CALIFORNIA STATE UNIVERSITY, BAKERSFIELD (CSUB)DEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERING & COMPUTER SCIENCE

ECE 423: DIGITAL COMMUNICATIONS

Homework 2Solution

QUESTION 1:(25 POINTS)(a) Use direct integration to find the Fourier transform G( f ) of signal

g (t ) = exp(−2|t −3|).

(a) Use direct integration to find the inverse Fourier transform of

G( f ) = δ(4π f )−δ( f −2).

Answer:(a)

G( f ) =∫ ∞

−∞g (t )e− j 2π f t d t =

∫ ∞

−∞e−2|t−3|e− j 2π f t d t

=∫ 3

−∞e2(t−3)e− j 2π f t d t +

∫ ∞

3e−2(t−3)e− j 2π f t d t

= e− j 6π f

2− j 2π f+ e− j 6π f

2+ j 2π f= e− j 6π f

1+π2 f 2

(b)

g (t ) =∫ ∞

−∞G( f )e j 2π f t d f =

∫ ∞

−∞δ(4π f )e j 2π f t d f −

∫ ∞

−∞δ( f −2)e j 2π f t d f

Let λ= 4π f in the integral involving δ(4π f ). Then∫ ∞

−∞δ(4π f )e j 2π f t d f =

∫ ∞

−∞δ(λ)e(λ/2)t dλ

4π= e0

4π= 1

4π

We also have∫ ∞−∞δ( f −2)e j 2π f t d f = e j 4πt . Hence

g (t ) = 1

4π−e j 4πt

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QUESTION 2:(25 POINTS)The Fourier transform of the triangular pulse g (t ) in the following figure is given by

G( f ) = 1

(2π f )2 (e j 2π f − j 2π f e j 2π f −1).

Use this information and the time shifting and time scaling properties to find the Fouriertransforms of the signals g 1(t )− g5(t ) shown bellow.

Answer:(b) g1(t ) = g (−t ). Then G1( f ) = 1

(2π f )2 (e− j 2π f + j 2π f e− j 2π f −1)

(c) g2(t ) = g (t −1)+ g1(t −1). Then

G2( f ) = e− j 2π f

(2π f )2 (e j 2π f − j 2π f e j 2π f −1)+ e− j 2π f

(2π f )2 (e− j 2π f + j 2π f e− j 2π f −1)

= 1

(2π f )2 (1− j 2π f −2e− j 2π f +e− j 4π f + j 2π f e− j 4π f )

(d) g3(t ) = g (t −1)+ g1(t +1). Then

G3( f ) = e− j 2π f

(2π f )2 (e j 2π f − j 2π f e j 2π f −1)+ e j 2π f

(2π f )2 (e− j 2π f + j 2π f e− j 2π f −1)

= 1

(2π f )2 (2−e− j 2π f −e j 2π f ) = 1

2(π f )2 (1−cos2π f )

(e) g4(t ) = g (t − 12 )+ g1(t + 1

2 ). Then

G4( f ) = e− jπ f

(2π f )2 (e j 2π f − j 2π f e j 2π f −1)+ e jπ f

(2π f )2 (e− j 2π f + j 2π f e− j 2π f −1)

= 1

(2π f )2 (− j 2π f e jπ f + j 2π f e− jπ f ) = − j

2π f(e jπ f −e− jπ f ) = 1

π fsin(π f )

(f) g5(t ) = 1.5g(1

2 (t −2)). Then

G5( f ) = 3e− j 4π f

(4π f )2 (e j 4π f − j 4π f e j 4π f −1) = 3

(4π f )2 (1− j 4π f −e− j 4π f )

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QUESTION 3:(25 POINTS)Signals g1(t ) = 104 ∏

(104t ) and g2(t ) = δ(t ) are applied at the inputs of the ideal low-pass fil-ters H1( f ) =∏

( f /20,000) and H2( f ) =∏( f /10,000) shown in the following figure. The output

y1(t ) and y2(t ) of these filters are multiplied to obtain the signal y(t ) = y1(t )y2(t ).

a) Sketch G1( f ) and G2( f ).

b) Sketch H1( f ) and H2( f ).

c) Sketch Y1( f ) and Y2( f ).

d) Find the bandwidths of y1(t ), y2(t ) and y(t ).

Answer: (d) We can read from the plots in the above figures that the bandwidth of y1(t ) =

10,000 Hz and the bandwidth of y2(t ) = 5000 Hz. To find the bandwidth of their producty(t ), we recall that the multiplication in the time domain is equivalent to convolution in thefrequency domain. Also recall the width property of convolution which states that the widthof the convolution of two signals is equal to the sum of their respective widths. Therefore, thebandwidth of y(t ) = 10,000+5000 = 15,000 Hz.

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QUESTION 4(25 POINTS)Using only time-shifting property and the Fourier transform Table given in the class, find theFT of the signals shown in the following figure.

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