Final answer:
The transfer function H(s) = 10 / (s + 10) + 4 / s represents a band-pass filter, which is a combination of a low-pass filter (first term) and a high-pass filter (second term), resulting in a filter that allows frequencies within a certain range to pass while attenuating those outside the range.
Step-by-step explanation:
We need to identify the type of filter represented by the transfer function H(s) = 10 / (s + 10) + 4 / s. To do this, let us examine the terms in the transfer function:
- The term 10 / (s + 10) is a first-order low-pass filter with a pole at s = -10, which allows low-frequency signals to pass while attenuating high frequencies.
- The term 4 / s is a first-order high-pass filter in the form of an integrator with a zero at s = 0, which allows high-frequency signals to pass while attenuating low frequencies.
Combining these two filters results in a band-pass filter, which allows frequencies in a certain range to pass while attenuating those outside that range. Typically, one would expect a band-pass filter to have both low-frequency and high-frequency cutoff points, but given the two terms provided and the lack of additional context, the filter can best be described as allowing a band of frequencies correlated with the poles and zeros to pass through.
Mathematically, the response of a band-pass filter will have gain at midrange frequencies, where neither the low-pass nor the high-pass filter components are effective at attenuation. The specific frequency range where the filter allows signals to pass is determined by the values of the components represented in the transfer function.