# Therefore since multiplication in the time domain is

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Therefore, since multiplication in the time domain is convolution ( ) in the frequency domain, the transform of equation ( a ) is H ( f ) = A ( f ) ⋆ B ( f ) ( d ) where A ( f ) and B ( f ) are the Fourier transforms of a ( t ) and b ( t ) given in ( b ) and ( c ). In b ( t ) we have f c = 4(10) 5 . Therefore, the function B ( f ) is given by B ( f ) = 0 . 5 δ parenleftBig f - 4(10) 5 parenrightBig + 0 . 5 δ parenleftBig f + 4(10) 5 parenrightBig ( e ) To find A ( f ), use the following Fourier transform pair: 2 BK sinc parenleftBig 2 Bt parenrightBig braceleftBigg K , - B f B 0 , else ( f ) Equating the sinc on the left side of ( f ) to the sinc given in ( b ) then gives 2 BK sinc parenleftBig 2 Bt parenrightBig = 40 sinc parenleftBig 2(10) 5 t parenrightBig ( g ) Equating the sinc arguments on both sides of ( g ) thus gives B : 2 B = 2(10) 5 B = 10 5 ( h ) Equating the sinc multiplying coefficient on both sides of ( g ) then gives K : 2 BK = 40 K = 20 B = 20 10 5 = 0 . 0002 ( i ) Substituting ( h ) and ( i ) into ( f ) then gives a ( t ) = 40 sinc parenleftBig 2(10) 5 t parenrightBig braceleftBigg 0 . 0002 , | f | ≤ 10 5 0 , else = A ( f ) ( j )

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ECE 301-001 Spring 2018 Test 3 w/Answers 5 Problem 2 (cont.) In the graph below, the top plot is A ( f ), the middle plot is the two-impulse frequency function in equation ( e ), and the bottom plot is their convolution, which is H ( f ) : A f ( ) B f ( ) H f ( ) 0 0 .0002 -0.1 0.1 0.5 0.4 -0.4 0.5 0 -0.4 0.3 0.4 0.5 f, MHz .0001 f, MHz f, MHz Now we can plot S ( f ). Since from the problem statement we have s ( t ) = h ( t ) ⋆ x ( t ), this implies that the spectrum S ( f ) is given by the multiplication S ( f ) = H ( f ) × X ( f ). A plot of X ( f ) from Part (a) is shown in the top plot in the figure below. The filter frequency response H ( f ) is shown in the middle plot in the figure below. The bottom plot is S ( f ), which is the multiplication of the top and middle plots. f , MHz X f ( ) f , MHz f , MHz H f ( ) 0 0.4 0.2 0.3 2 0.3 0.4 0 .0002 -0.3 0 -0.2 -0.4 -0.3 -0.4 S f ( ) .0001 0.5 -0.5 0.3 1.5 .00015
ECE 301-001 Spring 2018 Test 3 w/Answers 6 3. The analog signal x ( t ) = 10 e - 50 t cos(2 π 50 t ) is multiplied by a sampling waveform s δ ( t ) to give the “impulse sampled” signal x δ ( t ) = x ( t ) × s δ ( t ). The sampling frequency is f s = 250 Hz. This sampling waveform and h ( t ), the impulse response of an analog filter, are given by s δ ( t ) = m =+ summationdisplay m = - ∞ δ ( t - mT s ) , h ( t ) = u ( t ) - u ( t - 0 . 5 T s ) The analog signal p ( t ) is then created by the convolution p ( t ) = h ( t ) ⋆ x δ ( t ). (a) Plot x δ ( t ) over 0 t 15 ms. and compute the values of x ( t ) at the sampling times. (b) Plot p ( t ) over the time range 0 t 15 ms.

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