Consider this system equation where x(t) = input $$y(t)=x(t) \cos(3t)$$ Using the superposition theorem, we can prove that the system is linear. For input x1(t), the output is $$y_1(t)=x_1(t) \cos(3t)$$ For input x2(t), the output is $$y_2(t)=x_2(t) \cos(3t)$$ For input [ x1(t) + x2(t) ], the output is $$y(t)=[x_1(t)+x_2(t)] \cos(3t)$$ That is, $$y(t)=y_1(t)+y_2(t)$$ Hence the system is linear. But I can't get the meaning of this. y(t) is linear with respect to x(t) means when I plot a graph of y(t) v/s x(t), I should get a straight line passing through the origin. But for the above case, it's not a straight line. Please clarify this confusion. Also, if it is found to be linear, is the system linear for any x(t) or not? I mean, if we take x(t)=tu(t) or x(t)=t^2u(t), is the system linear in both cases?
9,280 23 23 gold badges 32 32 silver badges 42 42 bronze badges asked Oct 5, 2015 at 6:27 Ajay shifu Ajay shifu 41 1 1 gold badge 3 3 silver badges 9 9 bronze badgesDepending on the system equation behavior of the curve of input versus output can be of any shape.
System is said to be linear if it satisfies these two conditions:
\$\begingroup\$ Thanks for this information. how does the graph y(t) v/s x(t) look like? \$\endgroup\$
Commented Oct 5, 2015 at 16:30 \$\begingroup\$ Look at reference I linked. \$\endgroup\$ Commented Oct 5, 2015 at 16:32\$\begingroup\$ The website you linked to seems to contradict your answer. The second sentence below Fig. C says only a system who's input-output characteristic resembles a line like y=x is linear. \$\endgroup\$
Commented Aug 31 at 14:18 \$\begingroup\$According to my understanding, the linearity of the system is checked by fixing the time. The input signal x(t) is varied at fixed value of t (let 1 sec). Then see how the output y(t) is varying at the same value of t. If the relationship between y and x is linear (straight line) and crossing through origin then the system is linear. If you find any time t at which the system is not linear then the system is non-linear.
answered Mar 16, 2018 at 6:35 11 1 1 bronze badge \$\begingroup\$Linear does not mean, that you get straight lines for y(t) over x(t). Just think about about an RC low pass. A sinus as x(t) will cause a phase shifted y(t) and thereby you'll get an elipis for y(t) over x(t).
Lineararity means, that when you have a system \$\Gamma(input, state)\$ and now we have \$x_1\$ , \$x_2\$ as inputs and \$s_1\$ , and \$s_2\$ as states, we can write:
$$\Gamma(a x_1 + b x_2, c s_1 + d s_2) = a \Gamma(x_1, 0) + b \Gamma(x_2, 0) + c \Gamma(0, s_1) + d \Gamma(0, s_2)$$
Think about linearity in systems more like linearity in a story line. From A follows B. More A and I get more B.
Finally, many people say linear but actually mean linear time-invariant (LTI). The example you give is linear but time variant. When you are LTI, there are great benefits in the frequency domain. The special interest in sin, cos and exp functions come from the nice property, that exp functions are eigenfunctions of LTI systems. If you excite an LTI system with an exponential function \$x=e^t\$ , then y is proportional to x so y(t) over x(t) will plot as a straight line. But this is true only for eigenfunctions.