The transfer function of a real system, H ( s) , is given as :

H(s) = As + B/ s² +Cs +D

where A, B ,C and D are positive constants. This system cannot operate as

The correct options are an integrator and high pass filter.

To understand why the given real system cannot operate as an integrator or a high-pass filter, let us analyze its transfer function:

$$H\left(s\right)=\frac{As+B}{s^2+Cs+D}$$

where A, B, C, and D are positive constants.

1. Why the system cannot operate as a High-Pass Filter:
A high-pass filter must allow high-frequency signals to pass through with little to no attenuation. Mathematically, this means that as the frequency approaches infinity (s ⇒ ∞), the magnitude of the transfer function must approach a non-zero constant or infinity.
Let us evaluate the limit of the transfer function as s approaches infinity:

$$\underset{s\to \infty }{lim}H\left(s\right)=\underset{s\to \infty }{lim}\frac{As+B}{s^2+Cs+D}=0$$

Since the degree of the polynomial in the denominator (degree 2) is strictly greater than the degree of the polynomial in the numerator (degree 1), the system completely attenuates high-frequency signals. Therefore, it is impossible for this system to operate as a high-pass filter.

2. Why the system cannot operate as an Integrator:
An ideal integrator has a transfer function of the form:

$$H_{\text{ideal}}\left(s\right)=\frac{K}{s}$$

An ideal integrator has a pole at the origin (s = 0), which causes its gain to become infinitely large as the frequency approaches zero (DC condition).
For the given transfer function, let us find the value at s = 0:

$$H\left(0\right)=\frac{B}{D}$$

Since B and D are positive constants, the DC gain is a finite value, B/D. Furthermore, the poles of this system are the roots of the denominator quadratic equation:

$$s^2+Cs+D=0$$

Since C > 0 and D > 0, all poles lie strictly in the left half of the s-plane (having negative real parts). Because there is no pole at the origin, the system cannot perform true mathematical integration over all frequency ranges. Thus, it cannot operate as an integrator.