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Bounds of the derivative of a bounded band-limited function
Clash Royale CLAN TAG#URR8PPP
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Let $f(t)$ be a function with properties:
$$beginarrayll
tinmathbfR&ttext is in reals\
f(t)inmathbfRtext for all t&f(t)text is in reals\
|f(t)|<Atext for all t&textabsolute value of f(t)text is bounded above by A\
int_-infty^infty f(t) e^- i omega t rm dt = 0text for all |omega|ge B&f(t)text is band-limited by frequency B in radians
endarray$$
Given $A$ and $B,$ what is the tight upper bound for $|f'(t)|,$ the absolute value of the derivative of the function?
Nothing else shall be assumed about $f(t)$ than what has been stated above. The bound should accommodate for this uncertainty.
For a sinusoid of amplitude $A$ and frequency $B,$ the maximum absolute value of the derivative is $AB.$ I wonder if this is an upper bound, and in that case also the tight upper bound. Or maybe a non-sinusoidal function has a steeper slope.
bandwidth derivative
asked Aug 30 at 7:22
Olli Niemitalo
7,0831233
add a comment |Â
up vote
7
down vote
favorite
Let $f(t)$ be a function with properties:
$$beginarrayll
tinmathbfR&ttext is in reals\
f(t)inmathbfRtext for all t&f(t)text is in reals\
|f(t)|<Atext for all t&textabsolute value of f(t)text is bounded above by A\
int_-infty^infty f(t) e^- i omega t rm dt = 0text for all |omega|ge B&f(t)text is band-limited by frequency B in radians
endarray$$
Given $A$ and $B,$ what is the tight upper bound for $|f'(t)|,$ the absolute value of the derivative of the function?
Nothing else shall be assumed about $f(t)$ than what has been stated above. The bound should accommodate for this uncertainty.
For a sinusoid of amplitude $A$ and frequency $B,$ the maximum absolute value of the derivative is $AB.$ I wonder if this is an upper bound, and in that case also the tight upper bound. Or maybe a non-sinusoidal function has a steeper slope.
bandwidth derivative
asked Aug 30 at 7:22
Olli Niemitalo
7,0831233
Have you checked this?
– Tendero
Aug 30 at 13:24
@Tendero thanks. There, the signal energy is known, rather than the peak absolute value as in my question.
– Olli Niemitalo
Aug 30 at 13:39
1
See my answer for the bound you seek. It says More generally, a result due to Bernstein says that if the maximum frequency in a generic $x(t)$ bounded within $[-1,1]$ is $f_0$, that is, $X(f) = 0$ for $|f| > f_0$, then $$max left| fracmathrm dxmathrm dtright| leq 2pi f_0.$$
– Dilip Sarwate
Aug 30 at 14:50
1
Based on the sharp version of Bernstein's inequality, from Dilip's linked answers, MBaz's edited answer and the literature cited, $AB$ is indeed the sharp (I called it tight meaning the same) upper bound for the maximum absolute value of the derivative, a full-scale sinusoid at exactly the band limit (not strictly allowed by the constraints I give) making the inequality an equality.
– Olli Niemitalo
Aug 31 at 7:02
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Let $f(t)$ be a function with properties:
$$beginarrayll
tinmathbfR&ttext is in reals\
f(t)inmathbfRtext for all t&f(t)text is in reals\
|f(t)|<Atext for all t&textabsolute value of f(t)text is bounded above by A\
int_-infty^infty f(t) e^- i omega t rm dt = 0text for all |omega|ge B&f(t)text is band-limited by frequency B in radians
endarray$$
Given $A$ and $B,$ what is the tight upper bound for $|f'(t)|,$ the absolute value of the derivative of the function?
Nothing else shall be assumed about $f(t)$ than what has been stated above. The bound should accommodate for this uncertainty.
For a sinusoid of amplitude $A$ and frequency $B,$ the maximum absolute value of the derivative is $AB.$ I wonder if this is an upper bound, and in that case also the tight upper bound. Or maybe a non-sinusoidal function has a steeper slope.
bandwidth derivative
asked Aug 30 at 7:22
Olli Niemitalo
7,0831233
Let $f(t)$ be a function with properties:
$$beginarrayll
tinmathbfR&ttext is in reals\
f(t)inmathbfRtext for all t&f(t)text is in reals\
|f(t)|<Atext for all t&textabsolute value of f(t)text is bounded above by A\
int_-infty^infty f(t) e^- i omega t rm dt = 0text for all |omega|ge B&f(t)text is band-limited by frequency B in radians
endarray$$
Given $A$ and $B,$ what is the tight upper bound for $|f'(t)|,$ the absolute value of the derivative of the function?
Nothing else shall be assumed about $f(t)$ than what has been stated above. The bound should accommodate for this uncertainty.
For a sinusoid of amplitude $A$ and frequency $B,$ the maximum absolute value of the derivative is $AB.$ I wonder if this is an upper bound, and in that case also the tight upper bound. Or maybe a non-sinusoidal function has a steeper slope.
bandwidth derivative
asked Aug 30 at 7:22
Olli Niemitalo
7,0831233
edited Aug 30 at 14:32
asked Aug 30 at 7:22
Olli Niemitalo
7,0831233
asked Aug 30 at 7:22
Olli Niemitalo
7,0831233
asked Aug 30 at 7:22
Olli Niemitalo
7,0831233
7,0831233
Have you checked this?
– Tendero
Aug 30 at 13:24
@Tendero thanks. There, the signal energy is known, rather than the peak absolute value as in my question.
– Olli Niemitalo
Aug 30 at 13:39
1
See my answer for the bound you seek. It says More generally, a result due to Bernstein says that if the maximum frequency in a generic $x(t)$ bounded within $[-1,1]$ is $f_0$, that is, $X(f) = 0$ for $|f| > f_0$, then $$max left| fracmathrm dxmathrm dtright| leq 2pi f_0.$$
– Dilip Sarwate
Aug 30 at 14:50
1
Based on the sharp version of Bernstein's inequality, from Dilip's linked answers, MBaz's edited answer and the literature cited, $AB$ is indeed the sharp (I called it tight meaning the same) upper bound for the maximum absolute value of the derivative, a full-scale sinusoid at exactly the band limit (not strictly allowed by the constraints I give) making the inequality an equality.
– Olli Niemitalo
Aug 31 at 7:02
add a comment |Â
Have you checked this?
– Tendero
Aug 30 at 13:24
@Tendero thanks. There, the signal energy is known, rather than the peak absolute value as in my question.
– Olli Niemitalo
Aug 30 at 13:39
1
See my answer for the bound you seek. It says More generally, a result due to Bernstein says that if the maximum frequency in a generic $x(t)$ bounded within $[-1,1]$ is $f_0$, that is, $X(f) = 0$ for $|f| > f_0$, then $$max left| fracmathrm dxmathrm dtright| leq 2pi f_0.$$
– Dilip Sarwate
Aug 30 at 14:50
1
Based on the sharp version of Bernstein's inequality, from Dilip's linked answers, MBaz's edited answer and the literature cited, $AB$ is indeed the sharp (I called it tight meaning the same) upper bound for the maximum absolute value of the derivative, a full-scale sinusoid at exactly the band limit (not strictly allowed by the constraints I give) making the inequality an equality.
– Olli Niemitalo
Aug 31 at 7:02
Have you checked this?
– Tendero
Aug 30 at 13:24
Have you checked this?
– Tendero
Aug 30 at 13:24
@Tendero thanks. There, the signal energy is known, rather than the peak absolute value as in my question.
– Olli Niemitalo
Aug 30 at 13:39
@Tendero thanks. There, the signal energy is known, rather than the peak absolute value as in my question.
– Olli Niemitalo
Aug 30 at 13:39
1
1
See my answer for the bound you seek. It says More generally, a result due to Bernstein says that if the maximum frequency in a generic $x(t)$ bounded within $[-1,1]$ is $f_0$, that is, $X(f) = 0$ for $|f| > f_0$, then $$max left| fracmathrm dxmathrm dtright| leq 2pi f_0.$$
– Dilip Sarwate
Aug 30 at 14:50
See my answer for the bound you seek. It says More generally, a result due to Bernstein says that if the maximum frequency in a generic $x(t)$ bounded within $[-1,1]$ is $f_0$, that is, $X(f) = 0$ for $|f| > f_0$, then $$max left| fracmathrm dxmathrm dtright| leq 2pi f_0.$$
– Dilip Sarwate
Aug 30 at 14:50
1
1
Based on the sharp version of Bernstein's inequality, from Dilip's linked answers, MBaz's edited answer and the literature cited, $AB$ is indeed the sharp (I called it tight meaning the same) upper bound for the maximum absolute value of the derivative, a full-scale sinusoid at exactly the band limit (not strictly allowed by the constraints I give) making the inequality an equality.
– Olli Niemitalo
Aug 31 at 7:02
Based on the sharp version of Bernstein's inequality, from Dilip's linked answers, MBaz's edited answer and the literature cited, $AB$ is indeed the sharp (I called it tight meaning the same) upper bound for the maximum absolute value of the derivative, a full-scale sinusoid at exactly the band limit (not strictly allowed by the constraints I give) making the inequality an equality.
– Olli Niemitalo
Aug 31 at 7:02
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
You'll be interested in Bernstein's inequality, which I first learned about in Lapidoth, A Foundation in Digital Communication (page 92).
With a well-behaved signal $f(t)$ as you defined it above (in particular, $f(t)$ is integrable and bandlimited to $B,textHz$, and $textsup,|f(t)| = A$), then $$left|fractextdf(t)textdtright| leq 2ABpi.
$$
Note that the original result by Bernstein established a bound of $4ABpi$; later, that bound was tightened to $2ABpi$.
answered Aug 30 at 13:06
MBaz
8,25241432
1
@OlliNiemitalo As pointed out in MattL's answer, the sinusoid $sin(2pi Bt)$ has maximum derivative $2pi B$. This meets Bernstein's bound, as stated in my answer here on dsp.SE (cited in a comment on your question) and in my answer on math.SE that you found, with equality.
– Dilip Sarwate
Aug 30 at 15:11
1
@OlliNiemitalo I found the proof given by Pinksy here (I hope that link works!). He definitely uses $4ABpi$ as the bound, not $2ABpi$.
– MBaz
Aug 30 at 15:46
2
@MBaz Your link works indeed! At the end of the section 2.3.8 they say that the best known version of Bernstein's inequality has the factor 2 instead of 4, which is sharp, and that for details consult Zygmund (1959) Vol. 2, p. 276. I think that's Zygmund, A. Trigonometric series. 2nd ed. Vol. II. Cambridge University Press, New York 1959.
– Olli Niemitalo
Aug 30 at 19:14
2
RP Boas, Some theorems on Fourier transforms and conjugate trigonometric integrals, Transactions of the American Mathematical Society 40 (2), 287-308, 1936 cites the relevant articles by Bernstein, Szegö, and Zygmund, already with the sharp bound, as far as I can tell.
– Olli Niemitalo
Aug 30 at 20:26
2
@OlliNiemitalo Excellent! I had missed that note at the end of section 2.3.8. I'll update my answer. Also: that book by Zygmund is in my university's library, but it's not online. I'll take it out tomorrow and see what it says.
– MBaz
Aug 30 at 22:15
 |Â
show 4 more comments
up vote
2
down vote
In general you would get something like this, but it might not be tight:
$$beginalign|f'(t)|&=left|frac12piint_-infty^inftyjomega F(jomega)e^-jomega tdomegaright|\&le frac12piint_-infty^infty|omega||F(jomega)|domega\&=frac12piint_-omega_c^omega_c|omega||F(jomega)|domega\&le fracomega_c2piint_-omega_c^omega_c|F(jomega)|domega\&=fracomega_cpiint_0^omega_c|F(jomega)|domegatag1endalign$$
The upper bound on $|f(t)|$ is of course implicit in $|F(jomega)|$.
For a sinusoid $Asin(omega_ct)$, $(1)$ gives $Aomega_c$ as an upper bound, as expected.
answered Aug 30 at 12:51
Matt L.
45.2k13678
@Olli Niemitalo , I had derived the sinusoid case I think this is the general case we were looking at. Thanks Matt L.
– MimSaad
Aug 30 at 18:22
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
You'll be interested in Bernstein's inequality, which I first learned about in Lapidoth, A Foundation in Digital Communication (page 92).
With a well-behaved signal $f(t)$ as you defined it above (in particular, $f(t)$ is integrable and bandlimited to $B,textHz$, and $textsup,|f(t)| = A$), then $$left|fractextdf(t)textdtright| leq 2ABpi.
$$
Note that the original result by Bernstein established a bound of $4ABpi$; later, that bound was tightened to $2ABpi$.
answered Aug 30 at 13:06
MBaz
8,25241432
1
@OlliNiemitalo As pointed out in MattL's answer, the sinusoid $sin(2pi Bt)$ has maximum derivative $2pi B$. This meets Bernstein's bound, as stated in my answer here on dsp.SE (cited in a comment on your question) and in my answer on math.SE that you found, with equality.
– Dilip Sarwate
Aug 30 at 15:11
1
@OlliNiemitalo I found the proof given by Pinksy here (I hope that link works!). He definitely uses $4ABpi$ as the bound, not $2ABpi$.
– MBaz
Aug 30 at 15:46
2
@MBaz Your link works indeed! At the end of the section 2.3.8 they say that the best known version of Bernstein's inequality has the factor 2 instead of 4, which is sharp, and that for details consult Zygmund (1959) Vol. 2, p. 276. I think that's Zygmund, A. Trigonometric series. 2nd ed. Vol. II. Cambridge University Press, New York 1959.
– Olli Niemitalo
Aug 30 at 19:14
2
RP Boas, Some theorems on Fourier transforms and conjugate trigonometric integrals, Transactions of the American Mathematical Society 40 (2), 287-308, 1936 cites the relevant articles by Bernstein, Szegö, and Zygmund, already with the sharp bound, as far as I can tell.
– Olli Niemitalo
Aug 30 at 20:26
2
@OlliNiemitalo Excellent! I had missed that note at the end of section 2.3.8. I'll update my answer. Also: that book by Zygmund is in my university's library, but it's not online. I'll take it out tomorrow and see what it says.
– MBaz
Aug 30 at 22:15
 |Â
show 4 more comments
up vote
4
down vote
accepted
You'll be interested in Bernstein's inequality, which I first learned about in Lapidoth, A Foundation in Digital Communication (page 92).
With a well-behaved signal $f(t)$ as you defined it above (in particular, $f(t)$ is integrable and bandlimited to $B,textHz$, and $textsup,|f(t)| = A$), then $$left|fractextdf(t)textdtright| leq 2ABpi.
$$
Note that the original result by Bernstein established a bound of $4ABpi$; later, that bound was tightened to $2ABpi$.
answered Aug 30 at 13:06
MBaz
8,25241432
1
@OlliNiemitalo As pointed out in MattL's answer, the sinusoid $sin(2pi Bt)$ has maximum derivative $2pi B$. This meets Bernstein's bound, as stated in my answer here on dsp.SE (cited in a comment on your question) and in my answer on math.SE that you found, with equality.
– Dilip Sarwate
Aug 30 at 15:11
1
@OlliNiemitalo I found the proof given by Pinksy here (I hope that link works!). He definitely uses $4ABpi$ as the bound, not $2ABpi$.
– MBaz
Aug 30 at 15:46
2
@MBaz Your link works indeed! At the end of the section 2.3.8 they say that the best known version of Bernstein's inequality has the factor 2 instead of 4, which is sharp, and that for details consult Zygmund (1959) Vol. 2, p. 276. I think that's Zygmund, A. Trigonometric series. 2nd ed. Vol. II. Cambridge University Press, New York 1959.
– Olli Niemitalo
Aug 30 at 19:14
2
RP Boas, Some theorems on Fourier transforms and conjugate trigonometric integrals, Transactions of the American Mathematical Society 40 (2), 287-308, 1936 cites the relevant articles by Bernstein, Szegö, and Zygmund, already with the sharp bound, as far as I can tell.
– Olli Niemitalo
Aug 30 at 20:26
2
@OlliNiemitalo Excellent! I had missed that note at the end of section 2.3.8. I'll update my answer. Also: that book by Zygmund is in my university's library, but it's not online. I'll take it out tomorrow and see what it says.
– MBaz
Aug 30 at 22:15
 |Â
show 4 more comments
up vote
4
down vote
accepted
up vote
4
down vote
accepted
You'll be interested in Bernstein's inequality, which I first learned about in Lapidoth, A Foundation in Digital Communication (page 92).
With a well-behaved signal $f(t)$ as you defined it above (in particular, $f(t)$ is integrable and bandlimited to $B,textHz$, and $textsup,|f(t)| = A$), then $$left|fractextdf(t)textdtright| leq 2ABpi.
$$
Note that the original result by Bernstein established a bound of $4ABpi$; later, that bound was tightened to $2ABpi$.
answered Aug 30 at 13:06
MBaz
8,25241432
You'll be interested in Bernstein's inequality, which I first learned about in Lapidoth, A Foundation in Digital Communication (page 92).
With a well-behaved signal $f(t)$ as you defined it above (in particular, $f(t)$ is integrable and bandlimited to $B,textHz$, and $textsup,|f(t)| = A$), then $$left|fractextdf(t)textdtright| leq 2ABpi.
$$
Note that the original result by Bernstein established a bound of $4ABpi$; later, that bound was tightened to $2ABpi$.
answered Aug 30 at 13:06
MBaz
8,25241432
edited Aug 30 at 22:18
answered Aug 30 at 13:06
MBaz
8,25241432
answered Aug 30 at 13:06
MBaz
8,25241432
answered Aug 30 at 13:06
MBaz
8,25241432
8,25241432
1
@OlliNiemitalo As pointed out in MattL's answer, the sinusoid $sin(2pi Bt)$ has maximum derivative $2pi B$. This meets Bernstein's bound, as stated in my answer here on dsp.SE (cited in a comment on your question) and in my answer on math.SE that you found, with equality.
– Dilip Sarwate
Aug 30 at 15:11
1
@OlliNiemitalo I found the proof given by Pinksy here (I hope that link works!). He definitely uses $4ABpi$ as the bound, not $2ABpi$.
– MBaz
Aug 30 at 15:46
2
@MBaz Your link works indeed! At the end of the section 2.3.8 they say that the best known version of Bernstein's inequality has the factor 2 instead of 4, which is sharp, and that for details consult Zygmund (1959) Vol. 2, p. 276. I think that's Zygmund, A. Trigonometric series. 2nd ed. Vol. II. Cambridge University Press, New York 1959.
– Olli Niemitalo
Aug 30 at 19:14
2
RP Boas, Some theorems on Fourier transforms and conjugate trigonometric integrals, Transactions of the American Mathematical Society 40 (2), 287-308, 1936 cites the relevant articles by Bernstein, Szegö, and Zygmund, already with the sharp bound, as far as I can tell.
– Olli Niemitalo
Aug 30 at 20:26
2
@OlliNiemitalo Excellent! I had missed that note at the end of section 2.3.8. I'll update my answer. Also: that book by Zygmund is in my university's library, but it's not online. I'll take it out tomorrow and see what it says.
– MBaz
Aug 30 at 22:15
 |Â
show 4 more comments
1
@OlliNiemitalo As pointed out in MattL's answer, the sinusoid $sin(2pi Bt)$ has maximum derivative $2pi B$. This meets Bernstein's bound, as stated in my answer here on dsp.SE (cited in a comment on your question) and in my answer on math.SE that you found, with equality.
– Dilip Sarwate
Aug 30 at 15:11
1
@OlliNiemitalo I found the proof given by Pinksy here (I hope that link works!). He definitely uses $4ABpi$ as the bound, not $2ABpi$.
– MBaz
Aug 30 at 15:46
2
@MBaz Your link works indeed! At the end of the section 2.3.8 they say that the best known version of Bernstein's inequality has the factor 2 instead of 4, which is sharp, and that for details consult Zygmund (1959) Vol. 2, p. 276. I think that's Zygmund, A. Trigonometric series. 2nd ed. Vol. II. Cambridge University Press, New York 1959.
– Olli Niemitalo
Aug 30 at 19:14
2
RP Boas, Some theorems on Fourier transforms and conjugate trigonometric integrals, Transactions of the American Mathematical Society 40 (2), 287-308, 1936 cites the relevant articles by Bernstein, Szegö, and Zygmund, already with the sharp bound, as far as I can tell.
– Olli Niemitalo
Aug 30 at 20:26
2
@OlliNiemitalo Excellent! I had missed that note at the end of section 2.3.8. I'll update my answer. Also: that book by Zygmund is in my university's library, but it's not online. I'll take it out tomorrow and see what it says.
– MBaz
Aug 30 at 22:15
1
1
@OlliNiemitalo As pointed out in MattL's answer, the sinusoid $sin(2pi Bt)$ has maximum derivative $2pi B$. This meets Bernstein's bound, as stated in my answer here on dsp.SE (cited in a comment on your question) and in my answer on math.SE that you found, with equality.
– Dilip Sarwate
Aug 30 at 15:11
@OlliNiemitalo As pointed out in MattL's answer, the sinusoid $sin(2pi Bt)$ has maximum derivative $2pi B$. This meets Bernstein's bound, as stated in my answer here on dsp.SE (cited in a comment on your question) and in my answer on math.SE that you found, with equality.
– Dilip Sarwate
Aug 30 at 15:11
1
1
@OlliNiemitalo I found the proof given by Pinksy here (I hope that link works!). He definitely uses $4ABpi$ as the bound, not $2ABpi$.
– MBaz
Aug 30 at 15:46
@OlliNiemitalo I found the proof given by Pinksy here (I hope that link works!). He definitely uses $4ABpi$ as the bound, not $2ABpi$.
– MBaz
Aug 30 at 15:46
2
2
@MBaz Your link works indeed! At the end of the section 2.3.8 they say that the best known version of Bernstein's inequality has the factor 2 instead of 4, which is sharp, and that for details consult Zygmund (1959) Vol. 2, p. 276. I think that's Zygmund, A. Trigonometric series. 2nd ed. Vol. II. Cambridge University Press, New York 1959.
– Olli Niemitalo
Aug 30 at 19:14
@MBaz Your link works indeed! At the end of the section 2.3.8 they say that the best known version of Bernstein's inequality has the factor 2 instead of 4, which is sharp, and that for details consult Zygmund (1959) Vol. 2, p. 276. I think that's Zygmund, A. Trigonometric series. 2nd ed. Vol. II. Cambridge University Press, New York 1959.
– Olli Niemitalo
Aug 30 at 19:14
2
2
RP Boas, Some theorems on Fourier transforms and conjugate trigonometric integrals, Transactions of the American Mathematical Society 40 (2), 287-308, 1936 cites the relevant articles by Bernstein, Szegö, and Zygmund, already with the sharp bound, as far as I can tell.
– Olli Niemitalo
Aug 30 at 20:26
RP Boas, Some theorems on Fourier transforms and conjugate trigonometric integrals, Transactions of the American Mathematical Society 40 (2), 287-308, 1936 cites the relevant articles by Bernstein, Szegö, and Zygmund, already with the sharp bound, as far as I can tell.
– Olli Niemitalo
Aug 30 at 20:26
2
2
@OlliNiemitalo Excellent! I had missed that note at the end of section 2.3.8. I'll update my answer. Also: that book by Zygmund is in my university's library, but it's not online. I'll take it out tomorrow and see what it says.
– MBaz
Aug 30 at 22:15
@OlliNiemitalo Excellent! I had missed that note at the end of section 2.3.8. I'll update my answer. Also: that book by Zygmund is in my university's library, but it's not online. I'll take it out tomorrow and see what it says.
– MBaz
Aug 30 at 22:15
 |Â
show 4 more comments
up vote
2
down vote
In general you would get something like this, but it might not be tight:
$$beginalign|f'(t)|&=left|frac12piint_-infty^inftyjomega F(jomega)e^-jomega tdomegaright|\&le frac12piint_-infty^infty|omega||F(jomega)|domega\&=frac12piint_-omega_c^omega_c|omega||F(jomega)|domega\&le fracomega_c2piint_-omega_c^omega_c|F(jomega)|domega\&=fracomega_cpiint_0^omega_c|F(jomega)|domegatag1endalign$$
The upper bound on $|f(t)|$ is of course implicit in $|F(jomega)|$.
For a sinusoid $Asin(omega_ct)$, $(1)$ gives $Aomega_c$ as an upper bound, as expected.
answered Aug 30 at 12:51
Matt L.
45.2k13678
@Olli Niemitalo , I had derived the sinusoid case I think this is the general case we were looking at. Thanks Matt L.
– MimSaad
Aug 30 at 18:22
add a comment |Â
up vote
2
down vote
In general you would get something like this, but it might not be tight:
$$beginalign|f'(t)|&=left|frac12piint_-infty^inftyjomega F(jomega)e^-jomega tdomegaright|\&le frac12piint_-infty^infty|omega||F(jomega)|domega\&=frac12piint_-omega_c^omega_c|omega||F(jomega)|domega\&le fracomega_c2piint_-omega_c^omega_c|F(jomega)|domega\&=fracomega_cpiint_0^omega_c|F(jomega)|domegatag1endalign$$
The upper bound on $|f(t)|$ is of course implicit in $|F(jomega)|$.
For a sinusoid $Asin(omega_ct)$, $(1)$ gives $Aomega_c$ as an upper bound, as expected.
answered Aug 30 at 12:51
Matt L.
45.2k13678
@Olli Niemitalo , I had derived the sinusoid case I think this is the general case we were looking at. Thanks Matt L.
– MimSaad
Aug 30 at 18:22
add a comment |Â
up vote
2
down vote
up vote
2
down vote
In general you would get something like this, but it might not be tight:
$$beginalign|f'(t)|&=left|frac12piint_-infty^inftyjomega F(jomega)e^-jomega tdomegaright|\&le frac12piint_-infty^infty|omega||F(jomega)|domega\&=frac12piint_-omega_c^omega_c|omega||F(jomega)|domega\&le fracomega_c2piint_-omega_c^omega_c|F(jomega)|domega\&=fracomega_cpiint_0^omega_c|F(jomega)|domegatag1endalign$$
The upper bound on $|f(t)|$ is of course implicit in $|F(jomega)|$.
For a sinusoid $Asin(omega_ct)$, $(1)$ gives $Aomega_c$ as an upper bound, as expected.
answered Aug 30 at 12:51
Matt L.
45.2k13678
In general you would get something like this, but it might not be tight:
$$beginalign|f'(t)|&=left|frac12piint_-infty^inftyjomega F(jomega)e^-jomega tdomegaright|\&le frac12piint_-infty^infty|omega||F(jomega)|domega\&=frac12piint_-omega_c^omega_c|omega||F(jomega)|domega\&le fracomega_c2piint_-omega_c^omega_c|F(jomega)|domega\&=fracomega_cpiint_0^omega_c|F(jomega)|domegatag1endalign$$
The upper bound on $|f(t)|$ is of course implicit in $|F(jomega)|$.
For a sinusoid $Asin(omega_ct)$, $(1)$ gives $Aomega_c$ as an upper bound, as expected.
answered Aug 30 at 12:51
Matt L.
45.2k13678
answered Aug 30 at 12:51
Matt L.
45.2k13678
answered Aug 30 at 12:51
Matt L.
45.2k13678
answered Aug 30 at 12:51
Matt L.
45.2k13678
45.2k13678
@Olli Niemitalo , I had derived the sinusoid case I think this is the general case we were looking at. Thanks Matt L.
– MimSaad
Aug 30 at 18:22
add a comment |Â
@Olli Niemitalo , I had derived the sinusoid case I think this is the general case we were looking at. Thanks Matt L.
– MimSaad
Aug 30 at 18:22
@Olli Niemitalo , I had derived the sinusoid case I think this is the general case we were looking at. Thanks Matt L.
– MimSaad
Aug 30 at 18:22
@Olli Niemitalo , I had derived the sinusoid case I think this is the general case we were looking at. Thanks Matt L.
– MimSaad
Aug 30 at 18:22
add a comment |Â
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Have you checked this?
– Tendero
Aug 30 at 13:24
@Tendero thanks. There, the signal energy is known, rather than the peak absolute value as in my question.
– Olli Niemitalo
Aug 30 at 13:39
1
See my answer for the bound you seek. It says More generally, a result due to Bernstein says that if the maximum frequency in a generic $x(t)$ bounded within $[-1,1]$ is $f_0$, that is, $X(f) = 0$ for $|f| > f_0$, then $$max left| fracmathrm dxmathrm dtright| leq 2pi f_0.$$
– Dilip Sarwate
Aug 30 at 14:50
1
Based on the sharp version of Bernstein's inequality, from Dilip's linked answers, MBaz's edited answer and the literature cited, $AB$ is indeed the sharp (I called it tight meaning the same) upper bound for the maximum absolute value of the derivative, a full-scale sinusoid at exactly the band limit (not strictly allowed by the constraints I give) making the inequality an equality.
– Olli Niemitalo
Aug 31 at 7:02