 at 12:21












  • $begingroup$
    Blame computers and software and editors. That's not something a signal processing expert did.
    $endgroup$
    – Andy Walls
    Dec 2 '18 at 12:56










  • $begingroup$
    Fair enough I suppose. Now I’m really curious to know how that confusion could arise. Surely if you’re marking up in LaTeX there’s no proximity between the two.
    $endgroup$
    – Igid
    Dec 2 '18 at 17:11
















0












$begingroup$

It's a '$-$' sign.



$sin(pi F_s [t - nT_s]) =sin(pi F_s t)cos(pi F_s nT_s) - cos(pi F_s t)sin(pi F_s nT_s) = sin(pi F_s t)(-1)^n -cos(pi F_s t)(0)$



since $F_sT_s =1$ by definition.



The overall expression is a convolution.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you, that's a very helpful expansion. But why on earth would someone print that for a minus sign? That's just pure obfuscation.
    $endgroup$
    – Igid
    Dec 2 '18 at 12:21












  • $begingroup$
    Blame computers and software and editors. That's not something a signal processing expert did.
    $endgroup$
    – Andy Walls
    Dec 2 '18 at 12:56










  • $begingroup$
    Fair enough I suppose. Now I’m really curious to know how that confusion could arise. Surely if you’re marking up in LaTeX there’s no proximity between the two.
    $endgroup$
    – Igid
    Dec 2 '18 at 17:11














0












0








0





$begingroup$

It's a '$-$' sign.



$sin(pi F_s [t - nT_s]) =sin(pi F_s t)cos(pi F_s nT_s) - cos(pi F_s t)sin(pi F_s nT_s) = sin(pi F_s t)(-1)^n -cos(pi F_s t)(0)$



since $F_sT_s =1$ by definition.



The overall expression is a convolution.






share|cite|improve this answer











$endgroup$



It's a '$-$' sign.



$sin(pi F_s [t - nT_s]) =sin(pi F_s t)cos(pi F_s nT_s) - cos(pi F_s t)sin(pi F_s nT_s) = sin(pi F_s t)(-1)^n -cos(pi F_s t)(0)$



since $F_sT_s =1$ by definition.



The overall expression is a convolution.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 2 '18 at 12:10

























answered Dec 2 '18 at 11:56









Andy WallsAndy Walls

1,724139




1,724139












  • $begingroup$
    Thank you, that's a very helpful expansion. But why on earth would someone print that for a minus sign? That's just pure obfuscation.
    $endgroup$
    – Igid
    Dec 2 '18 at 12:21












  • $begingroup$
    Blame computers and software and editors. That's not something a signal processing expert did.
    $endgroup$
    – Andy Walls
    Dec 2 '18 at 12:56










  • $begingroup$
    Fair enough I suppose. Now I’m really curious to know how that confusion could arise. Surely if you’re marking up in LaTeX there’s no proximity between the two.
    $endgroup$
    – Igid
    Dec 2 '18 at 17:11


















  • $begingroup$
    Thank you, that's a very helpful expansion. But why on earth would someone print that for a minus sign? That's just pure obfuscation.
    $endgroup$
    – Igid
    Dec 2 '18 at 12:21












  • $begingroup$
    Blame computers and software and editors. That's not something a signal processing expert did.
    $endgroup$
    – Andy Walls
    Dec 2 '18 at 12:56










  • $begingroup$
    Fair enough I suppose. Now I’m really curious to know how that confusion could arise. Surely if you’re marking up in LaTeX there’s no proximity between the two.
    $endgroup$
    – Igid
    Dec 2 '18 at 17:11
















$begingroup$
Thank you, that's a very helpful expansion. But why on earth would someone print that for a minus sign? That's just pure obfuscation.
$endgroup$
– Igid
Dec 2 '18 at 12:21






$begingroup$
Thank you, that's a very helpful expansion. But why on earth would someone print that for a minus sign? That's just pure obfuscation.
$endgroup$
– Igid
Dec 2 '18 at 12:21














$begingroup$
Blame computers and software and editors. That's not something a signal processing expert did.
$endgroup$
– Andy Walls
Dec 2 '18 at 12:56




$begingroup$
Blame computers and software and editors. That's not something a signal processing expert did.
$endgroup$
– Andy Walls
Dec 2 '18 at 12:56












$begingroup$
Fair enough I suppose. Now I’m really curious to know how that confusion could arise. Surely if you’re marking up in LaTeX there’s no proximity between the two.
$endgroup$
– Igid
Dec 2 '18 at 17:11




$begingroup$
Fair enough I suppose. Now I’m really curious to know how that confusion could arise. Surely if you’re marking up in LaTeX there’s no proximity between the two.
$endgroup$
– Igid
Dec 2 '18 at 17:11


















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