3

Your transfer function looks correct. Note that you can rewrite $H(s)$ as $$H(s)=-\frac{3s}{5s+3}\tag{1}$$ or, equivalently, $$5sY(s)+3Y(s)=-3sX(s)\tag{2}$$ Since multiplication by $s$ corresponds to differentiation in the time domain you obtain from $(2)$ $$5y'(t)+3y(t)=-3x'(t)\tag{3}$$


1

I assume that you want to see that the transfer-function magnitude is inverse proportional to the frequency. In this case, you replace $S$ with $i\omega$ you get that the transfer function magnitude decays with the frequency$|G(\omega)|=\left|\frac{n}{ n+i\omega}\right|$


1

One answer is very simple: Information. An AC signal simply can't be quantified with a single number. Sum two 100hz 1V signals and you might get anything between 0 and 2 depending on phase. Complex numbers solve this problem by 'carrying around' two pieces of information all the time. Poles and zeros are similar - their frequency doesn't tell you ...


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