By considering wavelengths in air and propagation times, it seems
trivial to show that (c = speed of sound in air; s = speed of source
toward the observer; o = speed of observer toward the source; f =
original frequency; f' = measured frequency at the observer) for a
stationary observer,
c c + o
f' = f ����� and for stationary source, f' = f ����� . When both
move, c - s c
c + o
we have by superposition, f' = f ����� , given in many texts. When
the c - s
distance between source and observer remains constant while both move,
Numerator and denominator become equal (remember my sign convention), so
the shift becomes zero, as it must. So here's an example of division by
a variable yielding a valid superposition; it's a linear system. I don't
know why I'm sort of surprised; I've known this for a long time.
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Is the Doppler effect really linear?
Started by ●August 19, 2003
Reply by ●August 19, 20032003-08-19
In article 3F42C944.504DA48E@ieee.org, Jerry Avins at jya@ieee.org wrote on 08/19/2003 21:05:> By considering wavelengths in air and propagation times, it seems > trivial to show that (c = speed of sound in air; s = speed of source > toward the observer; o = speed of observer toward the source; f = > original frequency; f' = measured frequency at the observer) for a > stationary observer, > > c c + o > f' = f ����� and for stationary source, f' = f ����� . When both > move, c - s c > c + o > we have by superposition, f' = f ����� , given in many texts. When > the c - s > distance between source and observer remains constant while both move, > Numerator and denominator become equal (remember my sign convention), so > the shift becomes zero, as it must. So here's an example of division by > a variable yielding a valid superposition; it's a linear system. I don't > know why I'm sort of surprised; I've known this for a long time.i've always considered it linear, but not time-invariant. sound transmission usually is linear unless obscenely high SPL levels are used (like what you might get in an explosion). at least that's what i thunk. r b-j
Reply by ●August 20, 20032003-08-20
Hello Jerry,
Jerry,
For sound the linearity depends on what is moving.
Moving receiver -> linear
Moving source -> nonlinear
The general formula (handles both Tx and Rx moving)
1+- Vr/v (+ is for approaching receiver)
v' = -----------
1-+Ve/v (+ is for receeding source)
Vr -> velocity of receiver
Ve-> velocity of emitter
v -> speed of sound
And of course for the case you cite, where both move together, the fraction
just simply becomes one!
Clay
"Jerry Avins" <jya@ieee.org> wrote in message
news:3F42C944.504DA48E@ieee.org...
> By considering wavelengths in air and propagation times, it seems
> trivial to show that (c = speed of sound in air; s = speed of source
> toward the observer; o = speed of observer toward the source; f =
> original frequency; f' = measured frequency at the observer) for a
> stationary observer,
>
> c c + o
> f' = f ----- and for stationary source, f' = f ----- . When both
> move, c - s c
> c + o
> we have by superposition, f' = f ----- , given in many texts. When
> the c - s
>
> distance between source and observer remains constant while both move,
> Numerator and denominator become equal (remember my sign convention), so
> the shift becomes zero, as it must. So here's an example of division by
> a variable yielding a valid superposition; it's a linear system. I don't
> know why I'm sort of surprised; I've known this for a long time.
>
> Jerry
> --
> Engineering is the art of making what you want from things you can get.
> �����������������������������������������������������������������������
Reply by ●August 20, 20032003-08-20
Whoops! that didn't format right. I ought to have been:
By considering wavelengths in air and propagation times, it seems
trivial to show that (c = speed of sound in air; s = speed of source
toward the observer; o = speed of observer toward the source; f =
original frequency; f' = measured frequency at the observer) for a
stationary observer,
c c + o
f' = f ����� and for stationary source, f' = f ����� . When both
c - s c
c + o
move, we have by superposition, f' = f ����� , given in many texts.
c - s
When the distance between source and observer remains constant while
both move, numerator and denominator become equal (remember my sign
convention), so the shift becomes zero, as it must. So here's an example
of division by a variable yielding a valid superposition; it's a linear
system. I don't know why I'm sort of surprised; I've known this for a
long time.
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by ●August 20, 20032003-08-20
"Clay S. Turner" wrote:> > Hello Jerry, > > Jerry, > > For sound the linearity depends on what is moving. > > Moving receiver -> linear > > Moving source -> nonlinearWell, the source is in the denominator, which troubled me in the first place. How is the nonlinearity manifest?> > The general formula (handles both Tx and Rx moving) > > 1+-Vr/v (+ is for approaching receiver) > v' = ----------- > 1-+Ve/v (+ is for receeding source) > > Vr -> velocity of receiver > Ve-> velocity of emitter > v -> speed of soundThat's essentially what I wrote, leaving it to the reader to see that receding changed the signs. (The usual convention of physics of picking the sign according to whether x increases or not simplifies the case when both move in the same direction at different speeds.) I just factored it differently. Using your notation and assuming that v' is f'/f, v +- Vr f' = f ������� v -+ Ve> > And of course for the case you cite, where both move together, the > fraction just simply becomes one!I thought I wrote that too.> > Clay >... Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●August 20, 20032003-08-20
>>>>> "Clay" == Clay S Turner <physicsNOOOOSPPPPAMMMM@bellsouth.net> writes:Clay> Hello Jerry, Clay> Jerry, Clay> For sound the linearity depends on what is moving. Clay> Moving receiver -> linear Clay> Moving source -> nonlinear Clay> The general formula (handles both Tx and Rx moving) Clay> 1+- Vr/v (+ is for approaching receiver) Clay> v' = ----------- Clay> 1-+Ve/v (+ is for receeding source) Clay> Vr -> velocity of receiver Ve-> velocity of emitter Clay> v -> speed of sound What is v' here? I've forgotten a lot of my physics, but it seems that the frequency at the observer would not depend on whether the observer is fixed and the source moving or the observer is moving and the source is fixed, as long as the relative motion is the same. Ray
Reply by ●August 20, 20032003-08-20
Hello Ray,
Sorry about the typo. My fingers gor ahead of my brain. How about:
1+- Vr/v (+ is for approaching receiver)
f' = f -----------
1-+Ve/v (+ is for receeding source)
Clay
"Raymond Toy" <toy@rtp.ericsson.se> wrote in message
news:4nhe4c1dm3.fsf@edgedsp4.rtp.ericsson.se...
> >>>>> "Clay" == Clay S Turner <physicsNOOOOSPPPPAMMMM@bellsouth.net>
writes:
>
> Clay> Hello Jerry,
> Clay> Jerry,
>
> Clay> For sound the linearity depends on what is moving.
>
> Clay> Moving receiver -> linear
>
> Clay> Moving source -> nonlinear
>
> Clay> The general formula (handles both Tx and Rx moving)
>
> Clay> 1+- Vr/v (+ is for approaching receiver)
> Clay> v' = -----------
> Clay> 1-+Ve/v (+ is for receeding source)
>
> Clay> Vr -> velocity of receiver
> Ve-> velocity of emitter
> Clay> v -> speed of sound
>
> What is v' here?
>
> I've forgotten a lot of my physics, but it seems that the frequency at
> the observer would not depend on whether the observer is fixed and the
> source moving or the observer is moving and the source is fixed, as
> long as the relative motion is the same.
>
> Ray
Reply by ●August 20, 20032003-08-20
Raymond Toy wrote:> > >>>>> "Clay" == Clay S Turner <physicsNOOOOSPPPPAMMMM@bellsouth.net> writes: > > Clay> Hello Jerry, > Clay> Jerry, > > Clay> For sound the linearity depends on what is moving. > > Clay> Moving receiver -> linear > > Clay> Moving source -> nonlinear > > Clay> The general formula (handles both Tx and Rx moving) > > Clay> 1 +- Vr/v (+ is for approaching receiver) > Clay> v' = ----------- > Clay> 1 -+ Ve/v (+ is for receeding source) > > Clay> Vr -> velocity of receiver > Ve-> velocity of emitter > Clay> v -> speed of sound > > What is v' here? > > I've forgotten a lot of my physics, but it seems that the frequency at > the observer would not depend on whether the observer is fixed and the > source moving or the observer is moving and the source is fixed, as > long as the relative motion is the same. > > RayRay, v' is either a slip of the mind or it stands for f'/f. The received frequency depends only on the relative motion with light (relativity applies!) but not with sound. A stationary source produces a wavelength of c/f, and by moving, the observer crosses fewer (receding) or more (approaching) crests in a given time. This effect is in the numerator. A moving source produces a shorter wavelength in its direction of motion and a longer one in the direction it moves away from. If you work it out, you will see that the effect of this on the observer is in the denominator. For low speeds, the difference is second order: 1 + .01 is nearly equal to 1/.99. Even the difference between 1.1 and 1/.9 is hard to measure in the real world (1%). Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●August 20, 20032003-08-20
Jerry and all I always thought (similar to what Ray thought) that the frequency strictly depends on the relative motion (and not on who is moving). Do you mean to say that a moving source will "produce" different frequency in different directions? Also what do you mean when you say "relativity applies"? Abhijit "Jerry Avins" <jya@ieee.org> wrote in message news:3F43B762.524703D6@ieee.org...> Raymond Toy wrote: > > > > >>>>> "Clay" == Clay S Turner <physicsNOOOOSPPPPAMMMM@bellsouth.net>writes:> > > > Clay> Hello Jerry, > > Clay> Jerry, > > > > Clay> For sound the linearity depends on what is moving. > > > > Clay> Moving receiver -> linear > > > > Clay> Moving source -> nonlinear > > > > Clay> The general formula (handles both Tx and Rx moving) > > > > Clay> 1 +- Vr/v (+ is for approaching receiver) > > Clay> v' = ----------- > > Clay> 1 -+ Ve/v (+ is for receeding source) > > > > Clay> Vr -> velocity of receiver > > Ve-> velocity of emitter > > Clay> v -> speed of sound > > > > What is v' here? > > > > I've forgotten a lot of my physics, but it seems that the frequency at > > the observer would not depend on whether the observer is fixed and the > > source moving or the observer is moving and the source is fixed, as > > long as the relative motion is the same. > > > > Ray > > Ray, > > v' is either a slip of the mind or it stands for f'/f. > > The received frequency depends only on the relative motion with light > (relativity applies!) but not with sound. A stationary source produces a > wavelength of c/f, and by moving, the observer crosses fewer (receding) > or more (approaching) crests in a given time. This effect is in the > numerator. A moving source produces a shorter wavelength in its > direction of motion and a longer one in the direction it moves away > from. If you work it out, you will see that the effect of this on the > observer is in the denominator. For low speeds, the difference is second > order: 1 + .01 is nearly equal to 1/.99. Even the difference between 1.1 > and 1/.9 is hard to measure in the real world (1%). > > Jerry > -- > Engineering is the art of making what you want from things you can get. > �����������������������������������������������������������������������
Reply by ●August 20, 20032003-08-20
Abhijit wrote:> > Jerry and all > > I always thought (similar to what Ray thought) that the frequency strictly > depends on the relative motion (and not on who is moving).You were wrong in detail. For speeds that are low relative to the speed of sound, there is little observable difference.> > Do you mean to say that a moving source will "produce" different frequency > in different directions?The wavelength in the air around a moving source is different in different directions. You knew that, but you apparently didn't know you knew. Stationary observers to the front and rear of the moving source hear different frequencies.> > Also what do you mean when you say "relativity applies"?Lorenz contraction, e=mc�, and all that. Light travels at the same speed with respect to each observer, irrespective of an observer's or emitter's speed. Doppler shift with light depends only on relative motion.> > Abhijit...>Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
