Find the Fourier transform of the function f(t) = 2−t2, |t|<1 0, otherwise You will need to use integration by parts

228.223 Assignment 4. Solutions — 2012 1. Find the Fourier transform of the function

f(t) =

2−t², |t|<1 0, otherwise You will need to use integration by parts.

F(ω) = Z ∞

−∞

f(t)e^−iωtdt = Z 1

−1

(2−t²)e^−iωtdt

Now Z

t²e^−iωtdt=e^−iωt[it²/ω+ 2t/ω²−2i/ω³]

so Z 1

−1

(2−t²)e^−iωtdt=

e^−iωt(2i/ω³ −2t/ω²+ (2−t²)i/ω)1

−1

=e^−iω(2i/ω³−2/ω²+i/ω)−e^iω(2i/ω³+2/ω²+i/ω) = 4 sinω

ω³ −4 cosω

ω² +2 sinω ω 2. IfF(ω)is the Fourier transform off(t), show that

(a)

F(0) = Z ∞

−∞

f(t)dt

F(ω) = Z ∞

−∞

f(t)e^−iωtdt so

F(0) = Z ∞

−∞

f(t)e^0×tdt= Z ∞

−∞

f(t)dt

(b)

f(0) = 1 2π

Z ∞

−∞

F(ω)dω

Similarly:

f(t) = 1 2π

Z ∞

−∞

F(ω)e^iωtdω

and it follows.

(c) Show that iff(t)is even,

F(ω) = 2 Z ∞

0

f(t) cos (ωt)dt

F(ω) = Z ∞

−∞

f(t)e^−iωtdt= Z 0

−∞

f(t)e^−iωtdt+ Z ∞

0

f(t)e^−iωtdt=I₁+I₂

1

Puttingt=−sinI₁we have I₁ =

Z 0

∞

f(−s)e^iωs(−ds) = Z ∞

0

f(s)e^iωsds

sincef is even. SincesinI₁is a dummy variable, we can replace it byt, obtaining

F(ω) =I1+I2 = Z ∞

0

f(t)e^iωtdt+ Z ∞

0

f(t)e^−iωtdt= Z ∞

0

f(t)[e^iωt+e^−iωt]dt

= 2 Z ∞

0

f(t) cos (ωt)dt

3. Letf(t) =H(t)e^−2t, whereHis the Heaviside function, and

g(t) =

1, |t|<2 0, otherwise

(a) Find the convolution off(t)andg(t). Your answer will be defined piece- wise. We will find

h(t) = Z ∞

−∞

f(τ)g(t−τ)dτ

If t + 2 < 0, i.e. t < −2, there is no overlap of the functions, so the convolution is zero.

Whent−2<0andt+ 2>0, i.e. when−2< t <2, the functions overlap for0< τ < t+ 2, so

h(t) = Z t+2

0

e^−2τdτ = (1−e^−2(t+2))/2

Fort−2>0, i.e. t >2, the overlap is fort−2< τ < t+ 2, so h(t) =

Z t+2 t−2

e^−2τdτ =e^−2t(e⁴−e⁻⁴)/2

So

h(t) =







0, t <−2

(1−e^−2(t+2))/2, −2< t <2 e^−2t(e⁴−e⁻⁴)/2, 2< t (b) Find the Fourier transforms off(t)andg(t).

F(ω) = Z ∞

0

e^−2te^−iωtdt= 1 2 +iω

G(ω) = Z 2

−2

e^−iωtdt=

e^−iωt

−iω 2

−2

= (2 sin (2ω))/ω

2

(c) Using your results, write down the Fourier transform of the result in part (a).

It is the product:

F(ω)G(ω) = 2 sin (2ω) ω(2 +iω)

4. Find the discrete Fourier transform of the sequence {1,1,1,−1}. Then find the inverse DFT of the same sequence.

X[r] = 1 N

N−1

X

k=0

x[k]e^−i2πrk/N

X[0] = (1 + 1 + 1−1)/4 = 1/2

X[1] = (1 +e^−iπ/2+e^−iπ−e^−3iπ/2)/4 =−i/2 X[2] = (1 +e^−iπ+e^−2iπ−e^−3iπ)/4 = 1/2 X[3] = (1 +e^−3iπ/2+e^−3iπ−e^−9iπ/2)/4 =i/2 And

x[r] =

N−1

X

r=0

X[r]e^i2πrk/N

so

x[0] = 1 + 1 + 1−1 = 2 x[1] = 1 +e^iπ/2+e^iπ−e^3iπ/2 = 2i

x[2] = 1 +e^iπ+e^2iπ−e^3iπ = 2 x[3] = 1 +e^3iπ/2+e^3iπ−e^9iπ/2 =−2i

5. Write a matrix W such that X = W x, where x is a column vector with 4 values of a sequence in it and X is a column vector containing the DFT of that sequence. Use this matrix to find the DFT of{1,0,−3,−1}.

W = 1 4











1 1 1 1

1 e^−iπ/2 e^−iπ e^−3iπ/2 1 e^−iπ e^−2iπ e^−3iπ 1 e^−3iπ/2 e^−3iπ e^−9iπ/2











= 1 4











1 1 1 1

1 −i −1 i

1 −1 1 −1

1 i −1 −i











X = (−3,4−i,−1,4 +i)/4

6. For the following difference equations, state their order, and whether they are homogeneous or inhomogeneous.

3

(a)

2x[n]−nx[n−1] +x[n−2] + 2n² = 0 Order=n−(n−2) = 2. Rewrite as

2x[n]−nx[n−1] +x[n−2] = −2n²

Inhomogeneous.

(b)

(x[n+ 1]−x[n−1])/2 =x[n]

Order=n+ 1−(n−1) = 2. Rewrite as

(x[n+ 1]−x[n−1])/2−x[n] = 0

Homogeneous.

(c)

3x[n−1]

x[n−2] = n−1 n+ 1 Order=n−1−(n−2) = 1. Rewrite as

3(n+ 1)x[n−1]−(n−1)x[n−2] = 0 Homogeneous.

4