Complex Numbers Exponential Form

MrAl

Joined Jun 17, 2014
7,588
Hi,

Just one question, why are they giving the solution to Vm_dot if they are going to give the entire solution in terms of Vm_dot and not just plain Vm ?
 

Thread Starter

helloeveryone

Joined Apr 8, 2011
63
Hi,

Just one question, why are they giving the solution to Vm_dot if they are going to give the entire solution in terms of Vm_dot and not just plain Vm ?
I'm not sure. I double checked and that is what is written, perhaps a typo? Maybe the Vm_dot in the given answer should be just plain Vm?

Also I don't really understand the problem. Re means Real part and Im means Imaginary part, correct? How can the Real part have a complex exponential? Doesn't that make it imaginary automatically if it contains a "j" ? So shouldn't both have Im{ } brackets?
 
Last edited:

MrAl

Joined Jun 17, 2014
7,588
Hi,

When you take the real part of a complex number you get a real number, and when you take the imaginary part of a complex number you get another real number. However, the sum may have a complex representation.

A complex number is as:
a+b*j

where both a and b are real and j the imaginary operator.

e^(j*x) though may be represented by:
cos(x)+j*sin(x)

due to the Euler formula.

Thus for:
e^(j*x+j*a)

when i sum the real and imaginary parts i get:
sin(x+a)+cos(x+a)

and this can be simplified a little and that brings up factors of sqrt(2) and pi/4, but does not seem to relate directly to the given "known" result otherwise.

What you could do is try to convert the "known" result to a form like this and then compare to the form you get doing it directly.

Do you have any information on what work you did just before this? Maybe that will give us a clue.
 

WBahn

Joined Mar 31, 2012
25,745
I would appreciate some help with the following problem:

View attachment 125626
The first step is to ask if the proposed answer makes sense.

We know that v is a real number, since both Re{z} and Im{z} are real numbers.

So is the claimed answer a real number? This would require that the arg of Vm_dot be equal to exactly -pi/4. But alpha is an arbitrary angle, so this is not the case.

Hence we know the answer is wrong.
 

MrAl

Joined Jun 17, 2014
7,588
The first step is to ask if the proposed answer makes sense.

We know that v is a real number, since both Re{z} and Im{z} are real numbers.

So is the claimed answer a real number? This would require that the arg of Vm_dot be equal to exactly -pi/4. But alpha is an arbitrary angle, so this is not the case.

Hence we know the answer is wrong.
Hi there,

Maybe you can show some work to help show what you are suggesting. The possible forms come out a little varied from the straightfoward calculation idea.

For example, if we take the sum of both real and imag parts we get:
sin(x+a)+cos(x+a)

and if we define two quantities A1 and A2:
A1=e^(j*y)=cos(y)+j*sin(y)
A2=e^(-j*y)=cos(y)-j*sin(y)

then we can convert that original result. First:
cos(y)=(A1+A2)/2

then:
sin(y)=j*(A2-A1)/2

so we can get:
sin(y)+cos(y)=j*A2/2-j*A1/2+A1/2+A2/2=-(j*e^(j*y))/2+e^(j*y)/2+(j*e^(-j*y))/2+e^(-j*y)/2

If we first convert:
sin(y)+cos(y)

into:
sqrt(2)*sin(y+pi/4)

we MIGHT be able to get a simpler form for the result than the above with four exponentials.

When i did it it started to look like it was not possible to get the same result as given though, so maybe you can show a little more work.

The closest i can come so far is to convert into the form:
sqrt(2)*cos(x+a-pi/4)

and then take this to be a 'phasor' and get:
sqrt(2)*e^(j*(a-pi/4))

which still contains the 'a', and which would really be represented as:
sqrt(2) at angle of a-pi/4.
 
Last edited:

WBahn

Joined Mar 31, 2012
25,745
I don't see what work is needed.

If I claim that A = B and you can show that A is real and B is complex (meaning in this context that the imaginary part is not identically zero), then isn't that sufficient to show that my claim cannot be true?

v = v1 + v2 is real. Why? Because both the real and the imaginary coefficients of a complex number are real and the reals are closed under addition, meaning that the sum of two real numbers is a real number.

The proposed answer is the product of three terms. The first, sqrt(2), is real. The second and third are complex. The product of two complex numbers can only be real if the sum of their args is an integer multiple of pi. The arg of the first is a constant of pi/4, which means that the arg of Vm must be pi(k-1/4) where k is an integer. Since the angle is an arbitrary parameter, there is no reason for this to always be the case, and hence the proposed answer is not guaranteed to be real even though, to be correct, it must be.
 

WBahn

Joined Mar 31, 2012
25,745
Ignoring the answer that was given (since we've established that it can't be correct for the problem as stated), we can still find the answer.

Your work in your original power is just fine as far as you took it.

Now apply the Re{} and Im{} operators knowing that

Re{A·e^(jw)} = A·cos(jw)
Im{A·e^(jw)} = A·sin(jw)
 
Top