MHB Where Does the Convolution Theorem Come From?

[bugatti79]

Hi Folks,

I have an article which explains the modulation of 2 signals given by

X_1(f) e^{-j 2 \pi f t_i} and X_2(f) e^{-j 2 \pi f t_i}

The only difference between the 2 signals is a time delay, however i don't see a phase difference in either expression

It states the convolution of these 2 signals in the frequency domain is given as

X_1 (f) * X_2(f) e^{-j 2 \pi f t_i}

Where does this come from? I expected something of the form, from basic rules,

A e^{iat} B e^{iat}= A B e^{i2at}

 

[I like Serena]

Hi Folks,

I have an article which explains the modulation of 2 signals given by

X_1(f) e^{-j 2 \pi f t_i} and X_2(f) e^{-j 2 \pi f t_i}

The only difference between the 2 signals is a time delay, however i don't see a phase difference in either expression

It states the convolution of these 2 signals in the frequency domain is given as

X_1 (f) * X_2(f) e^{-j 2 \pi f t_i}

Where does this come from? I expected something of the form, from basic rules,

A e^{iat} B e^{iat}= A B e^{i2at}

Hi bugatti79!

It looks to me as if your formulas are somewhat incomplete.

It looks like $X_1(f)$ is the frequency spectrum of the first signal.
If I assume that to be true, then the corresponding signal would be:
$$x_1(t_i) = \int_0^\infty X_1(f)e^{-j 2 \pi f t_i} df$$
So it seems as if the integral was left out.The convolution (from Fourier theory) of $x_1(t_i)$ and $x_2(t_i)$ would be:
$$x_1(t_i) * x_2(t_i) = \mathscr{F}^{-1}(X_1(f)\cdot X_2(f)) = \int_0^\infty X_1(f)\cdot X_2(f)e^{-j 2 \pi f t_i} df$$

 

[bugatti79]

Hi bugatti79!

It looks to me as if your formulas are somewhat incomplete.

It looks like $X_1(f)$ is the frequency spectrum of the first signal.
If I assume that to be true, then the corresponding signal would be:
$$x_1(t_i) = \int_0^\infty X_1(f)e^{-j 2 \pi f t_i} df$$
So it seems as if the integral was left out.The convolution (from Fourier theory) of $x_1(t_i)$ and $x_2(t_i)$ would be:
$$x_1(t_i) * x_2(t_i) = \mathscr{F}^{-1}(X_1(f)\cdot X_2(f)) = \int_0^\infty X_1(f)\cdot X_2(f)e^{-j 2 \pi f t_i} df$$

Hi there!,

Thanks for the reply. The paper doesn't seem to evaluate the integral... i have attached the extract for your convenience. Its just one page. I am puzzled how he arrived at eqn 9...

 

[I like Serena]

Hi there!,

Thanks for the reply. The paper doesn't seem to evaluate the integral... i have attached the extract for your convenience. Its just one page. I am puzzled how he arrived at eqn 9...

Ah okay.

Suppose $\mathscr F(x_1(t)) = X_1(f)$.
And suppose $x_i(t)$ is the same signal shifted in time by $t_i$.
So $x_i(t) = x_1(t-t_i)$.

Then:
\begin{aligned}X_i(f) &= \int_{-\infty}^\infty x_i(t)e^{-j2\pi f t}dt \\
&=\int_{-\infty}^\infty x_1(t-t_i)e^{-j2\pi f t}dt \\
&=e^{-j2\pi f t_i}\int_{-\infty}^\infty x_1(t-t_i)e^{-j2\pi f (t-t_i)}d(t-t_i)\\
&=e^{-j2\pi f t_i}\int_{-\infty}^\infty x_1(\tau)e^{-j2\pi f \tau}d\tau \\
&=X_1(f)e^{-j2\pi f t_i}
\end{aligned}

Or in short:
$$X_i(f) = X_1(f) e^{-j2\pi f t}$$
That is, a shift in time in the signal $x_1(t)$ results in a factor $e^{-j2\pi f t}$ in the frequency spectrum.

 

[bugatti79]

Ah okay.

Suppose $\mathscr F(x_1(t)) = X_1(f)$.
And suppose $x_i(t)$ is the same signal shifted in time by $t_i$.
So $x_i(t) = x_1(t-t_i)$.

Then:
\begin{aligned}X_i(f) &= \int_{-\infty}^\infty x_i(t)e^{-j2\pi f t}dt \\
&=\int_{-\infty}^\infty x_1(t-t_i)e^{-j2\pi f t}dt \\
&=e^{-j2\pi f t_i}\int_{-\infty}^\infty x_1(t-t_i)e^{-j2\pi f (t-t_i)}d(t-t_i)\\
&=e^{-j2\pi f t_i}\int_{-\infty}^\infty x_1(\tau)e^{-j2\pi f \tau}d\tau \\
&=X_1(f)e^{-j2\pi f t_i}
\end{aligned}

Or in short:
$$X_i(f) = X_1(f) e^{-j2\pi f t}$$
That is, a shift in time in the signal $x_1(t)$ results in a factor $e^{-j2\pi f t}$ in the frequency spectrum.

Hi,

That is clear to me now however, the equation before (9), I am not sure how the exp term was handled. It seems only one of the exp terms was brought forward instead of the 2 exp terms...ie it seems

A e^{iat} B e^{iat}= A B e^{iat}

I was expecting it to be

A e^{iat} B e^{iat}= A B e^{2iat}...?

 

[I like Serena]

Hi,

That is clear to me now however, the equation before (9), I am not sure how the exp term was handled. It seems only one of the exp terms was brought forward instead of the 2 exp terms...ie it seems

A e^{iat} B e^{iat}= A B e^{iat}

I was expecting it to be

A e^{iat} B e^{iat}= A B e^{2iat}...?

That's a nice conundrum! ;)

But it is correct, since:
$$
\left(X(f)e^{-j2\pi ft_i}\right) * \left(Y(f)e^{-j2\pi ft_i}\right)
=\mathscr F\Big(x(t-t_i) \cdot y(t-t_i)\Big)
=\mathscr F\Big((x \cdot y)(t-t_i)\Big)
=\Big(X(f) * Y(f)\Big) e^{-j2\pi ft_i}
$$

 

[bugatti79]

Hi,

That is clear to me now however, the equation before (9), I am not sure how the exp term was handled. It seems only one of the exp terms was brought forward instead of the 2 exp terms...ie it seems

A e^{iat} B e^{iat}= A B e^{iat}
I was expecting it to be

A e^{iat} B e^{iat}= A B e^{2iat}...?

Hi,

Sorry if i seem ignorant on the matter, but how do you mean a conundrum?

 

[I like Serena]

Hi,

Sorry if i seem ignorant on the matter, but how do you mean a conundrum?

I mean that it's a nice puzzle showing an unexpected property of convolution.

What you expected would be correct if we'd be talking about multiplication. But we're not - it is about convolution. which behaves subtly different, as I've shown in my previous post.

 

[bugatti79]

I mean that it's a nice puzzle showing an unexpected property of convolution.

What you expected would be correct if we'd be talking about multiplication. But we're not - it is about convolution. which behaves subtly different, as I've shown in my previous post.

Ah...now i see the light! Thank you!

=e^{-j2\pi f t_i}\int_{-\infty}^\infty x_1(t-t_i)e^{-j2\pi f (t-t_i)}d(t-t_i)

Can you point to an online source that shows this kind of integration...ie d(x-y)

\int f(x) d(x-y)

Thanks

 

[bugatti79]

That's a nice conundrum! ;)

But it is correct, since:
$$
\left(X(f)e^{-j2\pi ft_i}\right) * \left(Y(f)e^{-j2\pi ft_i}\right)
=\mathscr F\Big(x(t-t_i) \cdot y(t-t_i)\Big)
=\mathscr F\Big((x \cdot y)(t-t_i)\Big)
=\Big(X(f) * Y(f)\Big) e^{-j2\pi ft_i}
$$

Ok, i guess you have used something along the fact that \frac{1}{\sqrt{2 \pi}} F(w)*G(w)= \mathscr F[(ft) g(t)[...?

Can we use the convolution theorem itself $$(f*g)(w)=\int_{0}^w f(\tau) g(w-\tau) d\tau$$

to show
\left(X(f)e^{-j2\pi ft_i}\right) * \left(Y(f)e^{-j2\pi ft_i}\right)=\Big(X(f) * Y(f)\Big) e^{-j2\pi ft_i}...?

Thanks

 

[I like Serena]

Ok, i guess you have used something along the fact that \frac{1}{\sqrt{2 \pi}} F(w)*G(w)= \mathscr F[(ft) g(t)[...?

Yes, but without the constant factor.
It's the Convolution Theorem, which you can find on wiki.

Can we use the convolution theorem itself $$(f*g)(w)=\int_{0}^w f(\tau) g(w-\tau) d\tau$$

to show

\left(X(f)e^{-j2\pi ft_i}\right) * \left(Y(f)e^{-j2\pi ft_i}\right)=\Big(X(f) * Y(f)\Big) e^{-j2\pi ft_i}...?

I was using the "convolution theorem". Presumably you meant the "definition of convolution", but what you're asking is to prove that the convolution theorem is true.
You can find that proof on the wiki page I've just mentioned.

 

[bugatti79]

Yes, but without the constant factor.
It's the Convolution Theorem, which you can find on wiki.
I was using the "convolution theorem". Presumably you meant the "definition of convolution", but what you're asking is to prove that the convolution theorem is true.
You can find that proof on the wiki page I've just mentioned.

Great, thanks!