Problem_13_Solution_1

# Problem_13_Solution_1 - t R . (c) Clearly, S 1 ( x )( t ) =...

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Solution to Midterm Problem 13 Someone asked me whether I’d mind posting a solution to this problem, so here it is. (a) You might just recognize this system as an averager, but if you don’t, you can still ﬁnd the impulse response by using h 1 = S 1 ( δ ). Accordingly, h 1 ( t ) = 1 2 Z t t - 2 δ ( τ ) = ± 1 2 if t - 2 < 0 < t 0 if t > 2 or t < 0 = ± 1 2 if 0 < t < 2 0 if t > 2 or t < 0 We really don’t care what happens right at t = 0 or t = 1 / 2, and I accepted any answer for those, such as 1 / 2 at each, or 0 at each, etc. (b) The value S 1 ( x )( t ) of the output of the system at any time t is the average of x ( τ ) over the interval t - 2 τ t . The square wave has period 2, as you can see when you graph it, and it has average value 3 / 2, so the system just puts out the constant signal with value 3 / 2 for all t . I.e. S 1 ( x )( t ) = 3 / 2 for all
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Unformatted text preview: t R . (c) Clearly, S 1 ( x )( t ) = 7 for all t (you can just plug in 7 as the input or you can remember that the system is an averager, so it puts out a constant output in response to a constant input signal). Now, S 2 ( S 1 ( x )) = h 2 * S 1 ( x ), so S 2 ( S 1 ( x ))( t ) = Z - h 2 ( ) S 1 ( x )( t- ) d = Z 7 e-3 d = 7 3 for all t R . (d) S 2 ( x ) = h 2 * x , so S 2 ( x )( t ) = Z - h 2 ( ) x ( t- ) d = Z e-3 e-( t- ) u ( t- ) d = e-t R t e-2 d if t if t &lt; = 1 2 ( e-t-e-3 t ) if t if t &lt; = 1 2 ( e-t-e-3 t ) u ( t ) for all t R . 1...
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