Use a 1 mH inductor to design a low-pass, R L, passive...

Use a 1 mH inductor to design a low-pass, $R L$, passive filter with a cutoff frequency of 5 kHz . a) Specify the value of the resistor. b) A load having a resistance of $68 \Omega$ is connected across the output terminals of the filter. What is the corner, or cutoff, frequency of the loaded filter in hertz? c) If you must use a single resistor from Appendix H for part (a), what resistor should you use? What is the resulting cutoff frequency of the filter? (FIGURE CAN'T COPY)

Use a 1 mH inductor to design a low-pass, $R L$, passive filter with a cutoff frequency of 5 kHz .
a) Specify the value of the resistor.
b) A load having a resistance of $68 \Omega$ is connected across the output terminals of the filter. What is the corner, or cutoff, frequency of the loaded filter in hertz?
c) If you must use a single resistor from Appendix H for part (a), what resistor should you use? What is the resulting cutoff frequency of the filter?
(FIGURE CAN'T COPY)

Key Concepts

RL Filter Design

An RL low?pass filter is constructed with a resistor in series with an inductor, and the output is taken across the resistor. Its frequency response is characterized by a cutoff frequency where the magnitude of the output falls to 70.7% (or –3 dB) of the passband value. This cutoff frequency is given by f?c? = R/(2?L), which makes it straightforward to choose the resistor value for a desired f?c? when the inductor value is known.

Inductive Reactance

Inductive reactance is the opposition an inductor presents to alternating current, and it increases linearly with frequency according to the relation X_L = ?L, where ? = 2?f. This frequency?dependent behavior of the inductor is fundamental in determining the filter’s frequency response because it sets the boundary between the passband and stopband.

Effect of Loading on Filter Performance

When a load resistor is connected across the filter’s output, it interacts with the filter’s components, effectively altering the impedance seen by the signal. The load in parallel with the resistor changes the effective resistance in the circuit, which in turn shifts the cutoff frequency. This behavior must be analyzed to determine the actual –3 dB (corner) frequency of the loaded filter.

Standard Resistor Values

In practical circuit design, resistor values are chosen from standard series available from vendor catalogs. Even when the calculated resistor value does not correspond exactly to a standard value, the designer must select the nearest available option, and then recalculate the cutoff frequency to account for any variation introduced by the resistor’s standard value tolerance.

Transcript

00:01 So in this problem we have a series lcr circuit.

00:03 So let's say omega r is equal to the resonance frequency.

00:18 So omega r must be equal to 1 by root over lc.

00:27 So we so that means 2 pi fr is equal to 1 by root over lc where fr is the resonance frequency and i mean omega is the angular resonance frequency so let's say this is angular and fr is the linear or the actual resonance frequency so if we put the values we will get 2 multiplied by 22 by 7 multiplied by 20 kilohertz is equal to 1 by root over of 10 multiplied by 10 to the power minus 10 minus 3 multiplied by c so if we solve this we will get c is equal to 6 .33 multiplied by 10 to the power minus 9 ferrat.

01:38 So this will be the value of the capacitor.

01:44 This is the answer for the first part.

01:46 Now for the second part we know that delta w, that is the bandwidth, is actually given by r by 2l, where r is the, let's say this is r total.

02:05 Where r total is the resistance of the total resistance.

02:11 R total is the total resistance of the circuit.

02:14 Total resistance of the circuit.

02:30 So if we put the values, so here in this book, delta w is assigned as r by 2 , generally we define bandwidth as r by l, but in this very book, delta w is assigned as r by 2l.

02:46 So if we put the values, we will get 2400 multiplied by 2 multiplied by 22 by 22 by 7 is equal to r plus 70 divided by 2 multiplied by 10 multiplied by 10 to the power minus 3.

03:14 Again if we solve it for r we will get r is equal to 231 .7.

03:27 So this is the maximum value of r.

03:32 Okay, if r is greater than this then bandwidth will be also greater than 2 .4 khz...

Please give Ace some feedback