Signal Processing

Signal Processing 

Lecture 1 

Sven Nordebo 

School of Computer Science, Physics and Mathematics 

Linnæus University, Sweden 

4ED044 Signal Processing, September 20, 2012

Chapter 1: Transform theory 

⎧ 

⎪⎨ 

⎪⎩ 

The Fourier transform: 

Continuous–time aperiodic signals x(t) 

∫ ∞ 

X(f) = x(t)e −i2πft dt 

−∞ 

∫ ∞ 

x(t) = X(f)e i2πft df 

−∞ 

⎧ 

⎪⎨ 

⎪⎩ 

The Fourier series: 

Continuous–time T -periodic signals ˜x(t) 

Discrete spectrum f k = k/T 

X k = 1 ∫ 

˜x(t)e −i 2π T kt dt 

T T 

∞∑ 

˜x(t) = X k e i 2π T kt 

k=−∞ 

⎧ 

The Discrete-Time Fourier transform: 

⎧ 

The Discrete Fourier Transform (DFT): 

Discrete–time aperiodic signals x n 

Discrete–time N-periodic signals ˜x n 

⎪⎨ 

1-periodic spectrum: 

˜X(ν) = 

∞∑ 

n=−∞ 

x ne −i2πνn 

˜X(ν + 1) = ˜X(ν) 

⎪⎨ 

N-periodic spectrum: 

˜Xk+N = ˜X k 

N−1 ∑ 

˜X k = ˜x ne −i 2π N kn , k = 0, . . . , N − 1 

n=0 

⎪⎩ 

∫ 1/2 

x n = ˜X(ν)e i2πνn dν 

−1/2 

⎪⎩ 

˜x n = 1 N 

N−1 ∑ 

k=0 

˜X k e i 2π N kn , n = 0, . . . , N − 1 

Sven Nordebo, School of Computer Science, Physics and Mathematics, Linnæus University, Sweden. 2(28)

Transform theory 

⎧ 

⎪⎨ 

⎪⎩ 

The Fourier transform: Alternative definition ω = 2πf [ rad/s] 

Continuous–time aperiodic signals 

∫ ∞ 

X(ω) = x(t)e −iωt dt 

−∞ 

x(t) = 1 ∫ ∞ 

X(ω)e i2πωt dω 

2π −∞ 

⎧ 

⎪⎨ 

⎪⎩ 

The Discrete-Time Fourier transform: Alternative definition Ω = 2πν 

Discrete–time aperiodic signals 

2π-periodic spectrum: X(Ω + 2π) = X(Ω) 

X(Ω) = 

∞∑ 

x(n)e −iΩn 

n=−∞ 

x(n) = 1 ∫ π 

X(Ω)e iΩn dΩ 

2π −π 



Exercise 1 (a): The Fourier transform for continuous–time aperiodic signals 

{ A 0 ≤ t ≤ Tp 

x(t) = 

0 otherwise 

∫ ∞ 

X(f) = x(t)e −i2πft dt = 

−∞ 

= A 1 − e−i2πfTp 

i2πf 

∫ Tp 

[ e 

Ae −i2πft −i2πft ] Tp 

dt = A 

0 

−i2πf 0 

= Ae −iπfTp eiπfTp − e −iπfTp 

i2πf 

−iπfTp sin(πfTp) 

= AT pe 

πfT p 

Zeros of X(f): 

X(f) = 0 ⇔ sin(πfT p) = 0 ⇔ πfT p = mπ ⇔ f = m T p 

, m = 0, ±1, ±2, . . . 

Draw a sketch of both the signal x(t) and its Fourier transform X(f). 



Exercise 1 (b): The Fourier series for continuous–time T -periodic signals 

⎧ 

A 0 ≤ t < T p 

⎪⎨ 

˜x(t) = 0 T p < t < T 

⎪⎩ 

˜x(t + T ) all t 

X k = 1 ∫ 

˜x(t)e −i 2π T kt dt = 1 ∫ T 

˜x(t)e −i 2π T kt dt = 1 ∫ Tp 

Ae −i 2π T kt dt 

T T 

T 0 

T 0 

{ 

= same integral as above with f k = k } 

= 1 T T ATpe−iπ T k sin(π k Tp T Tp) 

π k T Tp 

Draw a sketch of both the signal ˜x(t) and its Fourier series X k . 

Note that the Fourier series coefficients X k of ˜x(t) are equal to 1 times the Fourier 

T 

transform of one single period x(t) of the periodic signal ˜x(t), sampled at the 

frequencies f = k T 

X k = 1 ∫ 

˜x(t)e −i 2π T kt dt = 1 ∫ 

x(t)e −i 2π T kt dt = 1 ∫ ∞ 

x(t)e −i 2π T kt dt 

T T 

T T 

T −∞ 

= 1 T X(f)| f= k T 



Exercise 1 (c): The Discrete-Time Fourier transform for discrete–time aperiodic signals 

{ A 0 ≤ n ≤ Np − 1 

x n = 

0 otherwise 

˜X(ν) = 

∞∑ 

n=−∞ 

N 

∑ p−1 

x ne −i2πνn = Ae −i2πνn = 

n=0 

⎧ 

⎨ 

N p−1 

⎩ Geom. ser. ∑ 

n=0 

= A 1 − e−i2πνNp 

1 − e −i2πν = A e−iπνNp (e iπνNp − e −iπνNp ) 

e −iπν (e iπν − e −iπν ) 

= Ae −iπν(Np−1) (eiπνNp − e −iπνNp )/(2i) 

(e iπν − e −iπν )/(2i) 

q n = 1 − qNp 

1 − q 

−iπν(Np−1) sin(πνNp) 

= Ae 

sin(πν) 

Zeros of ˜X(ν): 

˜X(ν) = 0 ⇔ sin(πνN p) = 0 ⇔ πνN p = mπ ⇔ ν = m N p 

, m = 0, ±1, ±2, . . . 

⎫ 

⎬ 

⎭ 

Draw a sketch of both the signal x n and its Fourier transform ˜X(ν). 



Exercise 1 (d): The Discrete Fourier Transform (DFT) for discrete–time N-periodic 

signals 

⎧ 

A 0 ≤ n ≤ N p − 1 

⎪⎨ 

˜x n = 0 N p ≤ n ≤ N − 1 

⎪⎩ 

˜x n+N all n 

N−1 ∑ 

˜X k = 

n=0 

N 

˜x ne −i 2π ∑ p−1 

N kn = 

n=0 

Draw a sketch of both the signal ˜x n and its DFT ˜X k . 

{ 

Ae −i 2π N kn = same sum as above with ν k = k } 

N 

= Ae −iπ k N (Np−1) sin(π k N Np) 

sin(π k N ) 

Note that the DFT ˜X k of ˜x n is equal to the Fourier transform of one single period x n 

of the periodic signal ˜x n, sampled at the frequencies ν = k N 

˜X k = 

N−1 ∑ 

n=0 

˜x ne −i 2π N kn = 

N−1 ∑ 

n=0 

x ne −i 2π N kn = 

∞∑ 

n=−∞ 

x ne −i 2π N kn = ˜X(ν)| ν= k 

N 



The Fourier transform as the limit of the Fourier series 

Let x(t) be an aperiodic continuos-time signal with the property 

lim x(t) = 0 

|t|→∞ 

Extend this signal as a periodic signal with period T 

so that 

˜x(t) = 

∞∑ 

n=−∞ 

x(t − nT ) 

lim ˜x(t) = x(t) for t ∈ [−T/2, T/2] 

T →∞ 

It follows that 

X k = 1 ∫ T /2 

˜x(t)e −i2π T k t dt → 1 ∫ T /2 

x(t)e −i2π T k t dt → 1 T −T /2 

T −T /2 

T X( k T ) 

as T → ∞. Moreover, let f k = k/T → f and f k+1 − f k = 1/T → df, so that 

˜x(t) = 

as T → ∞. 

∞∑ 

k=−∞ 

X k e i2πf kt → 

∞∑ 

k=−∞ 

∫ 

1 

∞ 

T X(f k)e i2πfkt → X(f)e i2πft df = x(t) 

−∞ 



Classical Fourier transform: F : L 2 → L 2 

⎧ ∫ ∞ 

⎪⎨ Fx(t) = x(t)e −iωt dt 

⎪⎩ 

Important properties: 

⎧ 

Duality: 

⎪⎨ 

⎪⎩ 

−∞ 

F −1 X(ω) = 1 

2π 

Time-derivative: 

Frequency-derivative: 

Time-shift: 

∫ ∞ 

−∞ 

X(ω)e iωt dω 

FFx(t) = 2πx(−t) 

F∂ t x(t) = iωX(ω) 

Ftx(t) = i∂ ω X(ω) 

Fx(t − t 0 ) = e −iωt0 X(ω) 

Frequency-shift: Fe iω0t x(t) = X(ω − ω 0 ) 

Scaling: Fx(at) = 1 

|a| X(ω a ) 



Important properties: 

⎧ 

Parseval: 

⎪⎨ 

⎪⎩ 

Parseval: 

Time-convolution: 

∫ ∞ 

−∞ 

∫ ∞ 

−∞ 

x(t)y ∗ (t)dt = 1 

2π 

|x(t)| 2 dt = 1 

2π 

∫ ∞ 

−∞ 

∫ ∞ 

−∞ 

Fx(t) ∗ y(t) = X(ω)Y (ω) 

Freq.-convolution: Fx(t)y(t) = 1 X(ω) ∗ Y (ω) 

2π 

Conj. symmetry: x(t) real ⇔ X(ω) = X ∗ (−ω) 

Conj. symmetry: 

Conj. asymmetry: 

Conj. asymmetry: 

X(ω) real ⇔ x(t) = x ∗ (−t) 

X(ω)Y ∗ (ω)dω 

|X(ω)| 2 dω 

x(t) imaginary ⇔ X(ω) = −X ∗ (−ω) 

X(ω) imaginary ⇔ x(t) = −x ∗ (−t) 



The Parseval’s relations for Fourier series 

∫ 

1 

T T 

∫ 

1 

T T 

˜x(t)ỹ ∗ (t) dt = 

|˜x(t)| 2 dt = 

∞∑ 

k=−∞ 

∞∑ 

k=−∞ 

X k Y ∗ k 

|X k | 2 

Proof: Employ the orthogonality relationship of the basis functions 

We have 

∫ 

{ 

1 

e i 2π T (k−k′)t 1 k = k ′ 

dt = δ kk ′ = 

T T 

0 k ≠ k ′ 

∫ 

1 

˜x(t)ỹ ∗ (t) dt = 1 ∫ ∞∑ 

X k e i 2π ∑ ∞ 

T kt Y ∗ 2π k 

T T 

T 

′e−i T k′t dt 

T 

k=−∞ 

k ′ =−∞ 

∞∑ ∞∑ 

∫ 

= 

X k Y ∗ 1 

∞∑ ∞∑ 

k ′ e i 2π T (k−k′)t dt = 

X k Yk ∗ T 

′δ kk ′ = 

∑ ∞ X k Yk 

∗ 

k=−∞ k ′ =−∞ 

T 

k=−∞ k ′ =−∞ 

k=−∞ 



The Parseval’s relations for Fourier transforms 

∫ ∞ 

−∞ 

∫ ∞ 

∫ ∞ 

x(t)y ∗ (t) dt = 

−∞ 

−∞ 

X(f)Y ∗ (f) df 

∫ ∞ 

|x(t)| 2 dt = |X(f)| 2 df 

−∞ 


We have 

∫ ∞ 

e i2π(f−f ′ )t dt = δ(f − f ′ ) 

−∞ 

∫ ∞ 

∫ ∞ ∫ ∞ 

∫ ∞ 

x(t)y ∗ (t) dt = 

X(f)e i2πft df Y ∗ (f ′ )e −i2πf ′t df ′ dt 

−∞ 

−∞ −∞ 

−∞ 

∫ ∞ ∫ ∞ 

∫ ∞ 

= 

X(f)Y ∗ (f ′ ) e i2π(f−f ′ )t dt df df ′ 

−∞ −∞ 

−∞ 

∫ ∞ ∫ ∞ 

∫ ∞ 

= 

X(f)Y ∗ (f ′ )δ(f − f ′ ) df ′ df = X(f)Y ∗ (f) df 

−∞ −∞ 

−∞ 



The Parseval’s relations for the Discrete Fourier Transform (DFT) 

N−1 ∑ 

n=0 

N−1 ∑ 

n=0 

˜x nỹ ∗ n = 1 N 

|˜x n| 2 = 1 N 

N−1 ∑ 

k=0 

N−1 ∑ 

k=0 

˜X k Ỹ ∗ k 

| ˜X k | 2 


We have 

= 1 N 

N−1 ∑ 

n=0 

N−1 ∑ 

˜x nỹn ∗ = 

N−1 ∑ 

N−1 ∑ 

k=0 k ′ =0 

N−1 

1 ∑ 

{ 

e i 2π N (k−k′)n = ˜δ 1 k = k 

kk ′ = 

′ + mN 

N 

0 k ≠ k ′ + mN 

n=0 

n=0 

N−1 

1 ∑ 

˜X k e i 2π N kn 1 N−1 ∑ 

N 

N 

k=0 

N−1 

˜X k Ỹk ∗ 1 ∑ 

′ 

N 

n=0 

k ′ =0 

e i 2π N (k−k′ )n = 1 N 

Ỹ ∗ k ′e−i 2π N k′ n 

N−1 ∑ 

N−1 ∑ 

k=0 k ′ =0 

˜X k Ỹ ∗ k ′ ˜δ kk ′ = 1 N 

N−1 ∑ 

k=0 

˜X k Ỹ ∗ k 



The Parseval’s relations for Discrete-Time Fourier transforms 

∞∑ 

n=−∞ 

∞∑ 

n=−∞ 

x ny ∗ n = ∫ 1/2 

−1/2 

˜X(ν)Ỹ ∗ (ν) dν 

∫ 1/2 

|x n| 2 = | ˜X(ν)| 2 dν 

−1/2 


∞∑ 

e i2π(ν−ν′ )n = ˜δ(ν − ν ′ ) mod 1 

n=−∞ 

We have 

∞∑ 

x nyn ∗ = 

∞∑ 

∫ 1/2 

∫ 1/2 

˜X(ν)e i2πνn dν Ỹ ∗ (ν ′ )e −i2πν′n dν ′ 

n=−∞ 

n=−∞ −1/2 

−1/2 

∫ 1/2 ∫ 1/2 

= 

˜X(ν)Ỹ ∗ (ν ′ ) 

∞∑ 

e i2π(ν−ν′ )n dν dν ′ 

−1/2 −1/2 

n=−∞ 

∫ 1/2 ∫ 1/2 

∫ 1/2 

= 

˜X(ν)Ỹ ∗ (ν ′ )˜δ(ν − ν ′ ) dν ′ dν = ˜X(ν)Ỹ ∗ (ν) dν 

−1/2 −1/2 

−1/2 



Convolution theorem for the Fourier transform (linear convolution in 

continuous time) 

Let 

Then 

Proof: 

∫ ∞ 

z(t) = x(t) ∗ y(t) = x(t − τ)y(τ) dτ 

−∞ 

Z(ω) = X(ω)Y (ω) 

∫ ∞ 

∫ ∞ {∫ ∞ 

} 

Z(ω) = z(t)e −iωt dt = 

x(t − τ)y(τ) dτ e −iωt dt 

−∞ 

−∞ −∞ 

∫ ∞ ∫ ∞ 

= {(u, τ) = (t − τ, τ), du dτ = dt dτ} = 

x(u)y(τ)e −iω(u+τ) du dτ 

−∞ −∞ 

∫ ∞ 

∫ ∞ 

= x(u)e −iωu du y(τ)e −iωτ dτ = X(ω)Y (ω) 

−∞ 

−∞ 



Convolution theorem for the Fourier transform (linear convolution in 

continuous frequency) 

Let 

Then 

Proof: 

z(t) = x(t)y(t) 

Z(ω) = 1 

∫ 

1 ∞ 

X(ω) ∗ Y (ω) = X(ω − ϕ)Y (ϕ) dϕ 

2π 2π −∞ 

{ } 

1 

z(t) = F −1 2π X(ω) ∗ Y (ω) = 1 ∫ ∞ { ∫ 1 ∞ 

} 

X(ω − ϕ)Y (ϕ) dϕ e iωt dω 

2π −∞ 2π −∞ 

= {(υ, ϕ) = (ω − ϕ, ϕ), dυ dϕ = dω dϕ} = 1 ∫ ∞ ∫ ∞ 

(2π) 2 X(υ)Y (ϕ)e i(υ+ϕ)t dυ dϕ 

−∞ −∞ 

= 1 ∫ ∞ 

X(υ)e iυt dυ 1 ∫ ∞ 

Y (ϕ)e iϕt dϕ = x(t)y(t) 

2π −∞ 

2π −∞ 



Convolution theorem for the Fourier series (periodic convolution in 

continuous time) 

Let 

Then 

Proof: 

˜z(t) = 1 T ˜x(t) ⊗ ỹ(t) = 1 ∫ 

˜x(t − τ)ỹ(τ) dτ 

T T 

Z k = X k Y k 

Z k = 1 ∫ 

˜z(t)e −i 2π T kt dt = 1 ∫ T { ∫ 1 T 

} 

˜x(t − τ)ỹ(τ) dτ e −i 2π T kt dt 

T T 

T 0 T 0 

= {(u, τ) = (t − τ, τ), du dτ = dt dτ} = 1 ∫ T ∫ T −τ 

T 2 ˜x(u)ỹ(τ)e −i 2π T k(u+τ) du dτ 

0 −τ 

= 1 ∫ 

˜x(u)e −i 2π T ku du 1 ∫ 

ỹ(τ)e −i 2π T kτ dτ = X k Y k 

T T 

T T 



Convolution theorem for the Fourier series (linear convolution in discrete 

frequency) 

Let 

Then 

Proof: 

˜z(t) = ˜x(t)ỹ(t) 

Z k = X k ∗ Y k = 

∞∑ 

q=−∞ 

X k−q Y q 

⎧ 

⎫ 

∞∑ 

∞∑ ⎨ ∞∑ ⎬ 

˜z(t) = Z k e i 2π T kt = 

X 

⎩ 

k−q Y q 

⎭ ei 2π T kt = {(p, q) = (k − q, q)} 

k=−∞ 

k=−∞ q=−∞ 

∞∑ ∞∑ 

∞∑ 

= 

X pY qe i 2π T (p+q)t = X pe i 2π ∑ ∞ 

T pt Y qe i 2π T qt = ˜x(t)ỹ(t) 

p=−∞ q=−∞ 

p=−∞ 

q=−∞ 



Convolution theorem for the Discrete-Time Fourier transform (linear 

convolution in discrete time) 

Let 

Then 

z n = x n ∗ y n = 

˜Z(ν) = 

∞∑ 

q=−∞ 

˜X(ν)Ỹ (ν) 

x n−q y q 

Proof: 

⎧ 

⎫ 

∞∑ 

∞∑ ⎨ ∞∑ ⎬ 

˜Z(ν) = z ne −i2πνn = 

x n−q y q 

⎩ 

⎭ e−i2πνn 

n=−∞ 

n=−∞ q=−∞ 

∞∑ ∞∑ 

= {(p, q) = (n − q, q)} = 

x py qe −i2πν(p+q) 

= 

∞∑ 

p=−∞ 

p=−∞ q=−∞ 

x pe −i2πνp 

∞ ∑ 

q=−∞ 

y qe −i2πνq = 

˜X(ν)Ỹ (ν) 



Convolution theorem for the Discrete-Time Fourier transform (periodic 

convolution in continuous frequency) 

Let 

Then 

Proof: 

z n = x ny n 

˜Z(ν) = ˜X(ν) ⊗ Ỹ (ν) = ∫ 1/2 

−1/2 

˜X(ν − α)Ỹ (α) dα 

{ } ∫ { 1/2 ∫ } 1/2 

z n = F −1 ˜X(ν) ⊗ Ỹ (ν) = 

˜X(ν − α)Ỹ (α) dα e i2πνn dν 

−1/2 −1/2 

= {(β, α) = (ν − α, α), dβ dα = dν dα} 

∫ 1/2 ∫ 1/2−α 

= 

˜X(β)Ỹ (α)ei2π(β+α)n dβ dα 

α=−1/2 β=−1/2−α 

∫ 1/2 

∫ 1/2 

= ˜X(β)e i2πβn dβ Ỹ (α)e i2παn dα = x ny n 

−1/2 

−1/2 



Convolution theorem for the Discrete Fourier Transform (periodic 

convolution in discrete time) 

Let 

Then 

Proof: 

N−1 ∑ 

˜z n = ˜x n ⊗ ỹ n = ˜x n−q ỹ q 

q=0 

˜Z k = ˜X k Ỹ k 

⎧ 

⎫ 

N−1 ∑ 

N−1 

˜Z k = ˜z ne −i 2π ∑ ⎨N−1 

∑ ⎬ 

N kn = 

˜x n−q ỹ q 

⎩ 

⎭ e−i 2π N kn 

n=0 

n=0 q=0 

= {(p, q) = (n − q, q)} = 

N−1 ∑ 

q=0 

N−1−q ∑ 

p=−q 

˜x pỹ qe −i 2π N k(p+q) 

N−1 ∑ 

N−1 

= ˜x pe −i 2π ∑ 

N kp ỹ qe −i 2π N kq = ˜X k Ỹ k 

p=0 

q=0 



Convolution theorem for the Discrete Fourier Transform (periodic 

convolution in discrete frequency) 

Let 

Then 

Proof: 

{ 1 

˜z n = DFT −1 ˜X k ⊗ 

N Ỹk 

˜z n = ˜x nỹ n 

˜Z k = 1 ˜X k ⊗ 

N Ỹk = 1 N−1 ∑ 

˜X k−q Ỹ q 

N 

q=0 

} 

= 1 N 

⎧ 

⎫ 

N−1 ∑ ⎨ N−1 

1 ∑ ⎬ 

˜X 

⎩ 

k−q Ỹ q 

N 

⎭ ei 2π N kn 

k=0 q=0 

= {(p, q) = (k − q, q)} = 1 N−1 ∑ N−1−q ∑ 

N 2 

= 1 N 

q=0 

N−1 ∑ 

p=0 

p=−q 

˜X pe i 2π N pn 1 N 

˜X p Ỹ qe i 2π N (p+q)n 

N−1 ∑ 

q=0 

Ỹ qe i 2π N qn = ˜x nỹ n 


Time-shift in the Fourier transform 


Proof: 

F{x(t − t 0 )} = e −iωt 0 

X(ω) 

∫ ∞ 

F{x(t − t 0 )} = x(t − t 0 )e −iωt dt = {u = t − t 0 , du = dt} 

∫ ∞ 

−∞ 

∫ ∞ 

= x(u)e −iω(u+t0) du = e −iωt 0 

x(u)e −iωu du = e −iωt 0 

X(ω) 

−∞ 

−∞ 

Frequency-shift in the Fourier transform 

F{x(t)e iω0t } = X(ω − ω 0 ) 

Proof: 

∫ ∞ 

∫ ∞ 

F{x(t)e iω0t } = x(t)e iω0t e −iωt dt = x(t)e −i(ω−ω0)t dt = X(ω − ω 0 ) 

−∞ 

−∞ 



Time-shift in the Discrete-Time Fourier transform 

F{x n−n0 } = e −i2πνn 0 ˜X(ν) 

Proof: 

∞∑ 

F{x n−n0 } = x n−n0 e −i2πνn = {p = n − n 0 } 

n=−∞ 

∞∑ 

= 

p=−∞ 

x pe −i2πν(p+n0) = e −i2πνn 0 

∞∑ 

p=−∞ 

x pe −i2πνp = e −i2πνn 0 ˜X(ν) 

Frequency-shift in the Discrete-Time Fourier transform 

F{x ne i2πν0n } = ˜X(ν − ν 0 ) 

Proof: 

F{x ne i2πν0n } = 

∞∑ 

n=−∞ 

x ne i2πν0n e −i2πνn = 

∞∑ 

n=−∞ 

x ne −i2π(ν−ν0)n = ˜X(ν−ν 0 ) 



Time-derivative in the Fourier transform 

F 

{ } ∂x 

= iωX(ω) 

∂t 

Proof: 

∂x 

∂t = ∂ ∫ 

1 ∞ 

X(ω)e iωt dω = 1 ∫ ∞ ∂ 

∂t 2π −∞ 

2π −∞ ∂t {X(ω)eiωt } dω 

= 1 ∫ ∞ 

iωX(ω)e iωt dω = F −1 {iωX(ω)} 

2π −∞ 

Frequency-derivative in the Fourier transform 

Proof: 

F{tx(t)} = i ∂X(ω) 

∂ω 

i ∂X(ω) 

∂ω 

= i ∂ ∫ ∞ 

∫ ∞ 

x(t)e −iωt ∂ 

dt = i 

∂ω −∞ 

−∞ ∂ω {x(t)e−iωt } dt 

∫ ∞ 

∫ ∞ 

= i (−it)x(t)e −iωt dt = tx(t)e −iωt dt = F{tx(t)} 

−∞ 

−∞ 



Example: Fourier transform of the Gaussian pulse 

{ √ 

F e −at2} π ω2 

= e− 4a 

a 

Proof: Notice that x(t) = e −at2 is a solution to 

∂x(t) 

∂t 

+ 2atx(t) = 0 

Take the Fourier transform and use F{ ∂x(t) } = iωX(ω) and F{tx(t)} = i ∂X(ω) 

∂t 

∂ω 

iωX(ω) + 2ai ∂X(ω) 

∂ω 

Parseval’s theorem 

= 0 ⇒ ∂X(ω) ( ) 1 

∂ω + 2 ωX(ω) = 0 ⇒ X(ω) = Ce − 4a 1 ω2 

4a 

∫ ∞ 

−∞ 

∫ ∞ 

x 2 (t) dt = e −2at2 dt = 1 

−∞ 

2π 

= {ω = 2at, dω = 2a dt} = C2 

2π 

∫ ∞ 

−∞ 

∫ ∞ 

−∞ 

∫ 

X 2 (ω) dω = C2 ∞ 

e − ω2 

2a dω 

2π −∞ 

e −2at2 2a dt ⇒ 1 = C2 

2π 2a ⇒ C = √ π 

a 



Important symmetry properties of the Fourier Transform 

{ |X(ω)| = |X(−ω)| 

x(t) real ⇔ X(ω) = X ∗ (−ω) ⇔ 

arg X(ω) = − arg X(−ω) 

Proof: 

∫ ∞ 

∫ ∞ 

X(ω) = X ∗ (−ω) ⇔ x(t)e −iωt dt = x ∗ (t)e −iωt dt 

−∞ 

−∞ 

⇔ x(t) = x ∗ (t) ⇔ x(t) real 

Proof: 

X(ω) real ⇔ x(t) = x ∗ (−t) 

x(t) = x ∗ (−t) ⇔ 1 ∫ ∞ 

X(ω)e iωt dω = 1 ∫ ∞ 

X ∗ (ω)e iωt dω 

2π −∞ 

2π −∞ 

⇔ X(ω) = X ∗ (ω) ⇔ X(ω) real 



Important symmetry properties of the Discrete Fourier Transform (DFT) 

˜x n real ⇔ ˜X k = ˜X ∗ −k = ˜X ∗ N−k ⇔ { | ˜Xk | = | ˜X −k | = | ˜X N−k | 

arg ˜X k = − arg ˜X −k = − arg ˜X N−k 

Proof: Similar as above. 

˜X k real ⇔ ˜x n = ˜x ∗ −n = ˜x∗ N−n

Signal Processing

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