I know that frequency mixer $y(t)=x(t)cos(\omega_ct)$ is a nonlinear system since its output has frequency components that are not present in the input. To prove its nonlinearity, it shuldd be shown that the system either violates scaling or additivity. However, I could not do that. Can someone please help me?


If the system is nonlinear then if $y_1(t)$ is the response to the signal $x_1(t)$, and $y_2(t)$ is the output given input signal $x_2(t)$ then the response to the signal


with arbitrary constants $a_1$ and $a_2$ will generally not be equal to


However, for the given system an input signal (1) always results in an output signal (2). The conclusion is that the system is linear.

The confusion arises because you think that linear systems cannot add new frequencies to the input signal. This is generally not true. Only linear and time-invariant (LTI) systems have that property. Linear but time-varying systems can produce new frequencies in the output signal, and a modulator is a classic example of a linear time-varying system.

  • $\begingroup$ The answer makes sense. But what about the following sentence which I am quoting from a famous communications textbook, Digital Communications by Bernard Sklar: "Can the mixer be a linear device? No. The output signal of a LINEAR device will only consist of the same component frequencies as the input signal, differing only in amplitude and/or phase." $\endgroup$ – ArmenG Feb 20 '15 at 23:30
  • $\begingroup$ @ArmenG: If you consider both signals to be mixed as input signals, then it's a non-linear device, because it multiplies the two signals. If you see it as a device with one input signal plus a fixed signal, being part of the device (as in your example), then it's a linear system. So it's basically a matter of modelling. The system $y(t)=x(t)\cos(\omega_c t)$ with $x(t)$ as (only) input and output $y(t)$ is definitely linear. The system $y(t)=x_1(t)\cdot x_2(t)$ with two input signals $x_1(t)$ and $x_2(t)$ is non-linear. $\endgroup$ – Matt L. Feb 21 '15 at 9:23
  • $\begingroup$ So is the following sentence, which is quoted from the same textbook, incorrect and to correct it, linear device should be replaced with LTI system? "The output signal of a LINEAR device will only consist of the same component frequencies as the input signal" $\endgroup$ – ArmenG Feb 21 '15 at 16:54
  • $\begingroup$ @ArmenG: That's right, you have to add "time-invariant" for this sentence to be correct. Strange that Sklar is kind of sloppy about this. $\endgroup$ – Matt L. Feb 21 '15 at 21:19

The system is linear but time-variant, because $$ x(t-t_0)\cos(\omega_\mathrm{c}t) \neq y(t-t_0)\quad\text{for }\omega_\mathrm{c} \neq k2\pi $$ which can be seen by comparing the above equation with $$ y(t-t_0) = x(t-t_0)\cos(\omega_\mathrm{c}t - \omega_\mathrm{c}t_0) $$ Therefore, a frequency mixer is no linear time-invariant (LTI) system. One consequence of this property is, that the system can not be represented by a transfer function $H(\omega)$. More precisely, only an LTI system can be expressed in frequency domain by $$ Y(\omega)=X(\omega)H(\omega). $$ Interestingly, amplitude demodulation can also be implemented by a time-invariant, non-linear system (namely a diode).


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