(New page: Determine if the following are: #Memoryless #Time Invariant #Linear #Causal #Stable '''A)''' #y(t)=x(t-2) + x(2-t) ##Let x3=ax1(t) + bx2(t) ##then the output of the function is y(t)=x3(t-...) |
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| + | --[[User:Asan|Asan]] 03:48, 14 June 2008 (EDT)The instructor has added some comments. | ||
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Determine if the following are: | Determine if the following are: | ||
#Memoryless | #Memoryless | ||
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'''A)''' | '''A)''' | ||
| − | + | y(t)=x(t-2) + x(2-t) | |
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| − | + | Let x3=ax1(t) + bx2(t) | |
| − | + | then the output of the function is y(t)=x3(t-2) + x(2-t) = (ax1(t-2) + bx2(t-s)) + (ax1(2-t) + bx2(2-t)) --[[User:Asan|Asan]] 03:42, 14 June 2008 (EDT)(Check the details) | |
| + | Therefore the system is Linear. --[[User:Asan|Asan]] 03:42, 14 June 2008 (EDT)(It is the system... not the signal) | ||
| − | + | x(t-T)-->S-->y(t)=x(t-T-2) + x(2-t-T) <math>\neq</math> x(t-T) | |
| + | Since t-T-2 <math>\neq</math> 2-t-T the system is not TI. --[[User:Asan|Asan]] 03:42, 14 June 2008 (EDT)(Wrong. This system is not TI. Try again) | ||
| − | + | Assuming that x(t) is bounded then the output y(t) is also bound because it is the sum of two bound functions. | |
| + | Since there is a time shift in both the positive and negative direction, the function is neither memoryless or causal | ||
| − | + | This function is Linear, Time Invariant, and Stable. | |
'''B)''' | '''B)''' | ||
| − | + | y(t)=[cos(3t)]x(t) | |
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| − | + | Since there is no time shift in the output function it is both memoryless and causal. | |
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| − | + | Let x3=ax1(t) + bx2(t)-->S-->y(t)=cos(3t)x3(t)=cos(3t)[ax1(t) + bx2(t)]=ay(t) + by(t) | |
| − | + | Therefore the function is linear. | |
| − | + | Let x2(t)=x1(t-T)-->S-->y2(t)=cos(3t)x2(t)=cos(3t)x1(t-T) | |
| + | This is not equal to y1(t-T), therefore the function is not Time Invariant. | ||
| − | + | Assuming that x(t) is bound, the function y(t) is also bound since it is the multiple of two bound functions. | |
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| + | The function is Memoryless, Causal, Linear, Stable. | ||
'''C)''' | '''C)''' | ||
| − | + | y(t)=<math>\int_-\infty^{2T} x(\tau)\,d\tau</math> | |
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| − | + | Let x3(t)=ax1(t) + bx2(t)-->S--><math>\int_-\infty^{2T} x3(\tau)\,d\tau</math>=<math>a\int_-\infty^{2T} x1(\tau)\,d\tau</math> + <math>b\int_-\infty^{2T} x2(\tau)\,d\tau</math> | |
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| + | Therefore the function is linear. | ||
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| + | The function is not memoryless or causal since it takes into account past and future time with the integration going from <math>-/infty</math> to 2t. | ||
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| + | Let x2(t)=x1(t-T)-->S--><math>\int_\infty^{2T} x2(\tau)\,d\tau</math>--><math>\int_\infty^{2T} x1(\tau)\,d\tau</math> | ||
| + | The function is not time invariant because the integral will evaluate from negative infinity to twice the current time. This will cause the output to be shifted by more than the inputs time shift. --[[User:Asan|Asan]] 03:48, 14 June 2008 (EDT)(Wrong. Try again) | ||
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| + | The function is not stable since the integrand has no lower limit, therefore the sum can grow infinitely large without bound. | ||
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| + | This function is Linear and Time Invariant. | ||
| + | --[[User:Asan|Asan]] 03:42, 14 June 2008 (EDT)--[[User:Asan|Asan]] 03:42, 14 June 2008 (EDT)--[[User:Asan|Asan]] 03:42, 14 June 2008 (EDT)--[[User:Asan|Asan]] 03:42, 14 June 2008 (EDT)'''Bold text''' | ||
Latest revision as of 16:51, 16 June 2008
--Asan 03:48, 14 June 2008 (EDT)The instructor has added some comments.
Determine if the following are:
- Memoryless
- Time Invariant
- Linear
- Causal
- Stable
A) y(t)=x(t-2) + x(2-t)
Let x3=ax1(t) + bx2(t) then the output of the function is y(t)=x3(t-2) + x(2-t) = (ax1(t-2) + bx2(t-s)) + (ax1(2-t) + bx2(2-t)) --Asan 03:42, 14 June 2008 (EDT)(Check the details) Therefore the system is Linear. --Asan 03:42, 14 June 2008 (EDT)(It is the system... not the signal)
x(t-T)-->S-->y(t)=x(t-T-2) + x(2-t-T) $ \neq $ x(t-T) Since t-T-2 $ \neq $ 2-t-T the system is not TI. --Asan 03:42, 14 June 2008 (EDT)(Wrong. This system is not TI. Try again)
Assuming that x(t) is bounded then the output y(t) is also bound because it is the sum of two bound functions. Since there is a time shift in both the positive and negative direction, the function is neither memoryless or causal
This function is Linear, Time Invariant, and Stable.
B) y(t)=[cos(3t)]x(t)
Since there is no time shift in the output function it is both memoryless and causal.
Let x3=ax1(t) + bx2(t)-->S-->y(t)=cos(3t)x3(t)=cos(3t)[ax1(t) + bx2(t)]=ay(t) + by(t) Therefore the function is linear.
Let x2(t)=x1(t-T)-->S-->y2(t)=cos(3t)x2(t)=cos(3t)x1(t-T) This is not equal to y1(t-T), therefore the function is not Time Invariant.
Assuming that x(t) is bound, the function y(t) is also bound since it is the multiple of two bound functions.
The function is Memoryless, Causal, Linear, Stable.
C) y(t)=$ \int_-\infty^{2T} x(\tau)\,d\tau $
Let x3(t)=ax1(t) + bx2(t)-->S-->$ \int_-\infty^{2T} x3(\tau)\,d\tau $=$ a\int_-\infty^{2T} x1(\tau)\,d\tau $ + $ b\int_-\infty^{2T} x2(\tau)\,d\tau $
Therefore the function is linear.
The function is not memoryless or causal since it takes into account past and future time with the integration going from $ -/infty $ to 2t.
Let x2(t)=x1(t-T)-->S-->$ \int_\infty^{2T} x2(\tau)\,d\tau $-->$ \int_\infty^{2T} x1(\tau)\,d\tau $ The function is not time invariant because the integral will evaluate from negative infinity to twice the current time. This will cause the output to be shifted by more than the inputs time shift. --Asan 03:48, 14 June 2008 (EDT)(Wrong. Try again)
The function is not stable since the integrand has no lower limit, therefore the sum can grow infinitely large without bound.
This function is Linear and Time Invariant. --Asan 03:42, 14 June 2008 (EDT)--Asan 03:42, 14 June 2008 (EDT)--Asan 03:42, 14 June 2008 (EDT)--Asan 03:42, 14 June 2008 (EDT)Bold text
