| Line 11: | Line 11: | ||
Let us check for y[n] = x[n]^2 | Let us check for y[n] = x[n]^2 | ||
| − | <math>y[x[n-n0]] = x{[n-n0]^2}</math> | + | *<math>y[x[n-n0]] = x{[n-n0]^2}</math> |
Also, | Also, | ||
| − | <math>y[n-n0] = x{[n-n0]^2}</math> | + | *<math>y[n-n0] = x{[n-n0]^2}</math> |
| + | Thus the above system is time invariant | ||
| + | |||
| + | |||
| + | == Time Variance check == | ||
| + | |||
| + | Let us test for | ||
| + | '''y[n]=cos[nQ]*x[n]''' | ||
Revision as of 10:06, 12 September 2008
Time invariance
A system is called time invariant if the cascade
x[n]----->Time delay ----> System -----> z[n] yields the same output as x[n]----->system----->Time Delay-----> y[n]
Time Invariance check
Let us check for y[n] = x[n]^2
- $ y[x[n-n0]] = x{[n-n0]^2} $
Also,
- $ y[n-n0] = x{[n-n0]^2} $
Thus the above system is time invariant
Time Variance check
Let us test for
y[n]=cos[nQ]*x[n]
