Nyquist rate of a function convolved with itself
From Schaum's Outline, Digital Signal Processing, Second Edition, 2012, page 155:
If the Nyquist rate for x(t) is:
$$Omega_s$$
what is the Nyquist rate for:
$$ y(t) = x^{2}(t)$$
Solution:
$$y(t)= x(t)x(t)$$
Multiplication in time domain equals convulsion in frequency domain. Therefore:
$$Y(jOmega) = frac{1}{2pi}X(jOmega) * X(jOmega)$$
Then the book say because of this relationship the highest frequency in y(t) will be twice the frequency of x(t).
My question is:
How can you make this inference from the convulution of X with itself in the frequency domain?
A few definitions:
$$ F(jOmega) = int_{-infty}^{infty}f(t)e^{-jOmega t} dt$$
fourier-transform
asked Nov 23 at 19:05
Bill Moore
1176
add a comment |
From Schaum's Outline, Digital Signal Processing, Second Edition, 2012, page 155:
If the Nyquist rate for x(t) is:
$$Omega_s$$
what is the Nyquist rate for:
$$ y(t) = x^{2}(t)$$
Solution:
$$y(t)= x(t)x(t)$$
Multiplication in time domain equals convulsion in frequency domain. Therefore:
$$Y(jOmega) = frac{1}{2pi}X(jOmega) * X(jOmega)$$
Then the book say because of this relationship the highest frequency in y(t) will be twice the frequency of x(t).
My question is:
How can you make this inference from the convulution of X with itself in the frequency domain?
A few definitions:
$$ F(jOmega) = int_{-infty}^{infty}f(t)e^{-jOmega t} dt$$
fourier-transform
asked Nov 23 at 19:05
Bill Moore
1176
see my answer below. Does it answer your question?
– kodlu
Nov 24 at 22:37
add a comment |
From Schaum's Outline, Digital Signal Processing, Second Edition, 2012, page 155:
If the Nyquist rate for x(t) is:
$$Omega_s$$
what is the Nyquist rate for:
$$ y(t) = x^{2}(t)$$
Solution:
$$y(t)= x(t)x(t)$$
Multiplication in time domain equals convulsion in frequency domain. Therefore:
$$Y(jOmega) = frac{1}{2pi}X(jOmega) * X(jOmega)$$
Then the book say because of this relationship the highest frequency in y(t) will be twice the frequency of x(t).
My question is:
How can you make this inference from the convulution of X with itself in the frequency domain?
A few definitions:
$$ F(jOmega) = int_{-infty}^{infty}f(t)e^{-jOmega t} dt$$
fourier-transform
asked Nov 23 at 19:05
Bill Moore
1176
From Schaum's Outline, Digital Signal Processing, Second Edition, 2012, page 155:
If the Nyquist rate for x(t) is:
$$Omega_s$$
what is the Nyquist rate for:
$$ y(t) = x^{2}(t)$$
Solution:
$$y(t)= x(t)x(t)$$
Multiplication in time domain equals convulsion in frequency domain. Therefore:
$$Y(jOmega) = frac{1}{2pi}X(jOmega) * X(jOmega)$$
Then the book say because of this relationship the highest frequency in y(t) will be twice the frequency of x(t).
My question is:
How can you make this inference from the convulution of X with itself in the frequency domain?
A few definitions:
$$ F(jOmega) = int_{-infty}^{infty}f(t)e^{-jOmega t} dt$$
fourier-transform
fourier-transform
asked Nov 23 at 19:05
Bill Moore
1176
asked Nov 23 at 19:05
Bill Moore
1176
edited Nov 23 at 19:27
asked Nov 23 at 19:05
Bill Moore
1176
asked Nov 23 at 19:05
Bill Moore
1176
asked Nov 23 at 19:05
Bill Moore
1176
1176
see my answer below. Does it answer your question?
– kodlu
Nov 24 at 22:37
add a comment |
see my answer below. Does it answer your question?
– kodlu
Nov 24 at 22:37
see my answer below. Does it answer your question?
– kodlu
Nov 24 at 22:37
see my answer below. Does it answer your question?
– kodlu
Nov 24 at 22:37
add a comment |
1 Answer
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In the convolution integral
$$ Y(j Omega)=int_{-infty}^{infty}X(jOmega') X(j(Omega-Omega')) ,dOmega',$$
both arguments are nonzero and there is a nonzero contribution to the integral when $Omegain (2Omega_0-epsilon,Omega_0]$$Omega'in (Omega_0-epsilon,Omega_0],$ where $Omega_0$ is the Nyquist frequency of $X(jOmega).$
answered Nov 23 at 21:00
kodlu
3,380716
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
In the convolution integral
$$ Y(j Omega)=int_{-infty}^{infty}X(jOmega') X(j(Omega-Omega')) ,dOmega',$$
both arguments are nonzero and there is a nonzero contribution to the integral when $Omegain (2Omega_0-epsilon,Omega_0]$$Omega'in (Omega_0-epsilon,Omega_0],$ where $Omega_0$ is the Nyquist frequency of $X(jOmega).$
answered Nov 23 at 21:00
kodlu
3,380716
add a comment |
In the convolution integral
$$ Y(j Omega)=int_{-infty}^{infty}X(jOmega') X(j(Omega-Omega')) ,dOmega',$$
both arguments are nonzero and there is a nonzero contribution to the integral when $Omegain (2Omega_0-epsilon,Omega_0]$$Omega'in (Omega_0-epsilon,Omega_0],$ where $Omega_0$ is the Nyquist frequency of $X(jOmega).$
answered Nov 23 at 21:00
kodlu
3,380716
add a comment |
In the convolution integral
$$ Y(j Omega)=int_{-infty}^{infty}X(jOmega') X(j(Omega-Omega')) ,dOmega',$$
both arguments are nonzero and there is a nonzero contribution to the integral when $Omegain (2Omega_0-epsilon,Omega_0]$$Omega'in (Omega_0-epsilon,Omega_0],$ where $Omega_0$ is the Nyquist frequency of $X(jOmega).$
answered Nov 23 at 21:00
kodlu
3,380716
In the convolution integral
$$ Y(j Omega)=int_{-infty}^{infty}X(jOmega') X(j(Omega-Omega')) ,dOmega',$$
both arguments are nonzero and there is a nonzero contribution to the integral when $Omegain (2Omega_0-epsilon,Omega_0]$$Omega'in (Omega_0-epsilon,Omega_0],$ where $Omega_0$ is the Nyquist frequency of $X(jOmega).$
answered Nov 23 at 21:00
kodlu
3,380716
answered Nov 23 at 21:00
kodlu
3,380716
answered Nov 23 at 21:00
kodlu
3,380716
answered Nov 23 at 21:00
kodlu
3,380716
3,380716
add a comment |
add a comment |
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see my answer below. Does it answer your question?
– kodlu
Nov 24 at 22:37