# Finding the natural frequency of transfer function (2s) / (3s^2+5s+2)

• Engineering
• s3a
In summary, if I have something like (2s) / (3s^2+5s+2), I'm not sure what to do. Do I just ignore the s on the numerator and proceed like if it wasn't there or what?
s3a
Homework Statement
See attached PDF
Relevant Equations
(ω_n)^2 / [s^2 + 2ζ(ω_n) + (ω_n)^2]
In the context of control systems, if I have a vibratory second-order system, (ω_n)^2 / [s^2 + 2ζ(ω_n) + (ω_n)^2], I know how to get the natural frequency ω_n. So, if I have something like 2 / (3s^2+5s+2), I know how to get the natural frequency ω_n.

However, if I instead have something like (2s) / (3s^2+5s+2), I'm not sure what to do. Do I just ignore the s on the numerator and proceed like if it wasn't there or what?

Any input would be greatly appreciated!

Edit:
P.S.
For what it's worth, I'm doing this as part of a larger problem, so if I made a mistake and it's impossible for a vibratory second-order system to have an s on the numerator, just let me know. The problem is question 3 from the PDF document

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First,G(s)=y2/u1 it is not equal G1(s) multiplied by G2(s) since y1 in G1(s) is calculated when is no load after y1.In case there is a load [L2 series with r2] then y2/u1=2s/(3s^2+6s+2) and not 2s/(3s^2+5s+2).
Second:
However, since 6^2-4*3*2=12>0 sqrt(+12) is real and then no oscillation is expected [the damping factor is more than 1 and no ω]

If approximations don't work you can do the Laplace transform to verify. The equation you provided you can do do a partial fraction and use a table.

Thanks for your input, Babadag, but I still don't fully understand how to compute the connection of the two "sub-circuits". I also don't understand why multiplying the two transfer functions isn't good enough. Could you please elaborate on those?

Having said that, I figured out how to do the problem (as you can see the final answer is what you said), but I'm not sure if that's how the problem statement intended I do it. For what it's worth, here is what I did.:

Equation 1:
u_1 = (1 Ω)(I_A (t)) + (1 H) d [I_A (t) - I_B (t)] /dt
U_1 (s) = (1 Ω)(I_A (s)) + (1 H) s [I_A (s) - I_B (s)]
Removing the units for visual clarity,
U_1 (s) = I_A (s) + s I_A (s) - s I_B (s)

Equation 2:
0 = (3 H)(d/dt I_b (t)) + (2 Ω)(I_b (t)) + (1 H)[ d (I_b (t) - I_a (t) /dt]
0 = (3 H)(s I_B (s)) + (2 Ω)(I_B (s)) + (1 H)[ s (I_B (s) - s I_A (s)]
Removing the units for visual clarity,
0 = 3 s I_B (s) + 2 I_B (s) + s I_B (s) - s I_A (s)
0 = 4s I_B(s) + 2 I_B (s) - s I_A (s)
0 = (4s + 2) I_B (s) - s I_A (s)
I_A (s) = (4s + 2)/s I_B (s)

Equation 3:
y_2 (t) = (2 Ω)( I_b (t) )
Y_2 (s) = (2 Ω) I_b (s)
Removing the units for visual clarity,
Y_2 (s) = 2 I_B (s)

Then, we mix equations 1 and 2 to get the following, Equation 4,
Y_1 (s) / U_1 (s) = [2 I_B (s)] / [I_A (s) + s I_A (s) - s I_B (s)]

Then, mixing Equation 2 and Equation 4, we get
Y_1 (s) / U_1 (s) = [2 I_B (s)] / [{(4s + 2)/s I_B (s)} + s {(4s + 2)/s I_B (s)} - s I_B (s)]
Y_1 (s) / U_1 (s) = [2 I_B (s)] / [{(4s + 2)/s I_B (s)} + {(4s + 2) I_B (s)} - s I_B (s)]
Y_1 (s) / U_1 (s) = [2 I_B (s)] / [(4s + 2)/s I_B (s) + (4s + 2) I_B (s) - s I_B (s)]
Y_1 (s) / U_1 (s) = 2 / [(4s + 2)/s + (4s + 2) - s]
Y_1 (s) / U_1 (s) = 2s / [s(4s + 2)/s + s(4s + 2) - s^2]
Y_1 (s) / U_1 (s) = 2s / [4s + 2 + 4s^2 + 2s - s^2]
Y_1 (s) / U_1 (s) = 2s / [6s + 2 + 3s^2]
Y_1 (s) / U_1 (s) = 2s / [3s^2 + 6s + 2]

And, Joshy, thanks for the partial fractions and table tip; I later realized that while doing another problem.

What do you mean about an approximation, though?

I'm not too familiar with this stuff, but I've only ever seen stuff about a linear approximation, but that was in the time domain, not frequency domain. Were you suggesting the possibility of using a linear approximation in the frequency domain and then getting the inverse Laplace transform of that?

P.S.
Sorry for the late response!

## 1. What is the natural frequency of a transfer function?

The natural frequency of a transfer function is a measure of how quickly a system will oscillate or vibrate when disturbed. It is a characteristic of the system and is independent of any external forces or inputs.

## 2. How is the natural frequency calculated?

The natural frequency of a transfer function can be calculated by finding the square root of the ratio of the coefficient of the highest order term in the denominator to the coefficient of the highest order term in the numerator. In the case of (2s) / (3s^2+5s+2), the natural frequency would be √(2/3) ≈ 0.8165.

## 3. What is the significance of the natural frequency in a transfer function?

The natural frequency is an important characteristic of a system as it determines the rate at which the system will oscillate or vibrate when disturbed. It also affects the stability and response of the system to different inputs.

## 4. How does the natural frequency affect the behavior of a system?

A higher natural frequency indicates that the system will oscillate or vibrate at a faster rate, while a lower natural frequency indicates a slower rate. This can affect the response of the system to different inputs, as well as its stability and ability to reach a steady state.

## 5. How can the natural frequency be adjusted in a transfer function?

The natural frequency can be adjusted by changing the coefficients of the transfer function, particularly the coefficients of the highest order terms in the numerator and denominator. This can be done through system design or by adding external components to the system.

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