(→Linear system) |
(→Non Linear System) |
||
| Line 15: | Line 15: | ||
</pre> | </pre> | ||
| − | == Non Linear | + | ==Non-Linear== |
| − | SYSTEM: y = | + | SYSTEM: y = (x(t))^5 |
| + | a.) 1X1(t) --> SYSTEM --> (x(t))^5 | ||
| + | b.) 4X2(t) --> SYSTEM --> (4x(t))^5 | ||
| + | |||
| + | To prove that this system is non-linear, we multiply output "a" by 9, then multiply output "b" by 6. Then, we take their sum. The calculations are as follows: 9x(t)^5 + 24x(t)^5 = 33x(t)^5. | ||
| + | |||
| + | Now we try multiplying the input in part "a" by 9 and the input of part "b" by 6. This gives us 9x(t) + 6x(t). Now we take their sum. This gives us 15x(t). Now, we run 15x(t) through the system to obtain: 759375(X(t))^5. Therefore, the system is not linear. | ||
| + | </pre> | ||
Revision as of 17:31, 10 September 2008
Linear system
SYSTEM: y = 3x(t) - 10 a.) 1X1(t) --> SYSTEM --> 3Y1(t) - 10 b.) 4X2(t) --> SYSTEM --> 12Y2(t) - 10 We can do the following proof to show that the above system is linear. Take two random constant numbers such as 9 and 6. Now multiply the output from "a" by 9. Then multiply the output from "b" by 6. Now take their sum. (27Y(t) - 90) + (72Y(t)-60)) = 99Y(t)-150 Now we will multiply the original signals by the constants, take their sum, and then send them through the system. If we end up with 99Y(t)-150, then the system must be linear. So, (9*1X) + (6*4X) = 33x. This gives 99Y(t) - 150. Therefore the system is linear.
Non-Linear
SYSTEM: y = (x(t))^5 a.) 1X1(t) --> SYSTEM --> (x(t))^5 b.) 4X2(t) --> SYSTEM --> (4x(t))^5
To prove that this system is non-linear, we multiply output "a" by 9, then multiply output "b" by 6. Then, we take their sum. The calculations are as follows: 9x(t)^5 + 24x(t)^5 = 33x(t)^5.
Now we try multiplying the input in part "a" by 9 and the input of part "b" by 6. This gives us 9x(t) + 6x(t). Now we take their sum. This gives us 15x(t). Now, we run 15x(t) through the system to obtain: 759375(X(t))^5. Therefore, the system is not linear. </pre>
