Integral of Shifted Dirac Function
up vote
-1
down vote
favorite
How to integrate this function:
$$x[n] = frac{1}{2pi}int_{-pi}^{pi}delta(omega-omega_{0}) e^{jnomega}domega$$
Where:
$$delta[omega-omega_0] = begin{cases}1&n=omega_o\0&nneqomega_oend{cases}$$
The book says the answer is suppose to be:
$$x[n] = frac{1}{2pi}e^{jnomega_o} $$
Here's what i'm trying to do:
$$x[n] = frac{1}{2pi}int_{-pi}^{pi}delta(omega-omega_{0}) e^{jnomega}domega$$
$$x[n] = frac{1}{2pi} e^{jnomega_o} int_{-pi}^{pi}domega$$
$$x[n] = frac{1}{2pi} e^{jnomega_o} (pi - (-pi))$$
$$x[n] = frac{1}{2pi} e^{jnomega_o} (2pi)$$
$$x[n] = e^{jnomega_o} $$
integration dirac-delta
asked Nov 22 at 16:07
Bill Moore
1176
add a comment |
up vote
-1
down vote
favorite
How to integrate this function:
$$x[n] = frac{1}{2pi}int_{-pi}^{pi}delta(omega-omega_{0}) e^{jnomega}domega$$
Where:
$$delta[omega-omega_0] = begin{cases}1&n=omega_o\0&nneqomega_oend{cases}$$
The book says the answer is suppose to be:
$$x[n] = frac{1}{2pi}e^{jnomega_o} $$
Here's what i'm trying to do:
$$x[n] = frac{1}{2pi}int_{-pi}^{pi}delta(omega-omega_{0}) e^{jnomega}domega$$
$$x[n] = frac{1}{2pi} e^{jnomega_o} int_{-pi}^{pi}domega$$
$$x[n] = frac{1}{2pi} e^{jnomega_o} (pi - (-pi))$$
$$x[n] = frac{1}{2pi} e^{jnomega_o} (2pi)$$
$$x[n] = e^{jnomega_o} $$
integration dirac-delta
asked Nov 22 at 16:07
Bill Moore
1176
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
How to integrate this function:
$$x[n] = frac{1}{2pi}int_{-pi}^{pi}delta(omega-omega_{0}) e^{jnomega}domega$$
Where:
$$delta[omega-omega_0] = begin{cases}1&n=omega_o\0&nneqomega_oend{cases}$$
The book says the answer is suppose to be:
$$x[n] = frac{1}{2pi}e^{jnomega_o} $$
Here's what i'm trying to do:
$$x[n] = frac{1}{2pi}int_{-pi}^{pi}delta(omega-omega_{0}) e^{jnomega}domega$$
$$x[n] = frac{1}{2pi} e^{jnomega_o} int_{-pi}^{pi}domega$$
$$x[n] = frac{1}{2pi} e^{jnomega_o} (pi - (-pi))$$
$$x[n] = frac{1}{2pi} e^{jnomega_o} (2pi)$$
$$x[n] = e^{jnomega_o} $$
integration dirac-delta
asked Nov 22 at 16:07
Bill Moore
1176
How to integrate this function:
$$x[n] = frac{1}{2pi}int_{-pi}^{pi}delta(omega-omega_{0}) e^{jnomega}domega$$
Where:
$$delta[omega-omega_0] = begin{cases}1&n=omega_o\0&nneqomega_oend{cases}$$
The book says the answer is suppose to be:
$$x[n] = frac{1}{2pi}e^{jnomega_o} $$
Here's what i'm trying to do:
$$x[n] = frac{1}{2pi}int_{-pi}^{pi}delta(omega-omega_{0}) e^{jnomega}domega$$
$$x[n] = frac{1}{2pi} e^{jnomega_o} int_{-pi}^{pi}domega$$
$$x[n] = frac{1}{2pi} e^{jnomega_o} (pi - (-pi))$$
$$x[n] = frac{1}{2pi} e^{jnomega_o} (2pi)$$
$$x[n] = e^{jnomega_o} $$
integration dirac-delta
integration dirac-delta
asked Nov 22 at 16:07
Bill Moore
1176
asked Nov 22 at 16:07
Bill Moore
1176
asked Nov 22 at 16:07
Bill Moore
1176
asked Nov 22 at 16:07
Bill Moore
1176
asked Nov 22 at 16:07
Bill Moore
1176
1176
add a comment |
add a comment |
2 Answers
2
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oldest
votes
up vote
1
down vote
You've made a few mistakes, such as giving an $n$-dependent definition of $delta (omega-omega_0)$ instead of an $omega$-dependent one, and I'm sure your book wants you to work with Dirac deltas, whereas you've used a Kronecker delta. If $omega_0in [a,,b]$, $int_a^bdelta(omega-omega_0)f(omega)domega=f(omega_0)$ (otherwise the result is $0$); that defines the Dirac delta, as well as making the problem trivial. If the given formula were really meant to use a Kronecker delta, the integrand would be function-valued and vanish except at $omega=omega_0$, and the integral would be $0$. Note that the Dirac delta is a measure, not a true function.
answered Nov 22 at 16:14
J.G.
21.1k21933
That's a good point. I'm mixing up dirac with unit step function. its really: $$delta[omega-omega_0] = begin{cases}infty&n=omega_o\0&nneqomega_oend{cases}$$
– Bill Moore
Nov 22 at 16:28
Its a derivation for DTFT property... I guess its confusing because they use dirac delta in n-domain, and kronecker delta in the frequency domain. $$delta(omega-omega_0) = begin{cases}infty&n=omega_o\0&nneqomega_oend{cases}$$ $$delta[n-n_o] = begin{cases}1&n=n_o\0&nneqn_oend{cases}$$
– Bill Moore
Nov 22 at 16:37
sorry reverse that... dirac delta in frequence domain and kronecker delta in n-domain
– Bill Moore
Nov 22 at 16:43
add a comment |
up vote
1
down vote
Remember the translation property of the Dirac Delta. If $a < x_0 < b$
$$
int_{a}^b {rm d}x~ f(x) delta(x - x_0) = f(x_0)
$$
If you apply this to your problem you get
$$
frac{1}{2pi} int_{-pi}^pi {rm d}omega ~ delta (omega - omega_0) e^{j n omega} = frac{1}{2pi} e^{j n omega_0}
$$
answered Nov 22 at 16:15
caverac
12.8k21028
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
You've made a few mistakes, such as giving an $n$-dependent definition of $delta (omega-omega_0)$ instead of an $omega$-dependent one, and I'm sure your book wants you to work with Dirac deltas, whereas you've used a Kronecker delta. If $omega_0in [a,,b]$, $int_a^bdelta(omega-omega_0)f(omega)domega=f(omega_0)$ (otherwise the result is $0$); that defines the Dirac delta, as well as making the problem trivial. If the given formula were really meant to use a Kronecker delta, the integrand would be function-valued and vanish except at $omega=omega_0$, and the integral would be $0$. Note that the Dirac delta is a measure, not a true function.
answered Nov 22 at 16:14
J.G.
21.1k21933
That's a good point. I'm mixing up dirac with unit step function. its really: $$delta[omega-omega_0] = begin{cases}infty&n=omega_o\0&nneqomega_oend{cases}$$
– Bill Moore
Nov 22 at 16:28
Its a derivation for DTFT property... I guess its confusing because they use dirac delta in n-domain, and kronecker delta in the frequency domain. $$delta(omega-omega_0) = begin{cases}infty&n=omega_o\0&nneqomega_oend{cases}$$ $$delta[n-n_o] = begin{cases}1&n=n_o\0&nneqn_oend{cases}$$
– Bill Moore
Nov 22 at 16:37
sorry reverse that... dirac delta in frequence domain and kronecker delta in n-domain
– Bill Moore
Nov 22 at 16:43
add a comment |
up vote
1
down vote
You've made a few mistakes, such as giving an $n$-dependent definition of $delta (omega-omega_0)$ instead of an $omega$-dependent one, and I'm sure your book wants you to work with Dirac deltas, whereas you've used a Kronecker delta. If $omega_0in [a,,b]$, $int_a^bdelta(omega-omega_0)f(omega)domega=f(omega_0)$ (otherwise the result is $0$); that defines the Dirac delta, as well as making the problem trivial. If the given formula were really meant to use a Kronecker delta, the integrand would be function-valued and vanish except at $omega=omega_0$, and the integral would be $0$. Note that the Dirac delta is a measure, not a true function.
answered Nov 22 at 16:14
J.G.
21.1k21933
That's a good point. I'm mixing up dirac with unit step function. its really: $$delta[omega-omega_0] = begin{cases}infty&n=omega_o\0&nneqomega_oend{cases}$$
– Bill Moore
Nov 22 at 16:28
Its a derivation for DTFT property... I guess its confusing because they use dirac delta in n-domain, and kronecker delta in the frequency domain. $$delta(omega-omega_0) = begin{cases}infty&n=omega_o\0&nneqomega_oend{cases}$$ $$delta[n-n_o] = begin{cases}1&n=n_o\0&nneqn_oend{cases}$$
– Bill Moore
Nov 22 at 16:37
sorry reverse that... dirac delta in frequence domain and kronecker delta in n-domain
– Bill Moore
Nov 22 at 16:43
add a comment |
up vote
1
down vote
up vote
1
down vote
You've made a few mistakes, such as giving an $n$-dependent definition of $delta (omega-omega_0)$ instead of an $omega$-dependent one, and I'm sure your book wants you to work with Dirac deltas, whereas you've used a Kronecker delta. If $omega_0in [a,,b]$, $int_a^bdelta(omega-omega_0)f(omega)domega=f(omega_0)$ (otherwise the result is $0$); that defines the Dirac delta, as well as making the problem trivial. If the given formula were really meant to use a Kronecker delta, the integrand would be function-valued and vanish except at $omega=omega_0$, and the integral would be $0$. Note that the Dirac delta is a measure, not a true function.
answered Nov 22 at 16:14
J.G.
21.1k21933
You've made a few mistakes, such as giving an $n$-dependent definition of $delta (omega-omega_0)$ instead of an $omega$-dependent one, and I'm sure your book wants you to work with Dirac deltas, whereas you've used a Kronecker delta. If $omega_0in [a,,b]$, $int_a^bdelta(omega-omega_0)f(omega)domega=f(omega_0)$ (otherwise the result is $0$); that defines the Dirac delta, as well as making the problem trivial. If the given formula were really meant to use a Kronecker delta, the integrand would be function-valued and vanish except at $omega=omega_0$, and the integral would be $0$. Note that the Dirac delta is a measure, not a true function.
answered Nov 22 at 16:14
J.G.
21.1k21933
answered Nov 22 at 16:14
J.G.
21.1k21933
answered Nov 22 at 16:14
J.G.
21.1k21933
answered Nov 22 at 16:14
J.G.
21.1k21933
21.1k21933
That's a good point. I'm mixing up dirac with unit step function. its really: $$delta[omega-omega_0] = begin{cases}infty&n=omega_o\0&nneqomega_oend{cases}$$
– Bill Moore
Nov 22 at 16:28
Its a derivation for DTFT property... I guess its confusing because they use dirac delta in n-domain, and kronecker delta in the frequency domain. $$delta(omega-omega_0) = begin{cases}infty&n=omega_o\0&nneqomega_oend{cases}$$ $$delta[n-n_o] = begin{cases}1&n=n_o\0&nneqn_oend{cases}$$
– Bill Moore
Nov 22 at 16:37
sorry reverse that... dirac delta in frequence domain and kronecker delta in n-domain
– Bill Moore
Nov 22 at 16:43
add a comment |
That's a good point. I'm mixing up dirac with unit step function. its really: $$delta[omega-omega_0] = begin{cases}infty&n=omega_o\0&nneqomega_oend{cases}$$
– Bill Moore
Nov 22 at 16:28
Its a derivation for DTFT property... I guess its confusing because they use dirac delta in n-domain, and kronecker delta in the frequency domain. $$delta(omega-omega_0) = begin{cases}infty&n=omega_o\0&nneqomega_oend{cases}$$ $$delta[n-n_o] = begin{cases}1&n=n_o\0&nneqn_oend{cases}$$
– Bill Moore
Nov 22 at 16:37
sorry reverse that... dirac delta in frequence domain and kronecker delta in n-domain
– Bill Moore
Nov 22 at 16:43
That's a good point. I'm mixing up dirac with unit step function. its really: $$delta[omega-omega_0] = begin{cases}infty&n=omega_o\0&nneqomega_oend{cases}$$
– Bill Moore
Nov 22 at 16:28
That's a good point. I'm mixing up dirac with unit step function. its really: $$delta[omega-omega_0] = begin{cases}infty&n=omega_o\0&nneqomega_oend{cases}$$
– Bill Moore
Nov 22 at 16:28
Its a derivation for DTFT property... I guess its confusing because they use dirac delta in n-domain, and kronecker delta in the frequency domain. $$delta(omega-omega_0) = begin{cases}infty&n=omega_o\0&nneqomega_oend{cases}$$ $$delta[n-n_o] = begin{cases}1&n=n_o\0&nneqn_oend{cases}$$
– Bill Moore
Nov 22 at 16:37
Its a derivation for DTFT property... I guess its confusing because they use dirac delta in n-domain, and kronecker delta in the frequency domain. $$delta(omega-omega_0) = begin{cases}infty&n=omega_o\0&nneqomega_oend{cases}$$ $$delta[n-n_o] = begin{cases}1&n=n_o\0&nneqn_oend{cases}$$
– Bill Moore
Nov 22 at 16:37
sorry reverse that... dirac delta in frequence domain and kronecker delta in n-domain
– Bill Moore
Nov 22 at 16:43
sorry reverse that... dirac delta in frequence domain and kronecker delta in n-domain
– Bill Moore
Nov 22 at 16:43
add a comment |
up vote
1
down vote
Remember the translation property of the Dirac Delta. If $a < x_0 < b$
$$
int_{a}^b {rm d}x~ f(x) delta(x - x_0) = f(x_0)
$$
If you apply this to your problem you get
$$
frac{1}{2pi} int_{-pi}^pi {rm d}omega ~ delta (omega - omega_0) e^{j n omega} = frac{1}{2pi} e^{j n omega_0}
$$
answered Nov 22 at 16:15
caverac
12.8k21028
add a comment |
up vote
1
down vote
Remember the translation property of the Dirac Delta. If $a < x_0 < b$
$$
int_{a}^b {rm d}x~ f(x) delta(x - x_0) = f(x_0)
$$
If you apply this to your problem you get
$$
frac{1}{2pi} int_{-pi}^pi {rm d}omega ~ delta (omega - omega_0) e^{j n omega} = frac{1}{2pi} e^{j n omega_0}
$$
answered Nov 22 at 16:15
caverac
12.8k21028
add a comment |
up vote
1
down vote
up vote
1
down vote
Remember the translation property of the Dirac Delta. If $a < x_0 < b$
$$
int_{a}^b {rm d}x~ f(x) delta(x - x_0) = f(x_0)
$$
If you apply this to your problem you get
$$
frac{1}{2pi} int_{-pi}^pi {rm d}omega ~ delta (omega - omega_0) e^{j n omega} = frac{1}{2pi} e^{j n omega_0}
$$
answered Nov 22 at 16:15
caverac
12.8k21028
Remember the translation property of the Dirac Delta. If $a < x_0 < b$
$$
int_{a}^b {rm d}x~ f(x) delta(x - x_0) = f(x_0)
$$
If you apply this to your problem you get
$$
frac{1}{2pi} int_{-pi}^pi {rm d}omega ~ delta (omega - omega_0) e^{j n omega} = frac{1}{2pi} e^{j n omega_0}
$$
answered Nov 22 at 16:15
caverac
12.8k21028
answered Nov 22 at 16:15
caverac
12.8k21028
answered Nov 22 at 16:15
caverac
12.8k21028
answered Nov 22 at 16:15
caverac
12.8k21028
12.8k21028
add a comment |
add a comment |
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