Question 10.2: Figure 10.10 shows a low pass filter. Calculate the cut-off ......
Figure 10.10 shows a low pass filter. Calculate the cut-off frequency, and the magnitude of the output voltage if the frequency of the input voltage is 400 Hz.
Step-by-Step
The cut-off frequency is,
f_{c}={\frac{1}{2\pi R C}}={\frac{1}{2\pi\times10\times10^{3}\times0.03\times10^{-6}}}=530.52\,\mathrm{Hz} (10.55)
The output voltage is,
|V_{0}|={\frac{|V_{i}|}{\sqrt{\left({\frac{f}{f_{c}}}\right)^{2}+1}}}={\frac{120}{\sqrt{\left({\frac{400}{530.52}}\right)^{2}+1}}}=96\,\mathrm{V} (10.56)
Related Answered Questions
Question: 10.1
Verified Answer:
The output current can be determined as,
I_...
Question: 10.5
Verified Answer:
The resonant frequency is calculated as,
f_...
Question: 10.3
Verified Answer:
The cut-off frequency is,
f_{1}={\frac{1}{2...
Question: 10.4
Verified Answer:
The resonant frequency is,
f_{0}={\frac{1}{...
Question: 10.6
Verified Answer:
Substituting s = jω in the transfer function yield...