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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?
frequency linear-systems frequency-response math
edited Sep 3 at 7:47
Laurent Duval
14.8k31955
asked Sep 3 at 5:00
Carpetfizz
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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?
frequency linear-systems frequency-response math
edited Sep 3 at 7:47
Laurent Duval
14.8k31955
asked Sep 3 at 5:00
Carpetfizz
1224
add a comment |Â
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?
frequency linear-systems frequency-response math
edited Sep 3 at 7:47
Laurent Duval
14.8k31955
asked Sep 3 at 5:00
Carpetfizz
1224
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?
frequency linear-systems frequency-response math
edited Sep 3 at 7:47
Laurent Duval
14.8k31955
asked Sep 3 at 5:00
Carpetfizz
1224
edited Sep 3 at 7:47
Laurent Duval
14.8k31955
edited Sep 3 at 7:47
Laurent Duval
14.8k31955
edited Sep 3 at 7:47
Laurent Duval
14.8k31955
14.8k31955
asked Sep 3 at 5:00
Carpetfizz
1224
asked Sep 3 at 5:00
Carpetfizz
1224
asked Sep 3 at 5:00
Carpetfizz
1224
1224
add a comment |Â
add a comment |Â
2 Answers
2
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.)
answered Sep 3 at 7:46
Laurent Duval
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
 |Â
show 1 more comment
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.
answered Sep 3 at 7:50
Matt L.
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
add a comment |Â
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.)
answered Sep 3 at 7:46
Laurent Duval
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
 |Â
show 1 more comment
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.)
answered Sep 3 at 7:46
Laurent Duval
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
 |Â
show 1 more comment
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.)
answered Sep 3 at 7:46
Laurent Duval
14.8k31955
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.)
answered Sep 3 at 7:46
Laurent Duval
14.8k31955
edited Sep 3 at 9:30
answered Sep 3 at 7:46
Laurent Duval
14.8k31955
answered Sep 3 at 7:46
Laurent Duval
14.8k31955
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
 |Â
show 1 more comment
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
 |Â
show 1 more comment
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.
answered Sep 3 at 7:50
Matt L.
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
add a comment |Â
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.
answered Sep 3 at 7:50
Matt L.
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
add a comment |Â
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.
answered Sep 3 at 7:50
Matt L.
45.2k13678
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.
answered Sep 3 at 7:50
Matt L.
45.2k13678
answered Sep 3 at 7:50
Matt L.
45.2k13678
answered Sep 3 at 7:50
Matt L.
45.2k13678
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
add a comment |Â
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
add a comment |Â
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