using Fourier Transforms to solve the question.
$begingroup$
I am given a question of Fourier Transform:
$$ e^{2(t-1)}u(t-1) $$
My teacher solved it by using the formula which I couldn't understand so I tried to apply the properties on it.
Now I have solved it by the following method:
$$ e^{2(t)}u(t) rightarrow frac{1}{2+jomega} $$
Now we know that:
$$delta(t-t_0) rightarrow e^{-jomega t_0} $$
So I used the above property on $u(t-1)$ and got the following answer which is same as my teacher got, which is:
$$ frac{e^{-jomega}}{2+jomega} $$
Is my method correct?
fourier-transform
asked Dec 25 '18 at 17:39
Ahmad QayyumAhmad Qayyum
677
$endgroup$
add a comment |
$begingroup$
I am given a question of Fourier Transform:
$$ e^{2(t-1)}u(t-1) $$
My teacher solved it by using the formula which I couldn't understand so I tried to apply the properties on it.
Now I have solved it by the following method:
$$ e^{2(t)}u(t) rightarrow frac{1}{2+jomega} $$
Now we know that:
$$delta(t-t_0) rightarrow e^{-jomega t_0} $$
So I used the above property on $u(t-1)$ and got the following answer which is same as my teacher got, which is:
$$ frac{e^{-jomega}}{2+jomega} $$
Is my method correct?
fourier-transform
asked Dec 25 '18 at 17:39
Ahmad QayyumAhmad Qayyum
677
$endgroup$
add a comment |
$begingroup$
I am given a question of Fourier Transform:
$$ e^{2(t-1)}u(t-1) $$
My teacher solved it by using the formula which I couldn't understand so I tried to apply the properties on it.
Now I have solved it by the following method:
$$ e^{2(t)}u(t) rightarrow frac{1}{2+jomega} $$
Now we know that:
$$delta(t-t_0) rightarrow e^{-jomega t_0} $$
So I used the above property on $u(t-1)$ and got the following answer which is same as my teacher got, which is:
$$ frac{e^{-jomega}}{2+jomega} $$
Is my method correct?
fourier-transform
asked Dec 25 '18 at 17:39
Ahmad QayyumAhmad Qayyum
677
$endgroup$
I am given a question of Fourier Transform:
$$ e^{2(t-1)}u(t-1) $$
My teacher solved it by using the formula which I couldn't understand so I tried to apply the properties on it.
Now I have solved it by the following method:
$$ e^{2(t)}u(t) rightarrow frac{1}{2+jomega} $$
Now we know that:
$$delta(t-t_0) rightarrow e^{-jomega t_0} $$
So I used the above property on $u(t-1)$ and got the following answer which is same as my teacher got, which is:
$$ frac{e^{-jomega}}{2+jomega} $$
Is my method correct?
fourier-transform
fourier-transform
asked Dec 25 '18 at 17:39
Ahmad QayyumAhmad Qayyum
677
asked Dec 25 '18 at 17:39
Ahmad QayyumAhmad Qayyum
677
asked Dec 25 '18 at 17:39
Ahmad QayyumAhmad Qayyum
677
asked Dec 25 '18 at 17:39
Ahmad QayyumAhmad Qayyum
677
asked Dec 25 '18 at 17:39
Ahmad QayyumAhmad Qayyum
677
677
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Assuming that your definition of the Fourier transform is
$$
hat f(omega) = int_{mathbb R} f(t)e^{jomega t} mathsf dt,
$$
then yes, your answer is correct. We can use a change of variables $s=t+1$ to compute
begin{align}
hat f(omega) &= int_1^infty e^{-2(t+1)}e^{jomega t} mathsf dt\
&= e^{-jomega}int_0^infty e^{-2s}e^{jomega s} mathsf ds\
&= e^{-jomega}cdotfrac{1}{2+jomega}.
end{align}
answered Dec 25 '18 at 22:26
Math1000Math1000
19.3k31745
$endgroup$
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
$begingroup$
Assuming that your definition of the Fourier transform is
$$
hat f(omega) = int_{mathbb R} f(t)e^{jomega t} mathsf dt,
$$
then yes, your answer is correct. We can use a change of variables $s=t+1$ to compute
begin{align}
hat f(omega) &= int_1^infty e^{-2(t+1)}e^{jomega t} mathsf dt\
&= e^{-jomega}int_0^infty e^{-2s}e^{jomega s} mathsf ds\
&= e^{-jomega}cdotfrac{1}{2+jomega}.
end{align}
answered Dec 25 '18 at 22:26
Math1000Math1000
19.3k31745
$endgroup$
add a comment |
$begingroup$
Assuming that your definition of the Fourier transform is
$$
hat f(omega) = int_{mathbb R} f(t)e^{jomega t} mathsf dt,
$$
then yes, your answer is correct. We can use a change of variables $s=t+1$ to compute
begin{align}
hat f(omega) &= int_1^infty e^{-2(t+1)}e^{jomega t} mathsf dt\
&= e^{-jomega}int_0^infty e^{-2s}e^{jomega s} mathsf ds\
&= e^{-jomega}cdotfrac{1}{2+jomega}.
end{align}
answered Dec 25 '18 at 22:26
Math1000Math1000
19.3k31745
$endgroup$
add a comment |
$begingroup$
Assuming that your definition of the Fourier transform is
$$
hat f(omega) = int_{mathbb R} f(t)e^{jomega t} mathsf dt,
$$
then yes, your answer is correct. We can use a change of variables $s=t+1$ to compute
begin{align}
hat f(omega) &= int_1^infty e^{-2(t+1)}e^{jomega t} mathsf dt\
&= e^{-jomega}int_0^infty e^{-2s}e^{jomega s} mathsf ds\
&= e^{-jomega}cdotfrac{1}{2+jomega}.
end{align}
answered Dec 25 '18 at 22:26
Math1000Math1000
19.3k31745
$endgroup$
Assuming that your definition of the Fourier transform is
$$
hat f(omega) = int_{mathbb R} f(t)e^{jomega t} mathsf dt,
$$
then yes, your answer is correct. We can use a change of variables $s=t+1$ to compute
begin{align}
hat f(omega) &= int_1^infty e^{-2(t+1)}e^{jomega t} mathsf dt\
&= e^{-jomega}int_0^infty e^{-2s}e^{jomega s} mathsf ds\
&= e^{-jomega}cdotfrac{1}{2+jomega}.
end{align}
answered Dec 25 '18 at 22:26
Math1000Math1000
19.3k31745
answered Dec 25 '18 at 22:26
Math1000Math1000
19.3k31745
answered Dec 25 '18 at 22:26
Math1000Math1000
19.3k31745
answered Dec 25 '18 at 22:26
Math1000Math1000
19.3k31745
19.3k31745
add a comment |
add a comment |
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