R4 = 200 kΩ
Notch Filter
A notch lter attenuates signals within a narrow band of frequencies or removes a single frequency. The frequency we
want to remove in this case is 60 Hz because most noise produced by electronic devices is at that frequency. A Q factor is
the ratio of the center frequency to bandwidth, and it also helps describe the shape of the magnitude plot. A larger Q
factor results in a narrower stop band. For calculations, we will be using a Q value of 8.
We decided to choose capacitor values we had. So, C1 = C2 = 0.1 uF, and C2 = 0.2 uF.
The equations we will be using to calculate R1, R2, and R3 are as follows:
R1 = 1 / (4*pi*Q*f*C1) = 1 / (4*pi*8*60*0.1E-6) = 1.6 kΩ
R2 = (2*Q) / (2*pi*f*C1) = (2*8) / (2*pi*60*0.1E-6) = 424 kΩ
R3 = (R1*R2) / (R1 + R2) = (1.6*424) / (1.6 + 424) = 1.6 kΩ
Lowpass Filter
A low pass lter attenuates high frequencies while allowing lower frequencies to pass through. The cuto frequency will
have value of 150 Hz because that is the correct ECG value for adults. Also, the gain (K value) will be 1, and constants a
and b are 1.414214 and 1, respectively.
We chose C1 to equal 68 nF because we had that capacitor. To nd C2 we used the following equation:
C2 >= (C2*4*b) / [a^2 + 4*b(K-1)] = (68E-9*4*1) / [1.414214^2 + 4*1(1-1)] --> C2 >= 1.36E-7
Therefore, we chose C2 to equal 0.15 uF
To calculate the two resistor values, we had to use the following equations:
R1 = 2 / (2*pi*f*[a*C2 + sqrt([a^2 + 4*b(K-1)]*C2^2 - 4*b*C1*C2)] = 7.7 kΩ
R2 = 1 / (b*C1*C2*R1*(2*pi*f)^2) = 14.4 kΩ