Integral of Shifted Dirac Function











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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







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    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







    share|cite|improve this question
























      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







      share|cite|improve this question













      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked Nov 22 at 16:07









      Bill Moore

      1176




      1176






















          2 Answers
          2






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          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.






          share|cite|improve this answer





















          • 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


















          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



          $$
          frac{1}{2pi} int_{-pi}^pi {rm d}omega ~ delta (omega - omega_0) e^{j n omega} = frac{1}{2pi} e^{j n omega_0}
          $$






          share|cite|improve this answer





















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

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            active

            oldest

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            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.






            share|cite|improve this answer





















            • 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















            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.






            share|cite|improve this answer





















            • 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













            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.






            share|cite|improve this answer












            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.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            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


















            • 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










            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



            $$
            frac{1}{2pi} int_{-pi}^pi {rm d}omega ~ delta (omega - omega_0) e^{j n omega} = frac{1}{2pi} e^{j n omega_0}
            $$






            share|cite|improve this answer

























              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



              $$
              frac{1}{2pi} int_{-pi}^pi {rm d}omega ~ delta (omega - omega_0) e^{j n omega} = frac{1}{2pi} e^{j n omega_0}
              $$






              share|cite|improve this answer























                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



                $$
                frac{1}{2pi} int_{-pi}^pi {rm d}omega ~ delta (omega - omega_0) e^{j n omega} = frac{1}{2pi} e^{j n omega_0}
                $$






                share|cite|improve this answer












                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



                $$
                frac{1}{2pi} int_{-pi}^pi {rm d}omega ~ delta (omega - omega_0) e^{j n omega} = frac{1}{2pi} e^{j n omega_0}
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 22 at 16:15









                caverac

                12.8k21028




                12.8k21028






























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