# Homework 3 Solutions, ECE301 Spring 2011 Prof. Boutin

## 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.

Oops! This is incorrect. See the correction below. pm

The following cascades yield two different outputs, so the system is not time invariant.

$x(t) \to \Bigg[ time\ delay\ n_0 \Bigg] \to y(t)=x(t-t_0) \to \Bigg[ system \Bigg] \to z(t)=y(\sin(t))= y(\sin(t)-t_0)$

$x(t)\to \Bigg[ system \Bigg] \to y(t)=x(\sin(t)) \to \Bigg[ time\ delay\ n_0 \Bigg] \to z(t)=y(t-t_0)x(\sin(t-t_0))$

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: $y[n]=x[n]-y[n-10]-y[n-9]-\dots-y[n-1]-y[n+1]-y[n+2]-\dots-y[n+10]$.

Actually, it is not that simple. For example, consider the system y[n]=x[n]-x[n-1]. This is obviously not an invertible system, because several signals give the same output. (For example, any constant signal yields a zero signal.) However, one could use the same arguments as above to isolate x[n]=y[n]+x[n-1], and then switch x and y to get supposedly the inverse of this system: y[n]=x[n]+y[n-1]. So there must be something wrong with the argument.
The correct answer to this problem is "no, the system is not invertible." To prove this, you must find two different inputs that yield the same output. See below. -pm
Here are a few examples of pairs of input for which the system's response is the same.
1. $x_1[n]=\sum_{k=-\infty}^\infty \delta[n-21k]$ and $x_2[n]=\sum_{k=-\infty}^\infty \delta[n+1-21k]$ both yield y[n]=1.
2. $x_1[n]=\sum_{k=-\infty}^\infty \delta[n-7k]$ and $x_2[n]=\sum_{k=-\infty}^\infty \delta[n+1-7k]$ both yield y[n]=3.
3. $x_1[n]=\sum_{k=-\infty}^\infty \delta[n-3k]$ and $x_2[n]=\sum_{k=-\infty}^\infty \delta[n+1-3k]$ both yield y[n]=7.

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 time-invariant systems, $S$ 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.

## Question 5

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

## Question 6

\begin{align} y[n]&=\sum_{k=-\infty}^{\infty} \left(\frac{1}{3}\right)^{-k} u[-k-1]u[n-k-3] \\ &=\sum_{k=-\infty}^{-1} \left(\frac{1}{3}\right)^{-k} u[n-k-3] \\ &=\left\{\begin{array}{ll} \sum_{k=-\infty}^{n-3} \left(\frac{1}{3}\right)^{-k}& \mbox{ for } n<2\\ \sum_{k=-\infty}^{-1} \left(\frac{1}{3}\right)^{-k}& \mbox{ for } n\geq 2\\ \end{array}\right. \\ &=\left\{\begin{array}{ll} \sum_{k=3-n}^{\infty} \left(\frac{1}{3}\right)^{k}& \mbox{ for } n<2\\ \sum_{k=1}^{\infty} \left(\frac{1}{3}\right)^{k}& \mbox{ for } n\geq 2\\ \end{array}\right. \\ &=\left\{\begin{array}{ll} \frac{(\frac{1}{3})^{3-n}}{1-\frac{1}{3}}& \mbox{ for } n<2\\ \frac{\frac{1}{3}}{1-\frac{1}{3}}& \mbox{ for } n\geq 2\\ \end{array}\right. \\ &=\left\{\begin{array}{ll} \frac{3^n}{18}& \mbox{ for } n<2\\ \frac{1}{2}& \mbox{ for } n\geq 2\\ \end{array}\right. \\ \end{align}

## Question 7

\begin{align} y(t)&=\int_{-\infty}^{\infty} [u(\tau+5)-u(\tau-7)]u(\tau-t) d\tau \\ &=\int_{-5}^{7} u(\tau-t) d\tau \\ &=\left\{\begin{array}{ll} \int_{-5}^7 d\tau& \mbox{ for } t<-5 \\ \int_t^7 d\tau& \mbox{ for} -5<t<7 \\ 0& \mbox{ for } t>7 \end{array}\right. \\ &=\left\{\begin{array}{ll} 13& \mbox{ for } t<-5 \\ 7-t& \mbox{ for} -5<t<7 \\ 0& \mbox{ for } t>7 \end{array}\right. \end{align}

## Question 8

\begin{align} y(t)&=\int_{-\infty}^{\infty} e^\tau u(-\tau+5)u(\tau-t-8) d\tau \\ &=\int_{-\infty}^{5} e^\tau u(\tau-t-8) d\tau \\ &=\left\{\begin{array}{ll} \int_{t+8}^5 e^\tau d\tau& \mbox{ for } t<-3 \\ 0& \mbox{ for } t>-3 \end{array}\right. \\ &=\left\{\begin{array}{ll} e^5-e^{t+8}& \mbox{ for } t<-3 \\ 0& \mbox{ for } t>-3 \end{array}\right. \\ &=(e^5-e^{t+8})u(-t-3) \end{align}

## Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood