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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?

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

$$\big|H(s)\big|^2=\frac{1}{1+\left(\frac{-s^2}{\Omega_c^2}\right)^{N}}\tag{1}$$

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

$$H(s)H(-s)=\frac{1}{1-\left(\frac{s^2}{\Omega_c^2}\right)}=\frac{\Omega_c^2}{\Omega_c^2-s^2}=\frac{\Omega_c}{\Omega_c+s}\cdot\frac{\Omega_c}{\Omega_c-s}\tag{2}$$

and

$$H(s)=\frac{\Omega_c}{\Omega_c+s}\tag{3}$$

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