- #1
roam
- 1,271
- 12
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?
Any help would be greatly appreciated.