Design a passive R C high pass filter (see Fig. 14.10[a]) with a cutoff frequency of 300 Hz us

a) The cutoff frequency in rad/s can be calculated using the formula ωc = 2πf, where f is the fr

Design a passive R C high pass filter (see Fig. 14.10[a])...

Design a passive $R C \text { high pass filter (see Fig. } 14.10[\mathrm{a}])$ with a cutoff frequency of $300 \mathrm{Hz}$ using a $100 \mathrm{nF}$ capacitor. a) What is the cutoff frequency in rad/s? b) What is the value of the resistor? c) Draw your circuit, labeling the component values and output voltage. d) What is the transfer function of the filter in part $(\mathrm{c}) ?$ e) If the filter in part (c) is loaded with a resistor whose value is the same as the resistor in $(\mathrm{b})$ what is the transfer function of this loaded filter? f) What is the cutoff frequency of the loaded filter from part (e)? g) What is the gain in the pass band of the loaded filter from part (e)?

Transcript

00:01 Hi there, so for this problem we are asked to design a passive rc high pass filter with a q -tuff frequency of 300 hz using a capacitor with a capacitance of 100 nanofarets.

00:33 So for part a of this problem, we need to determine to calculate the code of frequency in radiance per second.

00:45 So to calculate this, we use the fact that the angular frequency is equal to 2 times pi the frequency.

00:56 So in here we just substitute that value that we are given, 300 hertz.

01:02 And from this we obtained a value of 1 ,884 .96 radiance per second.

01:27 So that's a solution for part a of this problem.

01:31 Now for part b, we are asked about what is the value of the resistor? so to calculate the value of the resistor, we use the fact that the angular frequency, the cutoff angular frequency, should be equal to the inverse of the product between the resistance and the capacitance.

01:55 So if we solve for the resistance from this expression, we will find that the resistance is equal to 1 over the angle, the cutoff angular frequency times the capacitance.

02:10 So we just now substitute those values, the value that we just have found, 1 ,884 .96 radiance per second.

02:24 And this times the capacitance that in this case is also given, and that is 100 nano, which means 10 to the minus 9 ferrets.

02:36 So from this, we obtain a resistance of 5305 .17.

02:55 Well, let's put 16 oms.

03:00 So that's a solution for part b of this problem.

03:04 Now, for par c, we are told to draw this circuit leveling the component value.

03:17 And output voltage.

03:20 So in this case, we are going to have the voltage in here, positive and minus.

03:35 This is the voltage.

03:40 And we also draw the capacitor and the resistor.

03:54 So this is the capacitance with its value that we are given 100 nanofarats.

04:03 The resistor 5 ,305 .16 oms.

04:13 And of course, in this case, is positive and minus in this case.

04:19 So that's a solution for part c of this problem.

04:23 Now for part d, we are asked, what is the transfer function of the filter in part c? so to calculate this, we use the following equation.

04:46 That is that the transfer function that depends on adds is equal to voltage divided by the initial voltage x...

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