Using circuit in (Figure 1) and only three components from...

Using circuit in (Figure 1) and only three components from the following table, design a high-pass filter with a cutoff frequency and passband gain as close as possible to 8 kHz and 14 dB respectively. Components 270 $Omega$ 330 $Omega$ 390 $Omega$ 1.8 k$Omega$ 2.7 k$Omega$ 3.3 k$Omega$ 18 k$Omega$ 100 $mu$F 47 $mu$F 22 nF 47 nF

Transcript

00:01 So for this question, the given data is, given data is, have the value of the cutoff frequency that will be noted by f0, which is 8 kilohertz, have the value of the passband gain, that is k, which is 14 d .b.

00:20 So to determine the percent error in cutoff frequency, the cutoff frequency is basically given by the omega knot will be equal to 1 divided by rc.

00:30 Let the value of the capacitance be 47 nm and the cut of frequency at 8 khz.

00:37 So the resistance will be calculated will be r will be equal to 1 divided by omega -0 c.

00:44 So here we are taking c as 47nf.

00:50 Let us see.

00:53 So with substitute the value, this will be equal to that is 1 divided by omega -0 will replace by 2 -5 f -0.

01:02 Will be remultiply by c this will be equal to 1 divided by 2 pi which is multiply by 8 which is multiplied by 10 to the power 3 will be multiplied by c that is 47 will be multiplied by 10 to the power minus 9 to became into f so this from this r will be equal to 4 to 3 .28 om now let the value of the resistance will be equal to we'll take here r i which is equal to 390 om.

01:37 So the cutoff frequency will be, that is omega -0, will be equal to 1 divided by rc.

01:42 This will be equal to 2 pi f -0, 2 -5 -f0, which is equal to 1 divided by rc.

01:50 So from this, f -0 will be equal to 1 divided by rc, which is multiplied by 2 -5.

01:58 So now let us substitute the value.

02:00 So f -0 will be equal to 1 divided by 2 -p, will be multiplied by 390 which is multiplied by 47 .47 which is multiplied by 10 to the power minus 9.

02:12 So from this, the cutoff frequency, that is, f0 will be equal to 8 .682 hertz.

02:20 Now consider the pass bend in k value as 14 db.

02:25 So the k is equal to 14 db.

02:28 So this will be equal to 10 to the power of 14 divided by 20, which will be equal to 5.

02:35 Let the value of k will be equal to 5 to rf divided by r i so r f divided by r i will be equal to 5 so rf will be equal to 390 which will be equal to 5 oms so from this we will get rf value that is here it is 5 ome so from this we are getting the rf value which is equal to 1950 now we'll choose the value of the resistance rf as 1 .8 kilo -oom.

03:13 Now the value of the new pass -bend gain will be equal to that is ka is equal to 20 log base 10...

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