How to get valid frequency for a transfer function?

[Effy]

Homework Statement

transfer function: H(s) = 1/(s^2+s+1).
question: compute half-power frequency

Homework Equations

none

The Attempt at a Solution

[/B]I got 4 answers: -0.605i, 0.065i, -1.169, 1.169
I know that only two of these are correct, one is 1.169 instead of -1.169 because frequency cannot be negative. Then I wrote the two imaginary answers into phaser form. I got 0.065<-90degree and 0.065<90 degree. Could someone tell me which phase I should choose and why?

 

[milesyoung]

Did you come up with those answers yourself, or were they given to you in the problem statement? If they're your work, then it would be much easier to help you if you showed how you came up with them.

The value of 's' is, in some contexts, called 'complex frequency', but I doubt that's what you're after. I think you probably just have to find the -3 dB point(s) of $H(j \omega)$.

 

[Effy]

Did you come up with those answers yourself, or were they given to you in the problem statement? If they're your work, then it would be much easier to help you if you showed how you came up with them.

The value of 's' is, in some contexts, called 'complex frequency', but I doubt that's what you're after. I think you probably just have to find the -3 dB point(s) of $H(j \omega)$.

Hi,

I wrote the transfer function H(s) into H(w)=1/(-w^2+iw+1)
so the magnitude would be |H(w)|=|1/(-w^2+iw+1)|=1/sqrt(1-w^2+w^4)
|H(w)|max would be sqrt(4/3) when 1-w^2+w^4 is at minimum when w=+ or -1/sqrt(2)
set |H(w)|=1/sqrt(1-w^2+w^4)=1/sqrt(2)*|H(w)|max=sqrt(2/3) and solve for w
since this is a 4th order equation, I got 4 answers. 1 imaginary positive, 1 imaginary negative, 1 real positive, 1 real negative.
Then I'm confused.

What do you mean by -3dB point? I don't know about that.

 

[lazyaditya]

@Effy , i think the answers you calculated are wrong. Please go through it once again.

 

[Effy]

Did you come up with those answers yourself, or were they given to you in the problem statement? If they're your work, then it would be much easier to help you if you showed how you came up with them.

The value of 's' is, in some contexts, called 'complex frequency', but I doubt that's what you're after. I think you probably just have to find the -3 dB point(s) of $H(j \omega)$.

Btw, the original question is a damper-spring-mass system. I got the transfer function from that and I'm pretty sure it's correct.

 

[lazyaditya]

-3db point is called as the cutoff frequency or half power frequency. It is a frequency at which magnitude of gain decreases by 3db from maximum gain.

 

[Effy]

@Effy , i think the answers you calculated are wrong. Please go through it once again.

I've done that several times and I'm still try

-3db point is called as the cutoff frequency or half power frequency. It is a frequency at which magnitude of gain decreases by 3db from maximum gain.

Does it mean that I shouldn't multiply the maximum magnitude by 1/sqrt(2) but subtract 3dB?

 

[lazyaditya]

I meant about the frequency values.

 

[lazyaditya]

I did it roughly .

 

[lazyaditya]

You can subtract 3 db only when the magnitude is also given in decibel .

 

[Effy]

I did it roughly .

I got the same answers and finally found out what I did wrong. Thank you so much!
But only the +1.27 is a valid frequency, right?

 

[milesyoung]

But only the +1.27 is a valid frequency, right?

That depends on how pedantic your instructor is. You can interpret the real-valued signal $\cos(\omega t)$ as a complex-valued signal:
$$
\cos(\omega t) = \frac{e^{j \omega t} + e^{-j \omega t}}{2}
$$
with frequency components at $-\omega$ and $\omega$, but I doubt that makes any difference for your assignment. The negative frequency components are redundant if you're only working with real-valued signals, so I think you can easily argue that it's correct to only include the positive frequency component in your answer.

 

[Effy]

That depends on how pedantic your instructor is. You can interpret the real-valued signal $\cos(\omega t)$ as a complex-valued signal:
$$
\cos(\omega t) = \frac{e^{j \omega t} + e^{-j \omega t}}{2}
$$
with frequency components at $-\omega$ and $\omega$, but I doubt that makes any difference for your assignment. The negative frequency components are redundant if you're only working with real-valued signals, so I think you can easily argue that it's correct to only include the positive frequency component in your answer.

Got it. Thanks!