Let's say that i want to find d xP(t)/dt for this equation: xP (t) = arctan [xQ(t), xI(t)] d xP(t)/dt = ? Does anyone know the solution? I'm just confused about taking the derivative of a two valued function. Since there are two functions (xQ(t) and xI(t)), i assume the chain rule would be involved.. Thanks, any help would be much appreciated
first derivative of the two valued arctan function?
Started by ●December 15, 2007
Reply by ●December 15, 20072007-12-15
maxplanck wrote:> Let's say that i want to find d xP(t)/dt for this equation: > > xP (t) = arctan [xQ(t), xI(t)] > > > d xP(t)/dt = ? > > Does anyone know the solution? I'm just confused about taking the > derivative of a two valued function. Since there are two functions (xQ(t) > and xI(t)), i assume the chain rule would be involved..There's a formula for the derivative of arctangent(U). Let P = I + jQ Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●December 15, 20072007-12-15
"maxplanck" <erik.bowen@comcast.net> writes:> Let's say that i want to find d xP(t)/dt for this equation: > > xP (t) = arctan [xQ(t), xI(t)] > > > d xP(t)/dt = ? > > Does anyone know the solution? I'm just confused about taking the > derivative of a two valued function. Since there are two functions (xQ(t) > and xI(t)), i assume the chain rule would be involved..The two argument arctan function arctan(x,y) is just arctan(y/x)+k*pi, where the k*pi term puts the result in the correct quadrant. Forget about the k*pi term--it's just a constant, so its derivative is zero.(*)) Anyway, your answer is just 1/(1+(y/x)^2) times the derivative of (y/x). Note that the two functions for your chain rule are not x and y, but arctan and (y/x). Does this help? (*) k*pi isn't constant in neighborhoods where x=0. But if you ignore this fact, you get the right answer anyway. Depending on what you're using this for, you may have to prove this. Scott -- Scott Hemphill hemphill@alumni.caltech.edu "This isn't flying. This is falling, with style." -- Buzz Lightyear
Reply by ●December 16, 20072007-12-16
On Dec 16, 10:48 am, "maxplanck" <erik.bo...@comcast.net> wrote:> Let's say that i want to find d xP(t)/dt for this equation: > > xP (t) = arctan [xQ(t), xI(t)] > > d xP(t)/dt = ? > > Does anyone know the solution? I'm just confused about taking the > derivative of a two valued function. Since there are two functions (xQ(t) > and xI(t)), i assume the chain rule would be involved.. > > Thanks, any help would be much appreciatedAre you demodulating a signal? If so this is not the best way. Use the formula QI'-IQ'/(I^2+Q^2) where ' is derivative which can be done with a simple difference or something more complicated (say an FIR filter approximating a derivative). Hardy
Reply by ●December 16, 20072007-12-16
>"maxplanck" <erik.bowen@comcast.net> writes: > >> Let's say that i want to find d xP(t)/dt for this equation: >> >> xP (t) = arctan [xQ(t), xI(t)] >> >> >> d xP(t)/dt = ? >> >> Does anyone know the solution? I'm just confused about taking the >> derivative of a two valued function. Since there are two functions(xQ(t)>> and xI(t)), i assume the chain rule would be involved.. > >The two argument arctan function arctan(x,y) is just arctan(y/x)+k*pi, >where the k*pi term puts the result in the correct quadrant. Forget >about the k*pi term--it's just a constant, so its derivative iszero.(*))>Anyway, your answer is just 1/(1+(y/x)^2) times the derivative of >(y/x). Note that the two functions for your chain rule are not x and >y, but arctan and (y/x). > >Does this help? > >(*) k*pi isn't constant in neighborhoods where x=0. But if you ignore >this fact, you get the right answer anyway. Depending on what you're >using this for, you may have to prove this. > >Scott >-- >Scott Hemphill hemphill@alumni.caltech.edu >"This isn't flying. This is falling, with style." -- Buzz Lightyear >this is just what i was looking for. Thanks a lot!!!
Reply by ●December 16, 20072007-12-16
>On Dec 16, 10:48 am, "maxplanck" <erik.bo...@comcast.net> wrote: >> Let's say that i want to find d xP(t)/dt for this equation: >> >> xP (t) = arctan [xQ(t), xI(t)] >> >> d xP(t)/dt = ? >> >> Does anyone know the solution? I'm just confused about taking the >> derivative of a two valued function. Since there are two functions(xQ(t)>> and xI(t)), i assume the chain rule would be involved.. >> >> Thanks, any help would be much appreciated > >Are you demodulating a signal? If so this is not the best way. Use the >formula > >QI'-IQ'/(I^2+Q^2) > >where ' is derivative which can be done with a simple difference or >something more complicated (say an FIR filter approximating a >derivative). > > >HardyThanks! This is in line with what Scott said. When i calculated the derivative of xP(t) as Scott suggested, the result is: QI'-IQ'/(I^2+Q^2). Thanks a lot for replying, i feel much more confident about the result now!
Reply by ●December 16, 20072007-12-16
maxplanck wrote:>> On Dec 16, 10:48 am, "maxplanck" <erik.bo...@comcast.net> wrote: >>> Let's say that i want to find d xP(t)/dt for this equation: >>> >>> xP (t) = arctan [xQ(t), xI(t)] >>> >>> d xP(t)/dt = ? >>> >>> Does anyone know the solution? I'm just confused about taking the >>> derivative of a two valued function. Since there are two functions > (xQ(t) >>> and xI(t)), i assume the chain rule would be involved.. >>> >>> Thanks, any help would be much appreciated >> Are you demodulating a signal? If so this is not the best way. Use the >> formula >> >> QI'-IQ'/(I^2+Q^2) >> >> where ' is derivative which can be done with a simple difference or >> something more complicated (say an FIR filter approximating a >> derivative). >> >> >> Hardy > > Thanks! This is in line with what Scott said. When i calculated the > derivative of xP(t) as Scott suggested, the result is: QI'-IQ'/(I^2+Q^2). > > Thanks a lot for replying, i feel much more confident about the result > now!Next time, if you write atan2(), I'll know what you mean. Words matter. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●December 16, 20072007-12-16
>Next time, if you write atan2(), I'll know what you mean. Words matter. > >Jerry >-- >Engineering is the art of making what you want from things you can get. >�����������������������������������������������������������������������Thanks Jerry, i will do that next time.
Reply by ●December 16, 20072007-12-16
On Sat, 15 Dec 2007 15:48:27 -0600, maxplanck wrote:> Let's say that i want to find d xP(t)/dt for this equation: > > xP (t) = arctan [xQ(t), xI(t)] > > > d xP(t)/dt = ? > > Does anyone know the solution? I'm just confused about taking the > derivative of a two valued function. Since there are two functions > (xQ(t) and xI(t)), i assume the chain rule would be involved.. > > > Thanks, any help would be much appreciatedd/dx f(x1(x), x2(x) should just be dx1/dx * d/dx1 f(x1(x), x2(x)) + dx2/dx * d/dx2 f(x1(x), x2(x)) Except in all those special circumstances that Calculus books warn about, but never happen unless you regularly visit the more seedy parts of Mathemagic land. -- Tim Wescott Control systems and communications consulting http://www.wescottdesign.com Need to learn how to apply control theory in your embedded system? "Applied Control Theory for Embedded Systems" by Tim Wescott Elsevier/Newnes, http://www.wescottdesign.com/actfes/actfes.html
Reply by ●December 17, 20072007-12-17
>>On Dec 16, 10:48 am, "maxplanck" <erik.bo...@comcast.net> wrote: >>> Let's say that i want to find d xP(t)/dt for this equation: >>> >>> xP (t) = arctan [xQ(t), xI(t)] >>> >>> d xP(t)/dt = ? >>> >>> Does anyone know the solution? I'm just confused about taking the >>> derivative of a two valued function. Since there are two functions >(xQ(t) >>> and xI(t)), i assume the chain rule would be involved.. >>> >>> Thanks, any help would be much appreciated >> >>Are you demodulating a signal? If so this is not the best way. Use the >>formula >> >>QI'-IQ'/(I^2+Q^2) >> >>where ' is derivative which can be done with a simple difference or >>something more complicated (say an FIR filter approximating a >>derivative). >> >> >>Hardy > >Thanks! This is in line with what Scott said. When i calculated the >derivative of xP(t) as Scott suggested, the result is:QI'-IQ'/(I^2+Q^2).> >Thanks a lot for replying, i feel much more confident about the result >now!woops, looks like it actually should be: IQ'-QI'/(I^2+Q^2)
