I want to take the square root of a transfer function

I am trying to plot a completely real transfer function

my current implementation is using:

s=tf('s')
w2=-(s)^2

Where w2 represents .

I am trying to implement the root with the sqrt function, however it is saying.

Incorrect number or types of inputs or outputs for function sqrt.

This transfer function needs to be multiplied by other (complex) transfer functions later in the project.

Is this the best way to implement it or is there another approach I could try?

[Rik]

11 years ago, someone else suggested you should be able to write your equation as a pure polynomial divided by another pure polynomial and that you can get there by approximating your function with a Taylor expansion. Have you tried that?

[Walter Roberson]

Does it even make sense for there to be negative frequencies? sqrt(omega^2) is just omega if omega is non-negative. (This does not help on the denominator sqrt though)

[Sam Chak]

Hi @Walter Roberson

The OP did not provide the background story regarding the issue. The transfer function is actually a function of a complex variable, s, which arises from performing the Laplace Transform on a function in the time domain. This complex variable has both a real and an imaginary component, typically denoted as , where j is the imaginary unit. For example, the ODE can be converted to the transfer function .

If the input u is a pure sinusoidal signal, the real part vanishes, leaving only the imaginary part, . The steady-state output can be found as follows:

and this is the origin for generating frequency response (in both magnitude and phase) that is typically depicted in Bode plots.

syms s omega

s = j*omega

s = 

s^2

ans = 

●

[Sam Chak]

I believe that the graph the OP wishes to plot is referred to as Displacement Transmissibility, which is commonly used in mechanical engineering to design structures that isolate vibrations, such as car suspension systems. I am not a mechanical engineer; perhaps @Yifeng Tang can assist in verifying this.

[Yifeng Tang]

Integrating the area under the Bode curve is indeed related to Transmissibility as Sam Chak pointed out. I don't think it's necessary to take the square root of the transfer function itself though. It's the norm/magnitude of the transfer function after replace "s" with , j being the imaginary number. Square root may show up when calculating the magnitude, but not on the transfer function itself.

Answer 1

[Sam Chak]

Hi @Calum

Unlike plotting time responses using the step() and lsim() functions, or first solving the ordinary differential equations using dsolve() and ode45(), plotting the steady-state response as a function of the driving frequency is relatively straightforward because the function defines a fixed relationship between the input and output. It is a form of what-you-see-is-what-you-get, where the output magnitude appears as it would when the input frequency is fed into the transfer function.

w   = linspace(0.1, 100, 10001);    % Angular Frequency
b   = (0.2*w)./sqrt((1 + w.^2).*(w.^4 - 1.51*w.^2 + 1));
semilogx(w, b), grid on
xlabel('Angular Frequency (rad/s)')
ylabel('Amplitude')
title ('Frequency Response')