Homework 3 Solutions

Question 1

a) Invertibility

Let $ x_1[n]=0 $ for all $ n $ be an input to the given system. Then, its response is $ y_1[n]=0 $ for all $ n $.

Let $ x_2[n]=\delta [n] $ be an input to the given system. Then, its response is $ y_2[n]=0 $ for all $ n $.

Since $ x_2[n]\neq x_1[n] $ and $ y_2[n]=y_1[n] $, then the system is not invertible.

Memory:

The output $ y[n] $ depends on past values of $ x[n] $, since we have $ x[n-1] $ in the system equation.

Hence, we deduce that system has memory.

Causality:

The output $ y[n] $ depends only on the current ($ x[n] $) and past ($ x[n-1] $) values of the input.

Hence, the given system is causal.

Stability

Let $ x[n] $ be a bounded signal by some number B, i.e. $ |x[n]|<B $ for all $ n $.

Then the response to $ x[n] $ is always bounded as such: $ |y[n]|<B^2 $ for all $ n $.

Thus the given system is stable.

Linearity

Let $ x_1[n] $ be an input to the given system. Then its response is $ y_1[n]=x_1[n]x_1[n-1] $.

Now, let $ x_2[n]=ax_1[n] $ be an input to the system, where $ a $ can be any number. Then its response is $ y_2[n]=(a^2x_1[n]x_1[n-1])\neq ay_1[n] $, then the system is not linear.

Time invariance

Let $ x_1[n] $ be an input to the system. Then $ y_1[n]=x[n]x[n-1] $ is its response.

Now, let $ x_2[n]=x_1[n-n_0] $ be an input to the system, where $ n_0 $ can be any number. Then, $ y_2[n]=x_1[n-n_0]x_1[n-1-n_0]=y_1[n-n_0] $.

Hence the system is time invariant.

b)