# High pass filter

Some one please help me with this question. I am getting -10.5 as the wrong final answer.I found the formula in the following slide.Since the result was coming out in complex number i got its modulus.That is how i ended up with a 10.5 (2 * 5.25).

• What is your working? Dec 8 '16 at 11:20
• As you're getting an answer, how are you getting that? The more information you give us, the more likely we are to help you. Dec 8 '16 at 11:29
• Your answer should be 5.25 volts - it's asking for amplitude and therefore you don't need the minus sign even though it is an inverting amplifier. Dec 8 '16 at 11:34
• Why should it be 5.25? @Andyaka Dec 8 '16 at 11:39
• Ahh now there's a question but first, what is the answer that you believe to be correct (in case it is wrong or mine is wrong) and how did you get to 10.5 as the amplitude? You need to explain that last bit in your question so please edit it. Dec 8 '16 at 11:42

For reference and in case the question gets changed, here is your circuit:

You say Rf = 10 kΩ, R1 = 1.5 kΩ, and C = 100 nF.

Right away, you should see that this circuit has two parts that can be analyzed separately. From the left end of R1 to Vo is just a ordinary inverting amplifier. It's gain is Rf/R1 = 6.7.

Looking into the left end of R1, it should be obvious (from understanding this kind of basic inverting amplifier) that the impedance seen will be just R1. Therefore we now have a high pass R-C filter with 100 nF and 1.5 kΩ. Whatever that does to the input signal will be multiplied by the 6.7 we determined previously.

The rolloff frequency of the R-C high pass filter is $1/2\pi RC$ = 1.1 kHz. I'll leave it to you to figure out what that does to a 1.36 kHz signal.

• One thing that's unclear to me: For C and R1 to form a high pass filter, shouldn't the output of the filter be the point between the C and R1? and not the point to the right of R1.. Dec 8 '16 at 12:13
• To clarify a bit: My concept of a filter (low-pass/high-pass) is that of a voltage divider formed by C and R, with the output between the two Dec 8 '16 at 12:14
• @m.Ali: It is. Look at the circuit from R1 to the right only. That's a basic amplifier. The input is the voltage at the left end of R1. As you say, that's also the output of the filter. This only works because the right side of R1 is held at ground potential, sometimes referred to as a virtual ground. Dec 8 '16 at 12:16
• It is a 1st-order highpass filter with gain (Rf/R1).
– LvW
Dec 8 '16 at 12:26

The feedback impedance divided by the input impedance is the gain magnitude of the circuit in the question.

• Feedback impedance is 10 kohm.
• Input impedance is $\sqrt{X^2_C + R^2_1}$ = $\sqrt{1172^2 + 1500^2}$ = 1903.6 ohms.

Gain magnitude is therefore 5.253 and output amplitude is 10.506 (and inverted).