Vextendsingle 1 j substituting s 0 into i and using

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vextendsingle = 1 ( j ) Substituting s = 0 into ( i ) and using the specification from ( j ) then gives H ( s = 0) = G 150 ( 150 2 ) = 1 G = 150 ( 150 2 ) = 3 . 375(10) 6 ( k ) Substituting ( k ) into ( i ) then gives the desired form for the transfer function: H ( s ) = 3 . 375(10) 6 parenleftBig s + 150 parenrightBig parenleftBig s 2 + 150 s + 2 . 25(10) 4 parenrightBig ( l ) (b) Compute the numerical value for vextendsingle vextendsingle vextendsingle A ( ω = 130) vextendsingle vextendsingle vextendsingle 2 . Use the form for | H ( ω ) | 2 given by equation (6) from Notes Set 10, repeated here for easy reference: | H ( ω ) | 2 = 1 1 + parenleftbigg ω ω B parenrightbigg 2 N ( m ) With N = 3 and ω B = 150, equation ( o ) becomes | H ( ω ) | 2 = 1 1 + parenleftbigg ω 150 parenrightbigg 6 ( n ) Substituting ω = 130 into ( n ) then gives | H ( ω = 130) | 2 = 1 1 + parenleftbigg 130 150 parenrightbigg 6 = 0 . 702 ( o )
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ECE 301-001 Spring 2018 Test 3 w/Answers 3 2. Let a filter have impulse response h ( t ) and a signal g ( t ) have spectrum G ( f ), where G ( f ) = braceleftBigg 4 - 10 - 5 vextendsingle vextendsingle vextendsingle f vextendsingle vextendsingle vextendsingle , | f | ≤ 10 5 0 , else , h ( t ) = 40 sinc parenleftBig 2(10) 5 t parenrightBig cos parenleftBig 8 π (10) 5 t parenrightBig (a) Plot G ( f ) over | f | ≤ 1 MHz. (b) Plot X ( f ) over | f | ≤ 1 MHz if x ( t ) = g ( t ) cos parenleftBig 6 π 10 5 t parenrightBig . (c) Plot S ( f ) over | f | ≤ 1 MHz if s ( t ) is the convolution s ( t ) = h ( t ) ⋆ x ( t ). DETAILED SOLUTION: Parts (a) and (b): Plot G ( f ) and X ( f ) over | f | ≤ 1 MHz. A plot of G ( f ) is shown in the top plot in the figure below. To find X ( f ), use the property that “ Multiplication in the time-domain is Convolution in the frequency domain ”. The spectrum of the modulating cosine cos( 6 π 10 5 t ) = cos parenleftBig 2 π 3(10) 5 t parenrightBig is shown in the middle plot below. The bottom plot is X ( f ), which is the convolution of the top and middle plots: f , MHz f , KHz X f ( ) 0 0 3 4 - 0.1 0.1 0 0.4 0.3 G f ( ) - 0.3 0.3 0.5 0.5 0.2 2 1.5 -0.3 Part (c): Plot S ( f ) over | f | ≤ 1 MHz. Since s ( t ) = h ( t ) ⋆ x ( t ), then the spectrum S ( f ) = H ( f ) × X ( f ). Therefore, we need to first find the frequency response H ( f ). Note that the impulse response h ( t ) in the problem statement is
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ECE 301-001 Spring 2018 Test 3 w/Answers 4 Problem 2 (cont.) actually the product of two simpler functions: h ( t ) = a ( t ) b ( t ) ( a ) where a ( t ) and b ( t ) are given by a ( t ) = 40 sinc parenleftBig 2(10) 5 t parenrightBig ( b ) b ( t ) = cos parenleftBig 8 π (10) 5 t parenrightBig = cos parenleftBig 2 π 4(10) 5 t parenrightBig ( c )
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