To determine the voltage across the capacitor in a First Order Low Pass Butterworth Filter circuit, we need to analyze the transfer function of the filter. In a low pass filter, the transfer function H(f)H(f)H(f) typically relates the input voltage VinV_{\text{in}}Vin to the output voltage VoutV_{\text{out}}Vout (which is across the capacitor).
For a First Order Low Pass Butterworth Filter, the general transfer function is given by:
H(f)=VoutVin=11+jffcH(f) = \frac{V_{\text{out}}}{V_{\text{in}}} = \frac{1}{1 + j\frac{f}{f_c}}H(f)=VinVout=1+jfcf1
where:
- fff is the frequency of the input signal,
- fcf_cfc is the cutoff frequency (the frequency at which the output power falls to half of its maximum value).
The voltage across the capacitor (output voltage) is then:
Vout=Vin×11+jffcV_{\text{out}} = V_{\text{in}} \times \frac{1}{1 + j\frac{f}{f_c}}Vout=Vin×1+jfcf1
If we compare this expression to the provided options:
Given Options:
- (a) V0=Vin1+0.0314jfV_0 = \frac{V_{\text{in}}}{1 + 0.0314jf}V0=1+0.0314jfVin
- (b) V0=Vin×(1+0.0314jf)V_0 = V_{\text{in}} \times (1 + 0.0314jf)V0=Vin×(1+0.0314jf)
- (c) V0=Vin+0.0314jf1+jfV_0 = \frac{V_{\text{in}} + 0.0314jf}{1 + jf}V0=1+jfVin+0.0314jf
- (d) None of the mentioned
From the structure of the transfer function, option (a) matches the form of the output voltage expression for a first-order low-pass filter, assuming 0.03140.03140.0314 is related to the cutoff frequency and frequency term.
Thus, the correct answer is:
(a) V0=Vin1+0.0314jfV_0 = \frac{V_{\text{in}}}{1 + 0.0314jf}V0=1+0.0314jfVin