Design an RC band-pass filter that passes a signal with frequency 5.00 kHz, has a ratio Vout /

Design an RC band-pass filter that passes a signal with...

Key Concepts

Impedance in AC Circuits

Impedance is the measure of a circuit’s opposition to alternating current (AC) and includes both resistive and reactive components. In RC circuits, the resistor provides a constant impedance while the capacitor’s impedance decreases with increasing frequency. Specifying an impedance at very high frequencies usually means that the capacitive contribution becomes negligible and the resistor defines the limiting impedance.

Phase Shift

Phase shift in RC filters refers to the difference in phase between the input and output signals. In RC circuits, capacitors introduce a phase shift that varies with frequency, causing the output signal to be delayed or advanced relative to the input. Accurately calculating the phase shift is essential for understanding the timing response of the filter at a given frequency.

Frequency Response and Cutoff Frequency

The frequency response of an RC circuit describes how the amplitude of the output signal varies with frequency. The cutoff frequency is a critical parameter that marks the point where the output falls to a specified level (commonly –3 dB) relative to the maximum. In band-pass filters, the relationship between the cutoff frequencies of the high-pass and low-pass sections determines the pass band and the filter’s overall gain at specific frequencies.

RC Band-Pass Filter

An RC band?pass filter is a circuit that allows signals within a specific frequency range to pass while attenuating frequencies outside that range. It is typically constructed by combining a high?pass filter (which blocks low frequencies) with a low?pass filter (which blocks high frequencies), resulting in a pass band where the desired signal is relatively unattenuated.

High-Pass and Low-Pass Filters

A high-pass filter is designed to allow frequencies above its cutoff frequency to pass while attenuating lower frequencies, and a low-pass filter does the opposite by allowing frequencies below its cutoff frequency to pass while attenuating higher frequencies. When these two are combined, they form a band-pass filter with a frequency window determined by the high-pass and low-pass cutoff frequencies.

Transcript

00:01 Hi there, so for this problem, we are asked to design an rc pan pass filter, high pass filter that passes a signal with a frequency equals to 5 kilohertz and has a radio between the output and input voltage equals to 0 ,000 ,000, and has an impedance of one kilo -ooms at a very high frequencies.

00:48 So for part a we are asked what components will you use for this? so the answer for that is we need to find the value for the capacitors.

01:07 So we know that the ratio of the voltages for the high pass rc filter is given by the following.

01:21 The voltage out divided by the voltage input is equal to 1 plus the angular frequency to the square, the resistants to the square, the capacitance to the square, the inverse of thought, and all of that elevated to minus 1 over 2.

01:42 Where we know that the angular frequency is defined as 2 times pi times the frequency.

01:52 Now, so from this, we can solve to obtain the ratio.

02:02 And, well, from this ratio, we can solve to obtain the capacitance.

02:06 So we have this expression.

02:08 So what we can do is to elevate both sides of this equation and take the inverse of thought, so we will obtain, remember that this in here is 0 .5.

02:20 So if we elevate that to the square, we will obtain 0 .25.

02:26 And if we take the inverse of thought, we will obtain 1 over 0 .25, which is equal to 4.

02:32 So we have 4 is equal to 1 plus 1 over the product between this.

02:44 And in here solving for the capacitance, we will find that that is equal to 1 over the square root of 3 times 1 over the angular frequency times the resistance.

03:02 Since we know that the angular frequency is 2 pi times the frequency, we will have this expression right here.

03:13 So we can, just substitute all of the values that we have from here.

03:20 So we have one over the square root of 3, and 2 pi times the frequency, which in this case is also given, that it is a value of 5 kilo, which means tens to the 3 hertz, times the resistance, which is also given, that is the impedance...

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