If we use a sinusoidal signal as an input signal to a linear transmission path, then we always get out a sine...
An Introduction to Information Theory: Symbols, Signals and Noise, by John R. Pierce, says the following:
With the very surprising property of linearity in mind, let us return to the transmission of signals over electrical circuits. We have noted that the output signal corresponding to most input signals has a different shape or variation with time from the input signal. Figures II-1 and II-2 illustrate this. However, it can be shown mathematically (but not here) that, if we use a sinusoidal signal, such as that of Figure II-4, as an input signal to a linear transmission path, we always get out a sine wave of the same period, or frequency. The amplitude of the output sine wave may be less than that of the input sine wave; we call this attenuation of the sinusoidal signal. The output sine wave, may rise to a peak later than the input sine wave; we call this phase shift, or delay of the sinusoidal signal.
I'm trying to find the aforementioned proof that, if we use a sinusoidal signal as an input signal to a linear transmission path, then we always get out a sine wave of the same period, or frequency.
During my research, the closest thing to this that I have come across is slide 30 of this presentation:
I would greatly appreciate it if people could please take the time to either prove this or redirect me somewhere that has the proof.
trigonometry fourier-analysis dynamical-systems mathematical-physics signal-processing
asked 17 hours ago
The Pointer
2,60021334
add a comment |
An Introduction to Information Theory: Symbols, Signals and Noise, by John R. Pierce, says the following:
With the very surprising property of linearity in mind, let us return to the transmission of signals over electrical circuits. We have noted that the output signal corresponding to most input signals has a different shape or variation with time from the input signal. Figures II-1 and II-2 illustrate this. However, it can be shown mathematically (but not here) that, if we use a sinusoidal signal, such as that of Figure II-4, as an input signal to a linear transmission path, we always get out a sine wave of the same period, or frequency. The amplitude of the output sine wave may be less than that of the input sine wave; we call this attenuation of the sinusoidal signal. The output sine wave, may rise to a peak later than the input sine wave; we call this phase shift, or delay of the sinusoidal signal.
I'm trying to find the aforementioned proof that, if we use a sinusoidal signal as an input signal to a linear transmission path, then we always get out a sine wave of the same period, or frequency.
During my research, the closest thing to this that I have come across is slide 30 of this presentation:
I would greatly appreciate it if people could please take the time to either prove this or redirect me somewhere that has the proof.
trigonometry fourier-analysis dynamical-systems mathematical-physics signal-processing
asked 17 hours ago
The Pointer
2,60021334
1
Work in the Fourier domain, where the input is a $delta$ function, so the output must also be a $delta$ function... i.e., an input sine wave yields and output sine wave.
– David G. Stork
17 hours ago
An LTI system operating on an input signal is a convolution in the time domain. A convolution in the time domain is just multiplication in the frequency domain. Multiplication of a signal spectrum by the frequency spectrum of LTI system impulse response, will never shift frequencies or create new frequencies.
– Andy Walls
16 hours ago
add a comment |
An Introduction to Information Theory: Symbols, Signals and Noise, by John R. Pierce, says the following:
With the very surprising property of linearity in mind, let us return to the transmission of signals over electrical circuits. We have noted that the output signal corresponding to most input signals has a different shape or variation with time from the input signal. Figures II-1 and II-2 illustrate this. However, it can be shown mathematically (but not here) that, if we use a sinusoidal signal, such as that of Figure II-4, as an input signal to a linear transmission path, we always get out a sine wave of the same period, or frequency. The amplitude of the output sine wave may be less than that of the input sine wave; we call this attenuation of the sinusoidal signal. The output sine wave, may rise to a peak later than the input sine wave; we call this phase shift, or delay of the sinusoidal signal.
I'm trying to find the aforementioned proof that, if we use a sinusoidal signal as an input signal to a linear transmission path, then we always get out a sine wave of the same period, or frequency.
During my research, the closest thing to this that I have come across is slide 30 of this presentation:
I would greatly appreciate it if people could please take the time to either prove this or redirect me somewhere that has the proof.
trigonometry fourier-analysis dynamical-systems mathematical-physics signal-processing
asked 17 hours ago
The Pointer
2,60021334
An Introduction to Information Theory: Symbols, Signals and Noise, by John R. Pierce, says the following:
With the very surprising property of linearity in mind, let us return to the transmission of signals over electrical circuits. We have noted that the output signal corresponding to most input signals has a different shape or variation with time from the input signal. Figures II-1 and II-2 illustrate this. However, it can be shown mathematically (but not here) that, if we use a sinusoidal signal, such as that of Figure II-4, as an input signal to a linear transmission path, we always get out a sine wave of the same period, or frequency. The amplitude of the output sine wave may be less than that of the input sine wave; we call this attenuation of the sinusoidal signal. The output sine wave, may rise to a peak later than the input sine wave; we call this phase shift, or delay of the sinusoidal signal.
I'm trying to find the aforementioned proof that, if we use a sinusoidal signal as an input signal to a linear transmission path, then we always get out a sine wave of the same period, or frequency.
During my research, the closest thing to this that I have come across is slide 30 of this presentation:
I would greatly appreciate it if people could please take the time to either prove this or redirect me somewhere that has the proof.
trigonometry fourier-analysis dynamical-systems mathematical-physics signal-processing
trigonometry fourier-analysis dynamical-systems mathematical-physics signal-processing
asked 17 hours ago
The Pointer
2,60021334
asked 17 hours ago
The Pointer
2,60021334
asked 17 hours ago
The Pointer
2,60021334
asked 17 hours ago
The Pointer
2,60021334
asked 17 hours ago
The Pointer
2,60021334
2,60021334
1
Work in the Fourier domain, where the input is a $delta$ function, so the output must also be a $delta$ function... i.e., an input sine wave yields and output sine wave.
– David G. Stork
17 hours ago
An LTI system operating on an input signal is a convolution in the time domain. A convolution in the time domain is just multiplication in the frequency domain. Multiplication of a signal spectrum by the frequency spectrum of LTI system impulse response, will never shift frequencies or create new frequencies.
– Andy Walls
16 hours ago
add a comment |
1
Work in the Fourier domain, where the input is a $delta$ function, so the output must also be a $delta$ function... i.e., an input sine wave yields and output sine wave.
– David G. Stork
17 hours ago
An LTI system operating on an input signal is a convolution in the time domain. A convolution in the time domain is just multiplication in the frequency domain. Multiplication of a signal spectrum by the frequency spectrum of LTI system impulse response, will never shift frequencies or create new frequencies.
– Andy Walls
16 hours ago
1
1
Work in the Fourier domain, where the input is a $delta$ function, so the output must also be a $delta$ function... i.e., an input sine wave yields and output sine wave.
– David G. Stork
17 hours ago
Work in the Fourier domain, where the input is a $delta$ function, so the output must also be a $delta$ function... i.e., an input sine wave yields and output sine wave.
– David G. Stork
17 hours ago
An LTI system operating on an input signal is a convolution in the time domain. A convolution in the time domain is just multiplication in the frequency domain. Multiplication of a signal spectrum by the frequency spectrum of LTI system impulse response, will never shift frequencies or create new frequencies.
– Andy Walls
16 hours ago
An LTI system operating on an input signal is a convolution in the time domain. A convolution in the time domain is just multiplication in the frequency domain. Multiplication of a signal spectrum by the frequency spectrum of LTI system impulse response, will never shift frequencies or create new frequencies.
– Andy Walls
16 hours ago
add a comment |
1 Answer
1
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$y=G(x)$ is translation invariant, $GBigl(T_sxBigr)(t)=(T_sy)(t)=y(t+s)$ Together with the linearity this has the consequence that also differential operators are preserved,
$$dot y(t)=lim_{sto 0}frac{(T_sy)(t)-y(t)}s=lim_{sto 0}frac{Gbigl(T_sxbigr)(t)-Gbigl(xbigr)(t)}{s}=Gleft(lim_{sto 0}frac{T_sx-x}sright)(t)=Gbigl(dot xbigr)(t).$$
Now you can also apply this to the oscillator equation, $G(ddot x+omega^2x)=ddot y+ω^2y$ and if $x$ is sinusoid with frequency $ω$, then so is $y$.
With $$G(cos(ω,cdot,))(t)=acos(ωt)+bsin(ωt)$$ you also get the shifted
$$
G(sin(ω,cdot,))=G(cos(ω,cdot,-tfracpi2))(t)=asin(ωt)-bcos(ωt)
$$
so that indeed there are only two free parameters per frequency. To get the attenuation and phase, you only need to compute the polar coordinates $(A,varphi)$ of the point $(a,-b)$.
answered 11 hours ago
LutzL
55.8k42054
Thanks for the answer. Shouldn't that be $lim_{s to 0} frac{Gbigl(T_sxbigr)(t)-Gbigl(xbigr)(t)}{s}$? And I'm confused as to how this is proof that, if we use a sinusoidal signal as an input signal to a linear transmission path, then we always get out a sine wave of the same period, or frequency?
– The Pointer
10 hours ago
Yes, that got missing in copying the parts of the expression. If $x$ is sinusoidal with frequency $ω$, then $ddot x+ ω^2x=0$. But then by LTI also $ddot y+ ω^2y=0$, which implies that also $y$ is sinusoidal with the same frequency. That the amplitude change and frequency shift is the only thing that happens at frequency $ω$ was the point of the second part, to show that, in still another formalism, $G({tmapsto e^{iωt}})={tmapsto (a-ib)e^{iωt}}$.
– LutzL
10 hours ago
ahh yes, I see. LTI meaning linear transformation, yes? And I think you mean phase-shift, not frequency-shift, since the entire point is that the period/frequency remains the same?
– The Pointer
10 hours ago
LTI is "Linear and Translation Invariant" (or V. Klemperers "Lingua Tertii Imperii"). Yes, phase shift.
– LutzL
10 hours ago
ok, this makes sense to me now. Thank you for taking the time to assist and clarify.
– The Pointer
10 hours ago
add a comment |
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$y=G(x)$ is translation invariant, $GBigl(T_sxBigr)(t)=(T_sy)(t)=y(t+s)$ Together with the linearity this has the consequence that also differential operators are preserved,
$$dot y(t)=lim_{sto 0}frac{(T_sy)(t)-y(t)}s=lim_{sto 0}frac{Gbigl(T_sxbigr)(t)-Gbigl(xbigr)(t)}{s}=Gleft(lim_{sto 0}frac{T_sx-x}sright)(t)=Gbigl(dot xbigr)(t).$$
Now you can also apply this to the oscillator equation, $G(ddot x+omega^2x)=ddot y+ω^2y$ and if $x$ is sinusoid with frequency $ω$, then so is $y$.
With $$G(cos(ω,cdot,))(t)=acos(ωt)+bsin(ωt)$$ you also get the shifted
$$
G(sin(ω,cdot,))=G(cos(ω,cdot,-tfracpi2))(t)=asin(ωt)-bcos(ωt)
$$
so that indeed there are only two free parameters per frequency. To get the attenuation and phase, you only need to compute the polar coordinates $(A,varphi)$ of the point $(a,-b)$.
answered 11 hours ago
LutzL
55.8k42054
Thanks for the answer. Shouldn't that be $lim_{s to 0} frac{Gbigl(T_sxbigr)(t)-Gbigl(xbigr)(t)}{s}$? And I'm confused as to how this is proof that, if we use a sinusoidal signal as an input signal to a linear transmission path, then we always get out a sine wave of the same period, or frequency?
– The Pointer
10 hours ago
Yes, that got missing in copying the parts of the expression. If $x$ is sinusoidal with frequency $ω$, then $ddot x+ ω^2x=0$. But then by LTI also $ddot y+ ω^2y=0$, which implies that also $y$ is sinusoidal with the same frequency. That the amplitude change and frequency shift is the only thing that happens at frequency $ω$ was the point of the second part, to show that, in still another formalism, $G({tmapsto e^{iωt}})={tmapsto (a-ib)e^{iωt}}$.
– LutzL
10 hours ago
ahh yes, I see. LTI meaning linear transformation, yes? And I think you mean phase-shift, not frequency-shift, since the entire point is that the period/frequency remains the same?
– The Pointer
10 hours ago
LTI is "Linear and Translation Invariant" (or V. Klemperers "Lingua Tertii Imperii"). Yes, phase shift.
– LutzL
10 hours ago
ok, this makes sense to me now. Thank you for taking the time to assist and clarify.
– The Pointer
10 hours ago
add a comment |
$y=G(x)$ is translation invariant, $GBigl(T_sxBigr)(t)=(T_sy)(t)=y(t+s)$ Together with the linearity this has the consequence that also differential operators are preserved,
$$dot y(t)=lim_{sto 0}frac{(T_sy)(t)-y(t)}s=lim_{sto 0}frac{Gbigl(T_sxbigr)(t)-Gbigl(xbigr)(t)}{s}=Gleft(lim_{sto 0}frac{T_sx-x}sright)(t)=Gbigl(dot xbigr)(t).$$
Now you can also apply this to the oscillator equation, $G(ddot x+omega^2x)=ddot y+ω^2y$ and if $x$ is sinusoid with frequency $ω$, then so is $y$.
With $$G(cos(ω,cdot,))(t)=acos(ωt)+bsin(ωt)$$ you also get the shifted
$$
G(sin(ω,cdot,))=G(cos(ω,cdot,-tfracpi2))(t)=asin(ωt)-bcos(ωt)
$$
so that indeed there are only two free parameters per frequency. To get the attenuation and phase, you only need to compute the polar coordinates $(A,varphi)$ of the point $(a,-b)$.
answered 11 hours ago
LutzL
55.8k42054
Thanks for the answer. Shouldn't that be $lim_{s to 0} frac{Gbigl(T_sxbigr)(t)-Gbigl(xbigr)(t)}{s}$? And I'm confused as to how this is proof that, if we use a sinusoidal signal as an input signal to a linear transmission path, then we always get out a sine wave of the same period, or frequency?
– The Pointer
10 hours ago
Yes, that got missing in copying the parts of the expression. If $x$ is sinusoidal with frequency $ω$, then $ddot x+ ω^2x=0$. But then by LTI also $ddot y+ ω^2y=0$, which implies that also $y$ is sinusoidal with the same frequency. That the amplitude change and frequency shift is the only thing that happens at frequency $ω$ was the point of the second part, to show that, in still another formalism, $G({tmapsto e^{iωt}})={tmapsto (a-ib)e^{iωt}}$.
– LutzL
10 hours ago
ahh yes, I see. LTI meaning linear transformation, yes? And I think you mean phase-shift, not frequency-shift, since the entire point is that the period/frequency remains the same?
– The Pointer
10 hours ago
LTI is "Linear and Translation Invariant" (or V. Klemperers "Lingua Tertii Imperii"). Yes, phase shift.
– LutzL
10 hours ago
ok, this makes sense to me now. Thank you for taking the time to assist and clarify.
– The Pointer
10 hours ago
add a comment |
$y=G(x)$ is translation invariant, $GBigl(T_sxBigr)(t)=(T_sy)(t)=y(t+s)$ Together with the linearity this has the consequence that also differential operators are preserved,
$$dot y(t)=lim_{sto 0}frac{(T_sy)(t)-y(t)}s=lim_{sto 0}frac{Gbigl(T_sxbigr)(t)-Gbigl(xbigr)(t)}{s}=Gleft(lim_{sto 0}frac{T_sx-x}sright)(t)=Gbigl(dot xbigr)(t).$$
Now you can also apply this to the oscillator equation, $G(ddot x+omega^2x)=ddot y+ω^2y$ and if $x$ is sinusoid with frequency $ω$, then so is $y$.
With $$G(cos(ω,cdot,))(t)=acos(ωt)+bsin(ωt)$$ you also get the shifted
$$
G(sin(ω,cdot,))=G(cos(ω,cdot,-tfracpi2))(t)=asin(ωt)-bcos(ωt)
$$
so that indeed there are only two free parameters per frequency. To get the attenuation and phase, you only need to compute the polar coordinates $(A,varphi)$ of the point $(a,-b)$.
answered 11 hours ago
LutzL
55.8k42054
$y=G(x)$ is translation invariant, $GBigl(T_sxBigr)(t)=(T_sy)(t)=y(t+s)$ Together with the linearity this has the consequence that also differential operators are preserved,
$$dot y(t)=lim_{sto 0}frac{(T_sy)(t)-y(t)}s=lim_{sto 0}frac{Gbigl(T_sxbigr)(t)-Gbigl(xbigr)(t)}{s}=Gleft(lim_{sto 0}frac{T_sx-x}sright)(t)=Gbigl(dot xbigr)(t).$$
Now you can also apply this to the oscillator equation, $G(ddot x+omega^2x)=ddot y+ω^2y$ and if $x$ is sinusoid with frequency $ω$, then so is $y$.
With $$G(cos(ω,cdot,))(t)=acos(ωt)+bsin(ωt)$$ you also get the shifted
$$
G(sin(ω,cdot,))=G(cos(ω,cdot,-tfracpi2))(t)=asin(ωt)-bcos(ωt)
$$
so that indeed there are only two free parameters per frequency. To get the attenuation and phase, you only need to compute the polar coordinates $(A,varphi)$ of the point $(a,-b)$.
answered 11 hours ago
LutzL
55.8k42054
edited 10 hours ago
answered 11 hours ago
LutzL
55.8k42054
answered 11 hours ago
LutzL
55.8k42054
answered 11 hours ago
LutzL
55.8k42054
55.8k42054
Thanks for the answer. Shouldn't that be $lim_{s to 0} frac{Gbigl(T_sxbigr)(t)-Gbigl(xbigr)(t)}{s}$? And I'm confused as to how this is proof that, if we use a sinusoidal signal as an input signal to a linear transmission path, then we always get out a sine wave of the same period, or frequency?
– The Pointer
10 hours ago
Yes, that got missing in copying the parts of the expression. If $x$ is sinusoidal with frequency $ω$, then $ddot x+ ω^2x=0$. But then by LTI also $ddot y+ ω^2y=0$, which implies that also $y$ is sinusoidal with the same frequency. That the amplitude change and frequency shift is the only thing that happens at frequency $ω$ was the point of the second part, to show that, in still another formalism, $G({tmapsto e^{iωt}})={tmapsto (a-ib)e^{iωt}}$.
– LutzL
10 hours ago
ahh yes, I see. LTI meaning linear transformation, yes? And I think you mean phase-shift, not frequency-shift, since the entire point is that the period/frequency remains the same?
– The Pointer
10 hours ago
LTI is "Linear and Translation Invariant" (or V. Klemperers "Lingua Tertii Imperii"). Yes, phase shift.
– LutzL
10 hours ago
ok, this makes sense to me now. Thank you for taking the time to assist and clarify.
– The Pointer
10 hours ago
add a comment |
Thanks for the answer. Shouldn't that be $lim_{s to 0} frac{Gbigl(T_sxbigr)(t)-Gbigl(xbigr)(t)}{s}$? And I'm confused as to how this is proof that, if we use a sinusoidal signal as an input signal to a linear transmission path, then we always get out a sine wave of the same period, or frequency?
– The Pointer
10 hours ago
Yes, that got missing in copying the parts of the expression. If $x$ is sinusoidal with frequency $ω$, then $ddot x+ ω^2x=0$. But then by LTI also $ddot y+ ω^2y=0$, which implies that also $y$ is sinusoidal with the same frequency. That the amplitude change and frequency shift is the only thing that happens at frequency $ω$ was the point of the second part, to show that, in still another formalism, $G({tmapsto e^{iωt}})={tmapsto (a-ib)e^{iωt}}$.
– LutzL
10 hours ago
ahh yes, I see. LTI meaning linear transformation, yes? And I think you mean phase-shift, not frequency-shift, since the entire point is that the period/frequency remains the same?
– The Pointer
10 hours ago
LTI is "Linear and Translation Invariant" (or V. Klemperers "Lingua Tertii Imperii"). Yes, phase shift.
– LutzL
10 hours ago
ok, this makes sense to me now. Thank you for taking the time to assist and clarify.
– The Pointer
10 hours ago
Thanks for the answer. Shouldn't that be $lim_{s to 0} frac{Gbigl(T_sxbigr)(t)-Gbigl(xbigr)(t)}{s}$? And I'm confused as to how this is proof that, if we use a sinusoidal signal as an input signal to a linear transmission path, then we always get out a sine wave of the same period, or frequency?
– The Pointer
10 hours ago
Thanks for the answer. Shouldn't that be $lim_{s to 0} frac{Gbigl(T_sxbigr)(t)-Gbigl(xbigr)(t)}{s}$? And I'm confused as to how this is proof that, if we use a sinusoidal signal as an input signal to a linear transmission path, then we always get out a sine wave of the same period, or frequency?
– The Pointer
10 hours ago
Yes, that got missing in copying the parts of the expression. If $x$ is sinusoidal with frequency $ω$, then $ddot x+ ω^2x=0$. But then by LTI also $ddot y+ ω^2y=0$, which implies that also $y$ is sinusoidal with the same frequency. That the amplitude change and frequency shift is the only thing that happens at frequency $ω$ was the point of the second part, to show that, in still another formalism, $G({tmapsto e^{iωt}})={tmapsto (a-ib)e^{iωt}}$.
– LutzL
10 hours ago
Yes, that got missing in copying the parts of the expression. If $x$ is sinusoidal with frequency $ω$, then $ddot x+ ω^2x=0$. But then by LTI also $ddot y+ ω^2y=0$, which implies that also $y$ is sinusoidal with the same frequency. That the amplitude change and frequency shift is the only thing that happens at frequency $ω$ was the point of the second part, to show that, in still another formalism, $G({tmapsto e^{iωt}})={tmapsto (a-ib)e^{iωt}}$.
– LutzL
10 hours ago
ahh yes, I see. LTI meaning linear transformation, yes? And I think you mean phase-shift, not frequency-shift, since the entire point is that the period/frequency remains the same?
– The Pointer
10 hours ago
ahh yes, I see. LTI meaning linear transformation, yes? And I think you mean phase-shift, not frequency-shift, since the entire point is that the period/frequency remains the same?
– The Pointer
10 hours ago
LTI is "Linear and Translation Invariant" (or V. Klemperers "Lingua Tertii Imperii"). Yes, phase shift.
– LutzL
10 hours ago
LTI is "Linear and Translation Invariant" (or V. Klemperers "Lingua Tertii Imperii"). Yes, phase shift.
– LutzL
10 hours ago
ok, this makes sense to me now. Thank you for taking the time to assist and clarify.
– The Pointer
10 hours ago
ok, this makes sense to me now. Thank you for taking the time to assist and clarify.
– The Pointer
10 hours ago
add a comment |
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Work in the Fourier domain, where the input is a $delta$ function, so the output must also be a $delta$ function... i.e., an input sine wave yields and output sine wave.
– David G. Stork
17 hours ago
An LTI system operating on an input signal is a convolution in the time domain. A convolution in the time domain is just multiplication in the frequency domain. Multiplication of a signal spectrum by the frequency spectrum of LTI system impulse response, will never shift frequencies or create new frequencies.
– Andy Walls
16 hours ago