Hello People In underwater communications, assume there is one direct path and one surface reflected path. And the receiver is moving at 0.1 m/s (either towards or away from transmitter). My question is: do both the paths have same doppler effect? Your guidance will be greatly appreciated. Regards, Chintan
Doppler effect
Started by ●July 8, 2008
Reply by ●July 8, 20082008-07-08
"cpshah99" <cpshah99@rediffmail.com> wrote in message news:ip-dnSHHVZ2J4u7VnZ2dnUVZ_rPinZ2d@giganews.com...> Hello People > > In underwater communications, assume there is one direct path and one > surface reflected path. And the receiver is moving at 0.1 m/s (either > towards or away from transmitter). > > My question is: do both the paths have same doppler effect? > > Your guidance will be greatly appreciated. > > Regards, > > ChintanThe surface acts like a mirror so the reflection looks like it's coming from a source that's above the surface. Motion would be the same but the geometry is not - whether it's negligible or not is something you can decide. If the source and the receiver are at the same depth then the radial velocity (which is what's important for Doppler) is the same as the horizontal velocity - which I assume is what you mean as "moving toward or moving away". The reflected source would have the same horizontal velocity but not the same radial velocity. Can you see why? Fred
Reply by ●July 8, 20082008-07-08
Fred Marshall wrote: ...> The reflected source would have the same horizontal velocity but not the > same radial velocity. Can you see why?Sure. Geometry. :-) Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●July 8, 20082008-07-08
> >"cpshah99" <cpshah99@rediffmail.com> wrote in message >news:ip-dnSHHVZ2J4u7VnZ2dnUVZ_rPinZ2d@giganews.com... >> Hello People >> >> In underwater communications, assume there is one direct path and one >> surface reflected path. And the receiver is moving at 0.1 m/s (either >> towards or away from transmitter). >> >> My question is: do both the paths have same doppler effect? >> >> Your guidance will be greatly appreciated. >> >> Regards, >> >> Chintan > >The surface acts like a mirror so the reflection looks like it's comingfrom>a source that's above the surface. Motion would be the same but the >geometry is not - whether it's negligible or not is something you can >decide. > >If the source and the receiver are at the same depth then the radial >velocity (which is what's important for Doppler) is the same as the >horizontal velocity - which I assume is what you mean as "moving towardor>moving away". > >The reflected source would have the same horizontal velocity but not the>same radial velocity. Can you see why? > >Fred > > >%%% HI Fred Thanks very much for replying. I am sorry but I am not getting your point. Just assume that there is no surface reflection. Also the TX and RX are at same depth. TX is steady and RX is moving. So the signal will be expnaded by the velocity. Is this correct? And as you are saying, 'The reflected source would have the same horizontal velocity but not the same radial velocity.' I am not getting this point. I have found all the angles and lenths of signal but I have no idea how to use them. Thanks again. Chintan
Reply by ●July 8, 20082008-07-08
cpshah99 wrote:> Hello People > > In underwater communications, assume there is one direct path and one > surface reflected path. And the receiver is moving at 0.1 m/s (either > towards or away from transmitter). > > My question is: do both the paths have same doppler effect? > > Your guidance will be greatly appreciated. > > Regards, > > ChintanNo they don't, because the reflected path-length is changing at a different rate from the direct path-length. Regards, John
Reply by ●July 8, 20082008-07-08
"cpshah99" <cpshah99@rediffmail.com> wrote in message news:S4udnV1RjK-nUe7VnZ2dnUVZ_hzinZ2d@giganews.com...> > >>"cpshah99" <cpshah99@rediffmail.com> wrote in message >>news:ip-dnSHHVZ2J4u7VnZ2dnUVZ_rPinZ2d@giganews.com... >>> Hello People >>> >>> In underwater communications, assume there is one direct path and one >>> surface reflected path. And the receiver is moving at 0.1 m/s (either >>> towards or away from transmitter). >>> >>> My question is: do both the paths have same doppler effect? >>> >>> Your guidance will be greatly appreciated. >>> >>> Regards, >>> >>> Chintan >> >>The surface acts like a mirror so the reflection looks like it's coming > from >>a source that's above the surface. Motion would be the same but the >>geometry is not - whether it's negligible or not is something you can >>decide. >> >>If the source and the receiver are at the same depth then the radial >>velocity (which is what's important for Doppler) is the same as the >>horizontal velocity - which I assume is what you mean as "moving toward > or >>moving away". >> >>The reflected source would have the same horizontal velocity but not the > >>same radial velocity. Can you see why? >> >>Fred >> >> >> > %%% > > HI Fred > > Thanks very much for replying. > > I am sorry but I am not getting your point. Just assume that there is no > surface reflection. Also the TX and RX are at same depth. TX is steady and > RX is moving. So the signal will be expnaded by the velocity. Is this > correct? > > And as you are saying, 'The reflected source would have the same > horizontal velocity but not the same radial velocity.' I am not getting > this point. > > I have found all the angles and lenths of signal but I have no idea how to > use them. > > Thanks again. > > ChintanPresumably you have the expression for Doppler shift as a function of speed of sound and radial velocity. If we ignore the reflected path for a moment then: - if the depths of source and receiver are the same then the radial velocity is the same as the horizontal velocity toward or away from. This resolves two ways: .. If the horizontal velocity V is 100% radial then you know the radial velocity. .. If the horizontal velocity V is at a right angle (90 degrees) from the line connecting the source and receiver, then the radial velocity is zero. .. If the horizontal velocity V is A degrees from the line connecting the source and receiver, then the radial velocity is V*cos(a) and the tangential velocity is V*sin(A), contributing zero Doppler as above. So, it's simpler to deal with the 100% case isn't it? Let's stick with that for now in the discussion below: If the source and the receiver are at different depths and the movement is purely a constant horizontal velocity V then we are motivated to calculate the vertical angle B between the source and the receiver. The radial velocity is V*cos(B) and is ever-changing as the distance between the source and receiver changes. In the long-distance limit, the angle goes to zero and the radial velocity is V. In the short-distance limit, the source and receiver are at the same x,y coordinate and the radial velocity is zero per V*cos(B). The reflected path situation is the same as the above where the depths are different. In this special case, the vertical "distance" between the source and the receiver is 2 times the depth of the observed object. Fred
Reply by ●July 9, 20082008-07-09
On 8 Jul, 22:21, "cpshah99" <cpsha...@rediffmail.com> wrote:> >"cpshah99" <cpsha...@rediffmail.com> wrote in message > >news:ip-dnSHHVZ2J4u7VnZ2dnUVZ_rPinZ2d@giganews.com... > >> Hello People > > >> In underwater communications, assume there is one direct path and one > >> surface reflected path. And the receiver is moving at 0.1 m/s (either > >> towards or away from transmitter). > > >> My question is: do both the paths have same doppler effect?....> Just assume that there is no > surface reflection.Underwater acoustics? Famous last words...> Also the TX and RX are at same depth. TX is steady and > RX is moving. So the signal will be expnaded by the velocity. Is this > correct?If you mean 'non-zero horizontal velocity might cause a Doppler shift' then yes, this is correct.> And as you are saying, 'The reflected source would have the same > horizontal velocity but not the same radial velocity.' I am not getting > this point.Then re-introduce the surface reflection. You need it to understand Fred's statement.> I have found all the angles and lenths of signal but I have no idea how to > use them.Then use a different geometrical setup (2D only): - Tx and Rx at different depths - No movement; no velocities 1) What is the horizontal distance between Tx and Rx? 2) What is the vertical distance between Tx and Rx? 3) What is the radial distance between Tx and Rx? - Let the Transmitter move in some direction. 1) What is the horizontal velocity of Tx relative to Rx? 2) What is the vertical velocity of Tx relative to Rx? 3) What is the radial velocity of Tx relative to Rx? - Go back and read Fred's post once more. Rune
Reply by ●July 9, 20082008-07-09
Hi Rune As you said>Then use a different geometrical setup (2D only): > >- Tx and Rx at different depths >- No movement; no velocities > >1) What is the horizontal distance between Tx and Rx? >2) What is the vertical distance between Tx and Rx? >3) What is the radial distance between Tx and Rx? >So I selected depth and range of the ocean as 100 meters each. Now lets say TX is at depth of 80 meters from surface and RX is at depth of 20 meters from surface. So: 1. The horizontal distance between TX and RX is 100 meters 2. The vertical distance between TX and RX is 60 meters 3. The radial distance between TX and RX is 116.61 meters. Is this correct? Hi Fred, thanks you very much. Now again I am keeping both TX and RX at same depth, 20 meters from surface. Starting with a simple case, both are steady: So no doppler. Now, TX is steady and RX is moving in straight line with velocity 0.1 m/s. So as the angle is 0, there for radial velocity is equal to horizontal velocity, i.e. v_r=v_h=0.1 m/s; So my doppler shift fd=v/c*f; Is this correct? Then I will move forward. Thanks again, Fred and Rune, for spending ur valueable time. Regards, Chintan
Reply by ●July 9, 20082008-07-09
On 9 Jul, 16:09, "cpshah99" <cpsha...@rediffmail.com> wrote:> Hi Rune > > As you said > > >Then use a different geometrical setup (2D only): > > >- Tx and Rx at different depths > >- No movement; no velocities > > >1) What is the horizontal distance between Tx and Rx? > >2) What is the vertical distance between Tx and Rx? > >3) What is the radial distance between Tx and Rx? > > So I selected depth and range of the ocean as 100 meters each. > > Now lets say TX is at depth of 80 meters from surface and RX is at depth > of 20 meters from surface. > > So: > 1. The horizontal distance between TX and RX is 100 meters > 2. The vertical distance between TX and RX is 60 meters > 3. The radial distance between TX and RX is 116.61 meters. > > Is this correct?Correct. Now apply the same line of reasoning to the path that is reflected at the surface. Rune
Reply by ●July 9, 20082008-07-09
Rune Allnor wrote:> On 9 Jul, 16:09, "cpshah99" <cpsha...@rediffmail.com> wrote: >> Hi Rune >> >> As you said >> >>> Then use a different geometrical setup (2D only): >>> - Tx and Rx at different depths >>> - No movement; no velocities >>> 1) What is the horizontal distance between Tx and Rx? >>> 2) What is the vertical distance between Tx and Rx? >>> 3) What is the radial distance between Tx and Rx? >> So I selected depth and range of the ocean as 100 meters each. >> >> Now lets say TX is at depth of 80 meters from surface and RX is at depth >> of 20 meters from surface. >> >> So: >> 1. The horizontal distance between TX and RX is 100 meters >> 2. The vertical distance between TX and RX is 60 meters >> 3. The radial distance between TX and RX is 116.61 meters. >> >> Is this correct? > > Correct. Now apply the same line of reasoning to the > path that is reflected at the surface.My first response in this thread (intended to be a hint) was "geometry". Consider a source and receiver 5 each meters below the surface and 25 meters apart. The reflected source will be 5 meters *above* the surface. How far is the phantom source from the receiver? If the source approaches to within 20 meters, how far will the phantom source be? 15 meters? 10? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
