help in qustion - the following system is:1-linear, 2-memory

[bleach118]

help in qustion

please can you help me answering this question
find out whether the following system is:
1-linear, 2-memoryless, 3-causel, 4-time invariant

y[n] = x[n] - x[n+1]

 

[AndyECE]

help in qustion

1: Yes - standard proof applies
2:No - output at time n relies on values of x other than at time n
3:No - Relies on future values of n
4:Yes - standard proof applies

 

[bleach118]

Re: help in qustion

1: Yes - standard proof applies
2:No - output at time n relies on values of x other than at time n
3:No - Relies on future values of n
4:Yes - standard proof applies

can you help me please by the way you improve the first and the fourth one so i can understand, i solved them already but i want to make sure

thanks

 

[AndyECE]

help in qustion

LINEAR:
For linear, there are two question:

1. If I multiply the output by a number, is that the same as multiplying the input by that number then feeding it through the system?

2. If I have two signals, and I feed them through one at a time then add the output later, is that the same as if I feed both signals through the system at the the same time?

We can check both simultaneously by the following:

Let there be two input signals: x1 and x2, each with their own output:

y1[n] = x1[n] - x1[n+1]
y2[n] = x2[n] - x2[n+1]

Now let there be a third signal that is a linear combination of the two input signals:
x3[n] = a*x1[n]+b*x2[n], where a and b are numbers

Then the output y3[n] = x3[n]-x[n+1] = ...
= (a*x1[n] + b*x2[n]) - (a*x1[n+1] + b*x2[n+1])
= a*x1[n] + b*x2[n] - a*x1[n+1] - b*x2[n+1]

Now then, suppose I set y3[n] aside for just a bit and look at y1[n] and y2[n]. I could make the same linear combination of these (call it y3'[n] as in "y three prime of n":
y3'[n] = a*y1[n]+b*y2[n]

which substituting from above is:
y3'[n] = a*(x1[n] - x1[n+1]) + b*(x2[n] - x2[n+1])
= a*x1[n] - a*x1[n+1]) + b*x2[n] - b*x2[n+1]

Note now that y3[n] = y3'[n]. In otherwords, scaling or summing (or both) of the different inputs is equivalent to scaling or summing (or both) the outputs. Therefore, it is linear.

TIME INVARIANT:

For time invariant, the question is: If I feed a given input x1[n] at time 0, I get an output y1[n]. If I feed a signal after time t, is the output of that signal the same as if I just shifted the output of the first signal by t?

so, let y1[n] = x1[n] - x1[n+1].

Now I feed x1[n+t] thorugh the system, and get
y2[n+t] = x1[n+t] - x1[n+t+1]

Is y2 (which has a time shifted input) the same as if I just shifted y1 by t? To tell, I just substitute in for n with n+t and get
y1[n+t] = x1[n+t] - x1[n+t+1]

since y1[n+t] = y2[n+t] it is time invariant.

---------------
This is a little trivial, and so maybe if I work another time invariance problem I can illustrate a little better:

LTI system: y[n] = n*x[n] + x[n+1]

So the first case is to time shift the output (i.e. take each 'n' and plug in 'n+t')
=> y1 = (n+t)*x[n+t] + x[n+t+1]

Now, we time shift the input and get:
=> y2 = n*x[n+t] + x[n+t+1]

y1 != y2 (NOT EQUAL)

Therefore NOT TIME INVARIANT

Hope this helps, and don't be afraid to click the helped me button!