Butterworth Filter frequency response is given as:

$$H_a(j\Omega)=\frac{1}{\sqrt{{1+\left(\frac{\Omega}{\Omega_c}\right)^{2N}}}}\quad \text{where $N$ is the order of the filter}$$

and for the transfer function, you could substitute $\Omega=\frac{s}{j}$

which implies $$H_a(s)=\frac{1}{\sqrt{{1+\left(\frac{-s^2}{\Omega_c^2}\right)^{N}}}}$$

But my professor skipped all this and directly evaluated all the poles and expressed the transfer function as a product of poles in the denominator.

For example Butterworth filter of order one according to me should be $$H_a(s)=\frac{1}{\sqrt{{1+\left(\frac{-s^2}{\Omega_c^2}\right)}}}=\frac{\Omega_c}{\sqrt{\Omega_c^2-s^2}}$$

but he got it as $$H_a(s)=\frac{\Omega_c}{s+\Omega_c}$$ considering order 1 has only one pole at $s=-\Omega_c$ which I don't deny.

But where am I going wrong in my expression of the transfer function? Why is my transfer function different from his?


Your first equation is the magnitude of the frequency response. So the squared magnitude of the transfer function becomes


Since $|H(s)|^2=H(s)H(-s)$, and since we want a causal and stable transfer function, we assign all poles in the left half-plane to $H(s)$. So for $N=1$ we obtain





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