midterm-sol

midterm-sol - MAE143 A - Signals and Systems - Winter 11...

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MAE143 A - Signals and Systems - Winter 11 Midterm, February 2nd Instructions (i) This exam is open book. You may use whatever written materials you choose, including your class notes and textbook. You may use a hand calculator with no communication capabilities (ii) You have 50 minutes (iii) Do not forget to write your name, student number, and instructor 1. Signals Consider the following mathematical description of a continuous-time signal x ( t ) = u ( t - 1) - (1 - e - ( t - 2) ) u ( t - 2) - δ ( t + 1) . Sketch the plot of the following derived signals: (a) (2 points) x ( t ) (b) (2 points) x (2 - t ) (c) (2 points) x ( t/ 2) Solution: NOTE: In this question we are not being picky about the value of u ( t ) at t = 0 ! (a) Start with the third summand, which is a negative impulse at time t = - 1 . This is the only nonzero value taken by the function for t < 1 . At t = 1 , the first summand, a unit step, which start adding 1 to the value of the function. Finally, the second summand is zero for t < 2 , so x ( t ) 1 for 1 < t < 2 . For t 2 , this summand is negative, with value 0 at t = 2 , and then smoothly approaching - 1 as t grows. A sketch of the plot is (+ 2 points) -2 0 2 4 6 8 -1 -0.5 0 0.5 1 x(t) -1
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(b) The function x ( - t ) has the same plot as the one above except that it is mirrored with respect to the vertical axis at 0 . For example, x ( - t ) at t = 1 is the impulse. The final function x (2 - t ) is then shifted in time by two seconds. For example, x (2 - t ) at t = 3 is the impulse. A sketch of the plot is (+ 2 points) -2 0 2 4 6 8 -1 -0.5 0 0.5 1 x(2-t) 3 (c) The function x ( t/ 2) has its time streched by a factor of 2 . For example, the impulse happens at
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This note was uploaded on 01/15/2012 for the course MAE 143A taught by Professor Miroslavkrstic during the Summer '08 term at UCSD.

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midterm-sol - MAE143 A - Signals and Systems - Winter 11...

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