How Do LTI Systems Use Convolution to Compute Outputs?

[lauritachris]

How would you interpretate this image about LTI systems and convolution?

 

[Runei]

h[X0] is the response due to the single pulse x[0].
h[X1] is the response due to the single pulse x[1].
h[X2] is the response due to the single pulse x[2].

And so forth.

In a discrete LTI system, the total response due to a signal is simply a sum of the responses due to each impulse that the signal is made up of.

In the above example, x[n] can be said to be made up of 4 impulses - x[0], x[1], x[2], x[3]. When passed through the system with impulse response h[n], each of the pulses would, on its own, create the shown response (h[X0] for example). When several impulses (the whole signal) is passed through the system at once, the output is the sum of the individual responses.

All the reponses are similar, they are, however, displaced in time.

Did this help? Otherwise I can try to elaborate :)

 

[lauritachris]

Thank you, what are we multiplying? At Y[3] for example.
x[3]h[0]+x[2]h[1]+x[1]h[2] - And why do we need to multiply it this way?

h[X0] is the response due to the single pulse x[0].
h[X1] is the response due to the single pulse x[1].
h[X2] is the response due to the single pulse x[2].

And so forth.

In a discrete LTI system, the total response due to a signal is simply a sum of the responses due to each impulse that the signal is made up of.

In the above example, x[n] can be said to be made up of 4 impulses - x[0], x[1], x[2], x[3]. When passed through the system with impulse response h[n], each of the pulses would, on its own, create the shown response (h[X0] for example). When several impulses (the whole signal) is passed through the system at once, the output is the sum of the individual responses.

All the reponses are similar, they are, however, displaced in time.

Did this help? Otherwise I can try to elaborate :)

 

[rbj]

looks like a case of superposition. that's how i would interpret it.

i would suggest making at least one sample of x[n] larger than the others, and then showing that in the delayed impulse response and the final sum.

 

[Runei]

Thank you, what are we multiplying? At Y[3] for example.
x[3]h[0]+x[2]h[1]+x[1]h[2] - And why do we need to multiply it this way?

h[0] is the impulse response. So it is the response you will see, if you give the system an input of the impulse function. Also, if you scale the input signal, the output signal will also be a scaled by the same amount (this is also a property of LTI systems).

Thus, x[3]h[0] is the scaled response of the system, due to the impulse x[3]. If x[3] is just 1, then the output is simply the normal impulse response, if x[3] is for example 4, then the input can be said to be a scaled impulse (scaled by 4), thus, the output will also be scaled.

So the multiplying comes from the fact that we scale the responses according to the input impulse that generated them.

 

[lauritachris]

Thank you very much! But why is y3:
x[3] multiplied with h[0]+ x[2] multiplied with h[1]+x[1] multiplied with h[2]
and not:

X[1] mutiplied with h[2] + X[2] multiplied with h[1] + X[3] multiplied with h[0]

for example?

Do you think that X is reversed, instead of the impulse response?

 

[Runei]

You are asking why is

y[3] = X[3]h[0] + X[2]h[1] + X[1]h[2]

and not

y[3] = X[1]h[2] + X[2]h[1] + X[3]h[0]

The two equations are the same.