let $ \omega_H>\omega_L>0$ be frequencies. I need to create an LTI non ideal band-pass filter so that$$ Y\left(s\right)=\begin{cases} 1 & \omega_{L}<\left|\omega\right|<\omega_{H}\\ 0 & else \end{cases} $$ The transfer function must be :
- Causal
- BIBO stable
- Rational
- Has a real valued impulse response.
- For every constant input $x(t)=c\Rightarrow y(x(t))=0 $
- Has minimal amount of poles and zeroes.
I'm not really sure how to approach all the details.
Because it is causal, we will have a right side ROC.
Because it is rational it means $ H\left(s\right)=\frac{\sum\limits _{i=1}^{k}\alpha_{i}s^{i}}{\sum\limits _{i=1}^{n}\beta_{i}s^{i}} $
Bibo stable means that the $ j\omega $ axis is included in the ROC.
Because it resets every constant, we know that it involves differentiating in the time domain, so $ H(s)=sX(s)$.
My problem is with the real valued impulse response. If $ h(t) $ is real valued then it's Laplace transform $ H(s) $ is an even function. But because $ H(s) $ is even and we have poles in $ \omega_L $ and $\omega_H$ it means we basically should have poles in $ -\omega_L$ and $-\omega_H$. The system is also $ 0 $ for every constant, so $ s=0 $ should be a zero of the transform, but in order to keep it an even function we must square it.
In the end my function looks like : $$ H\left(s\right)=\frac{s^{2}}{\left(s+\omega_{L}\right)\left(s+\omega_{H}\right)\left(s-\omega_{L}\right)\left(s-\omega_{H}\right)} $$
However this function contradicts the BIBO property since the ROC is jut right of $\omega_H$ and hence we don't include $j\omega $. Where did I go wrong?