The signal which is sampled at 500Hz has a power line artifact at 60 Hz. Find the location of zeros in the transfer function of the notch filter to remove the artifact at 60 Hz.
0.512+j0.66
0.512−j0.66
0.7289+j0.6845
0.7289−j0.6845
0.512+j0.66
0.7289+j0.6845
0.7289−j0.6845
To remove the power line artifact at 60 Hz, we need to design a notch filter with a transfer function that has zeros at the frequency of the artifact.
A zero in the transfer function corresponds to a value where the numerator of the transfer function becomes zero. In this case, we want the numerator to be zero at 60 Hz.
Given the transfer function of the notch filter as:
H(z) = (1 - 2cos(ω₀)z⁻¹ + z⁻²) / (1 - 2Re(ω₀)z⁻¹ + |ω₀|²z⁻²)
where ω₀ is the angular frequency of the zero, we can try to find the value of ω₀ that corresponds to 60 Hz.
ω₀ = 2πf₀
where f₀ is the frequency in Hz. In this case, f₀ = 60 Hz.
ω₀ = 2π * 60 = 120π
Now, we need to convert ω₀ to the complex number representation in the form of a + jb, where a is the real part and b is the imaginary part.
ω₀ = 120π
β = ω₀ / fs
where fs is the sampling frequency. In this case, fs = 500 Hz.
β = (120π) / 500
Next, we can calculate the real and imaginary parts of β.
a = cos(β) = cos((120π) / 500)
b = sin(β) = sin((120π) / 500)
Now, we can find the zeros of the transfer function by substituting the values of a and b into the complex number representation:
Zero = a + jb
Plugging in the values into the options given:
Option 1: 0.512 + j0.66
a = 0.512, b = 0.66
Option 2: 0.512 - j0.66
a = 0.512, b = -0.66
Option 3: 0.7289 + j0.6845
a = 0.7289, b = 0.6845
Option 4: 0.7289 - j0.6845
a = 0.7289, b = -0.6845
By comparing the calculated values of a and b with the values in the options, we can see that the correct answer is:
Option 4: 0.7289 - j0.6845
This means that this is the location of the zero in the transfer function of the notch filter to remove the power line artifact at 60 Hz.