The transfer-function: $$G(s) = \frac{\beta s}{s^2 + \beta s + \omega_0^2}$$

is to be used in an application that requires the magnitude of the frequency response to be of the band-pass form.

  • Show that this property is present by considering the cases when $\omega = 0, \omega = 1$ and show also that $\lvert G(j\omega_0)\rvert = 1$.
  • Determine also the equation that defines the frequencies where the filter magnitude is $−3\textrm{ dB}$.

I use $G(s)=G(j\omega)|_{s=j\omega}$, then $$ \lvert G(j\omega_0)\rvert= \frac{j\beta \omega_0}{-\omega_0^2 + j\beta\omega_0 + \omega_0^2}=1 $$ when $\omega=0$, the numerator equal to $0$ so $\omega=0$ is be filtered out.

My questions are:

  • Could I use the $G(s)=\lvert G(j\omega)\rvert$ directly or $G(s)=G(j\omega)=\lvert G(j\omega)\rvert e^{-jk\omega}$?
  • I have no idea about the case when $\omega=+\infty$?
  • The last question, I use $-20\log\lvert G(j\omega)\rvert=-3\textrm{ dB}$, so we can solve the $\omega$ from this equation, is that correct?
  • $\begingroup$ Please add the homework tag. Where are you stuck ? $\endgroup$
    – Gilles
    Commented Jan 8, 2017 at 12:03
  • $\begingroup$ this is not a homework, should i add the tag? $\endgroup$
    – Haoming Li
    Commented Jan 8, 2017 at 12:05
  • $\begingroup$ Now we know the exercise that you need to solve, but we still don't know what your problem is with that exercise. $\endgroup$
    – Matt L.
    Commented Jan 8, 2017 at 12:07

1 Answer 1


Even though this may not be homework, it is a basic textbook problem, and for this reason I just give you a few hints, which should nevertheless help you solve the problem if you put in a bit of effort.

  1. All questions refer to the filter's frequency response, which is simply obtained by evaluating $G(s)$ for $s=j\omega$.
  2. The first question should read "Show that this property ... $\omega=0$, $\omega={\infty}$" (not $\omega=1$).
  3. Your expression for $|G(j\omega)|$ is wrong; you wrote down $G(j\omega_0)$ (without magnitude). Of course, if $G(j\omega)=0$, or $G(j\omega)=1$, then also $|G(j\omega)|=0$ and $|G(j\omega)|=1$, respectively.
  4. In order to see what happens for $\omega\rightarrow\infty$, write down $|G(j\omega)|^2$ (that's maybe easier than the magnitude), and take the limit.
  5. For the 3 dB cut-off frequencies (there must be two positive solutions), use $|G(j\omega)|^2$ derived above and solve $|G(j\omega_c)|^2=\frac12$. Substitute $x=\omega^2$ and solve the resulting quadratic equation for $x$. From that solution, derive the two positive solutions for $\omega_c$.

All of this involves basic math skills, such as computing the (squared) magnitude of a complex number, and solving quadratic equations. If you find that difficult, read up on these basics because you'll need them all of the time when doing signal processing.


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