3015_Discussion_5_Solutions

3015_Discussion_5_Solutions - University of Minnesota Dept....

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Unformatted text preview: University of Minnesota Dept. of Electrical and Computer Engineering EE3015 Signals and Systems Discussion Session #5: CT convolution Consider the following RC circuit, with input voltage x ( t ) and output voltage y ( t ). ) ( ) ( ) ( t x t y dt t dy RC = + 1. Find and sketch the output voltage when the input is x ( t )= Vu ( t ). The circuit is assumed to be at rest for t <0. The homogeneous equation is ) ( ) ( = + t y dt t dy RC h h . The solution is of the form st h Ae t y = ) ( f o r t >0 Plugging it in the homogeneous equation, = + st st Ae RCAse . Dividing by st Ae we obtain RC s RCs 1 1 = = + So the homogeneous solution is RC t h Ae t y / ) ( = for t >0. Now, consider a particular solution of the form B t y p = ) ( f o r > t . Substituting in the differential equation, V B RCB = + . Hence, V B = . The complete solution is the sum of the homogeneous and particular solutions. V Ae t y t y t y RC t p h = + = / ) ( ) ( ) ( f o r t> 0 Using now the initial rest condition, V A V Ae y = = ) ( and the output is given by ( ) RC t p h e V t y t y t y / 1 ) ( ) ( ) ( = + = for t >0 Or, equivalently, ( ) ) ( 1 ) ( / t u e V t y RC t = . x ( t ) C R y ( t ) t V y ( t ) 2. Find and sketch the impulse response h ( t ) of the circuit. The circuit can be seen as an LTI system. As we know, we can find the impulse response of the circuit by differentiating the step response....
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3015_Discussion_5_Solutions - University of Minnesota Dept....

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