Convolution properteis and the imaginary unit

In summary, the given conversation discusses finding the Fourier transform of x(t)=sin(πt)sin(50πt) using the convolution operator. The resulting solution involves manipulating the signs and coefficients in the expression, which may have been a result of a typo made by the professor. The use of j2=-1 may have been applied to both factors instead of just the first. Additionally, the concept of constants coming outside integrals and only multiplying one factor at a time is also discussed.
  • #1
atrus_ovis
101
0
finding the FT of x(t)=sin(πt) sin(50πt) :

( '*' is the convolution operator)

its FT X(Ω)=(1/2π) F(sin(πt)) * F(sin(50πt))

= (1/2π) jπ(δ(Ω+π)-δ(Ω-π)) * (jπ (δ(Ω+π)-δ(Ω-π)) ) (a)

from my professor's solution it next goes:

= (π/2) (-δ(Ω+π)+δ(Ω-π)) * ((-δ(Ω+π)+δ(Ω-π)) ) (b)

1)How did the signs change?
2)Where did the j's go? ( i suspect 1,2 are related =P )

also,
3) doesn't the 1/π cancel out the π coefficients in each quantity to be convoluted?
from wiki:
a (f * g) = (a f) * g = f * (a g)

why is that? i mean, why isn't it:
a (f * g) = (a f) *(a g)

?
 
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  • #2
atrus_ovis said:
finding the FT of x(t)=sin(πt) sin(50πt) :

( '*' is the convolution operator)

its FT X(Ω)=(1/2π) F(sin(πt)) * F(sin(50πt))

= (1/2π) jπ(δ(Ω+π)-δ(Ω-π)) * (jπ (δ(Ω+π)-δ(Ω-π)) ) (a)

from my professor's solution it next goes:

= (π/2) (-δ(Ω+π)+δ(Ω-π)) * ((-δ(Ω+π)+δ(Ω-π)) ) (b)

1)How did the signs change?
2)Where did the j's go? ( i suspect 1,2 are related =P )
I think he is just using j2=-1 and inadvertently applied it to both factors instead of just the first.
also,
3) doesn't the 1/π cancel out the π coefficients in each quantity to be convoluted?
from wiki:
a (f * g) = (a f) * g = f * (a g)

why is that? i mean, why isn't it:
a (f * g) = (a f) *(a g)

?
Because constants come outside the integrals. a multiplies only one factor at a time.
 
  • #3
I think he is just using j2=-1 and inadvertently applied it to both factors instead of just the first.
But where does the 2nd j come from? :rolleyes:
 
  • #4
atrus_ovis said:
= (1/2π) jπ(δ(Ω+π)-δ(Ω-π)) * (jπ (δ(Ω+π)-δ(Ω-π)) ) (a)

atrus_ovis said:
But where does the 2nd j come from? :rolleyes:

Right there. I see two of them.
 
  • #5
Just a sec,i'm confused.
Because constants come outside the integrals. a multiplies only one factor at a time.
What integrals?

Also, the sign is inverted at both the operands of the convolution.
We had Aj*jB , we should have Aj*j*B = -A*B or A*-B , but it says -A*-B
 
  • #6
LCKurtz said:
I think he is just using j2=-1 and inadvertently applied it to both factors instead of just the first.

atrus_ovis said:
Just a sec,i'm confused.


Also, the sign is inverted at both the operands of the convolution.
We had Aj*jB , we should have Aj*j*B = -A*B or A*-B , but it says -A*-B

I agree; that's what I said in the first place above -- I think he made a typo.
 
  • #7
Oh, sorry.
Okay,thanks.
 

FAQ: Convolution properteis and the imaginary unit

1. What is convolution?

Convolution is a mathematical operation that combines two functions to produce a third function. It is used to describe the relationship between input and output signals in a system.

2. What are the properties of convolution?

The properties of convolution include commutativity, associativity, distributivity, and time shifting. Commutativity means that the order of the functions does not affect the result of the convolution. Associativity means that the order in which multiple convolutions are performed does not affect the final result. Distributivity means that convolution distributes over addition. Time shifting refers to the shifting of the input signal in time, which results in a corresponding shift in the output signal.

3. What is the imaginary unit in convolution?

The imaginary unit in convolution is represented by the letter j, and it is defined as the square root of -1. It is used to represent the imaginary part of a complex number in mathematics.

4. How is the imaginary unit used in convolution?

The imaginary unit is used in convolution to represent the phase shift of the input signal. It is multiplied by the frequency of the input signal to calculate the phase shift. This allows for the representation of complex functions and signals in convolution.

5. What are some real-world applications of convolution?

Convolution is widely used in signal processing and image processing applications. It is also used in areas such as physics, engineering, and statistics to describe the relationship between input and output signals. Some specific examples include noise reduction in audio signals, image filtering and edge detection, and frequency analysis of signals in communication systems.

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