What does it mean for a function to have frequencies?

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In a lecture, my professor mentioned that $cos$ has two frequencies. I see that using the inverse Euler's formula we can express $cos$ as a some linear combination of complex exponentials, each with a $omega$ term. But, I don't really have an intuitive understanding of what it means for a function to "have" a frequency. What other functions have a frequency, and why?







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    up vote
    0
    down vote

    favorite












    In a lecture, my professor mentioned that $cos$ has two frequencies. I see that using the inverse Euler's formula we can express $cos$ as a some linear combination of complex exponentials, each with a $omega$ term. But, I don't really have an intuitive understanding of what it means for a function to "have" a frequency. What other functions have a frequency, and why?







    share|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      In a lecture, my professor mentioned that $cos$ has two frequencies. I see that using the inverse Euler's formula we can express $cos$ as a some linear combination of complex exponentials, each with a $omega$ term. But, I don't really have an intuitive understanding of what it means for a function to "have" a frequency. What other functions have a frequency, and why?







      share|improve this question














      In a lecture, my professor mentioned that $cos$ has two frequencies. I see that using the inverse Euler's formula we can express $cos$ as a some linear combination of complex exponentials, each with a $omega$ term. But, I don't really have an intuitive understanding of what it means for a function to "have" a frequency. What other functions have a frequency, and why?









      share|improve this question













      share|improve this question




      share|improve this question








      edited Sep 3 at 7:47









      Laurent Duval

      14.8k31955




      14.8k31955










      asked Sep 3 at 5:00









      Carpetfizz

      1224




      1224




















          2 Answers
          2






          active

          oldest

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          up vote
          3
          down vote



          accepted










          I understand that "to have frequencies" can be misleading. This is a shorthand for "having non-zero energy or amplitude at two (specific) frequencies". Rephrased, for the cosine: when a cosine signal is represented in the frequency domain, it has two non-zero components at frequencies $pm omega$, and the other values anywhere on the frequency axis have zero amplitude. Or: its frequency spectrum has only two non-zero values.



          Frequency in signal processing often refers to a domain of analysis well-suited to certain signals and systems (linear systems for instance). With Fourier analysis, such signals can be represented as a weighted sum of sines or cosines, and essentially the frequencies a signal has is directly related to the number of sines/cosines with non-zero weight in the sum. One could be more precise, depending on which type of signal and Fourier transformation one is talking about (discrete/continuous).



          The concept of frequency is somehow independent on the signal, just like the time, and time and frequency are interlinked via Fourier transforms. So "having frequencies" by itself has no meaning for a signal, just as "having time". But a signal can be described with either amplitudes in the time domain, or amplitudes in a frequency domain.



          In that sense, whenever a Fourier transform is properly defined for a signal (integrability, etc.), and there is at least one non-zero value, the signal "has frequencies".



          So, most signals you will meet have frequencies, except the zero signal, and those whose Fourier transform cannot be defined correctly. As you might see later in signal processing lectures, one often uses "having frequencies" in a loose sense, meaning "above a certain threshold" instead of being simply non-zero, as in DSP practice, it is rare to have exactly zero values (finite signals, approximations, etc.)






          share|improve this answer






















          • Do you mind expanding on "having non-zero energy at two (specific) frequencies"?
            – Carpetfizz
            Sep 3 at 7:49






          • 1




            Thank you. Is the added sentence "when a cosine signal is represented in the frequency domain, it has two non-zero components at frequencies $pm omega$, and the other values anywhere on the frequency axis have zero amplitude." clearer?
            – Laurent Duval
            Sep 3 at 7:58






          • 1




            Matt's answer is IMO a version of mine with precise statements and formulae
            – Laurent Duval
            Sep 3 at 8:06






          • 1




            I hadn't seen your answer while I was writing up mine. I'll just leave it for the ones who want to see some formulas.
            – Matt L.
            Sep 3 at 8:23






          • 1




            That's why I mentioned that yours deserved attention
            – Laurent Duval
            Sep 3 at 9:20

















          up vote
          4
          down vote













          Let's first consider periodic signals. Under rather mild conditions, a periodic signal with period $T$ can be represented by its Fourier series:



          $$f(t)=sum_k=-infty^inftyc_ke^j2pi kt/Ttag1$$



          Eq. $(1)$ shows that the periodic function $f(t)$ is represented by a sum of weighted complex exponentials with frequencies $omega_k=2pi k/T$, which are integer multiples of the fundamental frequency $2pi/T$. So $f(t)$ has an infinite number of discrete frequencies, unless it is strictly band-limited, in which case the number of discrete frequencies is finite.



          For non-periodic functions, again under rather mild conditions, there exists a representation known as the Fourier integral:



          $$f(t)=frac12piint_-infty^inftyF(jomega)e^jomega tdomegatag2$$



          The weighting $F(jomega)$ is now a function defined for all frequencies $omega$, and it is known as the Fourier transform of $f(t)$. Eq. $(2)$ shows that $f(t)$ has all frequencies for which $F(jomega)$ is not equal to zero. So generally, a non-periodic signal contains a continuum of frequencies.






          share|improve this answer
















          • 1




            Thank you for your answer. While it has provided some new insights, I believe @LaurentDuval ‘s answer immediately answers the original question for new readers.
            – Carpetfizz
            Sep 3 at 8:11










          Your Answer




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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          3
          down vote



          accepted










          I understand that "to have frequencies" can be misleading. This is a shorthand for "having non-zero energy or amplitude at two (specific) frequencies". Rephrased, for the cosine: when a cosine signal is represented in the frequency domain, it has two non-zero components at frequencies $pm omega$, and the other values anywhere on the frequency axis have zero amplitude. Or: its frequency spectrum has only two non-zero values.



          Frequency in signal processing often refers to a domain of analysis well-suited to certain signals and systems (linear systems for instance). With Fourier analysis, such signals can be represented as a weighted sum of sines or cosines, and essentially the frequencies a signal has is directly related to the number of sines/cosines with non-zero weight in the sum. One could be more precise, depending on which type of signal and Fourier transformation one is talking about (discrete/continuous).



          The concept of frequency is somehow independent on the signal, just like the time, and time and frequency are interlinked via Fourier transforms. So "having frequencies" by itself has no meaning for a signal, just as "having time". But a signal can be described with either amplitudes in the time domain, or amplitudes in a frequency domain.



          In that sense, whenever a Fourier transform is properly defined for a signal (integrability, etc.), and there is at least one non-zero value, the signal "has frequencies".



          So, most signals you will meet have frequencies, except the zero signal, and those whose Fourier transform cannot be defined correctly. As you might see later in signal processing lectures, one often uses "having frequencies" in a loose sense, meaning "above a certain threshold" instead of being simply non-zero, as in DSP practice, it is rare to have exactly zero values (finite signals, approximations, etc.)






          share|improve this answer






















          • Do you mind expanding on "having non-zero energy at two (specific) frequencies"?
            – Carpetfizz
            Sep 3 at 7:49






          • 1




            Thank you. Is the added sentence "when a cosine signal is represented in the frequency domain, it has two non-zero components at frequencies $pm omega$, and the other values anywhere on the frequency axis have zero amplitude." clearer?
            – Laurent Duval
            Sep 3 at 7:58






          • 1




            Matt's answer is IMO a version of mine with precise statements and formulae
            – Laurent Duval
            Sep 3 at 8:06






          • 1




            I hadn't seen your answer while I was writing up mine. I'll just leave it for the ones who want to see some formulas.
            – Matt L.
            Sep 3 at 8:23






          • 1




            That's why I mentioned that yours deserved attention
            – Laurent Duval
            Sep 3 at 9:20














          up vote
          3
          down vote



          accepted










          I understand that "to have frequencies" can be misleading. This is a shorthand for "having non-zero energy or amplitude at two (specific) frequencies". Rephrased, for the cosine: when a cosine signal is represented in the frequency domain, it has two non-zero components at frequencies $pm omega$, and the other values anywhere on the frequency axis have zero amplitude. Or: its frequency spectrum has only two non-zero values.



          Frequency in signal processing often refers to a domain of analysis well-suited to certain signals and systems (linear systems for instance). With Fourier analysis, such signals can be represented as a weighted sum of sines or cosines, and essentially the frequencies a signal has is directly related to the number of sines/cosines with non-zero weight in the sum. One could be more precise, depending on which type of signal and Fourier transformation one is talking about (discrete/continuous).



          The concept of frequency is somehow independent on the signal, just like the time, and time and frequency are interlinked via Fourier transforms. So "having frequencies" by itself has no meaning for a signal, just as "having time". But a signal can be described with either amplitudes in the time domain, or amplitudes in a frequency domain.



          In that sense, whenever a Fourier transform is properly defined for a signal (integrability, etc.), and there is at least one non-zero value, the signal "has frequencies".



          So, most signals you will meet have frequencies, except the zero signal, and those whose Fourier transform cannot be defined correctly. As you might see later in signal processing lectures, one often uses "having frequencies" in a loose sense, meaning "above a certain threshold" instead of being simply non-zero, as in DSP practice, it is rare to have exactly zero values (finite signals, approximations, etc.)






          share|improve this answer






















          • Do you mind expanding on "having non-zero energy at two (specific) frequencies"?
            – Carpetfizz
            Sep 3 at 7:49






          • 1




            Thank you. Is the added sentence "when a cosine signal is represented in the frequency domain, it has two non-zero components at frequencies $pm omega$, and the other values anywhere on the frequency axis have zero amplitude." clearer?
            – Laurent Duval
            Sep 3 at 7:58






          • 1




            Matt's answer is IMO a version of mine with precise statements and formulae
            – Laurent Duval
            Sep 3 at 8:06






          • 1




            I hadn't seen your answer while I was writing up mine. I'll just leave it for the ones who want to see some formulas.
            – Matt L.
            Sep 3 at 8:23






          • 1




            That's why I mentioned that yours deserved attention
            – Laurent Duval
            Sep 3 at 9:20












          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          I understand that "to have frequencies" can be misleading. This is a shorthand for "having non-zero energy or amplitude at two (specific) frequencies". Rephrased, for the cosine: when a cosine signal is represented in the frequency domain, it has two non-zero components at frequencies $pm omega$, and the other values anywhere on the frequency axis have zero amplitude. Or: its frequency spectrum has only two non-zero values.



          Frequency in signal processing often refers to a domain of analysis well-suited to certain signals and systems (linear systems for instance). With Fourier analysis, such signals can be represented as a weighted sum of sines or cosines, and essentially the frequencies a signal has is directly related to the number of sines/cosines with non-zero weight in the sum. One could be more precise, depending on which type of signal and Fourier transformation one is talking about (discrete/continuous).



          The concept of frequency is somehow independent on the signal, just like the time, and time and frequency are interlinked via Fourier transforms. So "having frequencies" by itself has no meaning for a signal, just as "having time". But a signal can be described with either amplitudes in the time domain, or amplitudes in a frequency domain.



          In that sense, whenever a Fourier transform is properly defined for a signal (integrability, etc.), and there is at least one non-zero value, the signal "has frequencies".



          So, most signals you will meet have frequencies, except the zero signal, and those whose Fourier transform cannot be defined correctly. As you might see later in signal processing lectures, one often uses "having frequencies" in a loose sense, meaning "above a certain threshold" instead of being simply non-zero, as in DSP practice, it is rare to have exactly zero values (finite signals, approximations, etc.)






          share|improve this answer














          I understand that "to have frequencies" can be misleading. This is a shorthand for "having non-zero energy or amplitude at two (specific) frequencies". Rephrased, for the cosine: when a cosine signal is represented in the frequency domain, it has two non-zero components at frequencies $pm omega$, and the other values anywhere on the frequency axis have zero amplitude. Or: its frequency spectrum has only two non-zero values.



          Frequency in signal processing often refers to a domain of analysis well-suited to certain signals and systems (linear systems for instance). With Fourier analysis, such signals can be represented as a weighted sum of sines or cosines, and essentially the frequencies a signal has is directly related to the number of sines/cosines with non-zero weight in the sum. One could be more precise, depending on which type of signal and Fourier transformation one is talking about (discrete/continuous).



          The concept of frequency is somehow independent on the signal, just like the time, and time and frequency are interlinked via Fourier transforms. So "having frequencies" by itself has no meaning for a signal, just as "having time". But a signal can be described with either amplitudes in the time domain, or amplitudes in a frequency domain.



          In that sense, whenever a Fourier transform is properly defined for a signal (integrability, etc.), and there is at least one non-zero value, the signal "has frequencies".



          So, most signals you will meet have frequencies, except the zero signal, and those whose Fourier transform cannot be defined correctly. As you might see later in signal processing lectures, one often uses "having frequencies" in a loose sense, meaning "above a certain threshold" instead of being simply non-zero, as in DSP practice, it is rare to have exactly zero values (finite signals, approximations, etc.)







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Sep 3 at 9:30

























          answered Sep 3 at 7:46









          Laurent Duval

          14.8k31955




          14.8k31955











          • Do you mind expanding on "having non-zero energy at two (specific) frequencies"?
            – Carpetfizz
            Sep 3 at 7:49






          • 1




            Thank you. Is the added sentence "when a cosine signal is represented in the frequency domain, it has two non-zero components at frequencies $pm omega$, and the other values anywhere on the frequency axis have zero amplitude." clearer?
            – Laurent Duval
            Sep 3 at 7:58






          • 1




            Matt's answer is IMO a version of mine with precise statements and formulae
            – Laurent Duval
            Sep 3 at 8:06






          • 1




            I hadn't seen your answer while I was writing up mine. I'll just leave it for the ones who want to see some formulas.
            – Matt L.
            Sep 3 at 8:23






          • 1




            That's why I mentioned that yours deserved attention
            – Laurent Duval
            Sep 3 at 9:20
















          • Do you mind expanding on "having non-zero energy at two (specific) frequencies"?
            – Carpetfizz
            Sep 3 at 7:49






          • 1




            Thank you. Is the added sentence "when a cosine signal is represented in the frequency domain, it has two non-zero components at frequencies $pm omega$, and the other values anywhere on the frequency axis have zero amplitude." clearer?
            – Laurent Duval
            Sep 3 at 7:58






          • 1




            Matt's answer is IMO a version of mine with precise statements and formulae
            – Laurent Duval
            Sep 3 at 8:06






          • 1




            I hadn't seen your answer while I was writing up mine. I'll just leave it for the ones who want to see some formulas.
            – Matt L.
            Sep 3 at 8:23






          • 1




            That's why I mentioned that yours deserved attention
            – Laurent Duval
            Sep 3 at 9:20















          Do you mind expanding on "having non-zero energy at two (specific) frequencies"?
          – Carpetfizz
          Sep 3 at 7:49




          Do you mind expanding on "having non-zero energy at two (specific) frequencies"?
          – Carpetfizz
          Sep 3 at 7:49




          1




          1




          Thank you. Is the added sentence "when a cosine signal is represented in the frequency domain, it has two non-zero components at frequencies $pm omega$, and the other values anywhere on the frequency axis have zero amplitude." clearer?
          – Laurent Duval
          Sep 3 at 7:58




          Thank you. Is the added sentence "when a cosine signal is represented in the frequency domain, it has two non-zero components at frequencies $pm omega$, and the other values anywhere on the frequency axis have zero amplitude." clearer?
          – Laurent Duval
          Sep 3 at 7:58




          1




          1




          Matt's answer is IMO a version of mine with precise statements and formulae
          – Laurent Duval
          Sep 3 at 8:06




          Matt's answer is IMO a version of mine with precise statements and formulae
          – Laurent Duval
          Sep 3 at 8:06




          1




          1




          I hadn't seen your answer while I was writing up mine. I'll just leave it for the ones who want to see some formulas.
          – Matt L.
          Sep 3 at 8:23




          I hadn't seen your answer while I was writing up mine. I'll just leave it for the ones who want to see some formulas.
          – Matt L.
          Sep 3 at 8:23




          1




          1




          That's why I mentioned that yours deserved attention
          – Laurent Duval
          Sep 3 at 9:20




          That's why I mentioned that yours deserved attention
          – Laurent Duval
          Sep 3 at 9:20










          up vote
          4
          down vote













          Let's first consider periodic signals. Under rather mild conditions, a periodic signal with period $T$ can be represented by its Fourier series:



          $$f(t)=sum_k=-infty^inftyc_ke^j2pi kt/Ttag1$$



          Eq. $(1)$ shows that the periodic function $f(t)$ is represented by a sum of weighted complex exponentials with frequencies $omega_k=2pi k/T$, which are integer multiples of the fundamental frequency $2pi/T$. So $f(t)$ has an infinite number of discrete frequencies, unless it is strictly band-limited, in which case the number of discrete frequencies is finite.



          For non-periodic functions, again under rather mild conditions, there exists a representation known as the Fourier integral:



          $$f(t)=frac12piint_-infty^inftyF(jomega)e^jomega tdomegatag2$$



          The weighting $F(jomega)$ is now a function defined for all frequencies $omega$, and it is known as the Fourier transform of $f(t)$. Eq. $(2)$ shows that $f(t)$ has all frequencies for which $F(jomega)$ is not equal to zero. So generally, a non-periodic signal contains a continuum of frequencies.






          share|improve this answer
















          • 1




            Thank you for your answer. While it has provided some new insights, I believe @LaurentDuval ‘s answer immediately answers the original question for new readers.
            – Carpetfizz
            Sep 3 at 8:11














          up vote
          4
          down vote













          Let's first consider periodic signals. Under rather mild conditions, a periodic signal with period $T$ can be represented by its Fourier series:



          $$f(t)=sum_k=-infty^inftyc_ke^j2pi kt/Ttag1$$



          Eq. $(1)$ shows that the periodic function $f(t)$ is represented by a sum of weighted complex exponentials with frequencies $omega_k=2pi k/T$, which are integer multiples of the fundamental frequency $2pi/T$. So $f(t)$ has an infinite number of discrete frequencies, unless it is strictly band-limited, in which case the number of discrete frequencies is finite.



          For non-periodic functions, again under rather mild conditions, there exists a representation known as the Fourier integral:



          $$f(t)=frac12piint_-infty^inftyF(jomega)e^jomega tdomegatag2$$



          The weighting $F(jomega)$ is now a function defined for all frequencies $omega$, and it is known as the Fourier transform of $f(t)$. Eq. $(2)$ shows that $f(t)$ has all frequencies for which $F(jomega)$ is not equal to zero. So generally, a non-periodic signal contains a continuum of frequencies.






          share|improve this answer
















          • 1




            Thank you for your answer. While it has provided some new insights, I believe @LaurentDuval ‘s answer immediately answers the original question for new readers.
            – Carpetfizz
            Sep 3 at 8:11












          up vote
          4
          down vote










          up vote
          4
          down vote









          Let's first consider periodic signals. Under rather mild conditions, a periodic signal with period $T$ can be represented by its Fourier series:



          $$f(t)=sum_k=-infty^inftyc_ke^j2pi kt/Ttag1$$



          Eq. $(1)$ shows that the periodic function $f(t)$ is represented by a sum of weighted complex exponentials with frequencies $omega_k=2pi k/T$, which are integer multiples of the fundamental frequency $2pi/T$. So $f(t)$ has an infinite number of discrete frequencies, unless it is strictly band-limited, in which case the number of discrete frequencies is finite.



          For non-periodic functions, again under rather mild conditions, there exists a representation known as the Fourier integral:



          $$f(t)=frac12piint_-infty^inftyF(jomega)e^jomega tdomegatag2$$



          The weighting $F(jomega)$ is now a function defined for all frequencies $omega$, and it is known as the Fourier transform of $f(t)$. Eq. $(2)$ shows that $f(t)$ has all frequencies for which $F(jomega)$ is not equal to zero. So generally, a non-periodic signal contains a continuum of frequencies.






          share|improve this answer












          Let's first consider periodic signals. Under rather mild conditions, a periodic signal with period $T$ can be represented by its Fourier series:



          $$f(t)=sum_k=-infty^inftyc_ke^j2pi kt/Ttag1$$



          Eq. $(1)$ shows that the periodic function $f(t)$ is represented by a sum of weighted complex exponentials with frequencies $omega_k=2pi k/T$, which are integer multiples of the fundamental frequency $2pi/T$. So $f(t)$ has an infinite number of discrete frequencies, unless it is strictly band-limited, in which case the number of discrete frequencies is finite.



          For non-periodic functions, again under rather mild conditions, there exists a representation known as the Fourier integral:



          $$f(t)=frac12piint_-infty^inftyF(jomega)e^jomega tdomegatag2$$



          The weighting $F(jomega)$ is now a function defined for all frequencies $omega$, and it is known as the Fourier transform of $f(t)$. Eq. $(2)$ shows that $f(t)$ has all frequencies for which $F(jomega)$ is not equal to zero. So generally, a non-periodic signal contains a continuum of frequencies.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Sep 3 at 7:50









          Matt L.

          45.2k13678




          45.2k13678







          • 1




            Thank you for your answer. While it has provided some new insights, I believe @LaurentDuval ‘s answer immediately answers the original question for new readers.
            – Carpetfizz
            Sep 3 at 8:11












          • 1




            Thank you for your answer. While it has provided some new insights, I believe @LaurentDuval ‘s answer immediately answers the original question for new readers.
            – Carpetfizz
            Sep 3 at 8:11







          1




          1




          Thank you for your answer. While it has provided some new insights, I believe @LaurentDuval ‘s answer immediately answers the original question for new readers.
          – Carpetfizz
          Sep 3 at 8:11




          Thank you for your answer. While it has provided some new insights, I believe @LaurentDuval ‘s answer immediately answers the original question for new readers.
          – Carpetfizz
          Sep 3 at 8:11

















           

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