The causal realization of a system transfer function H( )s having poles at (2, –1), (–2, 1) and zeroes at (2, 1), (–2, –1) will be

The correct option is Unstable, complex, all pass.

Let us analyze the given transfer function step-by-step based on the locations of its poles and zeros in the complex s-plane.

1. Given Pole and Zero Locations:

The system transfer function $H\left(s\right)$ has poles at:

$p_1=2-j1=2-j$

$p_2=-2+j1=-2+j$

The zeros are located at:

$z_1=2+j1=2+j$

$z_2=-2-j1=-2-j$

2. Real vs. Complex System Coefficients:

For a system to have real coefficients in its transfer function, all complex poles and zeros must occur in complex conjugate pairs (i.e., if $a+jb$ is a pole, $a-jb$ must also be a pole).

Here, the conjugate of pole $p_1=2-j$ would be $2+j$, but $2+j$ is a zero ($z_1$), not a pole. Thus, the poles do not occur in conjugate pairs, which means the realization/system coefficients are complex.

3. Stability:

A causal LTI system is stable if and only if all of its poles lie strictly in the left half of the s-plane (i.e., real parts of all poles are strictly negative).

One of the poles is $p_1=2-j$, which has a positive real part ($\mathrm{Re}\left\{p_1\right\}=2>0$). Since a pole lies in the right-half of the s-plane, the causal system is unstable.

4. Frequency Response (All-Pass Feature):

Notice that for each pole $p_k$, there is a corresponding zero at $z_k=-p_k^{*}$ (reflection across the imaginary axis):

For $p_1=2-j$, $-p_1^{*}=-(2+j)=-2-j=z_2$.

For $p_2=-2+j$, $-p_2^{*}=-(-2-j)=2+j=z_1$.

Since the zeros are symmetric to the poles with respect to the imaginary axis, the magnitude response $\left|H\right(j\omega \left)\right|$ is constant for all frequencies $\omega$. Therefore, the system is an all-pass filter.

Combining all three observations, the causal realization of the system transfer function is Unstable, complex, all pass.