Signals & Systems - what's the output?

Thread starter Angello90
Start date Dec 6, 2010
Tags

Output Signals Systems

AI Thread Summary

The discussion revolves around determining the output of a system given the input x[n] and the impulse response h[n]. The input is defined as x[n] = r[n] - 2r[n - 1] + r[n - 2], and the impulse response is h[n] = δ[n - 2]. The proposed output is y[n] = u[n - 3], but there is confusion regarding the multiplication of the two signals, leading to the assertion that y[n] could be 0. Participants are seeking clarification and tips on the correct output calculation.

Dec 6, 2010

Angello90

Homework Statement

Input and system response is as follow:

x[n] = r[n] - 2r[n - 1] + r[n - 2]
h[n] = \delta[n - 2]

The output is:
y[n] = u[n - 3]

Is that true?

Homework Equations

The Attempt at a Solution

I draw the h[n] and x[n] and they looked as follows:

attachment.php?attachmentid=30426&stc=1&d=1291631179.jpg

As for me when these two are multiplied, it will become y[n]=0.
Am I right?

Thanks

Physics news on Phys.org

Dec 29, 2010

Angello90

Can anyone give me a tip maybe? Thanks!

Thread 'Find the distance given a function for v(t)'

My solution: v = 2*t + 5*t^2. dx/dt = 2*t + 5*t^2. Therefore dx = (2t + 5t^2)dt. After taking the anti derivative my equation is s = t^2/2 + 5/3(t^3).

View full post »

Thread 'PLL - How to find all the gains of a PI corrector and fix Ki ? MATLAB'

Hello everyone, it's not really for my homework (because I'm not at school anymore) but I'm new and I don't know how to start a new forum. I've never done automatic before and I need help. I have a 2nd ordre PLL on MATLAB, with an NCO and a PI corrector; the PLL works via the Mid/End algorithm (that's part is ok). I need to find the link between the theoretical formulas and their values in the code to find the value of the gain Ki, and possibly that of Kp if it's not correct. Unfortunately...

View full post »

Thread 'Have I solved this structural engineering equation correctly?'

Hi all, I have a structural engineering book from 1979. I am trying to follow it as best as I can. I have come to a formula that calculates the rotations in radians at the rigid joint that requires an iterative procedure. This equation comes in the form of: $$ x_i = \frac {Q_ih_i + Q_{i+1}h_{i+1}}{4K} + \frac {C}{K}x_{i-1} + \frac {C}{K}x_{i+1} $$ Where: ## Q ## is the horizontal storey shear ## h ## is the storey height ## K = (6G_i + C_i + C_{i+1}) ## ## G = \frac {I_g}{h} ## ## C...

View full post »