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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.
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 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.
Hence the answer is True.
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} $
