Rewriting complex number expressions

In summary: Just expand that expression for V0(t) using that trig identity and equate it to the real part that you found. The coefficients of cos(wt) and sin(wt) must be the same on both sides. Try to solve for A and θ. To get a notion of what the result should look like there is...In summary, the homework statement is that a turbine can deliver an infinitely large current without affecting the frequency. However, the mill (single phased) is now affected by a load from a resistor. We assume that the turbine can deliver an infinitely large current without affecting the frequency. Using trigonometric identities, we can expand the expression for V0(t) to the form V0(
  • #1
Tyrone123
10
0

Homework Statement


The mill (single phased) is now affected by a load from a resistor. We assume that the turbine can deliver an infinitely large current without affecting the frequency.

figure 1 illustrates the load:
http://puu.sh/19GGR

The voltage over the capacitor can can be expressed as

V0(t) = Re(vp * ei*ω*t)

where ω is the angular frequency, i is the complex number, and vp is a complex number (also called a phasor) and can be expressed as

vp = k * [itex]\frac{1}{1+iω*R*C}[/itex]

The values R and Ck er unknown, but can be assumed to be constant and real.

Make an expression for V0(t) where only the real part is described.

Rewrite the expression for V0(t) to the form V0(t) = A * cos(ωω*t + θω)

Where A is the amplitude



The Attempt at a Solution


As for the first part, my idea was to rewrite the exponential part ei*ω*t to cos(t*ω)+i*sin(t*ω) and then multiply it with vp as you would with two complex numbers, and then simply discard the imaginary part..
which gives me [itex]\frac{k*cos(t*w)}{1+w*R*C}[/itex]

But I am pretty sure I am way off.. and I could use some help!
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
I don't think you are doing the 'multiply by v_p' part correctly. If you have a complex number in the denominator and you want to take the real part you need to multiply numerator and denominator by the conjugate of the denominator. For example, 1/(a+bi)=(a-bi)/((a+bi)(a-bi))=(a-bi)/(a^2+b^2). Show the details of how you got your answer.
 
  • #3
I realize my mistake and I have now ended up with the following equation for vp by multiplying both the denominator and numerator by the conjugate of the denominator

http://puu.sh/19J1F (having issues with the fraction feature, so excuse me pasting an image instead)

the above expression for vp is then multiplied with ei*w*t (which equals cos(t*w)+ i*sin(t*w).. which leaves me with one hell of result, but as I am supposed to find the real part of the expression, I can simply discard the imaginary part of the result from the multiplication at this point?

if so, I end up with this expression for the real part of the complex number:
http://puu.sh/19J7t

which still isn't very pretty.. if this was the correct procedure, any thoughts on how to simplify it?
 
Last edited by a moderator:
  • #4
Tyrone123 said:
I realize my mistake and I have now ended up with the following equation for vp by multiplying both the denominator and numerator by the conjugate of the denominator

http://puu.sh/19J1F (having issues with the fraction feature, so excuse me pasting an image instead)

the above expression for vp is then multiplied with ei*w*t (which equals cos(t*w)+ i*sin(t*w).. which leaves me with one hell of result, but as I am supposed to find the real part of the expression, I can simply discard the imaginary part of the result from the multiplication at this point?

if so, I end up with this expression for the real part of the complex number:
http://puu.sh/19J7t

which still isn't very pretty.. if this was the correct procedure, any thoughts on how to simplify it?

That looks right. It doesn't get too much simpler than that, you could write it over the common denominator, but that's not a whole lot simpler.
 
Last edited by a moderator:
  • #5
Dick said:
That looks right. It doesn't get too much simpler than that, you could write it over the common denominator, but that's not a whole lot simpler.

If we can agree that http://puu.sh/19J7t is the result to the first part, namely finding an expression for the real part of the equation v_p * ei*w*t.. then so far so good.

The next part, which is to rewrite that expression to the form

V0(t) to the form V0(t) = A * cos(ωω*t + θω)

I am very unsure of how to proceed.. there is no given definition of A (which is the amplitude).. my teacher mentioned something about using trigonometric identities such as cos(x - y) = cos(x) cos(y) + sin(x) sin(y) to help rewrite it.. but I can't seem to grasp his idea..
 
Last edited by a moderator:
  • #6
Tyrone123 said:
If we can agree that http://puu.sh/19J7t is the result to the first part, namely finding an expression for the real part of the equation v_p * ei*w*t.. then so far so good.

The next part, which is to rewrite that expression to the form

V0(t) to the form V0(t) = A * cos(ωω*t + θω)

I am very unsure of how to proceed.. there is no given definition of A (which is the amplitude).. my teacher mentioned something about using trigonometric identities such as cos(x - y) = cos(x) cos(y) + sin(x) sin(y) to help rewrite it.. but I can't seem to grasp his idea..

Just expand that expression for V0(t) using that trig identity and equate it to the real part that you found. The coefficients of cos(wt) and sin(wt) must be the same on both sides. Try to solve for A and θ. To get a notion of what the result should look like there is a similar formula in http://en.wikipedia.org/wiki/List_of_trigonometric_identities under 'Linear combinations'.
 
Last edited by a moderator:
  • #7
"just" :3

Expanding the desired expression with the trig identity I mentioned equals

V0(t) = A * (cos(w*t) * cos(θ) - sin(w*t) * sin(θ))

and equating it to http://puu.sh/19J7t where the coefficient of cos(t*w) is [itex]\frac{k}{(W^2 *R^2 *C^2) }[/itex]
and the coefficient of sin(t*w) is [itex]\frac{w*R*C*k}{W^2 *R^2 *C^2)}[/itex]

from here I am not sure how to approach it.. since the coefficient on the left side is just A, but on the right side its different for cosine and sinus.

you said "solve for A and θ".. by that do you mean to take the two equations, equate them and isolate A from that? (and same for θ)
 
Last edited by a moderator:
  • #8
I mean that the two equations you then want to solve are [tex]A cos(\theta)=\frac{k}{W^2 R^2 C^2}[/tex] and [tex]-A sin(\theta)=\frac{w R C k}{W^2 R^2 C^2}[/tex] for [itex]A[/itex] and [itex]\theta[/itex]. It's not as hard it might look. First square both equations and add them and see if you can find [itex]A[/itex]. Then divide the two equations and see if you can solve for [itex]\theta[/itex].
 

Related to Rewriting complex number expressions

1. What are complex numbers?

Complex numbers are numbers that contain both a real part and an imaginary part. They are written in the form a + bi, where a is the real part and bi is the imaginary part (with i representing the square root of -1).

2. Why do we need to rewrite complex number expressions?

Rewriting complex number expressions allows us to simplify and manipulate the numbers in a way that makes them easier to work with in mathematical operations. It also helps to identify the real and imaginary parts of the numbers.

3. What are the basic rules for rewriting complex number expressions?

The basic rules for rewriting complex number expressions include combining like terms, distributing and factoring, and using the properties of exponents and radicals.

4. How do we add and subtract complex numbers?

To add or subtract complex numbers, we combine the real parts and the imaginary parts separately. For example, (3 + 2i) + (5 - 4i) = (3+5) + (2i - 4i) = 8 - 2i.

5. Can we rewrite all complex number expressions?

Yes, all complex number expressions can be rewritten using the basic rules mentioned above. However, the level of complexity may vary depending on the specific expression.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
390
  • Calculus and Beyond Homework Help
Replies
1
Views
378
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Differential Equations
Replies
3
Views
2K
  • Differential Equations
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
28
Views
3K
  • Calculus and Beyond Homework Help
Replies
7
Views
4K
Back
Top