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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)) | 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)) | ||
Therefore the signal is Linear. | Therefore the signal is Linear. | ||
| − | |||
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. | Therefore the signal is Time Invariant because the output will be shifted by the same amount that the input was shifted by. | ||
| − | |||
Assuming that x(t) is bounded then the output y(t) is also bound because it is the sum of two bound functions. | 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 | |
| − | 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. | This function is Linear, Time Invariant, and Stable. | ||
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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) | 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. | Therefore the function is linear. | ||
| − | |||
Let x2(t)=x1(t-T)-->S-->y(t)=cos(3t)x2(t)=cos(3t)x1(t-T) | Let x2(t)=x1(t-T)-->S-->y(t)=cos(3t)x2(t)=cos(3t)x1(t-T) | ||
This is not equal to y(t-T), therefore the function is not Time Invariant. | This is not equal to y(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. | 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. | The function is Memoryless, Causal, Linear, Stable. | ||
'''C)''' | '''C)''' | ||
| − | y(t)=<math> | + | y(t)=<math>\int_\infty^{2T} x(\tau)\,d\tau</math> |
| − | Let x3(t)=ax1(t) + bx2(t)-->S--><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> |
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> | + | |
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. | ||
| − | |||
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. | ||
Revision as of 15:45, 13 June 2008
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)) Therefore the signal is Linear. 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. 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-->y(t)=cos(3t)x2(t)=cos(3t)x1(t-T) This is not equal to y(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 $=$ \int_\infty^{2T} x1(\tau)\,d\tau $ + $ \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. 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.
