3dB Frequency of an LED Transfer Function

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Homework Statement


According to my textbook:

The LED transfer function ##H(\omega_m)## is defined as:

$$H(\omega_m) = \frac{1}{1+j\omega_{m}\tau_{c}}$$

The 3-dB modulation bandwidth ##f_{\text{3 dB}}## is defined as the modulation frequency at which ##H(\omega_m)## is reduced by 3 dB or by a factor of 2. The result is:

$$f_{\text{3 dB}}=\sqrt{3}(2\pi\tau_{c})^{-1} \tag{1}$$

I don't understand how they derived the last expression.

Homework Equations



The Attempt at a Solution



Let ##G(j\omega_{m})=\frac{1}{1+j\omega_{m}\tau_{c}}##.

##|G(j\omega_{m})|^{2}=\frac{1}{1+j\omega_{m}\tau_{c}}.\frac{1}{1-j\omega_{m}\tau_{c}}=\frac{1}{1+\omega_{m}^{2}\tau_{c}^{2}}##

##\therefore |G(j\omega_{m})|=\frac{1}{\sqrt{1+\omega_{m}^{2}\tau_{c}^{2}}}##

And ##|G(0)|=1##, therefore to find the 3 dB point we must solve:

$$\frac{|G(j\omega)|}{|G(0)|}=\frac{1}{\sqrt{1+\omega_{\text{3 dB}}^{2}\tau_{c}^{2}}}=\frac{1}{\sqrt{2}}$$

However when I solve this I get:

$$\omega_{\text{3 dB}}=\frac{1}{\tau_{c}}\ \text{or}\ f_{\text{3 dB}}=\frac{1}{2\pi\tau_{c}}$$

So where does the ##\sqrt{3}## factor in equation (1) come from? What is the mistake here? :confused:

Any help would be greatly appreciated.
 

Answers and Replies

  • #2
rude man
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I think this is problem as stated is a mess.

They're apparently thinking 3dB of power, not whatever the units of H(w) are (light power output/current input?). Your H(w) is down by a factor of √2, not 2, at w = 1/T. So the problem actually is a contradiction in terms since it gives H(w) as 1/(1+ jwT) which at 3dB down (i.e. 1/√2) gives w = 1/T as you found.

So they meant 3dB down in power which is 6dB down in H(w) which gives their posted answer. But the problem is misstated.
 
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