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Find the Energy of the signal x(n) = 0.5*u(n) + 8nu(-n-1)
1 Answer
written 8.3 years ago by |
The total energy of a sequence x[n] is defined by,
Ex=∞∑n=−∞|x[n]|2
∴Ex=∞∑−∞|u(n)2+8nu(−n−1)|2
∴Ex=∞∑−∞|u2(n)4|+∞∑−∞|2∗u(n)∗8nu(−n−1)|+∞∑−∞|8nu(−n−1)|2
We know, u(n)=1;n≥0=0;n<0
∴, u(-n-1) =1;n<1=0;n≥0
∴Ex=14∞∑0|u2|+0+−1∑−∞|82n|
Substituting n=−m∴,
Ex=14+0+∞∑1|64−m|
∴,Ex=∴14+0+∞∑1|164m|
m=∞∑m=1am=a1−a
∴,Ex=14+0+(1⁄64)(1−1⁄64)
∴,Ex=14+0+163=67252
Result: Energy of the given signal x(n) is, Ex=67252