To get a centre frequency if 1.5 MHz and a bandwidth of 300 kHz implies a Q factor of 1500/300 = 5. However, you will never get a centre frequency to bandwidth ratio as required by the question using a simple op-amp and high-pass and low-pass networks. If you'd have used 1 uF, R would be 0.096 ohms and that isn't practical for any op-amp. However, what you will find is that 1 uF is too high in value for an op-amp to work with at circa 1.5 MHz and you will need to choose a much lower default value.įor instance, using a 100 pF capacitor, it will form a 3 dB point at 1.65 MHz when R = 964 ohms. That allows you to calculate both resistors based on both capacitors being 1 uF. make the high-pass network cut-off at 1.5 MHz - 150 kHz (1.35 MHz).make the low-pass network cut-off at 1.5 MHz + 150 kHz (1.65 MHz) and.The two filter points need to be different to obtain a 300 kHz bandwidth so, It won't make a very sharp or precise filter using the configuration shown but, to make life simpler (given that it is a really sloppy filter) you can regard C1 and R1 as forming a high-pass network and C2 and R2 as forming a low-pass network.
Now I'm supposed to solve a system of 2 equations with 2 unknowns, but for some reason, it is not working with me I tried solving for R1 in eq 2, and the substitute in eq 1 and I got a negative R2 resistance, what could be wrong? is there another way to solve it that I am not aware of? I am allowed to assume 2 unknowns out of the four, so, I went with C1 = C2 = 1uF. The first step is to find the transfer function of the circuit, to compare it to the general TF of the bandpass filter: I have this question in Linear Circuit Analysis course where I need to find the values of R1, R2, C1, C2, of the given bandpass filter circuit in order to get a peak frequency wp = 1.5 MHz and a bandwidth Bw = 300 kHz.
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