FPGA Central

For minimum phase systems the magnitude uniquely specifys the phase
and the phase uniquely specifys the magnitude to within a scale
factor.

It is cited from the great textbook discrete time signal processing
by oppenheim.

How to prove the statement above? I am especially interested in the
latter.

HyeeWang

On Aug 21, 2:14 am, HyeeWang <hyeew...@gmail.com> wrote:
> For minimum phase systems the magnitude uniquely specifys the phase
> and the phase uniquely specifys the magnitude to within a scale
> factor.
>
> It is cited from the great textbook discrete time signal processing
> by oppenheim.
>
> How to prove the statement above?

this is a good and important question. one that i have thunked about
since i was in grad school in the '80s.

> I am especially interested in the latter.

i dunno what is the former vs. latter here.

do you want me to simply quote the proof for disrete-time systems (z-
plane) out of O&S?

for s-plane, continuous-time systems, i can give you a hint on one
method to prove it:

consider that the Hilbert transform (HT), or more specifically, the
hypothetical Hilbert transformer is a linear, time-invariant (LTI)
system with impulse response:

h(t) = 1/(pi*t)

and frequency response:

H(jw) = -(pi/2)*sgn(w)

so, if you input a delayed signal to a HT, the output of the HT is
identical to the original undelayed signal, except delayed by the same
amount. that's what the "TI" means in LTI.

now that is also true if the signal is not a function of time, but a
function of frequency. delayed in time means shifted in time. so the
Hilbert transformer acting on a function of frequency that is shifted
by a constant will have the same result as acting on the unshifted
function of frequency except it will be shifted by the same amount.

so let's examine the magnitude (actually the log magnitude) and phase
of a minimum-phase system. for *any* s-plane system with poles and
zeros, the transfer function is:

(s-q1)(s-q2)(s-q3)...(s-q_m)
H(s) = A ------------------------------
(s-p1)(s-p2)(s-p3)...(s-p_n)


for a minimum phase filter, the poles p_i and zeros q_i all have a
real part that is negative. for the poles, this is necessary for
stability and for the zeros, it is necessary for minimum phase. if
any of the zeros were "reflected" to the right half-plane (where their
real part is positive abs value of the minimum phase value, and the
imaginary part is unchanged) then the magnitude response is shown
below to be the same, but there is *more* negative phase. it's as if
you started with the so-called "minimum-phase filter" (where all the
zeros are in the left half-plane and have negative real part) and
cascaded that with an all-pass filter with a pole that lies on top of
the min-phase zero (and cancels it) and a zero at the reflected
location in the right half-plane. the all-pass filters only serve to
increase the (negative) phase response. the minimum phase filter has
none of these all-pass filters, since there are no zeros in the right
half-plane.

the complex frequency response is obtained by plugging in jw for s.


(jw-q1)(jw-q2)(jw-q3)...(jw-q_m)
H(jw) = A ----------------------------------
(jw-p1)(jw-p2)(jw-p3)...(jw-p_n)


the magnitude is


|jw-q1||jw-q2||jw-q3|...|jw-q_m|
|H(jw)| = |A| ----------------------------------
|jw-p1||jw-p2||jw-p3|...|jw-p_n|


the (natural) log magnitude is

m n
log|H(jw)| = log|A| + SUM{ log|jw-q_i| } - SUM{ log|jw-p_i| }
i=1 i=1

this is the real part of the complex log of H(jw)

the phase response is already in the log domain, it is the imaginary
part of the complex log of H(jw).


m n
arg[H(jw)] = arg[A] + SUM{ arg[jw-q_i] } - SUM{ arg[jw-p_i] }
i=1 i=1

so someone proposes that the negative of the Hilbert Transform (with
"w" as the independent variable instead of "t") of the log magnitude
is the phase response (measured in radians, if the log magnitude is
the natural log).

since the HT is LTI, we can look at each term above individually. we
know that the HT of the constant log|A| is zero, and like a
differentiator, the information regarding that constant term is lost.
for real and postive A, the arg[A] is also zero. so it doesn't matter
what A is, as long as it's real and >0.

now, consider any single zero or pole (all of which are in the left
half-plane):

log| jw - p | = log| jw - Re[p] - j*Im[p] |

= log| -Re[p] + j*(w-Im[p]) |

= (1/2)*log( (-Re[p])^2 + (w-Im[p])^2 )
and
arg[ jw - p ] = arg[ jw - Re[p] - j*Im[p] ]

= arg[ -Re[p] + j*(w-Im[p]) ]

= arctan( (w-Im[p])/(-Re[p]) )

(i'm leaving the subscript "i" off, for the time being.

if we can somehow show that the negative Hilbert transform of the top
expression (the log magnitude or the real part of the complex log) is
the bottom expression; if we can show that the two form a HT pair,
we're done, because the HT is linear, the HT of the summation above is
the same as the summation of the separate HTs.

now, remember that Re[p] < 0 so -Re[p] is a positive real number.

the Im[p] constant term only serves to shift both functions by the
same constant amount of Im[p] on the linear frequency axis. because
the HT is "time"-invariant, which means in the present frequency
domain, the HT operation is invariant to constant offsets in
frequency, then applying the HT to the constant offset expression for
log magnitude

log| jw - p | = (1/2)*log( (-Re[p])^2 + (w-Im[p])^2 )

will result in the same function of w as it would if there was no
offset, *except* that the result would also be offset (as it is in the
phase or arg equation). that means if we can show that these form an
HT pair:

log| jw - p | = (1/2)*log( (-Re[p])^2 + w^2 )

arg[ jw - p ] = arctan( w/(-Re[p]) )

then we're done, because the LTI property says if we offset both input
and output by the same value (namely Im(p)), nothing has changed. now
the real-valued scaler, -Re[p], comes out in the wash, it just scales
things:

log| jw - p | = (1/2)*log( 1 + (w/-Re[p])^2 ) + log(-Re[p])

arg[ jw - p ] = arctan( w/(-Re[p]) )


applying the HT to the top loses the constant log(-Re[p]) term just
like it loses the constant log|A| term. w/(-Re[p]) is just another
scaled version of w, so if we ditch it in the top expression, we can
ditch it in the bottom.

so if the minimum-phase relation works for a single real pole in the
left half-plane, it will work for any complex pole in the left-half
place.


so, the task finally (almost) comes down to showing that the negative
HT of (1/2)*log( 1 + w^2 ) is arctan(w). if we can show

HT{ (1/2)*log( 1 + w^2 ) } = -arctan(w)

we're done. but the integral on the left side is a bitch, so we're
not done yet. so even though we cannot directly show the following
pair are an HT pair:

(1/2)*log( 1 + w^2 ) <----> -arctan(w)

what we *can* show is that their derivatives are an HT pair.

HT{ w/(1 + w^2) } = -1/(1 + w^2)

it takes a little doing (and partial fraction expansion), but you can
prove that the above are a Hilbert pair.

i'm getting tired now. but, because both HTs and differentiators and
integrators are all LTI, we can switch around their order of
operation. HTing the output of the integrator is the same as
integrating the output of the HT.

so if you apply an integrator (w.r.t. "w" rather than "t") to both
sides of

HT{ w/(1 + w^2) } = -1/(1 + w^2)

you will get

HT{ (1/2)*log( 1 + w^2 ) } = -arctan(w)

or

-HT{ (1/2)*log( 1 + w^2 ) } = arctan(w)

or

-HT{ (1/2)*log( 1 + (w/-Re[p])^2 ) } = arctan( w/(-Re[p]) )

or

-HT{ (1/2)*log((-Re[p])^2 + (w-Im[p])^2) } = arctan((w-Im[p])/(-Re
[p]))

or

-HT{ log| jw - p | } = arg[ jw - p ]

or

-HT{ log| jw - p_i | } = arg[ jw - p_i ]
and
-HT{ log| jw - q_i | } = arg[ jw - q_i ]

and finally, after you sum up all of the terms,

-HT{ log|H(jw)| } = arg[ H(jw) ]


that's one way to look at it in the analog world. the other way uses
residue theory and is succinct, but more sophisticated. this is sorta
a brute force method.

r b-j

On Aug 21, 4:21*pm, robert bristow-johnson <r...@audioimagination.com>
wrote:
> On Aug 21, 2:14 am, HyeeWang <hyeew...@gmail.com> wrote:
>
> > For minimum phase systems the magnitude uniquely specifys the phase
> > and the phase uniquely specifys the magnitude to within a scale
> > factor.
>
> > It is cited from the great textbook *discrete time signal processing
> > by oppenheim.
>
> > How to prove the statement above?
>
> this is a good and important question. *one that i have thunked about
> since i was in grad school in the '80s.
>
> > I am especially interested in the latter.
>
> i dunno what is the former vs. latter here.
>
> do you want me to simply quote the proof for disrete-time systems (z-
> plane) out of O&S?
>
> for s-plane, continuous-time systems, i can give you a hint on one
> method to prove it:
>
> consider that the Hilbert transform (HT), or more specifically, the
> hypothetical Hilbert transformer is a linear, time-invariant (LTI)
> system with impulse response:
>
> * * h(t) = 1/(pi*t)
>
> and frequency response:
>
> * * H(jw) = -(pi/2)*sgn(w)
>
> so, if you input a delayed signal to a HT, the output of the HT is
> identical to the original undelayed signal, except delayed by the same
> amount. *that's what the "TI" means in LTI.
>
> now that is also true if the signal is not a function of time, but a
> function of frequency. *delayed in time means shifted in time. *so the
> Hilbert transformer acting on a function of frequency that is shifted
> by a constant will have the same result as acting on the unshifted
> function of frequency except it will be shifted by the same amount.
>
> so let's examine the magnitude (actually the log magnitude) and phase
> of a minimum-phase system. *for *any* s-plane system with poles and
> zeros, the transfer function is:
>
> * * * * * * * (s-q1)(s-q2)(s-q3)...(s-q_m)
> * * H(s) = A ------------------------------
> * * * * * * * (s-p1)(s-p2)(s-p3)...(s-p_n)
>
> for a minimum phase filter, the poles p_i and zeros q_i all have a
> real part that is negative. *for the poles, this is necessary for
> stability and for the zeros, it is necessary for minimum phase. *if
> any of the zeros were "reflected" to the right half-plane (where their
> real part is positive abs value of the minimum phase value, and the
> imaginary part is unchanged) then the magnitude response is shown
> below to be the same, but there is *more* negative phase. *it's as if
> you started with the so-called "minimum-phase filter" (where all the
> zeros are in the left half-plane and have negative real part) and
> cascaded that with an all-pass filter with a pole that lies on top of
> the min-phase zero (and cancels it) and a zero at the reflected
> location in the right half-plane. *the all-pass filters only serve to
> increase the (negative) phase response. *the minimum phase filter has
> none of these all-pass filters, since there are no zeros in the right
> half-plane.
>
> the complex frequency response is obtained by plugging in jw for s.
>
> * * * * * * * *(jw-q1)(jw-q2)(jw-q3)...(jw-q_m)
> * * H(jw) = A ----------------------------------
> * * * * * * * *(jw-p1)(jw-p2)(jw-p3)...(jw-p_n)
>
> the magnitude is
>
> * * * * * * * * * |jw-q1||jw-q2||jw-q3|...|jw-q_m|
> * *|H(jw)| = |A| ----------------------------------
> * * * * * * * * * |jw-p1||jw-p2||jw-p3|...|jw-p_n|
>
> the (natural) log magnitude is
>
> * * * * * * * * * * * * * *m * * * * * * * * * *n
> * * log|H(jw)| = log|A| + SUM{ log|jw-q_i| } - SUM{ log|jw-p_i| }
> * * * * * * * * * * * * * i=1 * * * ** * * * *i=1
>
> this is the real part of the complex log of H(jw)
>
> the phase response is already in the log domain, it is the imaginary
> part of the complex log of H(jw).
>
> * * * * * * * * * * * * * *m * * * * * * * * * *n
> * * arg[H(jw)] = arg[A] + SUM{ arg[jw-q_i] } - SUM{ arg[jw-p_i] }
> * * * * * * * * * * * * * i=1 * * * ** * * * *i=1
>
> so someone proposes that the negative of the Hilbert Transform (with
> "w" as the independent variable instead of "t") of the log magnitude
> is the phase response (measured in radians, if the log magnitude is
> the natural log).
>
> since the HT is LTI, we can look at each term above individually. *we
> know that the HT of the constant log|A| is zero, and like a
> differentiator, the information regarding that constant term is lost.
> for real and postive A, the arg[A] is also zero. *so it doesn't matter
> what A is, as long as it's real and >0.
>
> now, consider any single zero or pole (all of which are in the left
> half-plane):
>
> * * log| jw - p | *= *log| jw - Re[p] - j*Im[p] |
>
> * * * * * * * * * *= *log| -Re[p] + j*(w-Im[p]) |
>
> * * * * * * * * * *= *(1/2)*log( (-Re[p])^2 + (w-Im[p])^2 )
> and
> * * arg[ jw - p ] *= *arg[ jw - Re[p] - j*Im[p] ]
>
> * * * * * * * * * *= *arg[ -Re[p] + j*(w-Im[p]) ]
>
> * * * * * * * * * *= *arctan( (w-Im[p])/(-Re[p]) )
>
> (i'm leaving the subscript "i" off, for the time being.
>
> if we can somehow show that the negative Hilbert transform of the top
> expression (the log magnitude or the real part of the complex log) is
> the bottom expression; if we can show that the two form a HT pair,
> we're done, because the HT is linear, the HT of the summation above is
> the same as the summation of the separate HTs.
>
> now, remember that Re[p] < 0 so -Re[p] is a positive real number.
>
> the Im[p] constant term only serves to shift both functions by the
> same constant amount of Im[p] on the linear frequency axis. *because
> the HT is "time"-invariant, which means in the present frequency
> domain, the HT operation is invariant to constant offsets in
> frequency, then applying the HT to the constant offset expression for
> log magnitude
>
> * * log| jw - p | *= *(1/2)*log( (-Re[p])^2 + (w-Im[p])^2 )
>
> will result in the same function of w as it would if there was no
> offset, *except* that the result would also be offset (as it is in the
> phase or arg equation). *that means if we can show that these form an
> HT pair:
>
> * * log| jw - p | *= *(1/2)*log( (-Re[p])^2 + w^2 )
>
> * * arg[ jw - p ] *= *arctan( w/(-Re[p]) )
>
> then we're done, because the LTI property says if we offset both input
> and output by the same value (namely Im(p)), nothing has changed. *now
> the real-valued scaler, -Re[p], comes out in the wash, it just scales
> things:
>
> * * log| jw - p | *= *(1/2)*log( 1 + (w/-Re[p])^2 ) + log(-Re[p])
>
> * * arg[ jw - p ] *= *arctan( w/(-Re[p]) )
>
> applying the HT to the top loses the constant log(-Re[p]) term just
> like it loses the constant log|A| term. *w/(-Re[p]) is just another
> scaled version of w, so if we ditch it in the top expression, we can
> ditch it in the bottom.
>
> so if the minimum-phase relation works for a single real pole in the
> left half-plane, it will work for any complex pole in the left-half
> place.
>
> so, the task finally (almost) comes down to showing that the negative
> HT of (1/2)*log( 1 + w^2 ) is arctan(w). *if we can show
>
> * * HT{ (1/2)*log( 1 + w^2 ) } = -arctan(w)
>
> we're done. *but the integral on the left side is a bitch, so we're
> not done yet. *so even though we cannot directly show the following
> pair are an HT pair:
>
> * * (1/2)*log( 1 + w^2 ) * *<----> * -arctan(w)
>
> what we *can* show is that their derivatives are an HT pair.
>
> * * *HT{ w/(1 + w^2) } *= *-1/(1 + w^2)
>
> it takes a little doing (and partial fraction expansion), but you can
> prove that the above are a Hilbert pair.
>
> i'm getting tired now. *but, because both HTs and differentiators and
> integrators are all LTI, we can switch around their order of
> operation. *HTing the output of the integrator is the same as
> integrating the output of the HT.
>
> so if you apply an integrator (w.r.t. "w" rather than "t") to both
> sides of
>
> * * *HT{ w/(1 + w^2) } *= *-1/(1 + w^2)
>
> you will get
>
> * * *HT{ (1/2)*log( 1 + w^2 ) } = -arctan(w)
>
> or
>
> * * -HT{ (1/2)*log( 1 + w^2 ) } = *arctan(w)
>
> or
>
> * * -HT{ (1/2)*log( 1 + (w/-Re[p])^2 ) } = arctan( w/(-Re[p]) )
>
> or
>
> -HT{ (1/2)*log((-Re[p])^2 + (w-Im[p])^2) } = arctan((w-Im[p])/(-Re
> [p]))
>
> or
>
> * * -HT{ log| jw - p | } = * arg[ jw - p ]
>
> or
>
> * * -HT{ log| jw - p_i | } = * arg[ jw - p_i ]
> and
> * * -HT{ log| jw - q_i | } = * arg[ jw - q_i ]
>
> and finally, after you sum up all of the terms,
>
> * * -HT{ log|H(jw)| } = arg[ H(jw) ]
>
> that's one way to look at it in the analog world. *the other way uses
> residue theory and is succinct, but more sophisticated. *this is sorta
> a brute force method.
>
> r b-j

Thank you! Rbj.

It is so complex and make me study a lot.

Inferring from my original description,the former is : for minimum
phase systems the magnitude uniquely specifys the phase ,

the latter is : for minimum phase systems tthe phase uniquely specifys
the magnitude to within a scale factor.

HyeeWang

On Aug 21, 11:18*am, HyeeWang <hyeew...@gmail.com> wrote:
> On Aug 21, 4:21*pm, robert bristow-johnson <r...@audioimagination.com>
> wrote:
>
>
>
> > On Aug 21, 2:14 am, HyeeWang <hyeew...@gmail.com> wrote:
>
> > > For minimum phase systems the magnitude uniquely specifys the phase
> > > and the phase uniquely specifys the magnitude to within a scale
> > > factor.
>
> > > It is cited from the great textbook *discrete time signal processing
> > > by oppenheim.
>
> > > How to prove the statement above?
>
> > this is a good and important question. *one that i have thunked about
> > since i was in grad school in the '80s.
>
> > > I am especially interested in the latter.
>
> > i dunno what is the former vs. latter here.
>
> > do you want me to simply quote the proof for disrete-time systems (z-
> > plane) out of O&S?
>
> > for s-plane, continuous-time systems, i can give you a hint on one
> > method to prove it:
>
> > consider that the Hilbert transform (HT), or more specifically, the
> > hypothetical Hilbert transformer is a linear, time-invariant (LTI)
> > system with impulse response:
>
> > * * h(t) = 1/(pi*t)
>
> > and frequency response:
>
> > * * H(jw) = -(pi/2)*sgn(w)
>
> > so, if you input a delayed signal to a HT, the output of the HT is
> > identical to the original undelayed signal, except delayed by the same
> > amount. *that's what the "TI" means in LTI.
>
> > now that is also true if the signal is not a function of time, but a
> > function of frequency. *delayed in time means shifted in time. *so the
> > Hilbert transformer acting on a function of frequency that is shifted
> > by a constant will have the same result as acting on the unshifted
> > function of frequency except it will be shifted by the same amount.
>
> > so let's examine the magnitude (actually the log magnitude) and phase
> > of a minimum-phase system. *for *any* s-plane system with poles and
> > zeros, the transfer function is:
>
> > * * * * * * * (s-q1)(s-q2)(s-q3)...(s-q_m)
> > * * H(s) = A ------------------------------
> > * * * * * * * (s-p1)(s-p2)(s-p3)...(s-p_n)
>
> > for a minimum phase filter, the poles p_i and zeros q_i all have a
> > real part that is negative. *for the poles, this is necessary for
> > stability and for the zeros, it is necessary for minimum phase. *if
> > any of the zeros were "reflected" to the right half-plane (where their
> > real part is positive abs value of the minimum phase value, and the
> > imaginary part is unchanged) then the magnitude response is shown
> > below to be the same, but there is *more* negative phase. *it's as if
> > you started with the so-called "minimum-phase filter" (where all the
> > zeros are in the left half-plane and have negative real part) and
> > cascaded that with an all-pass filter with a pole that lies on top of
> > the min-phase zero (and cancels it) and a zero at the reflected
> > location in the right half-plane. *the all-pass filters only serve to
> > increase the (negative) phase response. *the minimum phase filter has
> > none of these all-pass filters, since there are no zeros in the right
> > half-plane.
>
> > the complex frequency response is obtained by plugging in jw for s.
>
> > * * * * * * * *(jw-q1)(jw-q2)(jw-q3)...(jw-q_m)
> > * * H(jw) = A ----------------------------------
> > * * * * * * * *(jw-p1)(jw-p2)(jw-p3)...(jw-p_n)
>
> > the magnitude is
>
> > * * * * * * * * * |jw-q1||jw-q2||jw-q3|...|jw-q_m|
> > * *|H(jw)| = |A| ----------------------------------
> > * * * * * * * * * |jw-p1||jw-p2||jw-p3|...|jw-p_n|
>
> > the (natural) log magnitude is
>
> > * * * * * * * * * * * * * *m * * * * * * * * * *n
> > * * log|H(jw)| = log|A| + SUM{ log|jw-q_i| } - SUM{ log|jw-p_i| }
> > * * * * * * * * * * * * * i=1 * * * * * * * * *i=1
>
> > this is the real part of the complex log of H(jw)
>
> > the phase response is already in the log domain, it is the imaginary
> > part of the complex log of H(jw).
>
> > * * * * * * * * * * * * * *m * * * * * * * * * *n
> > * * arg[H(jw)] = arg[A] + SUM{ arg[jw-q_i] } - SUM{ arg[jw-p_i] }
> > * * * * * * * * * * * * * i=1 * * * * * * * * *i=1
>
> > so someone proposes that the negative of the Hilbert Transform (with
> > "w" as the independent variable instead of "t") of the log magnitude
> > is the phase response (measured in radians, if the log magnitude is
> > the natural log).
>
> > since the HT is LTI, we can look at each term above individually. *we
> > know that the HT of the constant log|A| is zero, and like a
> > differentiator, the information regarding that constant term is lost.
> > for real and postive A, the arg[A] is also zero. *so it doesn't matter
> > what A is, as long as it's real and >0.
>
> > now, consider any single zero or pole (all of which are in the left
> > half-plane):
>
> > * * log| jw - p | *= *log| jw - Re[p] - j*Im[p] |
>
> > * * * * * * * * * *= *log| -Re[p] + j*(w-Im[p])|
>
> > * * * * * * * * * *= *(1/2)*log( (-Re[p])^2 + (w-Im[p])^2 )
> > and
> > * * arg[ jw - p ] *= *arg[ jw - Re[p] - j*Im[p] ]
>
> > * * * * * * * * * *= *arg[ -Re[p] + j*(w-Im[p])]
>
> > * * * * * * * * * *= *arctan( (w-Im[p])/(-Re[p]) )
>
> > (i'm leaving the subscript "i" off, for the time being.
>
> > if we can somehow show that the negative Hilbert transform of the top
> > expression (the log magnitude or the real part of the complex log) is
> > the bottom expression; if we can show that the two form a HT pair,
> > we're done, because the HT is linear, the HT of the summation above is
> > the same as the summation of the separate HTs.
>
> > now, remember that Re[p] < 0 so -Re[p] is a positive real number.
>
> > the Im[p] constant term only serves to shift both functions by the
> > same constant amount of Im[p] on the linear frequency axis. *because
> > the HT is "time"-invariant, which means in the present frequency
> > domain, the HT operation is invariant to constant offsets in
> > frequency, then applying the HT to the constant offset expression for
> > log magnitude
>
> > * * log| jw - p | *= *(1/2)*log( (-Re[p])^2 + (w-Im[p])^2 )
>
> > will result in the same function of w as it would if there was no
> > offset, *except* that the result would also be offset (as it is in the
> > phase or arg equation). *that means if we can show that these form an
> > HT pair:
>
> > * * log| jw - p | *= *(1/2)*log( (-Re[p])^2 + w^2 )
>
> > * * arg[ jw - p ] *= *arctan( w/(-Re[p]) )
>
> > then we're done, because the LTI property says if we offset both input
> > and output by the same value (namely Im(p)), nothing has changed. *now
> > the real-valued scaler, -Re[p], comes out in the wash, it just scales
> > things:
>
> > * * log| jw - p | *= *(1/2)*log( 1 + (w/-Re[p])^2 ) + log(-Re[p])
>
> > * * arg[ jw - p ] *= *arctan( w/(-Re[p]) )
>
> > applying the HT to the top loses the constant log(-Re[p]) term just
> > like it loses the constant log|A| term. *w/(-Re[p]) is just another
> > scaled version of w, so if we ditch it in the top expression, we can
> > ditch it in the bottom.
>
> > so if the minimum-phase relation works for a single real pole in the
> > left half-plane, it will work for any complex pole in the left-half
> > place.
>
> > so, the task finally (almost) comes down to showing that the negative
> > HT of (1/2)*log( 1 + w^2 ) is arctan(w). *if we can show
>
> > * * HT{ (1/2)*log( 1 + w^2 ) } = -arctan(w)
>
> > we're done. *but the integral on the left side is a bitch, so we're
> > not done yet. *so even though we cannot directly show the following
> > pair are an HT pair:
>
> > * * (1/2)*log( 1 + w^2 ) * *<----> * -arctan(w)
>
> > what we *can* show is that their derivatives are an HT pair.
>
> > * * *HT{ w/(1 + w^2) } *= *-1/(1 + w^2)
>
> > it takes a little doing (and partial fraction expansion), but you can
> > prove that the above are a Hilbert pair.
>
> > i'm getting tired now. *but, because both HTs and differentiators and
> > integrators are all LTI, we can switch around their order of
> > operation. *HTing the output of the integrator is the same as
> > integrating the output of the HT.
>
> > so if you apply an integrator (w.r.t. "w" rather than "t") to both
> > sides of
>
> > * * *HT{ w/(1 + w^2) } *= *-1/(1 + w^2)
>
> > you will get
>
> > * * *HT{ (1/2)*log( 1 + w^2 ) } = -arctan(w)
>
> > or
>
> > * * -HT{ (1/2)*log( 1 + w^2 ) } = *arctan(w)
>
> > or
>
> > * * -HT{ (1/2)*log( 1 + (w/-Re[p])^2 ) } = arctan( w/(-Re[p]) )
>
> > or
>
> > -HT{ (1/2)*log((-Re[p])^2 + (w-Im[p])^2) } = arctan((w-Im[p])/(-Re
> > [p]))
>
> > or
>
> > * * -HT{ log| jw - p | } = * arg[ jw - p ]
>
> > or
>
> > * * -HT{ log| jw - p_i | } = * arg[ jw - p_i ]
> > and
> > * * -HT{ log| jw - q_i | } = * arg[ jw - q_i ]
>
> > and finally, after you sum up all of the terms,
>
> > * * -HT{ log|H(jw)| } = arg[ H(jw) ]
>
> > that's one way to look at it in the analog world. *the other way uses
> > residue theory and is succinct, but more sophisticated. *this is sorta
> > a brute force method.
>
> > r b-j
>
> Thank you! Rbj.
>
> It is so complex and make me study a lot.

i tried to break it down to the smallest steps. but i did not show
the kernel:

HT{ w/(1 + w^2) } = -1/(1 + w^2)

or

+inf
integral{ 1/(pi*(w-v)) * v/(1 + v^2) dv } = -1/(1 + w^2)
-inf


you have to be the one to prove that. use partial fractions.

> Inferring from my original description,the former is : for minimum
> phase systems the magnitude uniquely specifys the phase ,
>
> the latter is : for minimum phase systems tthe phase uniquely specifys
> the magnitude to within a scale factor.

"to within a scale factor" is needed in both. in the former, the
scale factor becomes a constant when the log magnitude is done, and
the HT of a constant is zero, no matter what the constant (or scale
factor) is.

those relationships are:

for a minimum-phase system, the phase response (in radians) is equal
to the negative of the Hilbert transform of the (natural) log of the
magnitude response. since the phase response must be an odd-symmetry
function, there is no arbitrary offset term.

for a minimum-phase system, the log magnitude response (in "nepers"
which is proportional to dB) is equal to the Hilbert transform of the
phase response. since the log magnitude response is an even-symmetry
function, any constant can be added to it.

r b-j

On Aug 20, 11:14*pm, HyeeWang <hyeew...@gmail.com> wrote:
> For minimum phase systems the magnitude uniquely specifys the phase
> and the phase uniquely specifys the magnitude to within a scale
> factor.
>
> It is cited from the great textbook *discrete time signal processing
> by oppenheim.
>
> How to prove the statement above? I am especially interested in the
> latter.
>
> HyeeWang

This is proven in Bodes Theorem in the original bode paper.


hardy

On Aug 21, 8:36*pm, HardySpicer <gyansor...@gmail.com> wrote:
> On Aug 20, 11:14*pm, HyeeWang <hyeew...@gmail.com> wrote:
>
> > For minimum phase systems the magnitude uniquely specifys the phase
> > and the phase uniquely specifys the magnitude to within a scale
> > factor.
>
> > It is cited from the great textbook *discrete time signal processing
> > by oppenheim.
>
> > How to prove the statement above? I am especially interested in the
> > latter.
>
> > HyeeWang
>
> This is proven in Bodes Theorem in the original bode paper.
>
> hardy

Actually it is Bodes Second Theorem and is shown here

http://books.google.co.nz/books?id=3...heorem&f=false


Hardy

HardySpicer <[email protected]> writes:

> On Aug 21, 8:36Â*pm, HardySpicer <gyansor...@gmail.com> wrote:
>> On Aug 20, 11:14Â*pm, HyeeWang <hyeew...@gmail.com> wrote:
>>
>> > For minimum phase systems the magnitude uniquely specifys the phase
>> > and the phase uniquely specifys the magnitude to within a scale
>> > factor.
>>
>> > It is cited from the great textbook Â*discrete time signal processing
>> > by oppenheim.
>>
>> > How to prove the statement above? I am especially interested in the
>> > latter.
>>
>> > HyeeWang
>>
>> This is proven in Bodes Theorem in the original bode paper.
>>
>> hardy
>
> Actually it is Bodes Second Theorem and is shown here
>
> http://books.google.co.nz/books?id=3...heorem&f=false
>
>
> Hardy

Hardy - got a snapshot of that page? I can't accessit.
--
Randy Yates % "Midnight, on the water...
Digital Signal Labs % I saw... the ocean's daughter."
mailto://[email protected] % 'Can't Get It Out Of My Head'
http://www.digitalsignallabs.com % *El Dorado*, Electric Light Orchestra

"Randy Yates" <[email protected]> wrote in message
news:[email protected]..
> HardySpicer <[email protected]> writes:
>
>> On Aug 21, 8:36 pm, HardySpicer <gyansor...@gmail.com> wrote:
>>> On Aug 20, 11:14 pm, HyeeWang <hyeew...@gmail.com> wrote:
>>>
>>> > For minimum phase systems the magnitude uniquely specifys the phase
>>> > and the phase uniquely specifys the magnitude to within a scale
>>> > factor.
>>>
>>> > It is cited from the great textbook discrete time signal processing
>>> > by oppenheim.
>>>
>>> > How to prove the statement above? I am especially interested in the
>>> > latter.
>>>
>>> > HyeeWang
>>>
>>> This is proven in Bodes Theorem in the original bode paper.
>>>
>>> hardy
>>
>> Actually it is Bodes Second Theorem and is shown here
>>
>> http://books.google.co.nz/books?id=3...heorem&f=false
>>
>>
>> Hardy
>
> Hardy - got a snapshot of that page? I can't accessit.
> --
> Randy Yates % "Midnight, on the water...
> Digital Signal Labs % I saw... the ocean's daughter."
> mailto://[email protected] % 'Can't Get It Out Of My Head'
> http://www.digitalsignallabs.com % *El Dorado*, Electric Light Orchestra

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Best wishes,
--Phil Martel