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

• Engineering
• s3a

#### 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!