Electr. Engineering - Digital Sig. Processing

[DefaultName]

Determine if the CT systems are 1) casual or uncasual 2) memory or memoryless.

Definitions:

Casual: If for any time t1, the output response y(t1) at time t1 resulting from input x(t) does not depend on the values of the input x(t) for t > t1.
Memory: If the output at time t1 depends in general on the past values of the input x(t) for some range of values of t up to t=t1.

x(t) is random input and y(t) is the output of x(t)

For Eq1:

y(t) = |x(t)| = \left\{ \begin{array}{l}<br /> x(t)\; \mathrm{if}\, x(t) \geq 0 \\<br /> -x(t)\; \mathrm{if}\, x(t) &lt; 0<br /> \end{array}\right.<br />

I said this system is CASUAL and MEMORYLESS.

• Casual - because at time t, y(t) will depend only t from the input function x(t), not some other arbitrary t value.
• Memoryless - because the outputs at time t do not depend on previous inputs.

For Eq 2:


y(t) = \int_0^t\lambda x(\lambda)d\lambda

I said this system is CASUAL and has MEMORY.

• Casual - because at time t, it doesn't really depend on the future. It only depends on the past, so I'm guessing casual. *This I'm not too sure about*
• Memory - because the outputs at time t do depend on previous inputs since youre taking the integral from 0 to time t. *I'm almost sure about this one*

 

[learningphysics]

Both your answers look good to me. By the way, it's "causal" not "casual".

 

[DefaultName]

Both your answers look good to me. By the way, it's "causal" not "casual".

Hahha, I just realized that. Wow.

Thanks tho.

 

[DefaultName]

Now, Eq1 is obviously linear, but when I graph eq2, it seems to be nonlinear... does that make sense?

 

[chroot]

Yes.

- Warren

 

[user101]

Yes.

- Warren

is it because when you take derivatives and integrals, the terms will become nonlinear

also, in the one DefualtName posted, for equation 2, would that be a time varying or time invariant one. i would say varying because the actual output will be different from the input