Hi, I'm an electrical engineering undergraduate in my final year and I'm still struggling to get my head around complex signals. I would be really grateful if someone could give me an answer to something that has had me stumped for the couple of days. What would the signal x[n] = e^(j(pi/8)n) look like on a scope, compared to say, the signal x[n] = sin((pi/8)n)?
Basic question - what does the signal e^jwt look like?
Started by ●March 30, 2009
Reply by ●March 30, 20092009-03-30
On Mar 30, 10:02�am, "ty34" <minesa...@hotmail.com> wrote:> Hi, I'm an electrical engineering undergraduate in my final year and I'm > still struggling to get my head around complex signals. > > I would be really grateful if someone could give me an answer to something > that has had me stumped for the couple of days. What would the signal > > x[n] = e^(j(pi/8)n) > > look like on a scope, compared to say, the signal x[n] = sin((pi/8)n)?Well scopes display real valued signals. And the signal you propose is complex valued. Of course you may display the real part on one axis and the imaginary part of the other axis then the result is points on a circle. I doubt that is what you want ed though. Now you can do a three dimensional plot and then your points will be on a helix. IHTH, Clay
Reply by ●March 30, 20092009-03-30
>On Mar 30, 10:02=A0am, "ty34" <minesa...@hotmail.com> wrote: >> Hi, I'm an electrical engineering undergraduate in my final year andI'm>> still struggling to get my head around complex signals. >> >> I would be really grateful if someone could give me an answer tosomethin=>g >> that has had me stumped for the couple of days. What would the signal >> >> x[n] =3D e^(j(pi/8)n) >> >> look like on a scope, compared to say, the signal x[n] =3Dsin((pi/8)n)?> >Well scopes display real valued signals. And the signal you propose is >complex valued. Of course you may display the real part on one axis >and the imaginary part of the other axis then the result is points on >a circle. I doubt that is what you want ed though. Now you can do a >three dimensional plot and then your points will be on a helix. > >IHTH, > >Clay >Thanks very much, that is a great help - I can see what your saying about it forming a circle in the complex plane and that you'd need two separate mag vs time traces to represent it. The real source of my confusion is that i've somehow got it into my head that the 'j' operator simply represents a phase shift of 90deg. ie. sin wt = -jcos(wt) But if that was the case, then e^jx = cos x + j (-j cos x) = 2 cos x and this is a non-complex signal that can be displayed on a scope. Can anyone show me where I am going wrong? Tim
Reply by ●March 30, 20092009-03-30
On Mar 30, 10:49�am, "ty34" <minesa...@hotmail.com> wrote:> >On Mar 30, 10:02=A0am, "ty34" <minesa...@hotmail.com> wrote: > >> Hi, I'm an electrical engineering undergraduate in my final year and > I'm > >> still struggling to get my head around complex signals. > > >> I would be really grateful if someone could give me an answer to > somethin= > >g > >> that has had me stumped for the couple of days. What would the signal > > >> x[n] =3D e^(j(pi/8)n) > > >> look like on a scope, compared to say, the signal x[n] =3D > sin((pi/8)n)? > > >Well scopes display real valued signals. And the signal you propose is > >complex valued. Of course you may display the real part on one axis > >and the imaginary part of the other axis then the result is points on > >a circle. I doubt that is what you want ed though. Now you can do a > >three dimensional plot and then your points will be on a helix. > > >IHTH, > > >Clay > > Thanks very much, that is a great help - I can see what your saying about > it forming a circle in the complex plane and that you'd need two separate > mag vs time traces to represent it. > > The real source of my confusion is that i've somehow got it into my head > that the 'j' operator simply represents a phase shift of 90deg. > > ie. sin wt = -jcos(wt) > > But if that was the case, > > then e^jx = cos x + j (-j cos x) = 2 cos x > > and this is a non-complex signal that can be displayed on a scope. Can > anyone show me where I am going wrong? > > Tim- Hide quoted text - > > - Show quoted text -Tim, As you surmised there is an error. Euler's identity is e^(jx) = cos(x) +j*sin(x) However your comment about rotation is not totally incorrect. A complex phaser (two component vector written as a+jb) will be rotated by theta degrees when the phaser is multiplied by e^(j*theta). So when theta=90 degrees, e^(j*theta) is j. But don't confuse mutiplying a phaser with just multiplying one of its componets by a complex entity. IHTH, Clay
Reply by ●March 30, 20092009-03-30
"ty34" <minesadab@hotmail.com> wrote in message news:G62dnVmJXdx_Rk3UnZ2dnUVZ_tjinZ2d@giganews.com...> The real source of my confusion is that i've somehow got it into my head > that the 'j' operator simply represents a phase shift of 90deg.Yes, that is true, but only when applied to a complex phasor (e^jwt), or its conjugate e^(-jwt)> ie. sin wt = -jcos(wt) >That is not the case because cos wt = ( e^(jwt) + e^(-jwt) ) / 2 whereas sin wt = ( e^(jwt) - e^(-jwt) ) / 2j Now, j would represent a phase shift or 90 degrees anticlockwise, which would be OK for the e^(jwt) term, but to achieve the required effect on the e(-jwt) term, ie a phase advance in the clockwise direction, you'd have to multiply by -j Also, to convert from sin to cos, you'd have to somehow change the sign of the e^(-jwt) term. I had a similar confusion to you. if we take the cos term, ( e^(jwt) + e^(-jwt) ) / 2 and apply it to a circuit, we get two results, one for each of the exponential terms, but one is always the complex conjugate of the other. So why do the work twice? As mathematicians and engineers we are always lazy. So we apply only the term e(jwt) to a circuit to do our calculations, and if we want the result due to excitation by e^(-jwt) it is merely the complex conjugate of what we calculated in the first place. So, we cannot really visulise e^(jwt) which is where we came in. We can visualise any sinusoid as two counter-rotating phasors
Reply by ●March 30, 20092009-03-30
>On Mar 30, 10:49=A0am, "ty34" <minesa...@hotmail.com> wrote: >> >On Mar 30, 10:02=3DA0am, "ty34" <minesa...@hotmail.com> wrote: >> >> Hi, I'm an electrical engineering undergraduate in my final yearand>> I'm >> >> still struggling to get my head around complex signals. >> >> >> I would be really grateful if someone could give me an answer to >> somethin=3D >> >g >> >> that has had me stumped for the couple of days. What would thesignal>> >> >> x[n] =3D3D e^(j(pi/8)n) >> >> >> look like on a scope, compared to say, the signal x[n] =3D3D >> sin((pi/8)n)? >> >> >Well scopes display real valued signals. And the signal you proposeis>> >complex valued. Of course you may display the real part on one axis >> >and the imaginary part of the other axis then the result is points on >> >a circle. I doubt that is what you want ed though. Now you can do a >> >three dimensional plot and then your points will be on a helix. >> >> >IHTH, >> >> >Clay >> >> Thanks very much, that is a great help - I can see what your sayingabout>> it forming a circle in the complex plane and that you'd need twoseparate>> mag vs time traces to represent it. >> >> The real source of my confusion is that i've somehow got it into myhead>> that the 'j' operator simply represents a phase shift of 90deg. >> >> ie. sin wt =3D -jcos(wt) >> >> But if that was the case, >> >> then e^jx =3D cos x + j (-j cos x) =3D 2 cos x >> >> and this is a non-complex signal that can be displayed on a scope. Can >> anyone show me where I am going wrong? >> >> Tim- Hide quoted text - >> >> - Show quoted text - > >Tim, > >As you surmised there is an error. Euler's identity is e^(jx) =3D cos(x) >+j*sin(x) > >However your comment about rotation is not totally incorrect. A >complex phaser (two component vector written as a+jb) will be rotated >by theta degrees when the phaser is multiplied by e^(j*theta). So when >theta=3D90 degrees, e^(j*theta) is j. But don't confuse mutiplying a >phaser with just multiplying one of its componets by a complex entity. > >IHTH, > >ClayThanks again Clay. You're right - I saw x(n) = cos (n) + j sin(n) as a combination of two out-of-phase signals varying in the same axis. Its clear to me now just how wrong that was. Thank you very much - I'm very grateful.
Reply by ●March 30, 20092009-03-30
On Mon, 30 Mar 2009 09:49:06 -0500, ty34 wrote:>>On Mar 30, 10:02=A0am, "ty34" <minesa...@hotmail.com> wrote: >>> Hi, I'm an electrical engineering undergraduate in my final year and > I'm >>> still struggling to get my head around complex signals. >>> >>> I would be really grateful if someone could give me an answer to > somethin= >>g >>> that has had me stumped for the couple of days. What would the signal >>> >>> x[n] =3D e^(j(pi/8)n) >>> >>> look like on a scope, compared to say, the signal x[n] =3D > sin((pi/8)n)? >> >>Well scopes display real valued signals. And the signal you propose is >>complex valued. Of course you may display the real part on one axis and >>the imaginary part of the other axis then the result is points on a >>circle. I doubt that is what you want ed though. Now you can do a three >>dimensional plot and then your points will be on a helix. >> >>IHTH, >> >>Clay >> >> > Thanks very much, that is a great help - I can see what your saying > about it forming a circle in the complex plane and that you'd need two > separate mag vs time traces to represent it. > > The real source of my confusion is that i've somehow got it into my head > that the 'j' operator simply represents a phase shift of 90deg. > > ie. sin wt = -jcos(wt) > > But if that was the case, > > then e^jx = cos x + j (-j cos x) = 2 cos x > > and this is a non-complex signal that can be displayed on a scope. Can > anyone show me where I am going wrong? > > TimNo no no! Transfer functions that generate 90 degree phase shifts will have j's in them, but j is not an operator. It is, as oddball as it sounds, a number -- sqrt(-1). e^jx = cos x + j sin x. So if you traced the real part of your signal vs. the imaginary part, you'd have a pair of sine waves, in quadrature. You can do this in a math package like Matlab or Scilab -- make some x = exp(i * t), then plot Real(x) and Imag(x) vs. time. -- http://www.wescottdesign.com
Reply by ●March 30, 20092009-03-30
>"ty34" <minesadab@hotmail.com> wrote in message >news:G62dnVmJXdx_Rk3UnZ2dnUVZ_tjinZ2d@giganews.com... >> The real source of my confusion is that i've somehow got it into myhead>> that the 'j' operator simply represents a phase shift of 90deg. > >Yes, that is true, but only when applied to a complex phasor (e^jwt), >or its conjugate e^(-jwt) > >> ie. sin wt = -jcos(wt) >> > >That is not the case because cos wt = ( e^(jwt) + e^(-jwt) ) / 2 > >whereas sin wt = ( e^(jwt) - e^(-jwt) ) / 2j > >Now, j would represent a phase shift or 90 degrees anticlockwise, which >would be OK for the e^(jwt) term, but to achieve the required effect onthe>e(-jwt) term, ie a phase advance in the clockwise direction, you'd have >to multiply by -j > >Also, to convert from sin to cos, you'd have to somehow change the signof>the >e^(-jwt) term. > >I had a similar confusion to you. if we take the cos term, ( e^(jwt) + >e^(-jwt) ) / 2 >and apply it to a circuit, we get two results, one for each of the >exponential >terms, but one is always the complex conjugate of the other. > >So why do the work twice? As mathematicians and engineers we are >always lazy. > >So we apply only the term e(jwt) to a circuit to do our calculations, andif>we want the result due to excitation by e^(-jwt) it is merely the complex>conjugate >of what we calculated in the first place. > >So, we cannot really visulise e^(jwt) which is where we came in. We can >visualise any sinusoid as two counter-rotating phasorsThanks Alun - thats a big help too. Its all starting to come together now. Although I knew the equation cos wt = ( e^(jwt) + e^(-jwt) ) / 2 off by heart, I didn't understand it. I can now see that it represents two counter-rotating phasors which added end to end, produce a horizontally-varying point. And if you traced that point over time, it would form a cosinusiod. For a sine wave, the phasors are instead subtracted, and the result is then rotated by -90 deg. Thankyou very much. Until you helped out, I'd been stuck for two days trying to get to the bottom of all this.
Reply by ●March 30, 20092009-03-30
"ty34" <minesadab@hotmail.com> wrote in message news:coadnZD9DtmTaE3UnZ2dnUVZ_h6WnZ2d@giganews.com...> Although I knew the equation > > cos wt = ( e^(jwt) + e^(-jwt) ) / 2 > > off by heart, I didn't understand it.40 years ago, this was part of the school syllabus at age 17. Look up the Maclaurin infinite series expansions for e^(x), sin x and cos x e^(x) = 1 + x + x^2/2! + x^3/3! etc sin x = x - x^3/3! + x^5/5! - x^7/7! etc cos x = 1 - x^2/2! + x^4/4! - x^6/6! etc Note for sin and cos that you get every other term, but oscillating in sign. then replace x by jx in the above expansions (remembering that j^2 = -1 and that j^3 = -j etc), and you get the equivalents that give us e^jx = cos x + j sin x.
Reply by ●March 30, 20092009-03-30
On Mar 30, 10:02�am, "ty34" <minesa...@hotmail.com> wrote:> Hi, I'm an electrical engineering undergraduate in my final year and I'm > still struggling to get my head around complex signals. > > I would be really grateful if someone could give me an answer to something > that has had me stumped for the couple of days. What would the signal > > x[n] = e^(j(pi/8)n) > > look like on a scope, compared to say, the signal x[n] = sin((pi/8)n)?You can try looking at these flash programs. They might give you some insight: http://www.fourier-series.com/fourierseries2/complex_tutorial.html
