Search:
Search questions and share your knowledge with the world here on Questionpoint.

:: Categories ::

Interview Question in Phasors

Interview Question :: Signals help. Electrical engineers help

How would you write the signal for sin50pt as the real parts of rotating phasors. Also how would you write it as one-half the sum of a rotating phasors and its complex conjugate.

I'm not understanding what they are looking for. A little help.
?

Answers to "Signals help. Electrical engineers help"

RE: Signals help. Electrical engineers help?
I normally explain this in tutorials using a coil sitting on an overhead projector. The coil is obviously 3-dimensional - the "length" being time, and the other two dimensions (left/right and up/down) are the real and imaginary axes. This means it represents a complex phasor moving around in space - exp(jw) is a point on the unit circle which makes an angle of 'w' to the real axis, so exp(jwt) is the same point moving around the circle and thus forms a coil as it moves along the 't' axis. <see note 1 at the end of this answer> However, by looking at the image the projector makes on the screen, you don't see the imaginary part, and all you get is the "real" part of the coil, which is (if you can picture it in) a cosine wave. So, to answer your first question, the real part of a rotating phasor is a cosine wave. Hence cos(50\pi t) = Re[exp(j50\pi t)] and sin(50\pi t) = Re[exp(j(50\pi t + \pi/2)]. But of course that's cheating. What we need is a way to remove the imaginary axis mathematically. This is where the conjugate comes in. If you look at the phasor in terms of it's real and imaginary parts, we have: z = a + jb ...and the conjugate... z* = a - jb ...so the sum of the two is: z + z* = 2a + j(b - b) = 2a = 2 Re[z] In other words, add the phasor with its conjugate and you double its real part, but remove the imaginary part. Which is exactly what we want... well, almost. Ideally we'd write it as: (z + z*)/2 = Re[z] ...i.e. one-half the sum of the rotating phasor and its complex conjugate, where z = exp(jwt) = exp(j50\pi t). <note 1> I'm not sure it this made much sense when I read it back. Just swing a conker around in circles on the end of a peice of string. Now walk forwards. What path does the conker take? It should be a coil, imaginary axis points upwards, real axis points left and time axis points forwards. At any instant in time, the conker's position will be given by Aexp(jp), where p is the angle the string makes to horizontal and A is the length of the string.
Vote for this answer ::

RE: Signals help. Electrical engineers help?
We know from euler's equation that: exp(ix)=cos(x)+isin(x); There is a difference between exp(x) and exp(ix) that is exp(x) is a real number increasing with x, while exp(ix) is a complex number with absolute of 1 and phase x. Now the rotating phasor can be shown as: exp(iwt) where w is frequency in rad/sec and t is time. By t increasing, absolute remains 1 while phase increases (i.e. rotates). So, from euler's equation the real part is cos(wt) and imaginaty part is sin(wt). Complex conjugate of the phasor will be: cos(wt)-isin(wt), adding the two signals: [cos(wt)-isin(wt)] + [cos(wt)+isin(wt)] = 2cos(wt) so: exp(iwt) + [exp(iwt)]* = 2 cos(wt) and: cos(wt) = 1/2 [exp(iwt) + [exp(iwt)]* ] I hope my guidance would help
Vote for this answer ::

Notify me whenever an answer is posted for this question Update Alert Setting

Advertisement2

Top Tags

- Sharepoint Developer Interview Questions - Java Interview Questions - J2EE Interview Questions - ASP.NET Interview Questions - MOSS 2007 Interview Questions - WPF Interview Questions - Microsoft CRM Interview Questions - Technical Interview Questions for Programmers - SSRS Interview Questions - SSAS Interview Questions

Search questions and share your knowledge with the world here on Questionpoint.

:: Categories ::

Interview Question in Phasors

Interview Question :: Signals help. Electrical engineers help

Top Tags