Question

All the components in the bandpass filter given below are ideal. The lower -3 dB frequency of the filter is 1 MHz. The upper -3 dB frequency (in MHz, rounded off to the nearest integer) is _________. 

 

Hint:

For a bandpass filter, the upper -3 dB frequency can be calculated using the formula: \[ f_H = \frac{1}{2 \pi R C_2}, \] where \( R \) and \( C_2 \) are the resistor and capacitor values that define the higher frequency limit.

Answer 1

Correct Answer: 50

Step 1: Substitute the values for the lower -3 dB frequency and capacitors:
The given circuit consists of a bandpass filter with the following components:
\( R \) and \( 2R \) resistors.
\( 0.1C \) and \( 10C \) capacitors.
The lower -3 dB frequency \( f_L = 1 \, {MHz} \).
The general formula for the lower and upper -3 dB frequencies for an ideal bandpass filter is: \[ f_L = \frac{1}{2 \pi R C_1} \quad {and} \quad f_H = \frac{1}{2 \pi R C_2}, \] where \( C_1 \) and \( C_2 \) are the capacitors in the filter. Given that:
\( f_L = 1 \, {MHz} \),
\( C_1 = 0.1C \),
\( C_2 = 10C \).
From the lower frequency formula: \[ f_L = \frac{1}{2 \pi R \cdot 0.1C} = 1 \, {MHz}. \] Solving for \( R \cdot C \): \[ R \cdot C = \frac{1}{2\pi \cdot 1\,{MHz} \cdot 0.1} = \frac{1}{0.628 \times 10^{-6}} = 1.59 \times 10^6\,\Omega\cdot{F} \] Step 2: Use \( R \cdot C \) to calculate the upper -3 dB frequency:
Now, use this value for \( R \cdot C \) to calculate the upper -3 dB frequency \( f_H \): \[ f_H = \frac{1}{2 \pi R \cdot 10C} \] Substitute \( R \cdot C = 1.59 \times 10^6 \, \Omega\cdot{F} \): \[ f_H = \frac{1}{2\pi \cdot 1.59 \times 10^6 \cdot 10} = \frac{1}{10 \times 0.628 \times 10^{-6}} = 50\,{MHz} \] Thus, the upper -3 dB frequency is 50 MHz.