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 this 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_1[n]x_1[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 integer. 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) Invertibility
Let $ x_1(t)=0 $ for all $ t $ be an input to the given system. Then, its response is $ y_1(t)=0 $ for all $ t $.
Let $ x_2(t)=\delta (t-2) $ be an input to the given system. Then, its response is $ y_2(t)=0 $ for all $ t $ since $ -1\leq\sin(t)\leq 1 $.
Since $ x_2(t)\neq x_1(t) $ and $ y_2(t)=y_1(t) $, then the system is not invertible.
Memory:
For $ t=-\pi/2 $, we have $ y(-\pi/2)=x(-1) $. However, $ -\pi/2< -1 $.
Then the output $ y(t) $ depends on future values of the input $ x(t) $.
Hence, we deduce that this system has memory.
Causality:
Using the same example for the memory part, we can say that the system is non-causal.
Stability
Let $ x(t) $ be a bounded signal by some number B, i.e. $ |x(t)|<B $ for all $ t $.
Then the response to $ x(t) $ is obviously always bounded as such: $ |y(t)|<B $ for all $ t $.
Thus the given system is stable.
Linearity
Let $ x_1(t) $ be an input to the given system. Then its response is $ y_1(t)=x_1(\sin(t)) $.
Let $ x_2(t) $ be an input to the given system. Then its response is $ y_2(t)=x_2(\sin(t)) $.
Now, let $ x_3(t)=ax_1(t)+bx_2(t) $ be an input to the system, where $ a $ and $ b $ can be any numbers. Then its response is $ y_3(t)=ax_1(\sin(t))+bx_2(\sin(t))=ay_1(t)+by_2(t) $.
Hence the system is linear.
Time invariance
Let $ x_1(t) $ be an input to the system. Then $ y_1(t)=x_1(\sin(t)) $ is its response.
Now, let $ x_2(t)=x_1(t-t_0) $ be an input to the system, where $ t_0 $ can be any number. Then, $ y_2(t)=x_1(\sin(t-t_0))=y_1(t-t_0) $.
Hence the system is time invariant.
c) Invertibility
The system equation can be written as
$ y[n]=x[n-10]+x[n-9]+\dots+x[n]+x[n+1]+\dots+x[n+10] $.
Hence, the input $ x[n] $ can be written in terms of the output as such:
$ x[n]=y[n]-x[n-10]-x[n-9]-\dots-x[n-1]-x[n+1]-x[n+2]-\dots-x[n+10] $.
Hence, the system is invertible and the inverse system has the following equation: $ y[n]=x[n]-y[n-10]-y[n-9]-\dots-y[n-1]-y[n+1]-y[n+2]-\dots-y[n+10] $.
Memory:
The output $ y[n] $ depends on past and future values of $ x[n] $, since we have $ x[n-1] $ and $ x[n+1] $, for example, in the system equation.
Hence, we deduce that this system has memory.
Causality:
Since the output depends on $ x[n+1] $, for example, we deduce that the system depends on future values of the input and hence the system is non-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]|<21B $ 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]=\sum_{k=n-10}^{n+10} x_1[k] $. Let $ x_2[n] $ be an input to the given system. Then its response is $ y_2[n]=\sum_{k=n-10}^{n+10} x_2[k] $.
Now, let $ x_3[n]=ax_1[n]+bx_2[n] $ be an input to the system, where $ a $ and $ b $ can be any numbers. Then its response is $ y_3[n]=\sum_{k=n-10}^{n+10}(ax_1[k] + bx_2[k])= a\sum_{k=n-10}^{n+10} x_1[k] + b\sum_{k=n-10}^{n+10} x_2[k] = ay_1[n]+by_2[n] $.
Hence the system is linear.
Time invariance
Let $ x_1[n] $ be an input to the system. Then $ y_1[n]=\sum_{k=n-10}^{n+10} x_1[k] $ 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 integer. Then, $ y_2[n]=\sum_{k=n-10}^{n+10} x_1[k-n_0]=\sum_{k=n-n_0-10}^{n-n_0+10} x_1[k] = y_1[n-n_0] $.
Hence the system is time invariant.
d) Invertibility
Let $ x_1(t)=0 $ for all $ t $ be an input to the given system. Then, its response is $ y_1(t)=0 $ for all $ t $.
Let $ x_2(t)=\delta (t-1) $ be an input to the given system. Then, its response is $ y_2(t)=t^2\delta(t)=0 $ for all $ t $.
Since $ x_2(t)\neq x_1(t) $ and $ y_2(t)=y_1(t) $, then the system is not invertible.
Memory:
The output $ y(t) $ depends on future values of the input $ x(t) $ since we have $ x(t+1) $ in the system equation.
Hence, we deduce that this system has memory.
Causality:
Using the same reasoning for the memory part, we can say that the system is non-causal.
Stability
Let $ x(t)=1 $ for all $ t $ be an input to the given system.
Then the response to $ x(t) $ is not bounded since $ y(\infty)=(\infty)^2.1=\infty $.
Thus the given system is not stable.
Linearity
Let $ x_1(t) $ be an input to the given system. Then its response is $ y_1(t)=t^2x_1(t+1) $.
Let $ x_2(t) $ be an input to the given system. Then its response is $ y_2(t)=t^2x_2(t+1) $.
Now, let $ x_3(t)=ax_1(t)+bx_2(t) $ be an input to the system, where $ a $ and $ b $ can be any numbers. Then its response is $ y_3(t)=t^2(ax_1(t+1)+bx_2(t+1))=ay_1(t)+by_2(t) $.
Hence the system is linear.
Time invariance
Let $ x_1(t) $ be an input to the system. Then $ y_1(t)=t^2x_1(t+1) $ is its response.
Now, let $ x_2(t)=x_1(t-t_0) $ be an input to the system, where $ t_0 $ can be any number. Then, $ y_2(t)=t^2x_1(t+1-t_0)\neq y_1(t-t_0) $.
Hence the system is time invariant.
