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Chapter 5

Frequency Analysis: the Fourier Transform

5.1

Basic Problems

5.1 (a) The Laplace transforms are x1 (t) = e−2t u(t) x2 (t) = r(t)

⇔

x3 (t) = te−2t u(t)

⇔

X1 (s) =

X2 (s) = ⇔

1 s+2

1 s2

X3 (s) =

σ > −2

σ>0 1 (s + 2)2

σ > −2

(b) The Laplace transforms of x1 (t) and of x3 (t) have regions of convergence containing the jΩ-axis, and so we can find their Fourier transforms from their Laplace transforms by letting s = jΩ (c) The Fourier transforms of x1 (t) and x3 (t) are X1 (Ω) = X3 (Ω) =

1 2 + jΩ 1 (2 + jΩ)2

1

216

Chaparro — Signals and Systems using MATLAB

5.2

5.2 (a) In this case we are using the duality of the Fourier transforms so that the Fourier transform of the sinc is a pulse of magnitude A and cut-off frequency Ω0 which we will need to determine. The inverse Fourier transform is x(t) = = =

1 2π

Z

∞

−∞ Z Ω0

A[u(Ω + Ω0 ) − u(Ω − Ω0 )]ejΩt dΩ

A ejΩt dΩ 2π −Ω0 A sin Ω0 t πt

so that A = π and Ω0 = 1, i.e., sin(t) t

⇔

π[u(Ω + 1) − u(Ω − 1)]

(b) The Fourier transform of x1 (t) = u(t + 0.5) − u(t − 0.5) is   1 0.5s sin(0.5Ω) −0.5s [e − e ] X1 (Ω) = = 0.5Ω s s=jΩ Using the duality property we have: sin(Ω/2) Ω/2

x1 (t) = u(t + 0.5) − u(t − 0.5)

⇔

X1 (Ω) =

sin(t/2) t/2

⇔

2π[u(Ω + 0.5) − u(Ω − 0.5)]

X1 (t) =

using the fact that x1 (t) is even. Then using the scaling property X1 (2t) =

sin(t) t

⇔ ⇔

2π [u((Ω/2) + 0.5) − u((Ω/2) − 0.5)] 2 π[u(Ω + 1) − u(Ω − 1)]

so x(t) = X1 (2t) = sin(t)/t is the inverse Fourier transform of X(Ω) = π[u(Ω + 1) − u(Ω − 1)]

Copyright 2014, Elsevier, Inc. All rights reserved.

217

Chaparro — Signals and Systems using MATLAB

5.3

5.3 (a) The signal x(t) is even while y(t) is odd. (b) The Fourier transform of x(t) is X(Ω) = = =

Z

∞

e−|t| e−jΩt dt

−∞ Z ∞

2

e−|t| cos(Ωt)dt − j

−∞ Z ∞

Z

∞

e−|t| sin(Ωt)dt

−∞

e−|t| cos(Ωt)dt

0

this is because the imaginary part is the integral of an odd function which is zero. Since cos(.) is an even function X(−Ω) = X(Ω) The Fourier transform X(Ω) is X(Ω) = =

2 Z

0

=

ejΩt + e−jΩt dt 2 0 Z ∞ ∞ e−(1+jΩ)t dt e−(1−jΩ)t dt +

Z

∞

e−t

0

2 1 1 = + 1 − jΩ 1 + jΩ 1 + Ω 2

which is real-valued. (c) For y(t), odd function, its Fourier transform is Z ∞ Y (Ω) = y(t)e−jΩt dt −∞ Z ∞ = −j y(t) sin(Ωt)dt −∞

because y(t) cos(Ωt) is an odd function and its integral is zero. The Y (Ω) is odd since Y (−Ω) = =

−j

Z

∞

y(t) sin(−Ωt)dt

−∞

−Y (Ω)

since the sine is odd. (d) Let’s use the Laplace transform to find the Fourier transform of y(t): Y (s) =

1 1 − s + 1 −s + 1

with a region of convergence −1 < σ < 1, which contains the jΩ-axis. So Y (Ω) = Y (s) |s=jΩ =

1 1 −2jΩ − = 1 + Ω2 jΩ + 1 −jΩ + 1

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218

Chaparro — Signals and Systems using MATLAB

5.4

which as expected is purely imaginary. Check: Let z(t) = x(t) + y(t) = 2e−t u(t) which has a Fourier transform Z(Ω) =

2(1 − jΩ) 2 = = X(Ω) + Y (Ω) 1 + Ω2 1 + jΩ

(f) If a signal is represented as x(t) = xe (t) + xo (t) then X(Ω) = Xe (Ω) + Xo (Ω) where the first is a cosine transform and the second a sine transform.

Copyright 2014, Elsevier, Inc. All rights reserved.

219

Chaparro — Signals and Systems using MATLAB 5.5

5.6

(a) x1 (t) = −x(t + 1) + x(t − 1), time–shift property X1 (Ω) = X(Ω)(−ejΩ + e−jΩ ) = −2jX(Ω) sin(Ω) (b) x2 (t) = 2 sin(t)/t by duality X2 (Ω) = 2π[u(−Ω + 1) − u(−Ω − 1)] = 2π[u(Ω + 1) − u(Ω − 1)] by symmetry of x(t). (c) Compression x3 (t) = 2x(2t) = 2[u(2t + 1) − u(2t − 1)] = 2[u(t + 0.5) − u(t − 0.5)] X3 (Ω) = 2

X(Ω/2) = X(Ω/2) 2

(d) Modulation: x4 (t) = cos(0.5πt)x(t) so X4 (Ω) = 0.5[X(Ω + 0.5π) + X(Ω − 0.5π)] (e) x5 (t) = X(t) so that by duality X5 (Ω) = 2πx(−Ω) = 2π[u(−Ω + 1) − u(−Ω − 1)] = 2πx(Ω)

Copyright 2014, Elsevier, Inc. All rights reserved.

221

Chaparro — Signals and Systems using MATLAB 5.6

5.7

(a) x(t) = cos(t)[u(t) − u(t − 1)] = cos(t)p(t), so X(Ω) = 0.5[P (Ω + 1) + P (Ω − 1)] where P (Ω) =

e−s/2 (es/2 − e−s/2 ) sin(Ω/2) |s=jΩ = 2e−jΩ/2 s Ω

(b) y(t) = x(2t) = cos(2t)p(2t) = cos(2t)[u(t) − u(t − 0.5)], so Y (Ω) = 0.5[P1 (Ω + 2) + P1 (Ω − 2)] where P1 (Ω) = F [u(t) − u(t − 0.5)] = 2e−jΩ/4

sin(Ω/4) Ω

z(t) = x(t/2) = cos(t/2)p(t/2) = cos(t/2)[u(t) − u(t − 2)] = cos(t/2)p2 (t) so Z(Ω) = 0.5[P2 (Ω + 0.5) + P2 (Ω − 0.5)] P2 (Ω) = F [u(t) − u(t − 2)] = 2e−jΩ

sin(Ω) Ω

Using P1 (Ω) = 0.5P (Ω/2) P2 (Ω) = 2P (2Ω) we have X(Ω) =

0.5[P (Ω + 1) + P (Ω − 1)]

Y (Ω) =

0.5[0.5P ((Ω/2) + 1) + 0.5P ((Ω/2) − 1)] = 0.5X(Ω/2)

Z(Ω) =

0.5[2P (2Ω + 1) + 2P (2Ω − 1)] = 2X(2Ω)

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222

Chaparro — Signals and Systems using MATLAB

5.8

5.7 F [δ(t − τ )] = L[δ(t − τ )]s=jΩ = e−jΩτ so (a) By linearity and time–shift F [δ(t − 1) + δ (t + 1)] = 2 cos(Ω) (b) By duality 0.5[δ(t − τ ) + δ(t + τ )]

↔

cos(Ωτ )

cos(Ω0 t)

↔

π[δ(Ω + Ω0 ) + δ (Ω − Ω0 )]

by letting τ = Ω0 in the second equation. (c) Considering F [δ(t − 1) − δ (t + 1)] = 2j sin(Ω), by duality −0.5j[δ(t − τ ) + δ(t + τ )]

↔

sin(Ωτ )

sin(Ω0 t)

↔

− jπ[δ(Ω + Ω0 ) + δ(Ω + Ω0 )]

by letting τ = Ω0 in the second equation.

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223

Chaparro — Signals and Systems using MATLAB 5.14

5.15

(a) Let X(Ω) = A[u(Ω + Ω0 ) − u(Ω − Ω0 )] its inverse Fourier transform is x(t) =

1 2π

Z

Ω0

−Ω0

AejΩt dΩ =

A sin(Ω0 t) πt

so A = 1, Ω0 = 0.5 and X(Ω) = u(Ω + 0.5) − u(Ω − 0.5). (b) Y (Ω) = H (Ω)X(Ω) = X(Ω) so that y(t) = (x ∗ x)(t) = x(t), or convolution of a sinc function with itself is a sinc.

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230

Chaparro — Signals and Systems using MATLAB 5.17

5.18

(a) Impulse response h(t) = =

Z Z 2 Z 2 1 0 j(Ωt+π/2) 1 1 e dΩ ej(Ωt−π/2) dΩ + 1ej ∠H(j Ω) ejΩt dΩ = 2π −2 2π 0 2π −2 j 1 − cos(2t) −j j2t (e − 1) − (e−j2t − 1) = 2πjt 2πjt πt

(b) The frequency components of x(t) with harmonic frequencies bigger than 2 are filtered out so yss (t) = 2|H (j1.5)| cos(1.5t + ∠H (j 1.5)) = 2 cos(1.5t − π/2) = 2 sin(1.5t)

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233

Chaparro — Signals and Systems using MATLAB 5.18

5.19

(a) Plot of X(Ω) as function of Ω:

X(Ω)

−π

Ω

π Figure 5.3: Problem 18

(b) 1 x(0) = 2π

Z

π

−π

2 |Ω| dΩ = π 2π

Z

π 0

1 Ω dΩ = π 2

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234

Chaparro — Signals and Systems using MATLAB 5.19

5.20

(a) Poles are roots of D(s) = s2 + 2s + 2 = (s + 1)2 + 1 = 0 or s1,2 = −1 ± j1 the zero is s = 0. It is a band-pass filter with center frequency around 1. Its magnitude response is using vectors from the zero and the poles to the point in the jΩ–axis where are finding the frequency response: Ω |H (jΩ)| 0

0 (zero at zero) √ √ 5(1)/[(1)( 4 + 1)] = 1

1 ∞

0 (vectors of two poles and zero have infinite lengths)

(b) Impulse response √ √ 5(s + 1) 5 H (s) = − (s + 1)2 + 1 (s + 1)2 + 1 √ √ −t √ h(t) = 5e (cos(t) − sin(t)) u(t) = 5e−t 2 cos(t + π/4)u(t) √ −t = 10e cos(t + π/4)u(t) (c) The steady state response corresponding to x(t) = B + cos(Ωt) is y(t) = =

B|H (j 0)| + |H (j Ω0 )| cos(Ω0 + ∠H (jΩ0 )) |H (jΩ0 )| cos(Ω0 + ∠H (jΩ0 ))

for Ω0 to be determined by looking at frequencies for which √ 5Ω0 =1 or |H (jΩ0 )| = p (2 − Ω20 )2 + 4Ω20 5Ω02 = 4 − 4Ω20 + Ω40 + 4Ω20 ⇒

Ω04 − 5Ω20 + 4 = (Ω20 − 4)((Ω02 − 1) = 0

giving values of Ω0 = ±2, ± 1 so we have that when Ω = 1 or 2 the dc bias is filtered out and the cosine has a magnitude of 1. The corresponding phases are using the pole and zero vectors Ω0 = 1

⇒

∠H (jΩ0 ) = π/2 − 0 − tan−1 (2)

Ω0 = 2

⇒

∠H (jΩ0 ) = π/2 − tan−1 (1) − tan−1 (3)

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235

Chaparro — Signals and Systems using MATLAB 5.22

5.23

(a) According to the eigenvalue property for x(t) = ejΩt , −∞ < Ω < ∞, the output in the steady-state would be y(t) = ejΩt H (jΩ) so that the differential equation gives

jΩejΩt H (jΩ) = −ejΩt H (jΩ) + ejΩt 1 giving H (jΩ) = 1 + jΩ 1 |H (jΩ)| = √ , ∠H (jΩ) = − tan−1 (Ω) 1 + Ω2

|Y (Ω) 1

0.707 Ω

−1

1

Figure 5.4: Problem 22

(b) The magnitude response indicates the filter is a low-pass filter, in particular Ω 0 1 ∞

|H (jΩ)| ∠H (j Ω) 1 1 √ 2

0

0 −π/4

−π/2

(c) The Fourier transform of x(t) is X(Ω) = u(Ω + 1) − u(Ω − 1) so that the Fourier transform of the output is

Y (Ω) = X(Ω)H (jΩ) with magnitude response as in Fig. 5.4

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