signal and system

Thread Starter

micro1

Joined Feb 22, 2015
79
hello every one

signal and system

properties of system Linearity

I did my problem

you can check my solution it is true?
 

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WBahn

Joined Mar 31, 2012
32,823
I'm assuming that you mean

y[n] = x²[n-1]

If so, then this says that the present value of y is equal to the square of the previous value of x.

This is NOT the same as

y[n] = x²[n] - x²

This says that the current value of y is equal to the square of the current value of x minus the square of something that is not even at a defined time.
 

Thread Starter

micro1

Joined Feb 22, 2015
79
if change the equation to

X1 = X1 [ Xn - X ]

X2 = X2 [ Xn -X ]

X3 = X1 +X2 ==> ( X1 [ Xn - X ] + X2 [ Xn -X ] )

X3 = X3 [ Xn -X ]


X3 = X1 +X2 ( linearity)


thank you for reply
 

WBahn

Joined Mar 31, 2012
32,823
if change the equation to

X1 = X1 [ Xn - X ]

X2 = X2 [ Xn -X ]

X3 = X1 +X2 ==> ( X1 [ Xn - X ] + X2 [ Xn -X ] )

X3 = X3 [ Xn -X ]


X3 ≠ X1 +X2 (no linearity)


thank you for reply
So, if you had the continuous time signal

y(t) = x²(t-1sec)

are you saying that you would rewrite this as

y(t) = x(x·t - x·1sec)

Does that make ANY sense whatsoever?
 
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