(→Linearity and Time Invariance) |
|||
| (11 intermediate revisions by the same user not shown) | |||
| Line 1: | Line 1: | ||
=Linearity and Time Invariance= | =Linearity and Time Invariance= | ||
Given system: | Given system: | ||
| + | |||
Input Output | Input Output | ||
| + | |||
X0[n]=δ[n] -> Y0[n]=δ[n-1] | X0[n]=δ[n] -> Y0[n]=δ[n-1] | ||
| Line 10: | Line 12: | ||
X3[n]=δ[n-3] -> Y3[n]=16 δ[n-4] | X3[n]=δ[n-3] -> Y3[n]=16 δ[n-4] | ||
| − | ... | + | ... -> ... |
Xk[n]=δ[n-k] -> Yk[n]=(k+1)2 δ[n-(k+1)] -> For any non-negative integer k | Xk[n]=δ[n-k] -> Yk[n]=(k+1)2 δ[n-(k+1)] -> For any non-negative integer k | ||
| + | |||
| + | |||
| + | =Time Invariant System?= | ||
| + | Suppose the system is defined as the third line where input is <math>X_2[n]= dirac[n-2]</math> and output: <math>Y_2[n]=9 dirac[n-3]</math> with a time delay of 2 seconds. | ||
| + | |||
| + | |||
| + | Using the system then time delay method: | ||
| + | |||
| + | dirac[n-2] -> SYSTEM -> 9 dirac[n-3] -> Time Delay -> 9 dirac[n-5] | ||
| + | |||
| + | |||
| + | Using the time delay then system method: | ||
| + | |||
| + | dirac[n-2] -> Time Delay -> dirac[n-4] -> SYSTEM -> 25 dirac[n-6] | ||
| + | |||
| + | Because the two outputs are not the same, the system is not time-invariant. | ||
| + | |||
| + | |||
| + | =Necessary Input to Yield u[n-1]= | ||
| + | In order to obtain <math>u[n-1]</math> as the output instead of a delta function, the desired input would simply be <math>X_0 = u[n]</math>. | ||
Latest revision as of 13:56, 12 September 2008
Linearity and Time Invariance
Given system:
Input Output
X0[n]=δ[n] -> Y0[n]=δ[n-1]
X1[n]=δ[n-1] -> Y1[n]=4δ[n-2]
X2[n]=δ[n-2] -> Y2[n]=9 δ[n-3]
X3[n]=δ[n-3] -> Y3[n]=16 δ[n-4]
... -> ...
Xk[n]=δ[n-k] -> Yk[n]=(k+1)2 δ[n-(k+1)] -> For any non-negative integer k
Time Invariant System?
Suppose the system is defined as the third line where input is $ X_2[n]= dirac[n-2] $ and output: $ Y_2[n]=9 dirac[n-3] $ with a time delay of 2 seconds.
Using the system then time delay method:
dirac[n-2] -> SYSTEM -> 9 dirac[n-3] -> Time Delay -> 9 dirac[n-5]
Using the time delay then system method:
dirac[n-2] -> Time Delay -> dirac[n-4] -> SYSTEM -> 25 dirac[n-6]
Because the two outputs are not the same, the system is not time-invariant.
Necessary Input to Yield u[n-1]
In order to obtain $ u[n-1] $ as the output instead of a delta function, the desired input would simply be $ X_0 = u[n] $.
