# How to obtain 3db frequency from transfer function

How can I calculate the 3db Frequency of the following transfer function

$$H(w)=\frac{1}{1-j\frac{250}{w}}$$ I have thought of doing the inverse fourier of H(w) so I can find h(t) and from that the period T and then the frequency . But I think that frequency will not be the 3dB I am looking for . Can anyone help?

The j on the H(w) is the imaginary number $$j^{2}=-1$$

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

• We know what j is... – Eugene Sh. Jan 19 '18 at 17:19
• @EugeneSh. I actually commented before editing , I didn't mean the j – maverick98 Jan 19 '18 at 17:20
• I meant the other formula. How do you convert the amplification/transfer into dB? – Eugene Sh. Jan 19 '18 at 17:21
• @EugeneSh. I wasn't given any other formula , nor can I find what the one you say – maverick98 Jan 19 '18 at 17:22
• en.wikipedia.org/wiki/Bode_plot See the "Example" section and the following "Magnitude plot" for something very similar to your exercise. – Eugene Sh. Jan 19 '18 at 17:22

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

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

you can infer that

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

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

so

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

hence

$$\omega = 250$$

Otherwise you could bring the transfer function in a (at least to me) more recognizeable 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.

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.

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.