EE200_Weber_10-23

# EE200_Weber_10-23 - EE 200 Frequency Response Example Find...

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24 EE 200 Frequency Response Example: Find the frequency response of the circuit below with a resistor and capacitor. R C x(t) y(t) The relationship between the voltages x(t) and y(t) is given by the differential equation i = C dy dt x ( t ) = RC dy dt ( t ) + y ( t )

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25 EE 200 Frequency Response Assuming the input is a complex exponential, the output will also be a complex exponential scaled by the frequency response function. Putting these into the equation for x(t) gives Solving for H( ω ) gives e i " t = RC ( i ) H ( ) e i t + H ( ) e i t x ( t ) = e i t y ( t ) = H ( ) e i t H ( ) = 1 1 + iRC = 1 1 + ( RC ) 2 # i RC 1 + ( RC ) 2
26 EE 200 Frequency Response We can get a better picture of what this circuit is doing if we change the response function into polar coordinates. The magnitude and phase are given by As the frequency goes up, the magnitude of the response goes down so this is a low-pass filter. H ( " ) = re i # H ( ) = H ( ) H * ( ) = 1 + ( RC ) 2 = tan # 1 Im{ H ( \$ )} Re{ H ( )} % & ( ) * = tan # 1 ( # RC )

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27 EE 200 Frequency Response Example: A discrete-time system outputs a three point weighted average of the inputs according to For an input of e i ω n , this becomes y ( n ) = 1 2 x ( n ) + 1 3 x ( n " 1) + 1 6 x ( n " 2) y ( n ) = H ( " ) e i n = 1 2 e i n + 1 3 e i ( n # 1) + 1 6 e i ( n # 2) = e i n ( 1 2 + 1 3 e # i + 1 6 e # i 2 ) H ( ) = 1 2 + 1 3 e # i + 1 6 e # i 2 1/2 1/3 1/6 + y(t) x(t) D D
28 EE 200 Frequency Response Systems characterized by a high-order difference equation of the form with input x(n) = e i ω n will have an output y(n)= H( ω ) e i ω n a 0 y ( n ) + a 1 y ( n "

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## This note was uploaded on 12/03/2008 for the course EE 200 taught by Professor Zadeh during the Fall '08 term at USC.

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EE200_Weber_10-23 - EE 200 Frequency Response Example Find...

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