Amplitude response of Lynn's low pass filter at the 3dB cutoff frequency I am referring to the calculations in https://courses.cs.washington.edu/courses/cse474/18wi/labs/l8/QRSdetection.pdf for the second order Lynn's low pass filter.
On the third page, the author mentions the transfer function used as:
$$
H(z)=\frac{(1−z^{-6})^2}{ (1−z^{-1})^2 }
$$
The amplitude frequency response for this is given as:
$$
|H(e^{j\omega T})|=\frac{\sin^2(3\omega T)}{\sin^2(\omega T/2)}
$$
Now, he has calculated the 3dB cutoff frequency to be around 11 Hz for the sampling frequency of 200 Hz (i.e. sampling time = 0.005 seconds). Substituting these values in the above equation for amplitude response, I get:
$$
|H(e^{j\omega_c T})|=\frac{\sin^2(3 \cdot 11 \cdot 0.005)}{\sin^2(11 \cdot 0.005/2)} = \frac{0.026979}{0.000756} = 35.69
$$
At the 3dB cutoff frequency indicated, shouldn't the amplitude response work out to around 0.707 and not 35.69? Where is my understanding incorrect?
Here's your mistake, you forgot that $ \omega = 2\pi f$