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-varying.

Question 2

a) The response to a unit impulse is $ y[n]=\delta [n] \delta [n-1]=(0)\delta [n] = 0 $ for all $ n $.

Hence, the unit impulse response is $ h[n]=0 $ for all $ n $.

b) The response to a unit impulse is $ y(t)=\delta (\sin(t)) $.

Hence, the unit impulse response is $ h(t)=\delta (\sin(t)) $.

c) The response to a unit impulse is $ y[n]=\sum_{k=n-10}^{n+10} \delta [k] $.

Hence, the unit impulse response is $ h[n]=\sum_{k=n-10}^{n+10} \delta [k] $.

d) The response to a unit impulse is $ y(t)=(-1)^2 \delta (t+1)=\delta (t+1) $.

Hence, the unit impulse response is $ h(t)=\delta (t+1) $.

Question 3

$ \int_{-\infty}^\infty \delta (2t) dt= \int_{-\infty}^{\infty} \delta (\tau) \frac{d\tau}{2} = \frac{1}{2} \int_{-\infty}^{\infty} \delta (\tau) d\tau= \frac{1}{2} $,

where we have made the change of variable $ \tau=2t $.

Hence, $ \delta(2t)=\frac{1}{2} \delta(t) $.

Question 4

Assume we have two system $ S $ and $ T $.

Stimulate system $ S $ by input $ x_1(t) $, then the response is $ z_1(t)=S\left[x_1(t)\right] $.

Then stimulate system $ T $ by $ z_1(t) $, then the response of the cascade to $ x_1(t) $ is $ y_1(t)=T\left[z_1(t)\right]=T\left[S\left[x_1(t)\right]\right] $.

Now, let the input to the cascade be $ x_2(t)=x_1(t-t_0) $, where $ t_0 $ can be any number. Then the response to system $ S $ is

$ z_2(t)=S\left[x(t-t_0)\right]=z_1(t-t_0) $,

since system $ S $ is time-invariant.

The response of system $ T $ to $ z_2(t) $ is

$ y_2(t)=T\left[z_2(t)\right]=T\left[z_1(t-t_0)\right]=y_1(t-t_0) $,

since system $ T $ is time-invariant.

Hence the answer is True.

Question 5

$ \begin{align} y[n]&=h[n]* x[n] \\ &=(\delta [n+1] + 2\delta [n-1])* x[n] \quad \mbox{(using shifting property of the unit impulse)}\\ &=x[n+1]+2x[n-1] \\ &=u[n]-u[n-6]+2u[n-2]-2[n-8] \\ \end{align} $