How to calculate the magnitude of a function?


by seto6
Tags: function, magnitude
seto6
seto6 is offline
#1
Apr22-12, 11:25 PM
P: 251
1. The problem statement, all variables and given/known data
I have this function:

G(w)=[itex]\frac{1}{j\omega\tau+1}[/itex]
I want to find the magnitude

2. Relevant equations
S=[itex]\alpha[/itex]+j[itex]\beta[/itex]

magnitude(S)=[itex]\sqrt{[itex]\alpha[/itex]^{2}+[itex]\beta[/itex]^{2}}[/itex]


[b]3. The attempt at a solution[/b

What i did was carry out the division,

so i got
[itex]\frac{j\omega\tau-1}{-(j\omega+1)}[/itex]
then do i just split it into real and imaginary part and then take the magnitude using this?
S=[itex]\alpha[/itex]+j[itex]\beta[/itex]

magnitude(S)=[itex]\sqrt{[itex]\alpha[/itex]^{2}+[itex]\beta[/itex]^{2}}[/itex]

Can anyone help i am not so sure how to approach this?
Phys.Org News Partner Science news on Phys.org
Cougars' diverse diet helped them survive the Pleistocene mass extinction
Cyber risks can cause disruption on scale of 2008 crisis, study says
Mantis shrimp stronger than airplanes
failexam
failexam is offline
#2
Apr23-12, 12:56 AM
P: 338
Quote Quote by seto6 View Post
What i did was carry out the division,

so i got
[itex]\frac{j\omega\tau-1}{-(j\omega+1)}[/itex]
This is where you got it wrong. Try to multiply the numerator and the denominator of the original G(ω) by [itex] - j ω \tau + 1[/itex] and see what you get.
failexam
failexam is offline
#3
Apr23-12, 01:01 AM
P: 338
Then, try and think what you had to multiply by the expression [itex]- j ω \tau + 1 [/itex] to get your answer.

Ray Vickson
Ray Vickson is offline
#4
Apr23-12, 01:52 AM
HW Helper
Thanks
P: 4,672

How to calculate the magnitude of a function?


Quote Quote by seto6 View Post
1. The problem statement, all variables and given/known data
I have this function:

G(w)=[itex]\frac{1}{j\omega\tau+1}[/itex]
I want to find the magnitude

2. Relevant equations
S=[itex]\alpha[/itex]+j[itex]\beta[/itex]

magnitude(S)=[itex]\sqrt{[itex]\alpha[/itex]^{2}+[itex]\beta[/itex]^{2}}[/itex]


[b]3. The attempt at a solution[/b

What i did was carry out the division,

so i got
[itex]\frac{j\omega\tau-1}{-(j\omega+1)}[/itex]
then do i just split it into real and imaginary part and then take the magnitude using this?
S=[itex]\alpha[/itex]+j[itex]\beta[/itex]

magnitude(S)=[itex]\sqrt{[itex]\alpha[/itex]^{2}+[itex]\beta[/itex]^{2}}[/itex]

Can anyone help i am not so sure how to approach this?
Failexam's suggestions are all you need to do. However, if you are going to use LaTeX, why not do it properly? Your expression for "magnitude(S)" is ugly; here is what it should look like: [itex]\text{magnitude}(S)=\sqrt{\alpha^2 + \beta^2}.[/itex] To get this, just remove the "inner" [i t e x]-[/i t e x] pairs; furthermore, if you want the word "magnitude" to appear in nice text font, just include it inside the [i t e x] command, but say \text{magnitude}.

RGV
RoshanBBQ
RoshanBBQ is offline
#5
Apr23-12, 02:54 AM
P: 280
The magnitude of two vectors divided is the division of the magnitudes. What is the magnitude of the numerator? What is the magnitude of the denominator? What is the result of dividing those magnitudes?


Register to reply

Related Discussions
How to calculate vector angles and magnitude Introductory Physics Homework 12
Need to calculate the magnitude of the initial velocity? Calculus & Beyond Homework 1
Need to calculate the magnitude of the initial velocity? Introductory Physics Homework 1
Calculate magnitude of spring force Introductory Physics Homework 8
Calculate the magnitude of buoyant force Introductory Physics Homework 1