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| − | The instructor has added some comments. | + | --[[User:Asan|Asan]] 03:48, 14 June 2008 (EDT)The instructor has added some comments. |
Determine if the following are: | Determine if the following are: | ||
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Let x3=ax1(t) + bx2(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)) --[[User:Asan|Asan]] 03:42, 14 June 2008 (EDT)(Check the details) | 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 signal is Linear. --[[User:Asan|Asan]] 03:42, 14 June 2008 (EDT)(It is the system... not the signal | + | Therefore the signal 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) = x(t-T) | x(t-T)-->S-->y(t)=x(t-T-2) + x(2-t-T) = x(t-T) | ||
Therefore the signal is Time Invariant because the output will be shifted by the same amount that the input was shifted by. --[[User:Asan|Asan]] 03:42, 14 June 2008 (EDT)(Wrong. This system is not TI. Try again) | Therefore the signal is Time Invariant because the output will be shifted by the same amount that the input was shifted by. --[[User:Asan|Asan]] 03:42, 14 June 2008 (EDT)(Wrong. This system is not TI. Try again) | ||
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'''C)''' | '''C)''' | ||
y(t)=<math>\int_\infty^{2T} x(\tau)\,d\tau</math> | y(t)=<math>\int_\infty^{2T} x(\tau)\,d\tau</math> | ||
| − | Let x3(t)=ax1(t) + bx2(t)-->S--><math>\int_\infty^{2T} x3(\tau)\,d\tau</math>=<math>\int_\infty^{2T} x1(\tau)\,d\tau</math> + <math>\int_\infty^{2T} x2(\tau)\,d\tau</math> | + | 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> |
Therefore the function is linear. | 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 <math>-/infty</math> to 2t. | 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. | ||
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> | 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 time invariant because the output will be shifted by the same amount as the input. | + | The function is time invariant because the output will be shifted by the same amount as the input. --[[User:Asan|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. | 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. | 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''' | --[[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''' | ||
Revision as of 02:48, 14 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 signal 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) = x(t-T) Therefore the signal is Time Invariant because the output will be shifted by the same amount that the input was shifted by. --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 time invariant because the output will be shifted by the same amount as the input. --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
