(New page: = Homework 3 Solutions = == Question 1 == a) <u>Invertibility</u> Let <math>x_1[n]=0</math> for all <math>n</math> be an input to the given system. Then, its response is <math>y_1[n]=0<...) |
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<u>Causality:</u> | <u>Causality:</u> | ||
| − | The output <math>y[n]</math> depends only on the current (<math>x[n]</math>) and past (<math>x[n-1]</math>) values of the input. | + | The output <math>y[n]</math> depends only on the current ( <math>x[n]</math> ) and past ( <math>x[n-1]</math> ) values of the input. |
Hence, the given system is '''causal'''. | Hence, the given system is '''causal'''. | ||
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Let <math>x_1[n]</math> be an input to the given system. Then its response is <math>y_1[n]=x_1[n]x_1[n-1]</math>. | Let <math>x_1[n]</math> be an input to the given system. Then its response is <math>y_1[n]=x_1[n]x_1[n-1]</math>. | ||
| − | Now, let <math>x_2[n]=ax_1[n]</math> be an input to the system, where <math>a</math> can be any number. Then its response is <math>y_2[n]= | + | Now, let <math>x_2[n]=ax_1[n]</math> be an input to the system, where <math>a</math> can be any number. Then its response is <math class="inline">y_2[n]=a^2x_1[n]x_1[n-1]\neq ay_1[n]</math>, then the system is '''not linear'''. |
<u>Time invariance</u> | <u>Time invariance</u> | ||
| − | Let <math>x_1[n]</math> be an input to the system. Then <math>y_1[n]= | + | Let <math>x_1[n]</math> be an input to the system. Then <math>y_1[n]=x_1[n]x_1[n-1]</math> is its response. |
Now, let <math>x_2[n]=x_1[n-n_0]</math> be an input to the system, where <math>n_0</math> can be any number. Then, <math>y_2[n]=x_1[n-n_0]x_1[n-1-n_0]=y_1[n-n_0]</math>. | Now, let <math>x_2[n]=x_1[n-n_0]</math> be an input to the system, where <math>n_0</math> can be any number. Then, <math>y_2[n]=x_1[n-n_0]x_1[n-1-n_0]=y_1[n-n_0]</math>. | ||
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Hence the system is '''time invariant'''. | Hence the system is '''time invariant'''. | ||
| − | b) | + | b) <u>Invertibility</u> |
| + | |||
| + | Let <math>x_1(t)=0</math> for all <math>t</math> be an input to the given system. Then, its response is <math>y_1(t)=0</math> for all <math>t</math>. | ||
| + | |||
| + | Let <math>x_2(t)=\delta (t-2)</math> be an input to the given system. Then, its response is <math>y_2(t)=0</math> for all <math>t</math> since <math class="inline">-1\leq\sin(t)\leq 1</math>. | ||
| + | |||
| + | Since <math class="inline">x_2(t)\neq x_1(t)</math> and <math>y_2(t)=y_1(t)</math>, then the system is '''not invertible'''. | ||
| + | |||
| + | <u>Memory:</u> | ||
| + | |||
| + | For <math>t=-\pi/2</math>, we have <math>y(-\pi/2)=x(-1)</math>. However, <math>-\pi/2\leq -1</math>. | ||
| + | |||
| + | Then the output <math>y(t)</math> depends on future values of the input <math>x(t)</math>. | ||
| + | |||
| + | Hence, we deduce that system has '''memory'''. | ||
| + | |||
| + | <u>Causality:</u> | ||
| + | |||
| + | Using the same example for the memory part, we can say that the system is '''non-causal'''. | ||
| + | |||
| + | <u>Stability</u> | ||
| + | |||
| + | Let <math>x(t)</math> be a bounded signal by some number B, i.e. <math>|x(t)|<B</math> for all <math>t</math>. | ||
| + | |||
| + | Then the response to <math>x(t)</math> is obviously always bounded as such: <math>|y(t)|<B</math> for all <math>t</math>. | ||
| + | |||
| + | Thus the given system is '''stable'''. | ||
| + | |||
| + | <u>Linearity</u> | ||
| + | |||
| + | Let <math>x_1(t)</math> be an input to the given system. Then its response is <math>y_1(t)=x_1(\sin(t))</math>. | ||
| + | |||
| + | Let <math>x_2(t)</math> be an input to the given system. Then its response is <math>y_2(t)=x_2(\sin(t))</math>. | ||
| + | |||
| + | Now, let <math>x_3(t)=ax_1(t)+bx_2(t)</math> be an input to the system, where <math>a</math> and <math>b</math> can be any number. Then its response is <math>y_2(t)=ax_1(\sin(t))+bx_2(\sin(t))=ay_1(t)+by_2(t)</math>. | ||
| + | |||
| + | Hence the system is '''linear'''. | ||
| + | |||
| + | <u>Time invariance</u> | ||
| + | |||
| + | Let <math>x_1(t)</math> be an input to the system. Then <math>y_1(t)=x_1(\sin(t))</math> is its response. | ||
| + | |||
| + | Now, let <math>x_2(t)=x_1(t-t_0)</math> be an input to the system, where <math>t_0</math> can be any number. Then, <math>y_2(t)=x_1(\sin(t-t_0))=y_1(t-t_0)</math>. | ||
| + | |||
| + | Hence the system is '''time invariant'''. | ||
Revision as of 09:32, 8 February 2011
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_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 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) 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\leq -1 $.
Then the output $ y(t) $ depends on future values of the input $ x(t) $.
Hence, we deduce that 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 number. Then its response is $ y_2(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.
