# Can someone let me know linearity, time-invariance, causality, memory characteristic of the system?

The input/output system is $$\frac{dy(t)}{dt}+2y(t)=2x^2(t)$$

I want to know this is linear? causal? time-invariant? memoryless?

According to solution, the answer is 'Linear if zero initial conditions, causal, time-invariant, with memory'.

My trial:

$$y(t)=x^2(t)-\frac{1}{2}\frac{dy(t)}{dt}$$ Sorry for my poor drawing.

1. for input value: $x_1(t)=2x(t)$, \begin{align} y_1(t)&=\left(2x(t)\right)^2+\left(-\frac12\right)\frac{d}{dt}\left(y(t)\right)\\ &=4x^2(t)-\frac12\frac{dy(t)}{dt}\\ 2y(t)&=2x^2(t)-\frac{dy(t)}{dt}\\ \therefore y_1(t) &\ne 2y(t) \end{align}

## $\because$ not satisfying Homogeneity, Non-Linear.

1. The system doesn't have $x(t+k)$. (It means $y(t)$ is not made by $x(t+k)$)

## $\therefore$ The system is causal because

1. for input value: $x_1(t)=x(t-t_0)$, \begin{align} y_1(t)&=x^2(t-t_0)-\frac{1}{2}\frac{dy(t)}{dt}\\ y(t-t_0)&=x^2(t-t_0)-\frac{1}{2}\frac{dy(t-t_0)}{dt}\\ \therefore y_1(t) &\ne y(t-t_0) \end{align}

## $\therefore$ time-varying.

2. The system doesn't have any $\displaystyle\int$.

i.e.) $\displaystyle\int x(\tau) d\tau$.

## $\therefore$ memoryless

What is the real solution?

• whoa, the system is not linear (you're right about that), but it is time-invariant and it is not memoryless. – robert bristow-johnson Nov 5 '15 at 1:12
• @johnson Thanks for a comment. Can you tell me why it is? – Danny_Kim Nov 5 '15 at 5:10
• Where can I read about how to do this analysis on continuous systems? I already understand discrete systems. – MackTuesday Nov 5 '15 at 18:59
• This problem is in the text book, named "Continuous And Discrete Signals And Systems 2/e - Soliman and Srinath". – Danny_Kim Nov 5 '15 at 19:18

There's an error in your 3rd section (linearity) - you replace $y(t)$ with $y_1(t)$ on the left side of the equation but not on the right side. You should have $$y_1(t)=x^2(t-t_0)-\frac{1}{2}\frac{dy_1(t)}{dt}$$. Now you see that $y_1(t)=y(t-t_0)$.
As for memoryless - start with your initial equation and integrate both sides. Now you have $$\int y(t)dt=\int x^2(t)dt-\frac{1}{2} y(t)dt$$
and rearranging gives $$y(t) = 2 \int x^2(t)dt-2 \int y(t)dt$$ Does that help?
• Thank you for the answer. I got my fault, thanks to you. I should have to change $y(t)$ to $y_1(t)$. I think, I have to study differential equation first. I don't know how to find $y(t)$ from differential equation (I know a little.). I've booked introduction to differential equation textbook. – Danny_Kim Nov 5 '15 at 19:23