Hello All, Conceptually I understand complex conjugate but what is the application of complex conjugates? Thanks in advance ... --------------------------------------- Posted through http://www.DSPRelated.com

# complex conjugate

Started by ●January 6, 2016

Reply by ●January 6, 20162016-01-06

>Hello All, > >Conceptually I understand complex conjugate but what is the applicationof>complex conjugates? > >Thanks in advance ... >--------------------------------------- >Posted through http://www.DSPRelated.comI can think of three main "uses": 1) To find the magnitude of a complex number. z = a + bi z_bar = a - bi |z| = sqrt( z * z_bar ) = sqrt( a*a + b*b ) 2) To perform complex division: z = (c+di)/(a+bi) Multiply numerator and denominator by a-bi z = (c+di)(a-bi)/(a*a+b*b) 3) The complex roots of polynomials with real valued coefficients come in conjugate pairs. The quadratic formula gives a clear example Ax^2 + Bx + C = 0 x = -B/(2A) +/- sqrt(B^2-4AC)/(2A) When the argument of the sqrt is negative this takes the form x = c +/- di It seems to me you meant to say you know the meaning "definitionally". The mirror image with respect to the real axis comes to mind. Ced --------------------------------------- Posted through http://www.DSPRelated.com

Reply by ●January 7, 20162016-01-07

>>Hello All, >> >>Conceptually I understand complex conjugate but what is the application >of >>complex conjugates? >> >>Thanks in advance ... >>--------------------------------------- >>Posted through http://www.DSPRelated.com > >I can think of three main "uses": > >1) To find the magnitude of a complex number. > >z = a + bi >z_bar = a - bi > >|z| = sqrt( z * z_bar ) = sqrt( a*a + b*b ) > >2) To perform complex division: > >z = (c+di)/(a+bi) > >Multiply numerator and denominator by a-bi > >z = (c+di)(a-bi)/(a*a+b*b) > >3) The complex roots of polynomials with real valued coefficients comein>conjugate pairs. The quadratic formula gives a clear example > >Ax^2 + Bx + C = 0 > >x = -B/(2A) +/- sqrt(B^2-4AC)/(2A) > >When the argument of the sqrt is negative this takes the form > >x = c +/- di > >It seems to me you meant to say you know the meaning "definitionally". >The mirror image with respect to the real axis comes to mind. >Dear Cedron, Thanks. I see that in communication systems, receivers using matrix of complex conjugate values (as sort of filters I guess). Hence my question. A simple question on the first use-case,>|z| = sqrt( z * z_bar ) = sqrt( a*a + b*b )if Z is in Cartesian form then one can directly compute sqrt (a*a + b*b). In fact that is what is happening above. So, I am trying to understand how introduction of complex conjugate term in the above equation helped? --------------------------------------- Posted through http://www.DSPRelated.com

Reply by ●January 7, 20162016-01-07

> >Dear Cedron, > >Thanks. I see that in communication systems, receivers using matrix of >complex conjugate values (as sort of filters I guess). Hence myquestion.> >A simple question on the first use-case, > >>|z| = sqrt( z * z_bar ) = sqrt( a*a + b*b ) > >if Z is in Cartesian form then one can directly compute sqrt (a*a +b*b).>In fact that is what is happening above. So, I am trying to understandhow>introduction of complex conjugate term in the above equation helped? > >--------------------------------------- >Posted through http://www.DSPRelated.comComplex conjugates are used to find the magnitudes of complex vectors just like with complex scalars. When you're doing the math, as in this snippet I did in one of your other threads, dealing with a complex number or vector as a single symbol is a lot more convenient. D^(*) * D = S^(*) * T^(*) * T * S = S^(*) * I * S ( I is the identity matrix ) = S^(*) * S When you are implementing an algorithm in code it is generally more efficient to use the component values. On certain platforms, you have a complex type so your implementation can resemble your math closer if that is what you want. a = ( z + z_bar ) / 2 b = ( z - z_bar ) / (2i) These are the transition. They are frequently found in the form: cos( th ) = ( e^(i*th) + e^(-i*th) ) / 2 sin( th ) = ( e^(i*th) - e^(-i*th) ) / (2i) Where 'th' stands for the Greek letter theta. To understand why e^(i*th) is the complex conjugate of e^(-i*th), and to understand complex numbers a little better, I recommend you read my first blog article titled "The Exponential Nature of the Complex Unit Circle" which can be found here: http://www.dsprelated.com/showarticle/754.php Complex conjugates pop up in DFTs when you have a real valued signal. The top half (or negative half) of the DFT is the complex conjugate of the bottom half. This actually comes from the fact that the roots of unity for any N are symmetric about the real axis. The DFT transform matrix (T) in my proof snippet above has the property that its inverse is also its conjugate transpose. Matrices that have this property have a special name, they are called Unitary Matrices. They are special because they preserve the lengths of the transformed vectors as proven in the snippet. Ced --------------------------------------- Posted through http://www.DSPRelated.com