# How to obtain 3 dB frequency from transfer function?

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.

• We know what j is... Commented Jan 19, 2018 at 17:19
• @EugeneSh. I actually commented before editing , I didn't mean the j
– user170589
Commented Jan 19, 2018 at 17:20
• I meant the other formula. How do you convert the amplification/transfer into dB? Commented Jan 19, 2018 at 17:21
• @EugeneSh. I wasn't given any other formula , nor can I find what the one you say
– user170589
Commented Jan 19, 2018 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. Commented Jan 19, 2018 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

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

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.

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.