ase330m_fall07_ps6 - ASE 330M ‐ Homework # 6 Due November...

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Unformatted text preview: ASE 330M ‐ Homework # 6 Due November 16th, 2007. Problem #1 Consider a mass on a flat non‐smooth surface, subject to a force F ( t ) . The surface friction is directly proportional to the velocity of the particle, x , through a friction coefficient μ . Thus, the equation of motion for the particle can be expressed as, mx = − μ x + F ( t ) Let V ( t ) denote the velocity of the particle, V ( t ) = x ( t ) . Then, the equation of motion can also be written in first order form as V ( t ) + μkV ( t ) = u ( t ) Where μk = μ / m and u ( t ) = F ( t ) / m . The dynamical process can be represented as an input‐output system such that u ( t ) → V ( t ) + μkV ( t ) = u ( t ) → V ( t ) Suppose the system is initially at rest, V ( 0 ) = 0 . Find the velocity response of the system due to the following inputs using the convolution integral. Verify your results by modifying the numerical convolution script used in class. The script plots the input signal, the impulse response, the exact solution, and the solution approximated via numerical convolution. You will have to edit the functions get_input, get_impulse_response, and get_exact_solution to coincide reflect the changed input below. For each case, discuss which sections of the source code you had to change and why. • • u ( t ) = t ⋅ 1( t ) u ( t ) = e− t ⋅ 1( t ) Problem #2 Show that the convolution integral, f (t ) * g (t ) = ∞ ∫ f (τ ) g ( t − τ ) dτ , −∞ Satisfies the following properties: • Commutative: f ( t ) * g ( t ) = g ( t ) * f ( t ) • Distributive: f ( t ) * ⎡ g ( t ) + w ( t ) ⎤ = f ( t ) * g ( t ) + f ( t ) * w ( t ) ⎣ ⎦ • Convolution with an Impulse: f ( t ) * δ ( t ) = f ( t ) • Shift Property: if f ( t ) * g ( t ) = c ( t ) , then f ( t − T ) * g ( t ) = f ( t ) * g ( t − T ) = c ( t − T ) . ...
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This note was uploaded on 09/25/2011 for the course ASE 18510 taught by Professor Jeannefalcon during the Spring '10 term at University of Texas at Austin.

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