[Complex exponential] Solutions must be wrong

In summary, the conversation discusses the accuracy of a z-plot for various complex exponential values. The conversation also delves into the concept of phasors and how they can be plotted in the z-plane. The conversation ultimately concludes that the discrepancy in the plotted values may be due to evaluating the exponentials at different times.
  • #1
JJ91
41
0

Homework Statement


Hi,

I would like to get feedback if my z-plot is accurate for the following complex exponentiala:
Code:
a=2*exp(j*∏*t)
b=2*exp(j*∏*-1.25)
c=1*exp(j*∏*t)
d=-j*exp(j*∏*t)

Further analysis:
a= -2 because cos(∏) = -1 and sin(∏)=0
b= actual complex number A*cos(∅)+j*sin(∅)
c= -1 because cos(∏) = -1 and sin(∏)=0
d=+j because -j*cos(∏)= +j and -j*j*sin(∏)=0

However according to solutions from other source my analysis is wrong.

I've decided to plot both graphs in MatLab so you can see the difference:
(my solution on the left)/(solution from other source on the right)
2hgh20g.jpg


2.
A*exp(x)=A*cos(x)+A*j*sin(x)

3.
As stated in one, theory doesn't agree with solution found on the internet.
Also, in whatever form I put the exponential in the function I keep getting totally different results.
e.g.
[2*exp(j*pi), 2*exp(j*pi*-1.25), 1*exp(j*pi), -j*exp(j*pi)] would not give me similar amplitude as above for some reason.

P.S. Ignore the R{1} on the left graph.
 
Physics news on Phys.org
  • #2
tomek91 said:
I would like to get feedback if my z-plot is accurate for the following complex exponentiala:
Code:
a=2*exp(j*∏*t)
b=2*exp(j*∏*-1.25)
c=1*exp(j*∏*t)
d=-j*exp(j*∏*t)

This looks very confused. From the solution, the above looks like phasors and should be written as:

a = 2 ejwt
b = 2 ej(wt - 1.25∏)
c = 1 ejwt
d = -j ejwt

where w could be pi rad/s but I think is more likely a frequency variable.

First thing you need to notice is these are not fixed points in the Z plane -- they vary with t. So instead we view each phasor as a time-varying part multiplied by a fixed complex multiplier. Do this by factoring out ejwt and plotting the multiplier in the Z plane. For example, A would be 2 times the time varying part ejwt.

On your plot, each point is understood to be multiplied by ejwt and as time varies, they all get rotated counterclockwise by wt radians but maintain fixed spatial relationships between them so that you can imagine rotating the entire sheet of paper the plot is on clockwise wt radians to find the location of each complex number for all time.

The solution provided looks like a phasor diagram for the above.
 
  • #3
aralbrec said:
tomek91 said:
I would like to get feedback if my z-plot is accurate for the following complex exponentiala:
Code:
a=2*exp(j*∏*t)
b=2*exp(j*∏*-1.25)
c=1*exp(j*∏*t)
d=-j*exp(j*∏*t)

This looks very confused. From the solution, the above looks like phasors and should be written as:

a = 2 ejwt
b = 2 ej(wt - 1.25∏)
c = 1 ejwt
d = -j ejwt

where w could be pi rad/s but I think is more likely a frequency variable.

First thing you need to notice is these are not fixed points in the Z plane -- they vary with t. So instead we view each phasor as a time-varying part multiplied by a fixed complex multiplier. Do this by factoring out ejwt and plotting the multiplier in the Z plane. For example, A would be 2 times the time varying part ejwt.

On your plot, each point is understood to be multiplied by ejwt and as time varies, they all get rotated counterclockwise by wt radians but maintain fixed spatial relationships between them so that you can imagine rotating the entire sheet of paper the plot is on clockwise wt radians to find the location of each complex number for all time.

The solution provided looks like a phasor diagram for the above.

As written,
[tex] b = 2 e^{-1.25 j \pi}.[/tex] Is that really what is wanted?

RGV
 
  • #4
aralbrec said:
On your plot, each point is understood to be multiplied by ejwt and as time varies, they all get rotated counterclockwise by wt radians but maintain fixed spatial relationships between them

That make sense, however, so if e.g. 2 is multiplied by ejωt where ω is pi rad/sec then the result of this exponential is -1 (considering only real part of this complex exponential) thus the resulting value of this exponential is -2 but solution says 2.

The first graphs (subplot 1 and 2) shows a synthesis of those 4 exponential using MatLab function which is based on this formula (sinusoidal synthesis):
rc7qdh.jpg
The question or the form of this exponentials is represented as follows in textbook:
dls043.jpg
 
  • #5
tomek91 said:
however, so if e.g. 2 is multiplied by ejωt where ω is pi rad/sec then the result of this exponential is -1 (considering only real part of this complex exponential) thus the resulting value of this exponential is -2 but solution says 2.

At what time are you supposing ej∏t is equal to -1? I'd agree with you it equals -1 at time t=1, eg, but at time t=0 it equals +1. By factoring out the time component we can plot the constant part of the phasor in the Z plane. If A = 2 ejwt and we agree to draw a diagram for t=0, then A will be plotted at (2,0) as in the solution. To find A at other times you have to multiply this position by ejwt, the part we factored out. This has the effect of rotating the point A counter clockwise by wt radians. For time t=1 and w=∏, eg, we would multiply 2 by ej∏ to find at time t=1, A=2*(-1) = -2 as you have been saying.

On the phasor diagram, you can add those plotted points as vectors to find out what overall sum of the signal will be, then multiply by ej∏t and take the real part for the sinusoid.

But I think your discrepancy is coming from evaluating ej∏t at time t=1 instead of t=0 if I've understood correctly.
 
  • #6
Just realized right now that I was over-thinking about that. Very basic, thanks for answer.
 

Related to [Complex exponential] Solutions must be wrong

1. Why do complex exponential solutions sometimes appear to be wrong?

Complex exponential solutions may appear to be wrong because they involve the use of imaginary numbers, which can be difficult to understand and visualize. These solutions are valid and necessary in certain mathematical and scientific applications, but they may seem counterintuitive or incorrect to those not familiar with them.

2. Are complex exponential solutions always incorrect?

No, complex exponential solutions are not always incorrect. In fact, they are often necessary in solving complex mathematical equations and represent valid solutions in many scientific fields, such as quantum mechanics and electrical engineering.

3. How can I tell if a complex exponential solution is valid?

A complex exponential solution is valid if it satisfies the original equation or problem it was derived from. It may involve imaginary numbers or complex terms, but as long as it accurately solves the problem, it can be considered a valid solution.

4. Can complex exponential solutions be simplified or converted to real numbers?

In some cases, complex exponential solutions can be simplified or converted to real numbers. However, this depends on the specific equation or problem being solved and may not always be possible. In general, complex exponential solutions are considered valid and do not need to be simplified for them to be correct.

5. How important are complex exponential solutions in scientific research?

Complex exponential solutions are extremely important in scientific research, particularly in fields such as physics and engineering. They allow for the accurate modeling and prediction of complex systems and phenomena that cannot be described by real numbers alone. Without the use of complex exponential solutions, many scientific advancements and discoveries would not have been possible.

Similar threads

  • Calculus and Beyond Homework Help
Replies
5
Views
429
  • Calculus and Beyond Homework Help
Replies
7
Views
3K
  • Calculus and Beyond Homework Help
Replies
1
Views
982
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
940
  • Calculus and Beyond Homework Help
Replies
1
Views
613
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
301
Back
Top