How to obtain 3 dB frequency from transfer function?

Question

How can I calculate the 3 dB frequency of the following transfer function?

$$ H(j\omega)=\frac{1}{1-j\frac{250}{\omega}} $$ I have thought of doing the inverse Fourier transform of \$H(j\omega)\$ so I can find \$h(t)\$, and from that the period \$T\$ and then the frequency. However, I think that frequency will not be the 3 dB I am looking for.

The \$j\$ in \$H(j\omega)\$ is the imaginary number \$ j^{2}=-1 \$

I have found this formula: $$ H(f_\text{3 dB})=H_\text{max}(\text{dB}) - 3\text{ dB}$$ but I still can't find any solution.

Answer 1

What do you mean by the 3 dB frequency? The frequency at which your magnitude has the value 3 dB? Or are you confusing this with the corner frequency where the magnitude is 3 dB less?

This would be the frequency marked with the red line in the bode plot.

As -3 dB is corresponding to

$$20 \cdot \log\left(\frac{1}{\sqrt{2}}\right) = - 3.01$$

you can infer that

$$20 \cdot \log(|H(j \omega)|) = - 3.01$$

$$20 \cdot \log\left(\frac{1}{\sqrt{1+\left(\frac{250}{\omega}\right)^2}}\right) = - 3.01$$

so

$$\frac{250}{\omega} = 1$$

hence \$\omega = 250\$.

Otherwise you could bring the transfer function in a (at least to me) more recognizable form:

$$H(s) = \frac{s}{s+250}$$

with \$s = j \omega\$

rearranging to

$$H(s) = \frac{\frac{1}{250}s}{s\frac{1}{250}+1}$$

$$H(s) = \frac{\frac{1}{250}s}{s \cdot T+1}$$

where \$1/T\$ is the corner frequency (the one you are supposedly be looking for). This kind of notation can vary from textbook to textbook or your teacher. You may also keep writing \$j \omega\$ and the arrive at \$j \frac{\omega}{\omega_c}\$ where \${\omega_c}\$ again is the corner frequency and you can easily read that it's 250.

Answer 2

I can see you are not really progressing via the comments so take the example of an RL high pass filter like this: -

I can see from the position of \$\omega\$ in your formula that you have the equivalent of a high pass RL filter and the transfer function is: -

H(s) = \$\dfrac{sL}{R+sL}\$ = \$\dfrac{1}{1+\frac{R}{sL}}\$

In \$j\omega\$ terms it is: -

H(jw) = \$\dfrac{1}{1-j\frac{R}{\omega L}}\$

And is in the same form as the equation in the question.

I know from experience that the 3 dB point occurs when the denominator's real and imaginery terms are magnitude-equal so, in your example, the frequency of the 3 dB point is \$\omega\$ = 250.

Equating those terms is the same as equating the magnitude of R and the magnitude of \$\omega L\$ in my RL circuit.

For an RC circuit it would be when R = \$\dfrac{1}{\omega C}\$.

If you want to think of it another way you could vectorially add 1 and 250/w in the denominator and equate it to the 3 dB point amplitude (\$\dfrac{1}{\sqrt2}\$) denominator.

So \$\sqrt{1^2 + \frac{250^2}{\omega^2}}\$ = \$\sqrt2\$

If you follow it through to the end, \$\omega\$ = 250.

Answer 3

To obtain the 3-dB cutoff frequency, you determine what angular frequency \$\omega\$ makes the magnitude of your transfer function equal to \$\frac{1}{\sqrt2}\$. Solve the value of \$\omega\$ which leads to this value and you have the cutoff frequency you want. Your expression is unusual because if uses an inverted pole: you have a pole at the origin and then a zero in higher frequency. This is a nice and compact way - read low-entropy - to write transfer functions. The below Mathcad sheet shows the determination of the cutoff frequency in this particular configuration.