System Characterization: multiply imaginary component by scalar

Question

I tried solving a test question that frankly stumped me. If you could explain to me the solution I’d be really grateful.

Given $a, b \in \mathcal{R}$ the system $R_v$ takes the complex input $a + bj$ and returns $a + bvj$. For which values of v the system is linear?

This system accepts both discrete and continuous inputs.

The answer is $v=1$, but I don’t understand why for any other $v$ linearity is not fulfilled. It even seemed trivial to me that it is. Like if I take take two complex inputs I can characterize as $x_1 = a(t) + b(t)j$ and $x_2 = c(t) + d(t)j$ applying the system:
would result in $$\psi\{\alpha x_1+ \beta x_2\} = (\alpha a+\beta c)(t) + (\alpha b+ \beta d)vj$$, whereas applying the system to each one separately would do the same thing: $$ \psi\{\alpha x_1 \} + \psi \{\beta x_2 \} = (\alpha a + \alpha bvj) + (\beta c + \beta dvj) $$ so I think I must be missing something here

Answer 1

I think the problem here is that the system has complex inputs and outputs and you would have to extend the definition of linear to complex scale factors. Assuming that $F(a+jb) = a+jvb$, $y = F(x)$ you would have to proof that

$$F(\alpha x_1 + \beta x_2) = \alpha F(x_1) + \beta F(x_2)$$

You have successfully proven that this is the case if the scale coefficients are real ($\alpha,\beta \in \mathbb{R}$). However, for a complex system this must also hold for complex scale coefficients ($\alpha,\beta \in \mathbb{C}$) which is indeed not the case here.

The answer $v=1$ is trivial since the system just turns into an identity.

Answer 2

The fact that the test for linearity suggests that the system is linear if we only consider real valued scale factors shows that the system would be linear if viewed as a real-valued multiple-input multiple-output (MIMO) system. We could model the system as one with two inputs and two outputs. In that case, the transfer function matrix is just a constant diagonal matrix:

$$\mathbf{H}=\left[\begin{array}{ll}1 & 0 \\ 0 & v\end{array}\right]\tag{1}$$

However, if viewed as a single-input single-output complex-valued system - as implied in the test question - , the system is not linear because the linearity test fails if we consider complex-valued scale factors. This is correctly pointed out in Hilmar's answer.

I'd like to show another way to see that the given complex-valued system is not linear. First, note that the given system is cleary time-invariant. Consequently, if it were linear, it would need to be a linear time-invariant (LTI) system, and as such its output must be computable from its input via convolution with a (complex-valued) impulse response

$$h(t)=h_R(t)+jh_I(t)$$

With $x(t)=x_R(t)+jx_I(t)$ being the complex-valued input, the corresponding output is

$$y(t)=(x_R\star h_R)(t)-(x_I\star h_I)(t)+j\big[(x_R\star h_I)(t)+(x_I\star h_R)(t)\big]\tag{2}$$

From the specification of the system, we require the output to be

$$y(t)=x_R(t)+jvx_I(t)\tag{3}$$

Comparing $(2)$ with $(3)$ shows that there exists no impulse response $h(t)$ satisfying $(3)$. Hence, the system cannot be linear.

A complex-valued system can be viewed as a MIMO system. However, the transfer function matrix is very specific, and we can't choose the matrix entries independently. The transfer function matrix of a complex-valued system has the form

$$\mathbf{H}=\left[\begin{array}{ll}H_R(s) & -H_I(s) \\ H_I(s) & H_R(s)\end{array}\right]\tag{4}$$

where $H_R(s)$ and $H_I(s)$ are the transfer functions of the real and imaginary parts of the impulse response. Comparing $(4)$ with $(1)$ shows again that the given system cannot be implemented by a linear complex-valued system.