The transfer function for a filter is: 

$$H(jomega)=(jomega)/(1+jomega)$$

This filter is best classified as a:

A.   low-pass filter
B.   high-pass filter
C.   band-pass filter
D.   notch filter

A brief description of the different types of filters listed as possible answer choices:

1. A low-pass filter allows low frequencies to pass through while attenuating high frequency signals.
2. A high-pass filter allows high frequencies to pass through while attenuating low frequency signals.
3. A band-pass filters allows frequencies in a specified frequency range to pass through.
4. A notch filter attenuates frequencies in a specified frequency range and allows all others outside that range to pass through.

In order to figure out which type of filter we have, we want to evaluate the transfer function as the frequency approaches both 0 and infinity:

$$\begin{gathered} H\left(j\omega \right) =\frac{j\omega }{1+j\omega } \\ \underset{j\omega \to 0}{lim}H\left(j\omega \right)=\frac{0}{1+0} \\ \underset{j\omega \to 0}{lim}H\left(j\omega \right)=0 \end{gathered}$$

Now evaluating the transfer function as the frequency approaches infinity:

$$\begin{gathered} H\left(j\omega \right) =\frac{j\omega }{1+j\omega } \\ \underset{j\omega \to \infty }{lim}H\left(j\omega \right)=\frac{\infty }{1+\infty }=\frac{\infty }{\infty } \end{gathered}$$

We must use L'Hopital's rule since we get infinity over infinity:

$\underset{x\to c}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to c}{lim}\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}$

Taking the derivative of both the numerator and denominator:

$$\begin{gathered} \frac{j\omega }{1+j\omega }\to \frac{1}{{\left(j\omega \right)}^{-1}+1} \\ \frac{1}{{\left(j\omega \right)}^{-1}+1}=\frac{1}{1+{\displaystyle \frac{1}{j\omega }}} \end{gathered}$$

Solving once again for the limit as frequency approaches infinity:

$$\begin{gathered} \underset{j\omega \to \infty }{lim}H\left(j\omega \right)=\frac{1}{1+{\displaystyle \frac{1}{\infty }}} \\ \underset{j\omega \to \infty }{lim}H\left(j\omega \right)=\frac{1}{1+{\displaystyle 0}} \\ \underset{j\omega \to \infty }{lim}H\left(j\omega \right)= 1 \end{gathered}$$

Therefore, our transfer function approaches 1 as the frequency approaches infinity and 0 when the frequency approaches 0. This means that the filter allows higher frequencies, above a certain cut-off frequency, to pass through but rejects signals at the lower frequency end. This corresponds to the high-pass filter that was described previously.

Therefore, our answer is B, high-pass filter.